text:
\begin{document}
\maketitle

\begin{abstract}
We prove a formula relating the Hausdorff dimension of a subset of the unit interval
and the Hausdorff dimension of the same set with respect to a random path matric on the interval,
which is generated using a multiplicative cascade.
When the random variables generating the cascade are
exponentials of Gaussians,
the well known KPZ formula of  Knizhnik, Polyakov and Zamolodchikov from quantum gravity~\cite{MR947880} appears.
This note was inspired by the recent work of Duplantier and Sheffield~\cite{DSinprep} proving a somewhat different
version of the KPZ formula for Liouville gravity.
In contrast with the Liouville gravity setting,
the one dimensional multiplicative cascade framework facilitates the determination
of the Hausdorff dimension, rather than some expected box count dimension.
\end{abstract}

\section{Introduction}
There is growing interest in establishing a rigorous theory of two dimensional continuum quantum gravity.
Heuristically, quantum gravity is a metric chosen on the sphere uniformly among all possible metrics.
Although there are successful discrete mathematical quantum gravity models, we do not yet
have a satisfactory continuum definition.
A highlight of quantum gravity in the physics literature is the mysterious KPZ formula of
Knizhnik, Polyakov and Zamolodchikov~\cite{MR913671, MR947880}
relating the dimensions of fractals
in the random geometry to the corresponding dimension in Euclidean geometry.
More specifically, the KPZ formula is
\begin{equation}
\label{e.KPZ}
\Dbb-\Daa=\frac{\Dbb(1-\Dbb)}{k+2}\,,
\end{equation}
where $2-2\,\Dbb$ is the dimension of a set in quantum gravity and
$2-2\,\Daa$ is the dimension of the
corresponding set in Euclidean geometry\footnote{The parameter $k$ comes up since it is presumed that
there is essentially one free parameter in the construction of quantum gravity.
This parameter is intimately related to the central charge, and the various variants of
quantum gravity are believed to arise by weighting a uniform measure with the partition
function of a statistical physics model.}.

Recently Duplantier and Sheffield~\cite{DSinprep} were able to prove an expected box count dimension version of the KPZ formula
in Liouville gravity. In their setup, they avoid the very difficult issue of defining the random metric\footnote{We use the term \lq\lq metric\rq\rq\ to mean a \lq\lq distance function\rq\rq, not a Riemannian metric tensor.}, and instead define a random measure.

Multiplicative cascades is a well studied object and
defines naturally a random metric $\rho$ on $[0,1]$.
(The definition will be recalled in Section~\ref{s.bg}.)
The metric space $\bigl([0,1],\rho\bigr)$ is a path metric space, which in this
case just means that it is isometric with some interval $[0,\ell]$ with its Euclidean
metric $|x-y|$. The length $\ell=\rho(0,1)$ of $[0,1]$ in the metric $\rho$ is in general random.
In this note we  prove a formula
relating the Hausdorff dimensions of sets $K\subset [0,1]$ with respect to the random metric $\rho$ on the one hand and
with respect to the Euclidean metric on the other hand.
The KPZ relation appears precisely when the random variables defining the cascade are exponentials of
Gaussians.
Interestingly, this is hinted at in the introduction of~\cite{MR1835030}.

One goal of the note is to establish in this simple one-dimensional setup a Hausdorff dimension version of
the KPZ relation.  We also generalize the discussion to the non-Gaussian setting, mainly to gain
perspective on the underpinnings of the KPZ relation.

The proof in~\cite{DSinprep} is based on large deviations arguments,
which is not the case for the present paper. While in a very general sense one could
say that the ideas are similar, it can reasonably be claimed that our proof is substantially
different and is rather elegant. As we have not yet seen a draft of~\cite{DSinprep},
this comparison is based on lectures by Scott Sheffield and on conversations with him.

\medskip

Early versions of multiplicative cascades were introduced by
Kolmogorov already in 1941~\cite{MR1124922} and were developed by Yaglom~\cite{YaglomTurbulence} and by
Mandelbrot~\cite{MandelbrotCascades}.
Many fundamental properties of multiplicative cascades were first proved
in the remarkable paper~\cite{\KahanePeyrier} by Kahane and Peyri\`ere.
For further
background and references regarding the long history of multiplicative cascades see e.g.,~\cite{MR1835030}.

\medskip

In Section~\ref{s.bg}, we describe the basic setup of multiplicative cascades and
define the random metric $\rho$.
Our main result is stated and proved in Section~\ref{s.haus}.
An appendix follows in which we prove some
essentially known necessary background facts about
multiplicative cascades which we need to use. In some cases, the proofs
in the appendix are simpler than the proofs that we were able to find
in the literature, and in other cases, the results are slightly stronger.

\bigskip
\noindent
{\bf Acknowledgements}: We are obliged to Scott Sheffield for numerous discussions explaining his insights to us.
Yuval Peres and Ed Waymire have been very helpful in enlightening us
with regards to the published work on multiplicative cascades.

\section{Setup}\label{s.bg}

We now describe our setup.
Let $\II_n$ denote the set of dyadic subintervals of $[0,1]$ of length $2^{-n}$;
namely,
$$
\II_n:=\Bigl\{ \bigl[k\,2^{-n},(k+1)\,2^{-n}\bigr]: n\in\N,\,k\in \{0,1,\dots,2^{n}-1\}
\Bigr\}.
$$
Then each interval in $\II_n$ has precisely two subintervals in $\II_{n+1}$,
its left half and its right half.
Also set $\II =\bigcup_{n\in\N} \II_n$.

