TITLE: What Would be the Experimental Signature of Composite Leptons?
QUESTION [6 upvotes]: So far, the quarks and leptons appear to be fundamental particles. But they're complicated enough that there's always been some speculation that they might be composite.
What experimental evidence would be needed to show that a lepton is composite?

REPLY [1 votes]: Composite electrons, muons and taus ought to be much easier to detect in high energy conditions where an electomagnetic field influences decay product paths than composite neutrinos (which are damned hard to observe much of anything about without elaborate and not very statistically powerful purpose built experiments like those being discussed at Neutel11).

If components parts of electon-like leptons had a charge other than -1 (or +1 for antiparticles), the path that even briefly unconfined lepton components took ought to be possible to reverse engineer with great precision (and without a lot of the QCD background issues that make some of the other calculations harder to do -- because you'd be looking at the distribution pattern of where the decay products end up in space relative to the collision point, rather than how many there were).
IIRC, there has been some recent experimental signals that show these kinds of unexpected and unexplained spatial distribution patterns.

Another way to see composite electrons in confined state would be to detect events with signatures that are like mesons or exotic baryons, but much lighter that had previously been screened out of data since we weren't looking for anything like that in that mass range.  For example, suppose that you revised your decay data sorting software and suddenly saw several dozen decays of a particle that was behaving like a Delta plus plus baryon (spin 3/2, charge +2, ddd), but with a mass on the order of 123 eV instead of 1232 MEv.
A third possiblity would be that you could look at processes that appear to show beyond the Standard Model CP violation and do some kind of cluster analysis of the data that show one group of events that closely match the Standard Model and a separate group of events that has some pattern that distinguishes it and then show how a composite lepton model could explain the pattern common to the "excess group".
Strong evidence of B-L non-conservation that seems to be coming from something in the lepton sector.


TITLE: List of all measures
QUESTION [0 upvotes]: Let Ω be a four-element set (elements do not matter).
How can I find a list of all measures µ: P(Ω)→ N, for which µ(Ω) = 7, where P(Ω) is Power set of Ω.

REPLY [1 votes]: This is just equivalent to set of all $4$-tuples $(a,b,c,d)$ in $\Bbb N$ such that $a+b+c+d=7$. We only need to assign integer values to the singletons and the measure is determined uniquely. It's quite easy to enumerate these; we will have $\binom{10}{7}$ many of them, by classic combinatiorial facts.


TITLE: Exactness of the differential equation
QUESTION [0 upvotes]: Let $f$ be an arbitrary continuously differentiable function. If the ordinary differential equation $$(3y^2-x)f(x+y^2)+2y(y^2-3x)f(x+y^2)y'=0$$
is exact. Then which of the following are true
1) $xf'(x)-3f(x)=0$
2)  $xf'(x)+3f(x)=0$
3) $f'(x)+3f(x)=0$
4)  $f'(x)-3f(x)=0$
I tried using $Mdx+Ndy=0$, but I did not get any option. Please help

REPLY [1 votes]: The second option seems correct to me ..
I used f for $f(y^2+x)$ for simplicity except in the last line
$$(3y^2-x)f(x+y^2)dx+2y(y^2-3x)f(x+y^2)dy=0$$
$$Pdx+Qdy=0$$
$$\partial_y P= f\times 6y+2yf' \times (3y^2-x)$$
$$\partial_x Q= f\times (-6y)+f' \times 2y(y^2-3x)$$
Equate both expressions for exactness
$$\partial_y P=\partial_x Q$$
$$ f\times 6y+2yf' \times (3y^2-x)- f\times (-6y)-f' \times 2y(y^2-3x)=0$$
$$12yf+4y(y^2+x)f'=0$$
$$(y^2+x)f'=-3f$$
$$(y^2+x)f'(y^2+x)=-3f(y^2+x) \implies xf'(x)=-3f(x)$$


text:
\section{Inverter Models}\label{Methodology}

Before introducing the models of the four types of solar PV inverter studied in
this paper, it is useful to define notation for the network.
The network under study consists of a set $\mathcal{N}$ of $N$ nodes that have a
load and, in some cases, a PV inverter. We often consider an extended network
$\mathcal{N'} = \mathcal{N} \cup \{\mathbf{s}\}$, where $\mathbf{s}$ denotes
the secondary side of the distribution transformer (the slack bus).
Lines are represented as $(m,n) \in \mathcal{M} \subset \mathcal{N'} \times \mathcal{N'}$.

In our studies, the network initially hosts $L$ legacy and $A$ autonomous inverters (described below) located on the nodes of the sets $\mathcal{L}$ and $\mathcal{A}$, respectively. We increase PV penetration by adding new solar PV systems equipped with one of the three following inverter models: coordinated inverters, non-exporting inverters or more autonomous inverters. Subscripts $l\in\mathcal{L}, a\in\mathcal{A}, c\in\mathcal{C}$, and $e\in\mathcal{E}$ denote the quantities pertaining to each legacy inverter $l$, autonomous inverter $a$, coordinated inverter $c$, and non-exporting inverter $e$. All inverters have the same rated apparent power capacity $\textbf{S}$, which sets the limits on the maximum active power injection $P^\text{inj}$ and reactive power support $Q$.
Complex voltages are expressed per unit and denoted by $V$ with $\Re$ and $\Im$ as the real and imaginary components. Constants beyond our control are written in bold. The individual inverter models are as follows.

\subsection{Legacy Inverters}\label{sec:LEG}

Legacy inverters $l\in\mathcal{L}$ are not equipped with grid support functions and can operate only at unity power factor. The operating state of legacy inverters is fully captured by a binary variable $u_l$, where $u_l=1$ if the inverter is operational, and $u_l=0$ if it has tripped due to overvoltage; this is referred to as the status of the inverter. The injected active power $P^\text{inj}_l$ is
 \begin{equation}
   P^\text{inj}_l=\begin{cases}
     \textbf{P}^\text{av}_l, & \text{if } u_l=1,\quad \forall l \in \mathcal{L},\\
     0, & \text{otherwise}.
   \end{cases}
 \label{eq:LegacyInv}
 \end{equation}

\subsection{Autonomous Inverters}\label{sec:AUT}

The operating state and output of non-coordinated inverters are dependent only on the local voltage measured by the inverter. An autonomous inverter $a\in\mathcal{A}$ enables voltage regulation using a combination of Volt/VAr and Volt/Watt droop curves (Fig.~\ref{fig:droopcruves}). The feasible region of the full Volt/VAr droop curve is a piecewise linear function with three constant sections and two decreasing ones bounded by voltage setpoints. However, the Volt/VAr function considered here consists of three pieces since we neglect the control of undervoltage for the scenarios studied in this paper. Given the measured inverter output voltage $V_a$, the nominal inverter reactive power injection in the grid operating in Volt/VAr response mode is
\begin{equation}
    Q^{pu}_a =  \textbf{m}^\text{V}V_a+\textbf{c}^\text{V}, \quad \forall a \in \mathcal{A}.
    \label{eq:Q_VV}
\end{equation}
\begin{figure}[t]
\centerline{\includegraphics[width= \linewidth]{FiguresSec41/fig1_InvControl.PNG}}
\caption{Volt/VAr and Volt/Watt droop curves on a per-unit basis. The highlighted areas represent the feasible setpoint space for coordinated inverter control.}
\label{fig:droopcruves}
\end{figure}
The slope $\textbf{m}^\text{V}$ and intercept $\textbf{c}^\text{V}$ of the Volt/VAr droop curve are
\begin{align}
    \textbf{m}^\text{V}=- \textbf{Q}^{\min,pu}/({\textbf{V}^\text{Qmin} - \textbf{V}^\text{DB}}),
    \label{eq:slopeVV}
    \\
    \textbf{c}^\text{V}=\textbf{Q}^{\min,pu}/\textbf{m}{\textbf{V}^\text{Qmin}},
    \label{eq:intercVV}
\end{align}
where $\textbf{V}^\text{DB}$ is the upper reference point of the deadband, and $\textbf{V}^\text{Qmin}$ denotes the voltage level at which the maximum reactive power support is reached.

Recognising that inverters may use reactive power to regulate voltage without compromising the active power injection, autonomous inverters are operated in reactive power priority mode. That means that voltage regulation is first attempted through reactive power support. The Volt/Watt response mode is activated when inverter voltage cannot be maintained within operational limits using reactive power.
The Volt/Watt function linearly reduces active power output until the cut-off voltage $\textbf{V}^{\max}$ is reached, as
\begin{equation}
    P^{pu}_a= \textbf{m}^\text{W}V_a+\textbf{c}^\text{W},\quad \forall a \in \mathcal{A}.
    \label{eq:Q_VW}
\end{equation}
The slope $\textbf{m}^\text{W}$ and intercept $\textbf{c}^\text{W}$ are derived from the line going through the points $(\textbf{V}^{\textbf{Q}_{\min}}, \textbf{P}^\text{av})$ and $(\textbf{V}^{\max}, \textbf{P}^{\min})$. When operating in Volt/Watt response mode, $Q^{pu}_a = \textbf{Q}^{\min,pu}$. Converting to base units, this gives
\begin{equation}
    Q_a = \textbf{S}Q^{pu}_au_a,\quad \forall a \in \mathcal{A},
    \label{eq:Q_AUT}
\end{equation}
where $u_a\in\{0,1\}$ is the inverter ON/OFF state variable. Then, the inverter injected power $P^\text{inj}_a$ operating in Volt/Watt response mode is
\begin{equation}
  P^\text{inj}_a = \min\left\{\sqrt{\textbf{S}^2-Q_a^2},\textbf{P}^\text{av}_aP^{pu}_a\right\}u_a, \quad \forall a \in \mathcal{A}.
 \label{eq:Pinj_AUT}
\end{equation}

\subsection{Coordinated Inverters}\label{sec:CIC}

The optimum active and reactive power target values ($P^\text{inj}_c, Q_c$) for a set of coordinated inverters $c\in\mathcal{C}$ are found by solving the optimisation problem \eqref{eq:Objmain}, which will be introduced in later stages. First, it is subject to a capacity limit
\begin{equation}
(P_c^\text{inj})^2 + Q_c^2 \leq \textbf{S}^2, \quad \forall c \in \mathcal{C}.
\label{eq:CICconst1}
\end{equation}
The injected PV output $P^\text{inj}_c$ from each coordinated inverter is non-negative and limited by the available active power on the secondary side of the PV inverter $\textbf{P}^\text{av}_c$ as
 \begin{equation}
\textbf{0} \leq P^\text{inj}_c \leq \textbf{P}^\text{av}_c, \quad \forall c \in \mathcal{C}.
\label{eq:CICconst2}
\end{equation}
Since we are considering the periods with excess PV output leading to overvoltages, the inverter capability to provide reactive power is ignored. Thus, the maximum reactive power support is limited to
 \begin{equation}
-\textbf{Q}^{\min,pu}\textbf{S} \leq Q_c \leq \textbf{0}, \quad \forall c \in \mathcal{C},
\label{eq:CICconst3}
\end{equation}
as in \cite{Hall2019}, where $\textbf{Q}^{\min,pu}$ is the maximum reactive power that the inverter can absorb from the grid.

\subsection{Non-Exporting Inverters}\label{sec:NonExp}

The last inverter type considered in this study requires inverters $e\in\mathcal{E}$ to be connected in a ``self-supply'' mode. This non-exporting setup prevents reverse power flow at the point of common coupling of the customer premises. The non-exporting mode can be seen as the most neutral future direction for distribution networks since the inverters neither contribute to the voltage rise nor provide grid support functions. It is also arguably the fairest model to the customers as all PV systems are equally affected. The excess PV output has to be either stored locally for later or curtailed. For this study, we only consider curtailment, and leave the evaluation of energy storage for future work. The injected power from non-exporting inverters $P^\text{inj}_e$ is
\begin{equation}
  P^\text{inj}_e=\begin{cases}
  \min\left\{\textbf{P}^\text{av}_{e},\textbf{P}^\text{d}_{e}\right\}, & \text{if } u_e=1,\quad \forall e \in \mathcal{E}, \\
  0, & \text{otherwise},
  \end{cases}
  \label{eq:Non-Exp}
\end{equation}
where $u_{e}\in\{0,1\}$ denotes the status of the inverter.

config: arxiv


TITLE: Let A be some matrix and suppose x ∈kernel A(transpose) and y∈ Im(A). How to show that x and y are perpendicular?
QUESTION [1 upvotes]: I sort of has problem with this question. I assigned a matrix and calculated the span of ImA and kerA(transpose). But no idea what to do after this ?

REPLY [1 votes]: Since $y \in$ Im $A$, $\exists x_0$ s.t. $A x_{0}=y$.
We have,
$A^Tx = 0$
$\implies x^TA = 0$ (take transpose both sides)
$\implies x^TAx_0 = 0$ (right multiply both sides by $x_0$)
$\implies x^Ty = 0$


TITLE: Regression and percentile
QUESTION [1 upvotes]: On a midterm exam, the average is $50$ points (out of $100$) with an SD of $10$ points.  On the final exam, the average was $100$ points (out of $200$) with an SD of 30 points.  The correlation between the two is 0.6.
What percent of the people who got a 40 on the midterm improved their percentile on the final?

The solution given in this case is 31% but it doesn't seem right to me .
I started the question by first standardize the score of 40 on midterm:
$\frac{40-50}{10}=-1$
Thus the same percentile of score on final would be $-1*30+100=70$
I then solve for the new average for the final score of the people who got 40 in the midterm, I get: $-1*0.6*30+100=82$
Finally I solve for the z score of the people who were 1 standard deviation below the new average and use normal approximation to get the percentage of people who improved on their final grade:
$\frac{70-82}{\sqrt{1-0.6^2}*30} = -0.5$
-0.5 corresponds to 30.85% under the curve, we are interested in people who improved in their final score therefore we get: 100%-30.85%=69.15%
Is my solution correct in this case??

REPLY [0 votes]: I will solve this according to the least-squares regression line equation.
The linear relation between $2$ variables $x$ & $y$ is given by the equation for the regression line,$$ Y= \beta_o+\beta_1Χ+e.....(1)$$
Where $Y$ is the dependent variable
$Χ$ is the independent variabl
$\beta_o$ is the y-intercept.It is the value of Υ for Χ=0
$\beta_1$ is the slope of the line.It gives the amount of change in $Y$ for every unit change in $Χ$.
$e$ is the random error.
The predicted value of $Y$ for a given value of $Χ$ is denoted by $Y^{\prime}$
$$Y^{\prime}=a+bx.....(2)$$where $a= \beta_0^{\prime}=$ the estimated $y$-intercept of the regression line
and $b=\beta_1^{\prime}=$ the estimated slope of the regression line.
Regression line of $y$ on $x$ is $y-\overline{y}=byx(x-\overline{x}).....\overline{y}$ is the mean of $y,\overline{x}$ is the mean of $x.....(3)$
we consider $x$ is the midterm exam score
$y$ is the final exam score
Given midterm mean $=50=\overline{x},\overline{y}=100$
and the standard deviation for $x=10$ and $y=30$ $$byx=r\times\dfrac{\sigma}{\sigma x}=0.6\times\dfrac{30}{10}=1.8$$
Plug in these values in $(3)$, we get
$$y-50=1.8(x-50)$$
$$y=1.8x+10$$$\beta_0$ is the $y$-intercept $=10$
$\beta_1$ is the slope of the line $=1.8$
Now, plug these in $(1)$, we get
If $X=40$ then
$Y^{\prime}(40)=82$
Therefore, the predicted final exam score is $82$


TITLE: Laurent's series and finding the image of a complex function
QUESTION [0 upvotes]: Let $f$ be a complex function $f(z) = \frac{z^2 +1}{z(z-i)^3}$. It needs to be put jnto Laurent's series, so that it will converge in a set $D = \{ g(z) \mid Re(z) > 0 \}$, where  $g(z) = \frac{2iz}{z-1}$.
I've been struggling with finding this set $D$. So the singularities for $f$ are $z = 0$ and $z = i$. Where should I even have the center of the series?

REPLY [1 votes]: One option is to write  $g$ as a composition of “simple” transformations:
$$
 g(z) = 2i \left( 1 + \frac{1}{z-1}\right)
$$
and then successively determine the image of the right half-plane.
Another option is to use that fact that $g$ is a Möbius transformation: It maps the imaginary axis to a circle (or line), and that image is uniquely determined by the image of three distinct points. From
$$
 g(-i) = -1+i, g(0) = 0, g(+i) = 1+i
$$
you can conclude that the imaginary axis is mapped to the circle with center $i$ and radius $1$. It follows that the right half-plane is mapped to the inside or outside of that circle. Since $g(1) = \infty$, we have
$$
 D = \{ g(z) : \operatorname{Re}(z) > 0  \} = \{ w : |w-i| > 1 \} \, .
$$
That is an annulus with center $i$, inner radius $1$ and outer radius $\infty$. Your task is now to determine the Laurent series of $f$ in that annulus.


TITLE: Stationary distribution and limiting distribution of a random walk
QUESTION [2 upvotes]: Consider the random walk on $S={0, . . . , N}$, defined as  follows.   We  are  given $p, q, r >0$  with$ p+q+r=  1$.   The  walk  increases by 1 with probability $p$, decreases by 1 with probability $q$, and stays put with probabity $r$.What is its stationary and limiting distribution?
I writed out the transition matrix.
$$
    \begin{pmatrix}
    r & p & \cdots &⋯ &⋯&q\\
    q & r & p &\cdots&\cdots&\cdots \\
    0 & q & r &p &\vdots&\vdots\\
    0 & 0 & q &r &p &⋯\\
    p & 0 & 0 &0 &q & r \\
    \end{pmatrix}
$$
the stationary matrix satisfies $$\pi=\pi p$$
then it satisfies $$p_{i,j} \pi_i = p_{j,i} \pi_j$$
$$p\pi_1 = q\pi_2, p\pi_2 = q\pi_3, p\pi_3 = q\pi_4...$$
the stationary situation then becomes
$$
    \begin{pmatrix}
    1 & \frac{p}{q}\pi_1 & (\frac{p}{q})^2\pi_1 & (\frac{p}{q})^3\pi_1 &...&(\frac{p}{q})^N\pi_1\\
    \end{pmatrix}
$$
We can then get the distribution because it adds up to 1.
Is my method correct?
Then, for the limiting distribution, I think there is no limiting distribution because it seems not to be converging to a specific state. Is it true? Thank you!

REPLY [2 votes]: The stationary measure will be the uniform measure, which you could see by a soft argument: the transition matrix is irreducible and aperiodic, and thus has a stationary measure; furthermore, it is invariant under the map $k \mapsto k+1$, implying that each state must get the same measure.
You could also show this manually: let $\mu_j = \frac{1}{N+1}$ for each $j \in \{0,1,\ldots,N\}$.  Then $$(P \mu)_j = p \mu_{j-1} + r \mu_{j} + q \mu_{j+1} = \frac{p + r + q}{N+1} = \frac{1}{N+1} = \mu_j$$
where the indices are $\mod{N+1}$.

EDIT: some further comment on the invariance thing; let $A$ be the map that cycles the states, i.e. $A (x_0 ,\ldots, x_{N})^T = (x_N,x_0,x_1,\ldots,x_{N-1})^T\,.$
Then note that $PA = P$.  If you set $\mu$ to be the stationary measure of $P$ (i.e. $\mu P = \mu$), then it must be invariant under $A$: $$\mu PA = \mu P \implies \mu A = \mu\,.$$  Iteratively applying $A$ shows that $\mu$ must be uniform.


text:
\begin{document}

\title{On the concentration of the number of solutions\\
of random satisfiability formulas}

\author{Emmanuel Abbe\thanks{Department of Computer and Communication Sciences, EPFL, Switzerland. Email: emmanuel.abbe@epfl.ch} \;\;\; and\;\;\;
Andrea Montanari\thanks{Department of Electrical Engineering,
and Department of Statistics,
Stanford University. Email: montanari@stanford.edu}}

\date{}

\maketitle

\begin{abstract}
Let $Z(F)$ be the number of solutions of a random $k$-satisfiability
formula $F$
with $n$ variables and clause density $\alpha$.
Assume that the probability that $F$ is unsatisfiable is
$O(1/\log(n)^{1+\e})$ for $\e>0$.
We show that (possibly excluding a countable set of
`exceptional' $\alpha$'s) there exists a non-random function
$\phi(\alpha)$ such that, for any $\delta>0$,
$(1/n)\log Z(F)\in [\phi-\delta,\phi+\delta]$ with high probability.
In particular, the assumption holds for all $\alpha<1$,
which proves the above concentration claim in the whole
satisfiability regime of random $2$-SAT. We also extend
these results to a broad class of constraint satisfaction problems.

The proof is based on an interpolation technique from spin-glass
theory, and on an application of Friedgut's theorem on sharp thresholds for graph properties.
\end{abstract}

\section{Introduction and main results}

Over the last twenty years, a considerable effort
has been devoted to understanding the typical properties
of random $k$-satisfiability ($k$-SAT) instances.
This line of work was initially motivated by the empirical
discovery of a striking relation between the
running time of standard solvers and the
proximity to the `satisfiability threshold'.
Since then, an important motivation for these investigations
has been to develop better heuristics to cope
with hard constraint satisfaction problems.

Significant progress has been made along this path.
In particular, it was estabilished early on that, indeed, the probability
that a random instance is unsatisfiable increases sharply from $0$
to $1$ when the `clause density' (number of clauses per variable)
crosses a critical threshold \cite{Friedgut}.
This phenomenon is referred to as the satisfiability phase transition
or satisfiability threshold.
The critical density might a priori depend on the number of variables,
although upper and lower bounds are known to match up to a
multiplicative constant that goes to $1$ when $k$ increases \cite{AchlioetalNature}.

On the other hand, a significantly more detailed
picture has been conjectured, building on non-rigorous techniques
from statistical physics such as the replica and cavity methods
\cite{Monasson99Nature,MezParZec_science,OurPNAS,MMBook}.
In particular, not only an $n$-independent
critical density is conjectured to exist,
but explicit values (depending on $k$) were computed in
\cite{MezParZec_science}. This type of analysis also lead to
an intriguing picture of the geometry of the set of solutions
of a random satisfiability instance
\cite{MezParZec_science,OurPNAS,MMBook}. While only a small
subset of these results have been estabilished rigorously,
they provided guidance and stimulus for exciting rigorous
developments \cite{CojaAchlioptas}.

In this paper we explore the most basic property
of the set of solutions of a random $k$-SAT instance:
its cardinality.
The problem of computing the number of solution of a $k$-satisfiability
($k$-SAT) formula is well known to be $\#$P-complete for
any $k\ge 2$. Even worse, there is no fully polynomial
randomized approximation scheme (FPRAS) to approximapte
the number of solutions  unless NP = RP \cite{DyerComplexity}.

Here we are instead interested in asymptotic estimates of the number of
solutions of a random formula, in the following sense.
Let $F(n,\alpha)$ denote a formula on
$n$ variables with clause density $\alpha$ (each clause having $k$ literals) and let
$Z(F(n,\alpha))$ be the number of solution of $F(n,\alpha)$.
From a statistical physics viewpoint, one of the most basic conjectures
is that $Z(F(n,\alpha))$  concentrates on the exponential scale.
Namely, for each $\alpha<\alpha_{\rm s}(k)$
(the satisfiability threshold), there exists $\phi = \phi(\alpha)$ non-random
such that, for any $\delta>0$,
$2^{n(\phi-\delta)}\le Z(F(n,\alpha)) < 2^{n(\phi+\delta)}$ with high probability.
In formula

\begin{eqnarray}

\lim_{n\to\infty}
\pp\{2^{n(\phi-\delta)}\le Z(F(n,\alpha)) < 2^{n(\phi+\delta)}\} = 1\, .
\label{eq:Conjecture}

\end{eqnarray}

While for arbitrary $k \geq 3$ we cannot establish this conjecture for any $\alpha$ below the satisfiable threshold (which would imply
in particular the existence of an $n$-independent
satisfiability threshold), we are able to prove
that (\ref{eq:Conjecture}) holds for any $\alpha$ such that
 $\pp\{Z(F(n,\alpha))=0\}$ (the unsatisfiability probability)
is upper bounded by $1/(\log n)^{1+\e}$ for some $\e>0$ and all $n$ large enough.
In particular, this establishes the
conjecture for $k=2$ (in the entire satisfiable phase), and for $k\ge 3$ and $\alpha<1$.
A verification of previous assumption for arbitrary values of $k$ and $\alpha$ up to the so-called `dynamic threshold' is also discussed.

We further generalize these results to a broad family
of constraint satisfaction problems.

One basic difficulty in estabilishing (\ref{eq:Conjecture})
is that the concentration of $\log Z(F)$ cannot be proved using
standard martingale methods. Such an argument typically requires
to control the difference $|\log Z(F')-\log Z(F)|$ for
$F$ and $F'$ differing in a single clause \cite{Ledoux}.
Unfortunately, adding a single clause can change the value of
$Z(F)$ from exponentially large to $Z(F') = 0$.

In order to overcome this difficulty, we use Friedgut's theorem to
prove that the property $Z(F)\le 2^{n\phi}$ has a sharp threshold.
We then translate this result into a concentration statement by proving
that $(1/n)\E\log(1+Z(F))$ converges, using an interpolation technique.
This is where the condition on the decay of the unsatisfiability probability is used, since the interpolation technique does not lead to a formal superadditivity property in this setting.

The interpolation method we use was first developed by
Francesco Guerra and Fabio Toninelli in the context of spin glass theory
\cite{GuerraToninelliLimit} and then generalized to a number of problems
from statistical physics, computer science
\cite{FranzLeone,FranzLeoneToninelli,PanchenkoTalagrand}
and coding theory \cite{Montanari05,Kudekar}.
A nice suite of combinatorics applications were
recently presented in \cite{BayatiInterpolation}.

In particular, the interpolation technique was applied to random $k$-SAT
in \cite{FranzLeone,FranzLeoneToninelli}, to show that a suitable normalized log-partition function
has a limit for $n\to\infty$.
The same work implies that the minimum number of unsatisfied
clauses per variable has a limit as well.
Let us emphasize a crucial difference between such results and
the conjecture (\ref{eq:Conjecture}):
The quantities (normalized log-partition function or optimal cost)
considered in  \cite{FranzLeone,FranzLeoneToninelli}
and following-up work are easily proved to
concentrate via martingale methods. This considerably simplifies those proofs.

\vspace{0.2cm}

{\bf Acknowlegements.}
The present research was initiated as far back as 2006, and remained dormant
for a long period of time. Some results were presented at the
`DIMACS Working Group on Message-Passing Algorithms' in October 2008.
We were finally motivated to polish and publish the manuscript
after discussions with Mohsen Bayati, David Gamarnik and Prasad Tetali
regarding their recent paper  \cite{BayatiInterpolation}. It is a pleasure
to thank them. The first author also thanks Emre Telatar for a careful reading
of some parts of the manuscript and for useful suggestions.

This work was partially supported by a Terman fellowship,
the NSF CAREER award CCF-0743978 and the NSF grants DMS-0806211,
CCF-0915145.

\section{Random $k$-SAT}

\begin{definition}
A $k$-clause is a disjunction of $k$ Boolean variables or their negations.
Let $C_k(n)$ be the set of all $ N={n \choose k} 2^k $ possible
$k$-clauses on $n$ Boolean variables.

We denote by $F_k(n,\alpha)$ a random formula which is formed by selecting
independently each element in $C_k(n)$ with probability $p_k(n,\alpha)=
\alpha n/({n \choose k} 2^k)$, and by taking the conjunction of
the selected clauses.
\end{definition}

The number of clauses in the above model is a binomial random variable,
which concentrates exponentially fast around its expectation $\alpha n$.
Some of our computations prove to be simpler within slightly
different models, whereby the number of clauses is either
Poisson or deterministic with the same mean  $\alpha n$.
Standard monotonicity arguments can be used to show
the equivalence of these models for our purposes and we will hence
switch freely between these different models.

Unless specified, the value of $k$ will remain fixed throughout the paper
and the $k$ subscript is dropped.

We denote by $Z(F)$ the number of satisfying assignments   (solutions) of a
Boolean formula $F$ and by

\begin{eqnarray*}

P_n(\alpha,\phi) := \pp\{Z(F(n,\alpha)) < 2^{n\phi}\}\, ,

\end{eqnarray*}

the probability that a random formula has less than $2^{n\phi}$
satisfying assignments.

\begin{definition}
Note that $P_n(\alpha,0)=
\pp\{Z(F(n,\alpha)) =0 \}$ is the probability that $F(n,\alpha)$
is unsatisfiable (UNSAT). We define

\begin{eqnarray*}

\alpha^* := \sup \big\{ \,\alpha :\;
P_n(\alpha,0) = O(1/(\log n)^{1+\e}), \mbox{ for some }\e>0 \big\}\, .

\end{eqnarray*}

\end{definition}

\noindent
{\bf Remark 1.} the results in this paper still hold when defining
$\alpha^*$ to be $\sup \{ \alpha : \sum_n P_n(\alpha,0)/n
< \infty\}$. However,
the above definition simplifies the proofs without much loss of generality.

\noindent
{\bf Remark 2.} For any $k\ge 2$, we have $\alpha^* \geq 1$.
Indeed, considering the case of $2$-SAT, \cite{DeLaVega}
proves that, for $\alpha < 1$, $\pp \{Z(F_2(n,\alpha)) =0\}=O(1/ n)$.
Since for any $k \geq 3$, $\pp \{Z(F_k(n,\alpha)) =0\}
\leq \pp \{Z(F_2(n,\alpha)) =0\}$, we conclude
$$\alpha^* \geq 1$$
for $k\ge 3$ as well.

Unfortunately, the bounds on the satisfiability threshold
based on the second moment method
\cite{AchlioetalNature} do not imply any quantitative estimate
on the probability that $F_k(n,\alpha)$ is UNSAT.
It might be possible to prove such an estimate by a careful
analysis of specific solution algorithms. In particular,
a careful analysis of the recent algorithm \cite{Coja09}
might lead to a proof of
$\alpha^* \geq 2^k(1-\delta) \log k/k$, for $k$ large enough \cite{COPersonal}.
We expect that $\alpha^*$ does coincide with the satisfiability
threshold.

\vspace{0.2cm}

Our main result estabilishes the conjecture (\ref{eq:Conjecture})
for $\alpha<\alpha^*$, apart possibly for countably many
`special' values of $\alpha$.

\begin{thm}\label{res}
There exist a countable set $\Cble\subseteq[0,\alpha^*)$,
and $\phi_s: [0,\alpha^*)\to [0,1]$
such that the following holds.
For any $\e>0$ and any $\alpha \in [0,\alpha^*)\setminus\Cble$, we have

\begin{eqnarray*}
\lim_{n \rightarrow \infty} P_n(\alpha,\phi_s(\alpha)- \varepsilon) &=& 0, \\
\lim_{n \rightarrow \infty} P_n(\alpha,\phi_s(\alpha) + \varepsilon) &=& 1.
\end{eqnarray*}
\end{thm}

In order to prove this result, which is done in Section \ref{proof3}, we first show the following sharp
threshold result.
\begin{thm}\label{fried}
For any $\varepsilon>0$ and $\phi \in [0,1)$, there exists $\{\alpha_n(\phi)\}_{n \in \mZ_+}$ such that
\begin{eqnarray*}
\lim_{n \rightarrow \infty}P_n(\alpha_n(\phi)  - \varepsilon, \phi) &=& 0, \\
\lim_{n \rightarrow \infty}P_n(\alpha_n(\phi)+ \varepsilon, \phi) &=& 1.
\end{eqnarray*}
\end{thm}
In other words, for any fixed $\phi$, the property
$\{ Z(F(n,\alpha)) < 2^{n\phi} \}$ has a sharp threshold in $\alpha$.
In \cite{Friedgut}, this result is shown for $\phi=0$,
 i.e., for the property $\{F(n,\alpha)
\text{ is UNSAT} \}$. As for the $\phi=0$ case, Theorem \ref{fried} is proved by showing that the monotone property $\{ Z(F(n,\alpha)) < 2^{n\phi} \}$ cannot be approximated by a ``local property'', and hence must have a sharp threshold.
For greater convenience of the reader, and in order to introduce
relevant notations, we reproduce in Section \ref{sec:Friedgut} the Friedgut's Theorem on sharp thresholds for monotone properties
\cite{Friedgut}. The proof of Theorem \ref{fried} is differed to Section \ref{prooffried}.

In order to prove Theorem \ref{res}, we then would like
to transfer the threshold in $\alpha$ (Theorem \ref{fried}) into
a threshold in $\phi$. This step is however not straightforward because of the little knowledge we have about the function $\phi \mapsto \alpha_n(\phi)$.
In order to establish this threshold transfer,
we first prove the following result, which shows  the existence
of the  $n\to\infty$ limit of the normalized logarithm of
the number of solutions when $\alpha < \alpha^*$.

\begin{thm}\label{convpsat}
Let $$ \psi_n (\alpha) :=  \frac{1}{n} \E [ \log Z(F(n,\alpha)) | Z(F(n,\alpha)) \geq 1 ] .$$
We have that $\psi_n (\alpha)$ converges to a limit $\phi_s(\alpha)$, for every $\alpha < \alpha^*$.
\end{thm}

This theorem is proved in Section \ref{proof2} and the key step in the proof consists in establishing the following pseudo-superadditivity property.

\begin{lemma}\label{guerra}
For any $\alpha$, let $Z_{n}:= Z(F(n,\alpha))$, we then have for any $n_1,n_2 \geq k$
\begin{align*}
 \E  \log (1+Z_{n_1+n_2}) \geq  \E \log (1+Z_{n_1} Z_{n_2})\, .
\end{align*}

\end{lemma}

The proof of Lemma \ref{guerra} is differed to Section \ref{proof2} and is based on the interpolation technique by
Guerra and Toninelli \cite{GuerraToninelliLimit}, and Franz-Leone
\cite{FranzLeone}. However, while in those cases
one obtains superadditivity  of $\E\log Z$, in the present case we
get a weaker result because of the ``1+'' term.
Hence, in this case the existence of the infinite volume
limit is not a straightforward consequence of Lemma \ref{guerra}.
Notice that this problem is intimately related to the fact that
$Z=0$ with positive probability, and therefore $\E\log Z$
is not defined. We use the fact that $\alpha < \alpha^*$ to circumvent this problem in Theorem \ref{convpsat}.

Finally, Theorem \ref{convpsat} implies the following result, which allows the complete the threshold transfer.
\begin{lemma}\label{conva}
For each $n$ and $\phi$, let

$\alpha_n (\phi)$ be such that $P_n(\phi,\alpha_n (\phi)) =1/2$.

We have that $\alpha_n(\phi)$ converges for almost every $\phi \in \phi_s(\A)$ (where $\phi_s$ is defined in Theorem \ref{convpsat}).
\end{lemma}
Lemma \ref{conva} is proved in Section \ref{proof3}.

\section{A general family of random CSP's}

In this section, we extend the results of the previous section to a general
family of random constraint satisfaction problems (CSP) over binary variables.
An ensemble in this family is defined as follows.

\begin{definition}
 Let $\mu$ be a distribution over Boolean functions $\varphi:\{-1,+1 \}^k\to\{0,1\}$, which we call the
{\it clause type distribution}. Let $n$ be an integer and $\alpha \in \mR_+$.
A random formula from the ensemble $F_k(\alpha, n, \mu)$
is drawn as follows. For each $a\in\{1,\dots,m=\lfloor \alpha n \rfloor\}$
the $a$-th clause is drawn independently from previous
ones. For  clause $a$, $k$ indices $i_1(a),\ldots,i_k(a)$ are drawn
independently and uniformly at random in $[n]$. Further
$\vp_a: \{-1,+1 \}^k \rightarrow \{0,1\}$ is drawn under the distribution
$\mu$, producing the clause $\vp_a(x_{i_1(a)}, \ldots,x_{i_k(a)})$.

An assignment $x\in\{+1,-1\}^n$ is said to
\emph{satisfy} the formula $F_k(\alpha, n, \mu)$
if, for each $a\in [m]$,   we have $\vp_a(x_{i_1(a)}, \ldots,x_{i_k(a)})=1$.
\end{definition}

As in previous section, we will often drop the subscripts $k$ in the following.
Further, $Z (F)$ is the number of satisfying assignments of formula $F$
and we define

\begin{eqnarray*}

P_n(\alpha, \phi, \mu) := \pp\{ Z(F(\alpha, n, \mu)) < 2^{n\phi} \}\, .

\end{eqnarray*}

\begin{definition}
Note that $P_n(\alpha,0, \mu)=
 \pp\{Z(F(\alpha,n, \mu)) =0 \}$ is the probability that $F(\alpha,n, \mu)$
is UNSAT. We define

\begin{eqnarray*}

\alpha^*(\mu) := \sup \big\{ \alpha :\;
 P_n(\alpha,0, \mu) = O(1/(\log n)^{1+\e})
\mbox{ for some }  \e>0 \big\}\, .

\end{eqnarray*}

\end{definition}

\begin{definition}
For $\vp: \{-1,1\}^k \rightarrow \{0,1\}$ and $\theta \in [-1,1]$, let
$$\| \vp \|_\theta^2 = \sum_{x \in \{-1,1\}^k} \vp(x)^2 v_\theta (x)\quad
\mbox{ and }\quad \| \vp \| = \| \vp \|_0$$
where $$v_\theta (x) = \prod_{i=1}^k \frac{1+x_i\theta}{2}.$$
\end{definition}
Note that $\| \vp \|_\theta^2$ is the probability that $\vp=1$ under the measure $v_\theta$, which assigns probabilities $(1-\theta)/2$ and $(1+\theta)/2$ to $-1$ and $+1$ respectively.

Our CSP ensemble is specified by the distribution $\mu$ over clause types and

we now describe two set of hypotheses on this distribution.

\vspace{0.2cm}

\noindent
{\bf H1.} (a) \emph{Dominance of balanced assignments.} For every
$\theta \in [-1,1]$,
$\E_\vp \log \| \vp \|_\theta \leq \E_\vp \log \| \vp \|,$
with equality only if $\theta=0$. This condition implies that, in a typical random instance, most solutions have almost as many $1$'s as $-1$'s.\\
(b) \emph{Unsatisfiability of uniform assignments.} For every $s\in\{-1,+1\}$,
there is at least one clause $\vp$ with $\mu(\vp)>0$ such that $\vp
(s, \ldots, s)=0$.

\vspace{0.2cm}

\noindent{\bf H2.} \emph{Convexity of $\Gamma_l$.} Let $M_1(\{-1,1\}^l)$ be the set of probability measure on $\{-1,1\}^l$ and let
\begin{align*}
\Gamma_l: \, M_1(\{-1,1\}^l) &\to \mR\,

\end{align*}

be defined by

\begin{align}

 \Gamma_l(\nu):=
\E_\vp \E_{Z^{(r)}} \prod_{r=1}^l (1-\vp(Z^{(r)})) \label{cond1}

\end{align}

where $Z^{(r)}$ are Boolean random vectors of dimension $k$ such that $Z_i=(Z_i^{(1)},\ldots, Z_i^{(l)})$, $i=1,\ldots,k$, are i.i.d. with distribution $\nu$, whereas $\vp$ is a random clause type drawn under distribution $\mu$.
Under H2, $\mu$ is assumed to make $\Gamma_l$ convex for for any $l \geq 1$.

\vspace{0.2cm}

Notice that conditions H1.(a) and H1.(b) coincide with  conditions 4 and 5
in \cite{MRT09}.
Further, hypothesis H1 is satisfied by a number of interesting
random CSP ensembles. Such examples include

\begin{itemize}

\item $k$-NAE-SAT, where $\vp(x)=\vp_s(x)= \1(x \notin \{-s,s\})$ and $\mu(\vp_s)=2^{-k}$ for each $s \in \{-1,1\}^k$;
\item Hypergraph $2$-coloring, where
$\vp(x)= \1(x \notin \{\underline{-1};
\underline{+1}\})$ is the unique clause in the support of $\mu$, with $\underline{-1} = (-1,\dots,-1)$
and  $\underline{+1} = (+1,\dots,+1)$;
\item $k$-XOR-SAT, where $\vp(x) = \vp_s(x) = \1(\prod_{i=1}^k x_i =s)$ and
$\mu(\vp_s)=1/2$ for each $s\in \{-1,1\}$;
\item $k$-SAT, where $\vp(x)=\vp_s(x)= \1(x \neq s)$ and $\mu(\vp_s)=2^{-k}$ for each $s \in \{-1,1\}^k$.

\end{itemize}

For the first three examples above, it is checked in
\cite{MRT09} that hypothesis H1 is satisfied.
 Let us check that this is the case for $k$-SAT as well. Note that

\begin{align*}

\E_s \| \vp \|_\theta^2= 1-\E_s  \prod_{i=1}^k \frac{1-s_i \theta}{2} =1-2^{-k} = \E_s \| \vp \|^2,

\end{align*}

hence

\begin{align*}
\E_s \log \| \vp \|_\theta^2 \leq \log \E_s \| \vp \|_\theta^2 =\log \E_s \| \vp \|^2 =  \E_s  \log\| \vp \|^2.
\end{align*}

This verifies condition H1.(a). Condition H1.(b) holds trivially.

Hypothesis H2 is not straightforward to check.
The next definition characterizes a family of clause type distributions
satisfying it.

\begin{definition}[$k$-factorizing distributions]
A clause type distribution $\mu$ is said to {\it k-factorize} if it has the following structure. There exists an integer $J \geq 1$,
such that any $\vp \in \supp (\mu)$ is of the form

\begin{align}
\vp (x) = \1 (x \notin \{s^{(1)}, \ldots, s^{(J)}\}),
\end{align}

for some $s^{(1)}, \ldots, s^{(J)} \in \{-1,1\}^k$,
and
\begin{align}
\mu ( \vp) = \prod_{i=1}^k \bar{\mu} (s_i^{(1)}, \ldots, s_i^{(J)}) \label{cond2}
\end{align}
where $\bar{\mu}$ is a probability distribution on $\{-1,1\}^J$.
In other words, the vectors $(s_i^{(1)}, \ldots, s_i^{(J)})$, for $i=1,\ldots,k$, can have correlated components but are mutually i.i.d. with distribution $\bar{\mu}$.
\end{definition}

This definition can be generalized by letting  $J$
itself to be random, but we stick to the above case for the sake of simplicity.

The class of $k$-factorizing clause type distributions includes, among other problems:
\begin{itemize}

\item $k$-NAE-SAT: $\vp(x) = \1 (x \notin \{-s,s \})$
for $s\in\{-1,+1\}^k$ uniformly random. This is $k$-factorizing
with $\bar{\mu}(-1,1)  = \bar{\mu}(1,-1)=1/2$;

\item Hypergraph $2$-coloring: $\vp(x) = \1 (x \notin \{\underline{-1},\underline{+1}\})$ with $\bar{\mu}(-1,1)  = 1$;

\item $k$-SAT: $\vp(x) = \1 (x \notin \{s\})$ with $\bar{\mu} (1) = \bar{\mu}(-1) = 1/2$.

\end{itemize}

Condition H2 is satisfied by $k$-factorizing distributions as stated formally
below.

\begin{lemma}\label{ch1}
The mapping $\Gamma_l$ is convex for any $l \geq 1$ if the clause type distribution k-factorizes.
\end{lemma}

Note that $k$-XOR-SAT does not belong to this class of distributions, nevertheless, condition H2 holds in this case as well, as stated below.

\begin{lemma}\label{ch2}
The mapping $\Gamma_l$ is convex for any $l \geq 1$ for k-XOR-SAT with k even.
\end{lemma}
The proofs of Lemma \ref{ch1} and Lemma \ref{ch2} are differed to Section \ref{proof4}.
We now state the equivalent of Theorem \ref{res} for this general class of
CSPs.

\begin{thm}\label{res2}
Assume $\mu$ to satisfy conditions H1 and H2. Then there exists a countable
set $\Cble$ and a function $\alpha\mapsto \phi_s(\alpha)$
such that, for any  $\alpha \in [0,\alpha^* (\mu))\setminus\Cble$,
and  any $\e>0$,

\begin{eqnarray*}
\lim_{n \rightarrow \infty} P_n(\alpha,\phi_s(\alpha)- \varepsilon, \mu) &=& 0, \\
\lim_{n \rightarrow \infty} P_n(\alpha,\phi_s(\alpha) + \varepsilon, \mu) &=& 1.
\end{eqnarray*}
\end{thm}

As in previous section, the proof of this theorem is based on the following
two theorems.

\begin{thm}
For any $\mu$ satisfying H1 and $\phi \in [0,1)$,
there exists $\{\alpha_n(\phi)\}_{n \in \mZ_+}$ such that for any $\e>0$,
\begin{eqnarray*}
\lim_{n \rightarrow \infty}P_n(\alpha_n(\phi)  - \varepsilon, \phi, \mu) &=& 0, \\
\lim_{n \rightarrow \infty}P_n(\alpha_n(\phi)+ \varepsilon, \phi, \mu) &=& 1.
\end{eqnarray*}
\end{thm}

This theorem is proved in \cite{MRT09}. Although in that paper
a larger set of conditions was required in the formal statement,
by simple inspection of the proof it follows that only conditions
H1.(a) and H1.(b) were in fact used.

\begin{thm}\label{convcsp}
Let $$ \psi_n (\alpha) :=  \frac{1}{n} \E [ \log Z(F(\alpha,n, \mu)) | Z(F(\alpha,n, \mu)) \geq 1 ] .$$
For any $\mu$ satisfying H2 and for any $\alpha < \alpha^*(\mu)$,
$\psi_n (\alpha)$ converges to a limit $\phi_s(\alpha)$.

\end{thm}
The proof of this
theorem is based on the following pseudo-superaddittivity lemma.

\begin{lemma}\label{guerracsp}
For any $\alpha$ and $\mu$ satisfying H2, let $Z_n:=Z(F(\alpha,n, \mu))$, we then have for any $n_1,n_2 \geq k$,
\begin{align*}
 \E  \log (1+Z_{n_1 + n_2}) \geq  \E \log (1+Z_{n_1} Z_{n_2}).
\end{align*}
\end{lemma}
The proof of Lemma \ref{guerracsp} is differed to Section \ref{proof4}. The proofs of Theorem \ref{convcsp} and Theorem \ref{res2} follow the same analytical arguments as for the $k$-SAT case and hence we omit the details here and refer to the proofs of Theorem \ref{convpsat} and Theorem \ref{res}.

\section{Proofs}

\subsection{Friedgut's theorem: A reminder}
\label{sec:Friedgut}

Let $\prob_p$ be the product measure on
$\{0,1\}^N$, such that for $(z_1,\ldots,z_N) \in \{0,1\}^N$,
$\prob_p (z_1,\ldots,z_N) = p^{\sum_{i=1}^N z_i}  (1-p)^{N -
\sum_{i=1}^N z_i}$.
If $N= \binom{n}{k}2^k$, the space $\{0,1\}^N$ is isomorphic
to the space of $k$-CNF formulas on $n$ variables and the measure $\pp_p$ is the measure used in this paper for random $k$-SAT.
We say that $A \subseteq \{0,1\}^N$ is monotone if, for any
$x \in \{0,1\}^N, y \in A$, $x \leq y$ (component-wise) implies
$y \in A$. Such a collection of vectors $A$ is also called a monotone
property, and note that by monotone it is really meant here monotonically
increasing.
When $N= \binom{n}{k}2^k$, a property is called \emph{symmetric}, equivalently called a formula property, if it is invariant under $k$-CNF
formula automorphisms (invariant under the action of the wreath product of the symmetric group $S_n$ with $k$ copies of $\mZ_2$).
For instance, the UNSAT property of a  $k$-CNF formula
is monotone and symmetric.

Note that for $N$ fixed and a monotone symmetric property $A=A_N$, the
function $p \mapsto \pp_p (A)$ is a monotonically increasing
polynomial with $\pp_0 (A)=0$ and $\pp_1 (A)=1$
(strictly increasing if $A\neq \{0,1\}^N$).
We can therefore define for
$\e \in [0,1]$, $p_\e$ such that $\pp_{p_\e} (A) = \e$ and we call
$\delta(\e) = p_{1-\e} - p_\e$ the {\it critical interval}.
We then say that the property $A$ (or the sequence of properties $A_N$)
has a {\it sharp threshold} if for all $\e \in (0,1)$, $\delta(\e)/p_{1/2}$
tends to zero when $N$ increases.
Note that for a monotone property $A$ having a
sharp threshold, there exists $\hat{p}\in(p_{\e},p_{1-\e})$
such that $\pp_p(A) \rightarrow 0$ if $p \leq \hat{p}(1-\e)$ and
$\pp_p(A) \rightarrow 1$ if $p \geq \hat{p}(1+\e)$.

If instead, for $\e \in (0,1)$,
$\delta(\e)/p_{1/2}$ is bounded away from zero, we say that $A$ has a
{\it coarse threshold}.
In typical examples, $\delta(\e)/p_{1/2}$ is bounded and hence admits
subsequential limits. In order to prove a sharp threshold for monotone properties,
it is therefore sufficient to rule out the case of a coarse threshold
(modulo reducing to subsequences).
If $A$ has a coarse threshold, then there exists $p^*\in(p_{\e},p_{1-\e})$
such that  $p^* \cdot  \frac{\partial}{\partial p} \pp_p (A)\big|_{p^*}
\leq 1/C$ uniformly in $N$.

For a given formula $F$, we denote
by $|F|$ the number of clauses in $F$. The average degree of a formula $F$ is the ratio between the number of variables and the number of clauses in $F$, and a balanced formula is a formula whose average degree is no less than that of any sub-formula. We also denote by $E(F)$ the expected number of sub-formulas isomorphic to $F$ in a random formula.

We now state the result of Friedgut.
\begin{thm}[\cite{Friedgut}]\label{thm:GeneralSharp}
Let $0<\beta<1$. There exist functions $B=B(\epsilon,c)$, $b_1=b_1(\epsilon,c)$ and $b_2=b_2(\epsilon,c)$ such that for any $N$, $c>0$, $\e>0$, $p$ and any monotone symmetric family $A$ of $k$-CNF formulas with $n$ variables such that
$p \cdot \frac{\partial}{\partial q} \pp_q (A)\big|_p \leq c$ and $\beta < \pp_p(A) <1-\beta$,
there exists a formula $G$ satisfying:

\begin{itemize}
\item $G$ is balanced
\item $b_1 < E (G) < b_2$
\item $|G| \leq B$
\item $\pp_p \{ A|G \} \geq 1-\epsilon$, where $\pp_p \{ A|G \}$ denotes the probability that a random formula belongs to $A$ conditioned on the appearance of a specific copy of $G$ in the random formula.
\end{itemize}
\end{thm}
Note that conditioning of the appearance of a formula satisfying the above hypothesis is not the same as conditioning on a specific copy of such a formula (with specified variables).

\subsection{Proof of Theorem \ref{fried}}\label{prooffried}

In order to prove Theorem \ref{fried}, let us assume that the property $\{ Z(F(n,\alpha)) < 2^{n \phi} \}$ has a coarse threshold. Let us consider $p_{\rm s}$ in the critical interval of this property such that $p_{\rm s} \cdot \frac{\partial}{\partial p} \pp_p (A)\big|_{p_{\rm s}} \leq c$
and define $p_{\rm s}=\alpha_{\rm s} n/N$.
By compactness there exists a subsequence of size $n$
along which $P_n(\alpha_{\rm s},\phi )$ has a limit $P_{\infty}=[\e,1-\e]$.
For simplicity of notation, consider $P_{\infty}=1/2$ (the
general case is completely analogous).
We can hence assume w.l.o.g. that $\alpha_{\rm s}=\alpha_{\rm s}(n)$ is such
that $ P_n(\alpha_{\rm s},\phi )= 1/2$ (and $\alpha_{\rm s}$ is well defined since $P_n$ is continuous and strictly increasing in $\alpha$).

The proof consists in showing that there does not exists a specific ``short'' formula $G$ of the kind described in Theorem \ref{thm:GeneralSharp}, providing a contradiction with the coarse threshold assumption.

We proceed by showing that the third assumption on $G$, the bounded number of clauses, leads directly to a contraction with the fourth one.
This is shown with the following lemma.

\begin{lemma}\label{crit}

Let $G$ be a satisfiable formula with a finite number of variables and clauses. For any $\e>0$, we have for $n$ large enough

\begin{eqnarray*}
\pp \{ Z(F(n,\alpha_{\rm s})) < 2^{n \phi} | G\} &\leq& 1/2 +\e\, ,
\end{eqnarray*}

where $\pp \{\,\cdot\, | G\}$ is interpreted as in the
statement of Theorem \ref{thm:GeneralSharp}.

\end{lemma}

Before proceeding to the proof of Lemma \ref{crit}, we need the following technical estimates for the case $\phi >0$.

\begin{lemma}\label{lip1}
For any $n, \alpha, \phi, \psi$, with $\phi \geq \psi$,
\begin{align*}
& \text{(i)} \quad \frac{\partial\phantom{\alpha}}{\partial \alpha} P_n( \alpha, \phi) \leq C_1 \sqrt{n} ,\\
& \text{(ii)} \quad
P_n(\alpha, \phi) - P_n(\alpha, \psi) \leq C_2(\phi -\psi) \sqrt{n} \log(n) + o(1),
\end{align*}
where $C_1$ and $C_2$ do not depend on $n$.
\end{lemma}

\begin{proof}
$(i)$ By definition
\begin{align*}
P_n(\alpha, \phi)
=  \sum_{0 \leq m \leq N} {N \choose m} p^m (1-p)^{N-m} \pp \{ Z(F(n,\alpha)) <  2^{n \phi} | |F|=m\},
\end{align*}
and $p=(n\alpha)/N$, thus
\begin{align*}
\frac{\partial}{\partial \alpha}   p^m (1-p)^{N-m}

= \frac{n}{N}   p^{m-1} (1-p)^{N-m-1}(m-Np)
\end{align*}
and
\begin{eqnarray*}
\frac{\partial}{\partial \alpha} P_n(\alpha, \phi) &=&  \frac{n}{N}   p^{-1} (1-p)^{-1}\\
&&\sum_{0 \leq m \leq N} {N \choose m} p^{m} (1-p)^{N-m}(m-Np) \pp \{ Z(F(n,\alpha)) <  2^{n \phi} | |F|=m\}\\

&\leq&  \frac{n}{N}   p^{-1} (1-p)^{-1} \E[(|F|-Np),|F|\geq Np]\\
&\leq&   \frac{n}{N}   p^{-1} (1-p)^{-1} (\E(|F|-Np)^2)^{1/2}\\
&=&   \frac{n}{N}   p^{-1} (1-p)^{-1} (Np(1-p))^{1/2}\\
&\leq& C_1 \sqrt{n},
\end{eqnarray*}

$(ii)$ Note that because of Lemma \ref{lip1} $(i)$,
$$ \alpha \mapsto \pp \{ Z(F(n,\alpha)) < 2^{m} \}$$ is Lipschitz with constant $C_1 \sqrt{n}$. Therefore, it is sufficient to show that for any $a \geq 0$
\begin{eqnarray}
 \pp \{ Z(F(n,\alpha)) < 2^{a+1} \} \leq  \pp \{ Z(F(n,\alpha+A\log(n)/n)) < 2^{a}\}  + c \log(n)/\sqrt{n}, \label{l}
 \end{eqnarray}
for some constants $c$ and $A$ (possibly depending on $k$), since then
\begin{eqnarray*}
 &&\pp \{ Z(F(n,\alpha)) < 2^{a+1} \} - \pp \{ Z(F(n,\alpha)) < 2^{a} \} \\
 &\leq&  \pp \{ Z(F(n,\alpha+A\log(n)/n)) < 2^{a}\}- \pp \{ Z(F(n,\alpha)) < 2^{a} \}  + c\log(n)/\sqrt{n}\\
 &\leq& C_2\log(n)/\sqrt{n},
\end{eqnarray*}
for some constant $C_2$.
We first verify \eqref{l} for the model where the formulas have exactly $\alpha n$ clauses drawn uniformly at random, using in that case the notation $F_{\alpha n}$. The result for the Binomial model follows then from standard monotonicity arguments.
We will show that there exists $\theta \in (0,1)$ such that for any $a\geq 0$ and any integer $l\geq 1$,
\begin{eqnarray}
 \pp \{ Z(F_{\alpha n}) < 2^{a+1} \} -\pp \{ Z(F_{\alpha n+l}) < 2^{a}\} \leq 2 \theta^l. \label{finite}
 \end{eqnarray}
Note that for a deterministic formula $D$, and $C$ a $k$-clause uniformly drawn in $C_k(n)$,
$$\E_C Z( D \wedge C) = (1-\frac{1}{2^k}) Z (D),$$
hence for $l$ uniformly drawn $k$-clauses
$$\E_{C_1,\ldots,C_l} Z(D \wedge C_1 \wedge \ldots \wedge C_l) \leq (1-\frac{1}{2^k})^l Z (D).$$
Therefore, denoting by $\E_{l}= \E_{C_1,\dots,C_l}$ the expectation with
respect to $C_1,\dots,C_l$, we get
$$ \pp \{ Z(F_{\alpha n}) \leq 2^{a+1}\} \leq \pp \{\E_{l} Z (F_{\alpha n +l })  \leq 2^{a+1} (1-\frac{1}{2^k})^l  \}, $$
and defining $T := 2 (1-2^{-k})^l ,$ we have
\begin{eqnarray*}
\pp\{ Z(F_{\alpha n +l }) \leq 2^a \} &\geq& \pp\{ Z(F_{\alpha n +l }) \leq 2^a | \E_{l} Z (F_{\alpha n +l })  \leq T 2^{a} \} \pp\{ \E_{l} Z (F_{\alpha n +l })  \leq T 2^{a} \} \\
&\geq& (1-T) \pp\{ \E_{l} Z (F_{\alpha n +l })  \leq T 2^{a} \},
\end{eqnarray*}
where last inequality follows from Markov's inequality.
Thus,
$$ \pp \{ Z(F_{\alpha n}) < 2^{a+1} \} -\pp \{ Z(F_{\alpha n+l}) < 2^{a}\} \leq  2 (1-\frac{1}{2^k})^l.$$
By setting $l=A(k)\log(n)$, previous upper bound is $2 n^{A(k) \log (1-\frac{1}{2^k})}$, hence by taking $A(k)$ appropriately we get \eqref{l}.

\end{proof}

We are now in position to prove Lemma \ref{crit}.

\begin{proof}[Proof of Lemma \ref{crit}.]  Let $G$ be a satisfiable formula with a bounded number of clauses and, say, $r$ variables. And let us assume without loss of generality that $G$ contains only variables $x_1,\ldots,x_r$, and
that it is satisfied when all variables
are set to true. We then have
\begin{eqnarray*}
\pp \{ Z(F(n,\alpha_{\rm s})) < 2^{n \phi} | G\}
&\leq& \pp \{ Z(F(n,\alpha_{\rm s})|_{x_1=\ldots=x_r=T} ) < 2^{n \phi} \} ,
\end{eqnarray*}
where $T$ refers to the true assignment.
The random formula $F(n,\alpha_{\rm s})|_{x_1=\ldots=x_r=T}$ has now a different structure and probability distribution. Let us denote by $F_{*}$ an equivalently distributed random formula, which contains clauses of size 1 to $k$ and has only $n-r$ variables.
Since $|C_k(n)| ={n \choose k} 2^k$, the number of $l$-clauses appearing in $F_*$, denoted by $S_l$, satisfies
\begin{eqnarray}
\E S_k &=& \Theta(n),\\
\E S_{k-1} &=&  O(1),\label{O1}
\end{eqnarray}
and more generally

\begin{eqnarray}
\E S_{k-i} = O(n^{1-i}),\quad \forall 1 \leq i \leq k-1\, . \label{O2}
\end{eqnarray}

Hence, defining a random formula $F_{**}$, which contains only $k$-clauses and $(k-1)$-clauses, with the $k$-clauses selected independently in $C_k(n-r)$ with probability $p=\frac{\alpha_{\rm s} n}{{n \choose k} 2^k}$, and exactly $d$
clauses of size $(k-1)$ uniformly selected in $C_{k-1}(n-r)$,
we get that for any $\tau >0$ and $n$ large enough
\begin{eqnarray*}
\pp \{ Z(F(n,\alpha_{\rm s})) < 2^{n \phi} |  G\}
&\leq& \pp \{ Z(F_{**}) < 2^{n \phi} \} +  \tau\, ,
\end{eqnarray*}
provided $d$ is large enough.
Since 1-clauses are more constraining than $(k-1)$-clauses, we can upper bound our estimate by replacing each $k$-clause by the disjunction of its $k$
literals.

Moreover, the clauses on the $n-r$ variables are drawn with probability $p(n,\alpha_{\rm s})=\alpha_{\rm s} n/\{ {n \choose k} 2^k\}$, hence, by drawing them from $p(n-r,\alpha_{\rm s})$ instead we get, for
$D=d(k-1)$,
\begin{align}
& \pp \{ Z(F(n,\alpha_{\rm s})) < 2^{n \phi} |  G\}
\leq \pp \{ Z(F(n-r,\alpha_{\rm s})\wedge_{i=1}^D C^{(1)}_i)  < 2^{n \phi} \} +  \tau. \label{D}
\end{align}

We now prove a useful fact.

\begin{remark}\label{rem1}
Let $f$ be a Boolean formula on $n$ variables, $C^{(l)}_i$ be $l$-clauses independently and uniformly selected in $C_l(n)$ and $y \in \mR$.
Then for any unbounded increasing sequence $h(n)$ and for any $\varepsilon>0$, we can take $n$ large enough such that,
$$ \pp \{ Z (f\wedge_{i=1}^D C^{(1)}_i) < y \} \leq \pp \{ Z (f\wedge_{i=1}^{h(n)} C^{(k)}_i )< k^D y \}  + \varepsilon.$$
\end{remark}
\begin{proof}
In order to prove this fact, we check that for $n$ large enough,
\begin{align}
\pp \{ Z ( f\wedge C^{(1)}_1 ) < y \} \leq \pp \{ Z ( f\wedge_{i=1}^{h(n)/D} C^{(k)}_i )< ky \}  + \varepsilon/D. \label{g}
\end{align}
Note that for any $g \geq 1$
\begin{align}
\pp \{ Z (f\wedge_{i=1}^{g} C^{(k)}_i ) < ky\} \geq \pp \{ \min_{1 \leq i \leq g}Z (f\wedge C^{(k)}_i )< k y \} \label{1}
\end{align}
Moreover,
\begin{align}
\pp \{ Z ( f\wedge C^{(k)}_1) < ky \} &= \pp \{ Z (f\wedge \vee_{i=1}^{k} C^{(1)}_i )< ky \} \notag \\

 &\geq \pp \{ \sum_{i=1}^{k} Z ( f\wedge C^{(1)}_i )< ky \}\notag\\
 &\geq \pp \{ Z (f\wedge C^{(1)}_i) < y,\,\forall 1\leq i \leq k \}\notag\\
 &= \pp \{ Z (f\wedge C^{(1)}_1 )< y\}^k. \label{2}
\end{align}
Therefore, putting \eqref{1} and \eqref{2} together, we get
\begin{eqnarray}
\pp \{ Z (f\wedge_{i=1}^{g} C^{(k)}_i) < ky\} &\geq&1- (1-  \pp \{ Z ( f\wedge C^{(1)}_1 )< y\}^k )^g, \notag
\end{eqnarray}
but this implies that we can take $g$ large enough, such that
\begin{align}
 \pp \{ Z ( f\wedge C^{(1)}_1 )< y\} \leq \pp \{ Z( f\wedge_{i=1}^{g} C^{(k)}_i) < ky\} +\varepsilon/D, \notag
\end{align}
which proves the remark.
\end{proof}

We now can use our remark to upper bound the estimate in Eq.~\eqref{D}
\begin{align}
\pp \{ Z(F(n,\alpha_{\rm s})) < 2^{n \phi} |  G\}
\leq \pp \{ Z(F(n-r,\alpha_{\rm s})\wedge_{i=1}^{h(n)} C^{(k)}_i)  < k^D 2^{n \phi} \} + 2\tau \label{t1}
\end{align}
and by taking $t$ large enough
\begin{align}
\pp \{ Z(F(n-r,\alpha_{\rm s})\wedge_{i=1}^{h(n)} C^{(k)}_i)  < k^D 2^{n \phi} \} \leq
\pp \{ Z(F(n-r,\alpha_{\rm s}+th(n)/(n-r))  < k^D 2^{n \phi} \}  +\tau. \label{t2}
\end{align}

Defining $H(n):=th(n)/(n-r)$, we get from Lemma \ref{lip1} $(ii)$
\begin{align}
\pp \{ Z(F(n-r ,\alpha_{\rm s} +H(n) ) ) < k^D 2^{r \phi} 2^{(n -r) \phi} \}
\leq \pp \{ Z(F(n-r ,\alpha_{\rm s}+H(n)) )< 2^{(n-r)  \phi} \} +\tau. \label{t3}
\end{align}
Note that by Lemma \ref{lip1} $(ii)$, the inequality in \eqref{t3} holds for a clause density which is independent of $n$, although here the clause density is $\alpha_{\rm s} (n)+H(n)$. Since we will pick $H(n)$ to be $o(1/\sqrt{n})$, hence the variation of $H(n)$ can be neglected. Regarding the variation of $\alpha_{\rm s} (n)$, note that this sequence fluctuates on a compact interval (for a fixed $\phi<1$, on an interval contained in $(0,A]$ where $A$ is an upper bound on the critical threshold of Friedgut \cite{Friedgut}), then one can check that the gap in this inequality, i.e., $P_n(\alpha,\phi) - P_n(\alpha,\phi + 1/n)$, tends to zero uniformly in $\alpha$, leading to the claimed inequality.

Putting \eqref{t1}, \eqref{t2} and \eqref{t3} together, we get
\begin{align}
\pp \{ Z(F(n,\alpha_{\rm s})) < 2^{n \phi} | G\}& \leq
\pp \{ Z(F(n-r ,\alpha_{\rm s}+H(n)) )< 2^{(n-r)  \phi} \} + 4 \tau \notag \\
& = P_{n-r}(\alpha_{\rm s}+H(n),\phi)  +4\tau \notag \\
&\leq P_n(\alpha_{\rm s}+2H(n),\phi)  +5\tau. \label{var}
\end{align}
To see that the inequality \eqref{var} holds, let us check that for $\alpha,\phi$ fixed, $g(n)$ increasing and $n$ large enough
\begin{align}
P_{n-1}(\alpha,\phi) \leq P_n(\alpha + g(n)/n,\phi) + \tau.\label{var2}
\end{align}
The inequality \eqref{var} can then be verified by an appropriate choice of $g(n)$ and by using a similar argument as discussed previously regarding the dependence in $n$ of the clause density.
Recall that a random formula in $F(n-1,\alpha)$ is drawn by picking each clause in $C_k(n-1)$ with probability $p(n-1,\alpha)= \frac{\alpha (n-1)}{{n-1 \choose k}2^k}$. By a coupling argument, since $p(n-1,\alpha) > p(n,\alpha)$, one can equivalently draw a first formula $F_1$ by picking each clause in $C_k(n-1)$ with probability $p(n,\alpha)$ and a second formula $F_2$ by picking each clause in $C_k(n-1)$ with probability $p(n-1,\alpha)-p(n,\alpha)$; creating the formula $F_1 \wedge F_2$. Note that a random formula $F(n,\alpha)$ picks each clause in $C_k(n-1)$ and also in $\{C_k(n)- C : C \in C_k(n-1)\}$ with probability $p(\alpha,n)$. Hence
\begin{align}
& P_{n-1}(\alpha,\phi) =  \pp\{ Z(F(n-1,\alpha)) < 2^{(n-1) \phi}\}\notag \\
&= \pp\{ Z(F_1 \wedge F_2) < 2^{(n-1) \phi}\} \notag \\
& \leq \pp\{ Z(F(n,\alpha) \wedge F_2) < 2^{(n-1) \phi}\}. \label{up1}
\end{align}
The expected number of clauses in $F_2$ is given by $\E |F_2| = |C_k(n-1)| (p(n-1,\alpha)-p(n,\alpha)) = O(1)$, hence, we can upper bound \eqref{up1} by replacing $F_2$ with a constant number of random 1-clauses and use Remark \ref{rem1} (as done above) to conclude that
\begin{align}
& P_{n-1}(\alpha,\phi) \leq \pp\{ Z(F(n,\alpha + g(n)/n)) < K 2^{(n-1) \phi}\}
\end{align}
for a constant $K$ and an increasing function $g(n)$.
(Note that the 1-clauses are drawn within the set of $n-1$ variables instead of $n$ variables, but this does not change the conclusion).
Finally, using Lemma \ref{lip1} $(ii)$, we get
\begin{align}
& P_{n-1}(\alpha,\phi)  \leq \pp\{ Z(F(n,\alpha + g(n)/n) < 2^{n \phi}\} + \tau = P_{n}(\alpha + g(n)/n,\phi) + \tau,
\end{align}
which proves \eqref{var2}. Hence, we have
\begin{align*}
\pp \{ Z(F(n,\alpha_{\rm s})) < 2^{n \phi} | G\}
&\leq P_n(\alpha_{\rm s}+2H(n),\phi)  +5\tau
\end{align*}
and by choosing $h(n)=o(\sqrt{n})$ (increasing), i.e., $H(n)=o(1/\sqrt{n})$, and using Lemma \ref{lip1} $(i)$, we get
\begin{align*}
\pp \{ Z(F(n,\alpha_{\rm s})) < 2^{n \phi} |  G\}  \leq P_n(\alpha_{\rm s},\phi ) +6 \tau.
\end{align*}

\end{proof}

\subsection{Proofs of Theorem \ref{convpsat} and Lemma \ref{guerra}}\label{proof2}

\begin{proof}[Proof of Lemma \ref{guerra}.] We refer to the proof of Lemma \ref{guerracsp} which is more general.
\end{proof}

In order to prove Theorem \ref{convpsat}, we first need the following technical lemma.

\begin{lemma}\label{anal}
Let $\Delta(n) = O(n/(\log n)^{1+\e})$ for some $\e>0$, and $t(n)=o(n)$.
Let $f(\cdot)$ be positive, such that $f(n)/n$ is bounded above and
$$f(n_1 + n_2) + \Delta(n_1 + n_2 ) \geq  f(n_1) + f(n_2) ,\quad \forall n_1, n_2 \geq t(n_1+n_2).$$
Then $f(n)/n$ converges.
\end{lemma}

\noindent
{\bf Remark 3.} This lemma still holds if $\Delta(n)$ is such that $\sum_{n} \frac{\Delta(n)}{n^2} < \infty$.

\begin{proof}
Let $\e >0$.
Since $f(n)/n$ is bounded above, we have $S := \lim \sup_n f(n)/n < \infty$.
Let $n_0$ large enough such that $f(n)/n < S + \e$ for any $n \geq n_0$.
Let $r$ be an integer (to be chosen at our convenience but to be kept fixed) and let $\gamma(n)$ such that $\gamma(n) \geq t(n)$ and $\gamma(n)=o(n)$. Define $$m(n) = \inf \{ r 2^k : r 2^k \geq \gamma (n) \}$$
and note that $ \gamma(n) \leq m(n) \leq 2 \gamma(n) \vee r$.
For any $n$, we then have
$$n= \lfloor n/m(n) -1 \rfloor m(n) + q(n),$$
where $q(n) \in [m(n), 2m(n)]$, with $\lfloor n/m(n) -1 \rfloor \geq 0$ for $n$ large enough.
Hence, using the property of $f$, we have
\begin{align*}
f(n) \geq  \lfloor n/m(n) -1 \rfloor f(m(n)) + f(q(n)) - \sum_{k=1}^{\lfloor n/m(n) -1 \rfloor} \Delta (q(n)+ k m(n)),
\end{align*}
and since $f$ is positive,
\begin{align}
\frac{f(n)}{n} \geq  \frac{m(n)}{n} \lfloor n/m(n) -1 \rfloor \frac{f(m(n))}{m(n)}   -\frac{1}{n} \sum_{k=1}^{\lfloor n/m(n) -1 \rfloor} \Delta (q(n)+ k m(n)). \label{tot}
\end{align}
Since $\gamma (n) = o(n)$,  for $n$ large enough, we have $$\frac{m(n)}{n} \lfloor n/m(n) -1 \rfloor > 1-\e.$$
We now show that for $n$ large enough, we also have $\frac{f(m(n))}{m(n)} > S - \e$. First, note that we can take $r$ large enough such that $r 2^k / 2 \geq \gamma (r 2^k)$ for all $k \geq 0$, since $\gamma (n) = o(n)$.
Hence, using the property of $f$, we have
$$\frac{f(r 2^{k+1})}{r 2^{k+1}} \geq  \frac{f(r 2^k )}{r 2^{k}} - \frac{1}{r 2^{k+1}} \Delta (r 2^{k+1}) $$
and
\begin{align}
 \frac{f(r 2^{k})}{r 2^{k}} \geq  \frac{f(r  )}{r } - \sum_{j=1}^k \frac{1}{r 2^{j}} \Delta (r 2^{j}) . \label{ugly}
\end{align}

Since we can pick $r$ at our convenience, note that if $r$ is a power of 2,
$$\sum_{j=1}^k \frac{1}{r 2^{j}} \Delta (r 2^{j}) = \sum_{j = \log_2 r+1}^{\log_2 r +k} \frac{1}{2^{j}} \Delta ( 2^{j}),$$
which is, when $r$ increases, tending to zero uniformly in $k$, provided that
$$\sum_{j=1}^\infty \frac{1}{ 2^{j}} \Delta ( 2^{j}) < \infty.$$
Since previous condition follows from our hypothesis on $\Delta$,
and since we can always take $r$ large enough to ensure that $f(r)/r > S-\e$,
we can take $r$ large enough such that, from \eqref{ugly}, we have for any $k$
$$ \frac{f(r 2^{k})}{r 2^{k}} \geq  S -  \e$$
and since $m(n)$ is of the form $r2^k$, for any $n$
$$ \frac{f(m(n))}{m(n)} \geq  S -  \e.$$
Finally, we need to show that the last term in \eqref{tot} is vanishing, i.e., that
$$\frac{1}{n} \sum_{k=1}^{\lfloor n/m(n) -1 \rfloor} \Delta (q(n)+ k m(n)) \stackrel{n\rightarrow \infty}{\longrightarrow} 0.$$
For this, we pick $\gamma(n)$ to be large enough. For example, if $\Delta = O(n/(\log n)^{1+\e})$, we have
\begin{align}
\frac{1}{n} \sum_{k=1}^{\lfloor n/m(n) -1 \rfloor} \Delta (q(n)+ k m(n)) \leq \frac{1}{n} \frac{n}{m(n)} \Delta (n),
\end{align}
and since $m(n) \geq \gamma (n)$, if $\gamma (n)= O(n/(\log n)^{1+\nu})$ with $\nu > \e$, we conclude the proof. In general, we pick $\gamma(n)$ such that $ \Delta (n)/\gamma(n)=o(1)$.
\end{proof}

\begin{proof}[Proof of Theorem \ref{convpsat}.]
Let $F_n = F(n,\alpha)$.
Note that
\begin{align*}
& \E \log (1 + Z( F_{n})) = \E [\log (1 + Z( F_{n})) ,Z (F_{n}) \geq 1]
\end{align*}
and
\begin{align*}

&\E [\log (1 + Z( F_{n})) ,Z (F_{n}) \geq 1] \\
& = \E [\log Z (F_{n}), Z( F_{n} )\geq 1]+\E [\log (1+Z(F_n)^{-1} ) , Z (F_{n}) \geq 1].
\end{align*}

Let $c>0$, we have
\begin{align*}
& \E [\log (1+Z(F_n)^{-1} ) , Z (F_{n}) \geq 1] \leq \E [Z(F_n)^{-1}  , Z (F_{n}) \geq 1] \\
 & \leq \E [Z(F_n)^{-1}  , Z (F_{n} )\geq 1, Z_{FV} \geq c n  ] + \pp\{ Z_{FV} < c n ,   Z (F_{n}) \geq 1 \}
\end{align*}
where $Z_{FV}$ is the number of free variables in $F_n$. Therefore,
$\E [Z(F_n)^{-1}  , Z (F_{n} )\geq 1, Z_{FV} \geq c n  ] \leq e^{-cn}$. Moreover, there exists $c, c_2>0$ such that
$$ \pp\{ Z_{FV} < c n, Z(F_{n}) \geq 1  \} =O(e^{-c_2 n}) $$

hence there exists $\xi>0$ such that
\begin{align}
& \tau(n) := \E [\log (1+Z(F_n)^{-1} ) , Z (F_{n}) \geq 1] = O(e^{- \xi n}). \label{claim}

\end{align}
On the other hand, we have
(denoting by $F_{n_1}$, $F_{n_2}$ two independent formulas and letting $n=n_1+n_2$)
\begin{align*}
& \E \log (1 + Z( F_{n_1}) Z(F_{n_2})) =\E[ \log(1+ Z (F_{n_1}) Z(F_{n_2})) , Z (F_{n_1}) Z(F_{n_2}) \geq 1]
\end{align*}
and
\begin{align*}
&\E[ \log(1+ Z (F_{n_1}) Z(F_{n_2})) , Z( F_{n_1})Z(F_{n_2}) \geq 1] \geq \E[ \log( Z (F_{n_1})Z(F_{n_2})) , Z (F_{n_1}) Z(F_{n_2}) \geq 1].
\end{align*}
Hence, using Lemma \ref{guerra}, we get the following inequality
\begin{align*}
& \E [\log  Z (F_{n}), Z (F_{n}) \geq 1]+ \tau(n) \geq \E[ \log( Z( F_{n_1}) Z(F_{n_2})) , Z (F_{n_1}) Z(F_{n_2}) \geq 1]
\end{align*}
or equivalently
\begin{align}
& g(n)+ \tau(n) \geq g(n_1) + g(n_2) - g(n_1) \e (n_2) - g(n_2) \e (n_1) , \quad \forall n_1,n_2 \geq k \label{start}
\end{align}
where
\begin{eqnarray*}
 g(n) &=& \E [\log Z (F_{n}), Z (F_{n}) \geq 1], \\
  \e (n) &=& \pp \{Z( F_n) = 0 \}.
\end{eqnarray*}
Note that $0 \leq g(n) \leq n$. Therefore \eqref{start} implies

\begin{align}
& g(n)+ \tau(n) \geq g(n_1) + g(n_2) - n_1 \e (n_2) - n_2 \e (n_1), \quad \forall n_1,n_2 \geq k. \label{gb}
\end{align}
Since $\alpha < \alpha^*$, we have that $\e(n) = O(1/(\log n)^{1+\e})$, for some $\e >0$.

We then restrict ourself to $$n_1,n_2 \geq t(n) := n/(\log n)^\eta, \quad \text{with } \eta = \e/3.$$

This implies that $n^{1-\eta} \leq n_2$ and $(1-\eta) \log n \leq \log n_2$.

So, for $n_1, n_2$ large enough, we have
\begin{align}
& n_1 \leq n_2 (\log n_2)^{2 \eta} \\
& n_2 \leq n_1 (\log n_1)^{2 \eta}.
\end{align}
Going back to \eqref{gb}, we get
\begin{align*}
& g(n)+ \tau(n) \geq g(n_1) + g(n_2) - n_1 (\log n_1)^{2 \eta} \e (n_1) - n_2 (\log n_2)^{2 \eta}  \e (n_2), \quad \forall n_1,n_2 \geq n/(\log n)^{2 \eta}
\end{align*}
or equivalently
\begin{align}
& f(n)+\Delta(n) \geq f(n_1) + f(n_2) , \quad \forall n_1,n_2 \geq t(n)
\end{align}
where
\begin{eqnarray*}
f(n)&=&g(n) -n (\log n)^{2 \eta} \e (n) \\
\Delta(n)&=&n (\log n)^{2 \eta} \e (n)+ \tau(n)\\
t(n)&=& n/(\log n)^{\eta}.
\end{eqnarray*}
But $ \e - 2 \eta = \e/3 >0$, hence $$\Delta (n) \leq O\left(\frac{n}{(\log n)^{1 + \e/3}}\right),$$

and we satisfy the hypothesis of Lemma \ref{anal}, which implies that $f(n)/n$ converges, hence $g(n)/n$ converges too.
\end{proof}

\subsection{Proofs of Lemma \ref{conva} and Theorem \ref{res}}\label{proof3}

\begin{proof}[Proof of Lemma \ref{conva}.]
From Theorem \ref{convpsat},
for every $\alpha < \alpha^*$, $\psi_n(\alpha)$ converges to a limit $\phi_s(\alpha)$.
Note that $\phi_s(\,\cdot\,)$ is a non-increasing function on $\A$, hence, it has a countable number of plateaus and discontinuities.
Let $\alpha_0 \in \A$
and denote $\phi_0 = \phi_s(\alpha_0)$.
If $\alpha_n (\phi_0)$ does not converge, define
$$\underline{\alpha}_0=\lim \inf_{n \rightarrow \infty} \alpha_n (\phi_0)  ,$$
$$n_k \nearrow \infty \text{ s.t. } \lim_{k \rightarrow \infty} \alpha_{n_k} (\phi_0) = \underline{\alpha}_0,$$
$$\bar{\alpha}_0=\lim \sup_{n \rightarrow \infty} \alpha_n (\phi_0) ,$$
$$m_k \nearrow \infty \text{ s.t. } \lim_{k \rightarrow \infty} \alpha_{m_k} (\phi_0) = \bar{\alpha}_0.$$
Then, for any $\alpha \in (\underline{\alpha}_0,\bar{\alpha}_0)$, there exists $\e>0$ such that
\begin{align}
P_{m_k}(\alpha,\phi_0) \leq P_{m_k} ( \alpha_{m_k} (\phi_0)-\e, \phi_0)
\stackrel{k \nearrow \infty}{\rightarrow} 0 \label{lim1}

\end{align}
and
\begin{align}
P_{n_k}(\alpha, \phi_0) \geq P_{n_k} (\alpha_{n_k} (\phi_0)+\e, \phi_0)
\stackrel{k \nearrow \infty}{\rightarrow} 1 \label{lim2}

\end{align}
Moreover, if $\alpha < \alpha^*$,
$$

P_{n_k}(\alpha, \phi_0)
= \pp \{ \frac{1}{n_k} \log Z(F(n_k,\alpha)) < \phi_0 | Z(F(n_k,\alpha)) \geq 1 \} + o(1),$$
hence
$$\lim_{k \rightarrow \infty} \E [ \frac{1}{n_k} \log Z(F(n_k,\alpha))  | Z(F(n_k,\alpha)) \geq 1 ] \le \phi_0, $$
i.e., since $\psi_n(\alpha)$ converges to $\phi_s(\alpha)$ from Lemma \ref{convpsat}, $$\phi_s (\alpha) \le \phi_0.$$ Similarly, we have
$$\lim_{k \rightarrow \infty} \E [ \frac{1}{m_k} \log Z(F(m_k,\alpha))  | Z(F(m_k,\alpha)) \geq 1 ] \ge \phi_0, $$
and
$$\phi_s (\alpha) \ge \phi_0.$$ Therefore, $\phi_0$ is a plateau of $\phi_s(\cdot)$, and since $\phi_s(\cdot)$ has countably many plateaus, there are countably many $\phi_0 \in \phi_s(\A)$, for which $\alpha_n(\phi_0)$ does not converge.
\end{proof}

\begin{proof}[Proof of Theorem \ref{res}]
From Theorem \ref{convpsat}, there exists a function $\phi_s(\cdot)$, such that for any $\alpha \in [0,\alpha^*)$, we have $\f(\alpha) = \lim_{n \rightarrow \infty} \psi_n(\alpha)$, where $\psi_n(\cdot)$ is defined in Theorem \ref{convpsat}. Let $I:=\f ([0,\alpha^*))$. From Lemma \ref{conva}, there exists a countable set $\mathcal{C} \subseteq I$ and a function $A: \,I \setminus \mathcal{C} \rightarrow [0,\alpha^*)$ such that for any $\phi \in I \setminus \mathcal{C}$, we can define the limit $A(\phi)=\lim_{n \rightarrow \infty} \alpha_n (\phi)$. Note that for any $\phi \in I \setminus \mathcal{C}$, Theorem \ref{fried} implies $\f(A(\phi))=\phi$.

Now, for any $\alpha \in [0,\alpha^*)$ which is not a discontinuity point of $\f$ (this holds except on a countable subset of $[0,\alpha^*)$),
and for any $\e >0$, there exists $\e^\prime <\e$ such that $\phi_* :=\f(\alpha) - \e^\prime \in I \setminus \mathcal{C}$ and hence $\alpha_n(\phi_*)$ tends to a limit $A_*$. Note that $A_*>\alpha$, since $\alpha$ is not a discontinuity point of $\f$ and since $\f(A_*) = \phi_*$. Therefore, there exists $\delta >0$ such that
$$P_n(\alpha, \phi_s (\alpha) - \e) \leq P_n(\alpha, \phi_*) \leq P_n(\alpha_n(\phi_*) - \delta, \phi_*)$$
and we conclude from Theorem \ref{fried} that $P_n(\alpha, \phi_s (\alpha) - \e) \rightarrow 0$ when $n \rightarrow \infty$. With a similar argument, we conclude that $P_n(\alpha, \phi_s (\alpha) + \e) \rightarrow 1$ when $n \rightarrow \infty$.
\end{proof}

\subsection{Proofs of Lemma \ref{ch1}, Lemma \ref{ch2} and Lemma \ref{guerracsp}}\label{proof4}

\begin{proof}[Proof of Lemma \ref{guerracsp}.] In this proof, we keep $\alpha$ fixed and split the $n$ variables into two sets of $n_1$ and $n_2=n-n_1$ variables, such as $\{1,\ldots,n_1\}$ and $\{n_1+1,\ldots,n\}$.
For convenience, we now work with the interpolated Poisson model.
We construct a random Boolean formula as follows: we first draw independently the integers $M$, $M_1$ and $M_2$ under Poisson distributions of parameters $\alpha n t$, $\alpha n_1 (1-t)$ and $\alpha n_2 (1-t)$ respectively. We then draw independently $M$ clauses from the full system, by picking for each clause the indices of the variables appearing in it independently and uniformly at random within the set of $n$ variables and by picking $\vp$ under $\mu$. We also draw independently $M_i$ clauses from each sub-systems, by picking for each clause the indices of the variables appearing in it independently and uniformly at random within the set of $n_i$ variables and by picking $\vp_i$ under $\mu$. Finally, we take the conjunction of all clauses to create the formula $F_n (t)$.

Note that the claim of the lemma is equivalent to
\begin{align}
\E \log (1 + Z (F_n(1))) \geq \E \log (1 + Z( F_{n}(0))) , \label{qsub}
\end{align}
which is proved by showing that
$$\frac{\de\phantom{t}}{\de t} \E \log (1 + Z (F_n(t))) \geq 0.$$
An elementary calculation yields
\begin{align}
& \frac{\de\phantom{t}}{\de t} \frac{1}{n} \E \log (1 + Z (F_n(t))) = \alpha  \left[\E_{\vp,I} \E \log (1 + Z (F_n(t) \wedge \vp(x_I))) - \E \log (1 + Z (F_n(t))) \right]  \notag \\
& - \alpha \frac{n_1}{n} \left[\E_{\vp_1,I_1} \E \log (1 + Z (F_n(t) \wedge \vp_1 (x_{I_1}))) - \E \log (1 + Z (F_n(t))) \right] \notag \\& - \alpha \frac{n_2}{n} \left[\E_{\vp_2,I_2} \E \log (1 + Z (F_n(t) \wedge \vp_2(x_{I_2}) )) - \E \log (1 + Z (F_n(t))) \right], \notag
\end{align}
where $\vp, \vp_1, \vp_2 \iid \mu$, $I \sim U^k$, $I_1 \sim U_1^k$, $I_2 \sim U_2^k$, all independent, and where $U^k$, respectively $U_i^k$, denotes the $k$-th product measure of $U$, respectively $U_i$ (where $U$, resp. $U_i$, denotes the uniform measure on the $n$ variables, resp. $n_i$ variables). Hence, $x_I = (x_{i_1}, \ldots, x_{i_k})$ with $i_1,\ldots, i_k$ iid uniform over the $n$ variables.

We then have
\begin{align}
&\E_{\vp, I} \E \log (1 + Z (F_n(t) \wedge \vp(x_I))) - \E \log (1 + Z (F_n(t)))    = \E_{\vp, I} \E \log \langle   \vp(X_I) \rangle  \notag
\end{align}
where $X$ is uniformly drawn within the augmented solution space $S(F_{n^*}(t))= S(F_{n}(t))\cup \{*\}$, where $*$ is an assignment which returns true on any Boolean functions, and  $\langle\, \cdot\, \rangle$ denotes the expectation with respect to $X$. Note that
\begin{align}
\E_{\vp, I} \E \log \langle   \vp(X_I) \rangle  = - \E_{\vp, I} \E \sum_{l=1}^\infty  \frac{ \langle \tilde{\vp}(X_I) \rangle^l }{l} . \label{serie}
\end{align}
where $\tilde{\vp} = 1 - \vp$.
We now introduce the `replicas' $X^{(r)}$, which are independent and
identically distributed copies of $X$. We then have
\begin{align*}
& \langle  \tilde{\vp}(X_I) \rangle^l = \langle \prod_{r=1}^l \tilde{\vp}(X^{(r)}_I) \rangle, \quad \forall l \geq 1.
\end{align*}
We are done if we can show that for any realizations of the $X^{(r)}$'s and for any $l \geq 1$,
\begin{align}
& \E_{\vp, I} \prod_{r=1}^l \tilde{\vp}(X^{(r)}_I) - \frac{n_1}{n} \E_{\vp, I_1} \prod_{r=1}^l \tilde{\vp}(X^{(r)}_{I_1}) - \frac{n_2}{n} \E_{\vp, I_2} \prod_{r=1}^l \tilde{\vp}(X^{(r)}_{I_2}) \geq 0. \label{targ}
\end{align}
Note that
\begin{align}
\E_{\vp, I} \prod_{r=1}^l \tilde{\vp}(X^{(r)}_I) = \E_\vp \E_{\hat{P}} \prod_{r=1}^l \tilde{\vp}(\xi^{(r)})
\end{align}
where $\xi^{(1)}, \ldots, \xi^{(l)} \iid \hat{P}$ and where $\hat{P}$ is the empirical distribution of $X^{(1)}_I, \ldots, X^{(l)}_I$, i.e. the distribution on $\{-1,1\}^{kl}$ given by
$$\hat{P}(x_1^{(1)},\ldots, x_k^{(1)}, \ldots, x_1^{(l)}, \ldots, x_k^{(l)})=\prod_{i=1}^k\bar{P}(x_i^{(1)},\ldots, x_i^{(l)}) $$
with
$$ \bar{P}(x_i^{(1)},\ldots, x_i^{(l)}) = \frac{\# \{ i \in \{1,\ldots,n\}: (X_{i}^{(1)},\ldots, X_{i}^{(l)})=(x_i^{(1)},\ldots, x_i^{(l)}) \}}{n}$$
and similarly
\begin{align}
\E_{\vp, I_s} \prod_{r=1}^l \tilde{\vp}(X^{(r)}_{I_i}) = \E_\vp \E_{\hat{P}_s} \prod_{r=1}^l \tilde{\vp}(\xi_s^{(r)}), \quad s =1,2
\end{align}
where $\xi_s^{(1)}, \ldots, \xi_s^{(l)} \iid \hat{P}_s$ and where $\hat{P}_s$ is the empirical distribution of $X^{(1)}_{I_s}, \ldots, X^{(l)}_{I_s}$, i.e. the distribution on $\{-1,1\}^{kl}$ given by
$$\hat{P}_s(x_1^{(1)},\ldots, x_k^{(1)}, \ldots, x_1^{(l)}, \ldots, x_k^{(l)})=\prod_{i=1}^k\bar{P}_s(x_i^{(1)},\ldots, x_i^{(l)}), \quad s=1,2 $$
with
\begin{align*}
& \bar{P}_1(x_i^{(1)},\ldots, x_i^{(l)}) = \frac{\# \{ i \in \{1,\ldots,n_1\}: (X_{i}^{(1)},\ldots, X_{i}^{(l)})=(x_i^{(1)},\ldots, x_i^{(l)}) \}}{n_1}, \\
& \bar{P}_2(x_i^{(1)},\ldots, x_i^{(l)}) = \frac{\# \{ i \in \{n_1+1,\ldots,n\}: (X_{i}^{(1)},\ldots, X_{i}^{(l)})=(x_i^{(1)},\ldots, x_i^{(l)}) \}}{n_2}.
 \end{align*}
Now, using the operator $\Gamma_l$ defined by \eqref{cond1} in H2, i.e.,
\begin{align*}
\Gamma_l: \, M_1(\{-1,1\}^l) \ni \nu \mapsto \E_\vp \E_{Z^{(r)}} \prod_{r=1}^l (1-\vp(Z^{(r)}))
\end{align*}
where $Z^{(r)}$ are Boolean random vectors of dimension $k$ such that $Z_i=(Z_i^{(1)},\ldots, Z_i^{(l)})$, $i=1,\ldots,k$, are i.i.d. with distribution $\nu$, note that \eqref{targ} is equivalent to
$$\Gamma_l (\bar{P}) - \frac{n_1}{n} \Gamma_l (\bar{P}_1)- \frac{n_2}{n} \Gamma_l (\bar{P}_2) \geq 0,$$
which holds by convexity of $\Gamma_l$, since
$$ \bar{P} = \frac{n_1}{n} \bar{P}_1 + \frac{n_2}{n} \bar{P}_2.$$
\end{proof}

\begin{proof}[Proof of Lemma \ref{ch1}.]
We have
\begin{align*}
\Gamma_l ( \nu ) &=\E_\vp \E_{Z^{(r)}} \prod_{r=1}^l \tilde{\vp}(Z^{(r)}) \\
&=\E_{\vp} \sum_{z^{(1)}, \ldots, z^{(l)} \in \{-1,1\}^k}  \prod_{r=1}^l \tilde{\vp}(z^{(r)}) \nu^k (z^{(1)},\ldots,z^{(l)})\\
& =\E_{s^{(j)}} \sum_{z^{(1)}, \ldots, z^{(l)} \in \{s^{(1)}, \ldots, s^{(J)}\} }  \prod_{i=1}^k \nu ((z^{(1)})_i,\ldots,(z^{(l)})_i) \\
& = \sum_{i_1,\ldots, i_l \in \{1,\ldots, J\}} [\E_{s^{(j)}_1} \nu(s_1^{(i_1)}, \ldots, s_1^{(i_l)})]^k
\end{align*}
and $\Gamma_l$ is convex for any $l\geq 1$.
\end{proof}

\begin{proof}[Proof of Lemma \ref{ch2}.]
We need to check the convexity of
\begin{align}
\nu \mapsto \E_s \E_{Z^{(r)}} \prod_{r=1}^l (1-\vp_s(Z^{(r)}))
\end{align}
where $Z^{(r)}$ are Boolean random vectors of dimension $k$ such that $Z_i=(Z_i^{(1)},\ldots, Z_i^{(l)})$, $i=1,\ldots,k$, are i.i.d. with distribution $\nu$,
$$\vp_s (x) = \1 (\prod_{i=1}^k x_i = s)$$
and
$$\mu(\vp_1) = \mu(\vp_{-1})=1/2.$$
Note that
\begin{align}
\E_{Z^{(r)}} \prod_{r=1}^l (1-\vp_s(Z^{(r)})) = \pp \{ \prod_{i=1}^k Z_i = -s^l \}
\end{align}
where $-s^l$ denotes the vector $(-s,\ldots,-s)$ with $l$ components and where $ \prod_{i=1}^k Z_i$ denotes the component-wise product of the vectors $Z_i$.
Since the $Z_i$ are i.i.d. under $\nu$ and valued in $\{-1,1\}$, and since we are interested in their product, we now work with their Fourier transform.
For any $Q \subseteq \{1,\ldots, l\}$, let
$$f(Q)=f_{Z_1}(Q) =  \E \prod_{r \in Q} Z_1^{(r)}.$$
Note that we can recover the distribution of $Z_1$ by knowing $f(Q)$ for any $Q$, in particular
$$\pp \{Z_1 = 1^l\} = \sum_{Q \in 2^{[l]}} f(Q)$$
and
$$\pp \{Z_1 = -1^l\} = \sum_{Q \in 2^{[l]}} (-1)^{|Q|} f(Q).$$
Moreover,
$$f_{\prod_{i=1}^k Z_i}(Q) =  f(Q)^k,$$
hence,
\begin{align*}
\E_s \pp \{ \prod_{i=1}^k Z_i = -s^l \} &= 1/2 \sum_{Q \in 2^{[l]}} f(Q)^k +1/2 \sum_{Q \in 2^{[l]}} (-1)^{|Q|} f(Q)^k \\
& = \sum_{Q \in 2^{[l]} \atop{|Q| \text{ even}}}  f(Q)^k.
\end{align*}
Since $f(Q)$ is linear in $\nu$ (it is the expectation of $\prod_{r \in Q} Z_1^{(r)}$ where $(Z_1^{(1)},\ldots,Z_1^{(l)}) \sim \nu$), the above summation is clearly convex in $\nu$ if $k$ is even.
\end{proof}

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text:
\begin{document}

\title{A c-holomorphic effective Nullstellensatz with parameter}

\author{Maciej P. Denkowski
}\address{Jagiellonian University, Faculty of Mathematics and Computer Science, Institute of Mathematics, \L ojasiewicza 6, 30-348 Krak\'ow, Poland}\email{maciej.denkowski@uj.edu.pl}\date{June 3rd 2014}
\keywords{Nullstellensatz, c-holomorphic functions}
\subjclass{32B15, 32C25, 14Q20}

\begin{abstract}
We prove a local Nullstellensatz with parameter for a continuous family of c-holomorphic functions with an effective exponent independent of the parameter: the local degree of the cycle of zeroes of the central section section. We assume that this central section defines a proper intersection and we show that we can omit this assumption in case of isolated zeroes.
\end{abstract}

\maketitle

\section{Introduction}

The idea of writing this note comes from an observation made in our recent paper on the \L ojasiewicz inequality with parameter \cite{D4}. Using the methods of \cite{PT} that led to the c-holomorphic effective Nullstellensatz presented in \cite{D2} and some intersection theory results introduced in \cite{T}, we obtain an effective Nullstellensatz for a continuous family of c-holomorphic functions.

We shall briefly recall the notion of \textit{c-holomorphic functions} introduced by Remmert in \cite{R} (see also \cite{Wh}). These are complex continuous functions defined on an analytic set (or more generally, analytic space) $A$ that are holomorphic at its regular points $\mathrm{Reg} A$. We denote by $\mathcal{O}_c(A)$ their ring for a fixed $A$. They have similar properties to those of holomorphic functions. Nevertheless, they form a larger class and do not allow the use of methods based on differentiability. Their main feature is the fact that they are characterized among all the continuous functions $A\to {\C}$ by the analycity of their graphs (see \cite{Wh}) which makes geometric methods appliable. Some effective results obtained for this class of functions from the geometric point of view are presented in \cite{D1}--\cite{D4}.  In particular, we have an identity principle on irreducible sets and a Nullstellensatz --- see \cite{D2}.

 Throughout the paper we are working with a topological, locally compact space $T$ that in addition is \textit{1st countable}. We fix also a pure $k$-dimensional analytic subset $A$ of an open set $\Omega\subset{\C}^m$ with $0\in A$ and consider a continuous function
$f=f(t,x)\colon T\times A\to{\C}^n$. We will use the notation $f_t(x)=f(t,x)$ and $f_t=(f_{t,1},\dots, f_{t,n})$. We assume that each $f_t$ is c-holomorphic. Of course, this happens to be true iff all the components $f_{t,j}\in\mathcal{O}_c(A)$.

Let $I_t(U)\subset\mathcal{O}_c(A\cap U)$ denote the ideal generated by $f_{t,1},\dots, f_{t_n}$ restricted to $U\cap A\neq\varnothing$ where $U$ is an open set.

If a mapping $h\in\mathcal{O}_c(A,{\C}^n)$ defines a proper intersection, i.e. $h^{-1}(0)$ (which corresponds to the interesection of the graph $\Gamma_h$ with $\Omega\times\{0\}^n$) has pure dimension $k-n$ (\footnote{By the identity principle presented in \cite{D2}, neither of the components of $h$ can vanish identically on any irreducible component of $A$.}), then we introduce the \textit{cycle of zeroes} for $h$ as the Draper proper intersection  cycle (\cite{Dr})
$$
Z_h:=\Gamma_h\cdot (\Omega\times\{0\}^n).
$$
In other words, $Z_h$ is a formal sum $\sum \alpha_j S_j$ where $\{S_j\}$ is the locally finite family of the irreducible components of $h^{-1}(0)$ and $\alpha_j=i(\Gamma_h\cdot(\Omega\times\{0\});S_j)$ denotes the Draper intersection multiplicity along $S_j$ (cf. \cite{Dr} and \cite{Ch}). We define the local degree (or Lelong number) of $Z_h$ at a point $a$ usually as $\mathrm{deg}_a Z_h:=\sum\alpha_j \deg_a S_j$ with the convention that $\deg_a S_j=0$ if $a\notin S_j$.

Note that $\Gamma_h$ is a pure $k$-dimensional analytic set and $k-n$ is the minimal possible dimension for the interesection $\Gamma_h\cap(\Omega\times\{0\}^n)$ which means this interesection is what we call a proper one.

Now we are ready to state our main result:

\begin{thm}\label{main}
Let $T,A,f$ be as above. Assume moreover that $f_{t}(0)=0$ for any $t$, and that $f_{t_0}^{-1}(0)$ has pure dimension $k-n$. Then there is a neighbourhood $T_0\times U$ of $(t_0,0)\in T\times A$ such that \begin{enumerate}
\item for all $t\in T_0$, the sets $f_t^{-1}(0)$ have pure dimension $k-n$, too;
\item for all $t\in T_0$, for any $g\in \mathcal{O}_c(A\cap U)$ vanishing on $f_t^{-1}(0)\cap U$, $g^\delta\in I_t(U)$ where $\delta=\mathrm{deg}_0 Z_{f_{t_0}}$ is independent of $t\in T_0$;
\item if $g\colon T_0\times (A\cap U)\to {\C}$ is continuous and such that each $g_t\in\mathcal{O}_c(A\cap U)$ vanishes on $f_t^{-1}(0)\cap U$, then there is a continuous function $h\colon T_0\times (A\cap U)\to {\C}^n$ with c-holomorphic $t$-sections and such that $g^\delta=\sum_{j=1}^n h_{j}f_{j}$ where $\delta$ is as above.
\end{enumerate}
\end{thm}

This is somehow related to some results of \cite{GGVL}. In the particular case when $\dim f_{t_0}^{-1}(0)=0$, we relax the assumptions in the last Section.

\section{Proof of the main result}

The proof of Theorem \ref{main} will be derived in several steps from some results presented in \cite{D4}.

First, we easily observe that \textsl{all} the results from \cite{D4} Section 2 hold true (with exactly the same proofs) for the mapping $f\colon T\times A\to{\C}^n$ provided $f_{t_0}^{-1}(0)$ has pure dimension $k-n$, i.e. the intersection $\Gamma_{f_{t_0}}\cap(\Omega\times\{0\}^n)$ is proper.

We will briefly state in clear the main arguments. But first let us recall that a family $Z_t=\sum_s \alpha_{t,s} S_{t,s}$ ($t\in T$) of positive (\footnote{i.e. with non-negative integer coefficients $\alpha_{t,s}$}) analytic $k$-cycles (\footnote{i.e. each irreducible set $S_{t,s}$ is pure $k$-dimensional and their family is locally finite for $t$ fixed.}) in $\Omega$ converges to a positive $k$-cycle $Z_{t_0}$ when $t\to t_0$ in the sense of Tworzewski \cite{T} (see the Introduction in \cite{D4}) if\begin{itemize}
\item the analytic sets $|Z_t|:=\bigcup_s S_{t,s}$ converge to $|Z_{t_0}|$ in the sense of Kuratowski, i.e. on the one hand, for any $x\in |Z_{t_0}|$ and any sequence $t_\nu\to t_0$ there are points $|Z_{t_\nu}|\ni x_\nu\to x$, while on the other any limit point $x$ of a sequence $x_\nu\in |Z_{t_\nu}|$ chosen for $t_\nu\to t_0$, belongs to $|Z_{t_0}|$ (cf. \cite{D4} Lemma 1.4);
\item the multiplicities are presevered: for regular point $a\in\mathrm{Reg} |Z_{t_0}|$ and any relatively compact submanifold $M\Subset \Omega$ of codimension $k$ transversal to $|Z_{t_0}|$ at $a$ and such that $\overline{M}\cap |Z_{t_0}|=\{a\}$, there is $\deg(Z_t\cdot M)=\deg (Z_{t_0}\cdot M)$ for all $t$ in a neighbourhood of $t_0$, where $\deg$ denotes here the total degree of the intersection cycle (\footnote{By \cite{TW} the intersections $|Z_t|\cap M$ are finite and proper; the \textit{total degree} is the sum of the intersection multiplicites computed at the intersection points.}).
\end{itemize}
We write then $Z_t\stackrel{T}{\longrightarrow} Z_{t_0}$.

\begin{proof}[Proof of Theorem \ref{main}]
Since $f$ is continuous, we obtain as in \cite{D4} Lemma 2.5, the Kuratowski convergence of the graphs $\Gamma_{f_t}$ to $\Gamma_{f_{t_0}}$ which together with the remark that this in fact is the local uniform convergence --- which on $\mathrm{Reg} A$ implies also the convergence of the differentials --- allows us to conclude, as in Proposition 2.6 from \cite{D4}, that the graphs converge in the sense of Kuratowski.

Now, since $f_{t_0}^{-1}(0)$ has pure dimension $k-n$, then \cite{D4} Proposition 1.7 based on the main result of \cite{TW} shows that $f_t^{-1}(0)$ have pure dimension $k-n$ for all $t$ sufficiently close to $t_0$, which gives (1). Moreover, these zero-sets converge to $f_{t_0}^{-1}(0)$ in the sense of Kuratowski, and so applying Lemma 3.5 from \cite{T} we obtain the Tworzewski convergence $Z_{f_{t}}\stackrel{T}{\longrightarrow} Z_{f_{t_0}}$, just as in Theorem 2.7 from \cite{D4}.

We can choose coordinates in ${\C}^m$ in such a way that $f_{t_0}^{-1}(0)$ projects properly onto the first $k-n$ coordinates and we have for the intersection multiplicity the following equality:
$$
i((\{0\}^{k-n}\times{\C}^{m-k+n})\cdot Z_{f_{t_0}};0)=\mathrm{deg}_0 Z_{f_{t_0}}.
$$
Define $\ell\colon {\C}^m\to {\C}^{k-n}$ as the linear epimorphism for which $\mathrm{Ker}\ell=\{0\}^{k-n}\times{\C}^{m-k+n}$ and take $$\varphi_t\colon A\ni x\mapsto (f_t(x),\ell(x))\in{\C}^n\times{\C}^{k-n}$$ for $t\in T$. We can find a polydisc $V\times W\subset{\C}^{k-n}\times{\C}^{m+k-n}$ centred at zero such that $$(\{0\}^{k-n}\times\overline{W})\cap f_{t_0}^{-1}(0)=\{0\}$$
and $f_{t_0}^{-1}(0)$ projects properly onto $V$.

Clearly,  $(\{0\}^{k-n}\times\overline{W})\cap f_{t_0}^{-1}(0)$ seen in  $V\times W\times \{0\}^k$ is exactly   $$(\overline{V\times W}\times \{0\}^n)\cap\Gamma_{\varphi_{t_0}}$$ where we note that $\Gamma_{\varphi_{t_0}}$ is a pure $k$-dimensional analytic set. Thus, there is a polydisc $P\subset{\C}^k$ such that $(V\times W\times P)\cap \Gamma_{\varphi_{t_0}}$ projects properly \textit{onto} $P$ along $V\times W$. This means that $\varphi_{t_0}|_{(V\times W)\cap A}$ is a proper mapping and its image is $P$.

Obviously, the mapping $$\Phi\colon T\times A\ni (t,x)\mapsto \varphi_t(x)\in{\C}^k$$ is continuous which implies (again as in \cite{D4} Proposition 2.6) that the graphs $\Gamma_{\varphi_t}$ converge in the sense of Kuratowski to $\Gamma_{\varphi_{t_0}}$ as $t\to t_0$. But the type of convergence implies that for all $t$ sufficiently close to $t_0$, the natural projection
$(V\times W\times P)\cap \Gamma_{\varphi_{t}}\to P$
is a branched covering over $P$. In particular, all these $\varphi_t$ have the same image $P$. Let $q_t$ denote the multiplicity of the branched covering $\varphi_t|_{A\cap (V\times W)}$.

In order to give a precise value of $q_{t}$ we remark first that actually it is the multiplicity of the projection
$$
\pi\colon {\C}^{k-1}\times{\C}^{m-k+1}\times{\C}\ni(u,v,w)\mapsto (w,u)\in{\C}\times {\C}^{k-1}
$$
over $P$ when restricted to $\Gamma_t:=\Gamma_{f_t}\cap (V\times W\times {\C})$. The classical Stoll Formula says that this multiplicity is the sum of the local multiplicities of the projection at the points of any fibre. On the other hand, as already noted in \cite{Dr}, this local multiplicity of the projection is nothing else but the isolated proper intersection multiplicity of the set we are projecting and the fibre of the linear projection. Hence, $q_t$ is just the total degree of the intersection cycle $\pi^{-1}(0)\cdot \Gamma_t$, i.e. $q_t=\deg((V\times W\times\{0\}^k)\cdot \Gamma_{t})$ (cf. the Stoll Formula).
Moreover, in view of \cite{TW2} Theorem 2.2, we can write
\begin{align*}
(V\times W\times\{0\}^k)\cdot \Gamma_{t}&=(\{0\}^{k-n}\times W)\cdot_{V\times W\times\{0\}^k}((V\times W\times\{0\}^k)\cdot \Gamma_t)=\\
&=(\{0\}^{k-n}\times W)\cdot Z_{f_t|_{A\cap (V\times W)}}=\\
&=(\{0\}^{k-n}\times W)\cdot Z_{f_t},
\end{align*}
and so
$$
q_t=\deg((\{0\}^{k-n}\times W)\cdot Z_{f_t}).
$$
Besides, thanks to the convergence $Z_{f_{t}}\stackrel{T}{\longrightarrow} Z_{f_{t_0}}$, we have for all $t$ in a neighbourhood $T_0$ of $t_0$
$$
\deg((\{0\}^{k-n}\times W)\cdot Z_{f_t})=\deg((\{0\}^{k-n}\times W)\cdot Z_{f_{t_0}}).
$$
From this, eventually, we obtain $q_t=\deg_0 Z_{f_{t_0}}$, $t\in T_0$.

Once we have established that, we may directly use Lemma 3.1 from \cite{D2} (which is the c-holomorphic counterpart of Lemma 1.1 in \cite{PT}) getting precisely the statement (2) for $T_0$ and $U:=V\times W$.

In order to obtain (3) it is enough to take a closer look at how \cite{D2} Lemma 3.1 is proved. Write $A_0=A\cap U$ and consider
$\varphi:=\Phi|_{T_0\times A_0}$.
This is a continuous mapping with c-holomorphic sections $\varphi_t=(\varphi_{t,1},\dots, \varphi_{t_k})$ that are proper mappings $A\to P\subset {\C}^k$ of multiplicity $\delta=\deg_0 Z_{f_{t_0}}$. Take a continuous $g(t,x)$ with c-holomorphic $t$-sections vanishing on $\{\varphi_{t_1}=\ldots=\varphi_{t,n}=0\}$, i.e. on $f_t^{-1}(0)\cap U$. The properness of $\varphi_t$ allows us to define for each $t_\in T_0$ the characteristic polynomial of $g_t$ with respect to $\varphi_t$, setting
$$
p_t(w,s)=\prod_{j=1}^\delta (s-g_t(x^{(j)}))=s^\delta+\sum_{j=1}^\delta a_j(t,w)t^{\delta-j}, \quad (w,s)\in P\times{\C},
$$
where $\varphi_t^{-1}(w)=\{x^{(1)},\dots, x^{(\delta)}\}$ consists of exactly $\delta$ points and $a_j(t,w)=(-1)^j\sum_{1\leq i_1<\ldots<i_j\leq\delta}\prod_{r=1}^j g_t(x^{(r)})$ are continuous, and extend the coefficients $a_j(t,\cdot)$ through the critical locus of the branched covering $\varphi_t$ thanks to the Riemann Theorem. We obtain thus $p_t\in\mathcal{O}(P)[s]$ and, clearly, the mapping $$p\colon T_0\times A_0\times {\C}\ni (t,w,s)\mapsto p_t(w,s)\in{\C}$$ is continuous.

Observe that
$$
p^{-1}(0)=\{(t,\varphi_t(x),g_t(x))\in T_0\times P\times {\C}\mid x\in A_0\}.
$$
The set $P\subset{\C}^k$ is a polydisc centred at zero, hence we can write $P=P_1\times\ldots\times P_k=(P_1\times\ldots\times P_n)\times P''$. Now, by the assumptions,
$$p^{-1}(0)\cap (T_0\times \{0\}^n\times P''\times{\C})=T_0\times \{0\}^n\times P''\times\{0\},
$$
which implies that $a_j|_{T_0\times \{0\}^n\times P''}\equiv 0$. We have the following parameter version of the Hadamard Lemma:
\begin{lem}
Let $h=h(t,x,y)\colon T\times D\times U\to {\C}$ be a continuous function where $T$ is a locally compact, 1st countable topological space, $D\subset{\C}^n$ is a convex neighbourhood of the origin and $U\subset{\C}^r$ is an open set. If $h_t\in\mathcal{O}(D\times U)$ and $h|_{T\times\{0\}^n\times U}\equiv 0$, then
$$
h(t,x,y)=\sum_{j=1}^n x_jh_j(t,x,y),\quad (t,x,y)\in T\times D\times U
$$
for some continuous functions $h_j$, holomorphic in $(x,y)$.
\end{lem}
\begin{proof}
Since $h$ is continuous, then for each $t_0$ and $t_\nu\to t_0$, we see as in \cite{D4} Lemma 2.5 that $h_{t_\nu}$ converge to $h_{t_0}$ locally uniformly. But htese are holomorphic functions, hence $\frac{\partial h_{t_\nu}}{\partial x_j}$ converge locally uniformly to $\frac{\partial h_{t_0}}{\partial x_j}$. This in turn implies that that the functions $\tilde{h}_j(t,x,y)=\frac{\partial h_{t}}{\partial x_j}(x,y)$ are continuous (and holomorphic in $(x,y)$).

Then we proceed as Hadamard: thanks to the convexity of $D$ we can write
$$
h(t,x,y)=\int_{[0,1]}\sum_{j=1}^n x_j\tilde{h}_j(t,sx,y)\ ds,\quad (t,x,y)\in T\times D\times U.
$$
Therefore, $h_j(t,x,y):=\int_{[0,1]}\tilde{h}_j(t,sx,y)\ ds$ are the functions looked for. Their continuity follows from classical analysis.
\end{proof}

Applying this Lemma to $a_j$ with $w=(w_1,\dots, w_n, w'')\in P_1\times\ldots\times P_n\times P''$ we obtain $a_j(t,w)=\sum_{i=1}^n w_i a_{ji}(t,w)$ with $a_{ji}$ continuous functions with holomorphic $t$-sections.

Finally, (3) follows from the identity $p(t,\varphi(t,x),g(t,x))\equiv 0$. This ends the proof of Theorem \ref{main}.
\end{proof}

\section{A particular case: isolated zeroes.}

Basing on \cite{D2} Theorem 4.1 (cf. \cite{S} and \cite{D1}) we can obtain a counterpart of the main theorem in the case of improper isolated intersection (a c-holomorphic parameter version of the main result of \cite{Cg}).

Observe that for a c-holomorphic mapping $h\colon A\to{\C}^n$ on a pure $k$-dimensional analytic subset of some open set in ${\C}^m$ containing zero and such that $h^{-1}(0)=\{0\}^m$ we have $$\mathrm{deg}_0 Z_h=i(\Gamma_h\cdot ({\C}^m \times\{0\}^n);0)$$
where the isolated intersection multiplicity $i_0(h):=i(\Gamma_h\cdot ({\C}^m \times\{0\}^n);0)$ is computed according to Draper \cite{Dr} (when $n=k$ i.e. the proper interesection case), or according to the extension of  Achilles-Tworzewski-Winiarski \cite{ATW} in the improper case (i.e. $n>k$; note that always $n\geq k$, for $h$ is necessarily proper in a neighbourhood of zero). In the proper intersection case, this multiplicity is just the geometric multiplicity (covering number) $m_0(f)$.

\begin{thm}
Let $T,A,f$ be as earlier. Assume moreover that $f_{t}(0)=0$ for any $t$, and that $f_{t_0}^{-1}(0)=\{0\}^m$. Then there is a neighbourhood $T_0\times U$ of $(t_0,0)\in T\times A$ and for each $t\in T_0$ there is a neighbourhood $V_t\subset U$ such that for $\delta=i_0(f_{t_0})$, \begin{enumerate}
\item for all $t\in T_0$, $\#\overline{U}\cap f_t^{-1}(0)=\# {U}\cap f_t^{-1}(0)\leq \delta$, $\overline{U}\cap f_{t_0}^{-1}(0)=\{0\}^m$;
\item for all $t\in T_0$, $\overline{V_t}\cap f_t^{-1}(0)=\{0\}$ and for any $g\in \mathcal{O}_c(A\cap V_t)$ such that $g(0)=0$,

$g^\delta\in I_t(V_t)$ where $\delta$ is independent of $t\in T_0$;
\item if $f^{-1}(0)\cap (T_0\times U)=T_0\times\{0\}^m$ and if $g\colon T_0\times (A\cap U)\to {\C}$ is continuous and such that each $g_t\in\mathcal{O}_c(A\cap U)$ vanishes at the origin, then there is a continuous function $h\colon T_0\times (A\cap U)\to {\C}^n$ with c-holomorphic $t$-sections and such that $g^\delta=\sum_{j=1}^n h_{j}f_{j}$.
\end{enumerate}
\end{thm}
\begin{proof}
Fix a neighbourhood $V\subset{\C}^m$ of zero such that $f_{t_0}$ is a proper mapping on $V\cap A$ over some neighbourhood $G\subset{\C}^n$ of the origin. then $X_0:=f_{t_0}(V\cap A)$ is an analytic, pure $k$-dimensional subset of $G$, by the Remmert Theorem. Consider a linear epimorphism $L\colon {\C}^n\to {\C}^k$ such that $\mathrm{Ker} L\cap C_0(X_0)=\{0\}^n$. Then $F:=(L\circ f)\colon T\times A\to {\C}^k$ satisfies the assumptions of our Main Theorem. Let $T_0, U$ and $\delta$ be as in the Main Theorem applied to $F$. Write $A_0:=A\cap U$. We restrict our considerations to this set.

Observe that $F_t=L\circ f_t$. Therefore, since $\deg_0 Z_{F_{t_0}}=i(\Gamma_{F_{t_0}}\cdot(U\times \{0\}^k);0)$ and we check as in the proof of Theorem 2.6 in \cite{D1} that the latter is equal to $i(\Gamma_{f_{t_0}}\cdot(U\times\{0\}^n);0)$, we obtain $\deg_0 Z_{F_{t_0}}=i_0(f_{t_0})$.

Now, as we have $Z_{F_{t}}\stackrel{T}{\longrightarrow} Z_{F_{t_0}}=i_0(f_{t_0})\{0\}^m$ (by the previous proof), the type of convergence implies that --- possibly after shrinking $U$ so that the conclusions of the Main Theorem hold true in a neighbourhood of the compact set $\overline{U}$ --- we have (1). Indeed, since $\deg Z_{F_{t}}=\deg Z_{F_{t_0}}$ whenever $t$ is sufficiently close to $t_0$, we have $\#F_t^{-1}(0)=\#|Z_{F_t}|\leq \deg Z_{F_t}=\delta$. But
$$
F_t(x)=0\ \Leftrightarrow\ x\in f_t^{-1}(\mathrm{Ker} L)
$$
and obviously, $f_t^{-1}(0)\subset f_t^{-1}(\mathrm{Ker} L)$.

Assume for the moment that for all $t\in T_0$ we have $\mathrm{Ker} L\cap f_t(A)=\{0\}^k$ so that $f_t(x)=0$ iff $F_t(x)=0$. Hence, if $g$ vanishes on the zeroes of $f_t$, it does so on the zeroes of $F_t$. Property (2) follows now from the obvious fact that for the components of the mapping we have $(L\circ f_t)_j=L_j\circ f_t$ and as $L_j$ is a linear form, we can write it in the form $L_j(w)=\sum_{i=1}^na_{j,i}w_i$ for some $a_{j,i}\in{\C}$. Therefore, if $g_t^\delta$ is a combination of $(L\circ F)_{t,j}$ with some coefficients $h_{t,i}\in\mathcal{O}_c(A_0)$, $i=1,\dots, n$, we have \begin{align*}
g_t(x)^\delta&=\sum_{i=1}^n h_{t,i}(x) (L_j\circ f_t)(x)=\\
&=\sum_{i=1}^n h_{t,i}(x)\sum_{\kappa=1}^n a_{j,\kappa} f_{t,\kappa}(x).
\end{align*}
It remains to put $\tilde{h}_{t,j}:=\sum_{i=1}^n a_{j,i} h_{t,i}\in\mathcal{O}_c(A_0)$ to get (2).

In the general case, since $\dim f_t^{-1}(0)=0$ and $f_t(0)=0$, we obviously can find neighbourhoods $V_t$ isolating the origin in the fibre for $t\in T_0$. The previous arguments applied for a fixed $t\in T_0$ assure that for each $g\in\mathcal{O}_c(A\cap V_t)$ vanishing at zero, we have $g^{\delta_t}\in I_t(V_t)$ with $\delta_t=\mathrm{deg}_0 Z_{F_t}=i_0(f_t)$. But $\mathrm{deg}_0 Z_{F_t}\leq \deg Z_{F_t}$ and we have already shown that the latter does not exceed $\delta$, whence $\delta_t\leq\delta$. It remains to observe that $g^\delta=g^{\delta-\delta_t}\cdot g^{\delta_t}\in I_t(V_t)$ and (2) is proved.

Finally, (3) follows easily from the above and Theorem \ref{main} (3).
\end{proof}
It should be pointed out that the Kuratowski convergence of the images $X_t:=f_t(A)\stackrel{K}{\longrightarrow} f_{t_0}(A)=:X_{t_0}$ (\footnote{Which we have by Lemma 4.4 from \cite{DD}.}) together with $f_t(0)=0$ for all $t$, does not imply the Kuratowski convergence of the tangent cones. In particular if $\Lambda$ is a linear subspace transversal to $C_0(X_0)$, it need not be transversal to any other $X_t$.
\begin{ex}
Let $f(t,x)=(x^2,tx)$ for $(t,x)\in{\C}^2$ and $t_0:=0$. It satisfies the assumptions of our Theorem, but $X_t=\{t^2u=v^2\}$ are parabol{\ae} converging to the $u$-axis $X_0=\{v^2=0\}$ counted twice for the Tworzewski convergence, by the way). Clearly $\Lambda:=C_0(X_t)=\{u=0\}$ ($t\neq 0$) is transversal to $C_0(X_0)=\{v=0\}$.
\end{ex}

Note also that $L\circ f_t$ can indeed produce some extra zeroes which explains why we were not able to find a neighbourhood of zero independent of the parameter in the proof above:
\begin{ex}
Let $A=\{xy=0\}$ be the union of the two axes in ${\C}^2$. Put for $(t,x,y)\in {\C}\times A$,
$$
f(t,x,y)=\begin{cases}
(x^2, tx), &\textrm{if}\ y=0,\\
(y^2+y, ty), &\textrm{if}\ x=0.
\end{cases}
$$
The assumptions of our Theorem are satisfied for $t_0=0$. Now, $X_0$ is simply the $x$-axis in ${\C}^2$, while $X_t$ ($t\neq 0$) consists of two parabol{\ae} of equations $t^2x=y^2$ and $t^2x=y(y+t)$.

Now, the $y$-axis $\Lambda$ is transversal to $X_0$, but intersects $X_t$ in two points: the origin and $(0,-t)$. Hence for $L(x,y)=x$ we have $(L\circ f_t)^{-1}(0)=\{(0,0), (0,-t)\}$, whereas $f_t$ itself vanishes only at the origin.
\end{ex}

\section{Acknowledgements}

During the preparation of this paper the author was partially supported by Polish Ministry of Science and Higher Education grant  1095/MOB/2013/0.

This article was written during the author's stay at the University Lille 1 whose staff he thanks for excellent working conditions.