Let $W$ be some positive random variable with mean $1$,
and let $W_{I}$, $I\in\II$, be an independent collection
of random variables, each of which has the distribution of $W$.
We now define inductively a sequence of random measures on $[0,1]$.
Let $\mu_0$ denote Lebesgue measure on $[0,1]$,
and let $\mu_1:= W_{[0,1]}\,\mu_0$.
Let $\mu_2$ denote the measure which agrees with $W_{[0,1/2]}\,\mu_1$
on $[0,1/2]$ and agrees with $W_{[1/2,1]}\,\mu_1$ on $[1/2,1]$.

Inductively, define $\mu_{n+1}$ as the measure

that on every $I\in \II_{n}$ agrees with $W_I\,\mu_n$.
Alternatively,

$$
\mu_{n}:= w_n\,\mu_0\,,
\qquad\text{ where }\qquad
w_n(x):=\prod_{j=0}^{n-1} W_{I_j(x)}\,,
$$
and $I_j(x)$ denotes the interval $I\in\II_j$ that contains $x$
(and if there is more than one, the one whose maximum is $x$, say).

We will need the following general result regarding multiplicative
cascades.

\begin{theorem}\label{t.bg}
The weak limit $\mu:=\lim_{n\to\infty}\mu_n$ exists almost surely.
Moreover, if $\Eb{W\log_2 W}<1$ then $\mu[0,1]>0$ a.s.\ and $\mu$
has no atoms a.s.
\end{theorem}

This theorem is entirely or almost entirely proved in~\cite{\KahanePeyrier}, but
we present in the appendix a different perhaps simpler proof.
Since $\mu[0,s]$ is a positive martingale for every $s\in[0,1]$, the first
claim is very easy to verify.
In~\cite{\KahanePeyrier} they show that
$\mu[0,1]>0$ a.s.\ if and only if $\Eb{W\log_2 W}<1$.
The claim about the non-existence of atoms would follow from the last remark
in~\cite{\KahanePeyrier}, but we were unable to verify the justification
of that remark (though it does follow under additional moment assumptions).

\smallskip

Henceforth, we will be assuming that
\begin{equation}
\label{e.WlogW}
\Eb{W\log_2W}<1\,.
\end{equation}
By the result of~\cite{\KahanePeyrier} mentioned above, this assumption is
necessary for the limit $\mu$ to be nonzero.

On $[0,1]$, define the random metric $\rho$ by
$$
\rho(x,y):=\mu[x,y]\,,\qquad \text{ for all }0\le x\le y\le 1\,.
$$
If $F(x):=\mu[0,x]$, then $\rho$ is just the pullback of the
Euclidean metric on $[0,F(1)]$ under $F$.
We also set
\begin{equation}
\label{e.ell}
\ell_n:=\mu_n[0,1]\,,\qquad\ell:=\mu[0,1]\,.
\end{equation}
Clearly, $\Eb{\ell_n}=1$ and $\Eb{\ell}\le 1$.
In fact, $\Eb{\ell}=1$ by~\cite{\KahanePeyrier}, but we do not
need this result.

\section{Hausdorff dimension}\label{s.haus}

\begin{theorem}\label{t.haus}
Suppose that
\begin{equation}\label{e.negmom}

\Eb{W^{-s}}<\infty
\qquad
\text{for all $s\in[0,1)$}
\end{equation}

(in addition to the standard
assumptions
$\Eb{W\log_2W}<1$,
$\Eb{W}=1$,  and $\Pb{W>0}=1$.)
Let $K\subset [0,1]$ be some
(deterministic) nonempty set, let $\Da$ denote its Hausdorff
dimension with respect to the Euclidean metric, and let
$\Db$ denote its Hausdorff dimension with respect to
the random metric $\rho$.
Then a.s.\ $\Db$  is the unique solution of the equation
\begin{equation}
\label{e.Da}
2^{\Da}=\frac{2^{\Db}}{\Eb{W^\Db}}
\end{equation}
in $[0,1]$.
\end{theorem}

As our proof shows, the assumption~\eqref{e.negmom}
may be significantly relaxed.
See Theorems \ref{t.upperbound} and \ref{t.lowerbd}.

\medskip

Now consider the case in which $\log W$ is a Gaussian random variable. Since $\Es{W}=1$,
this implies $W=\exp(\sigma\,Y-\sigma^2/2)$, where $Y$ is a standard Gaussian
of zero mean and unit variance and $\sigma\ge 0$. The assumption~\eqref{e.WlogW}
is then equivalent to the requirement $\sigma^2<\log 4$. In this case,
the moments $\Eb{W^s}$ are easily evaluated and~\eqref{e.Da} gives
\begin{equation}
\label{e.gaussian}
\Da - \Db = \frac{\sigma^2}{\log 4}\,\Db \,(1-\Db)\,,
\end{equation}
in agreement with~\eqref{e.KPZ}.
\footnote{In~\eqref{e.KPZ}, $2\,(1-\Daa)$ is the dimension and similarly for $\Dbb$.
The factor of $2$ comes from the fact that the ambient space is two dimensional,
and that the dimension is defined in terms of the measure, not the distance function.
The transition from $\Daa$ to $1-\Daa$ is a passage from the dimension to the
co-dimension, and does not change the form of~\eqref{e.KPZ}.}

For comparison, suppose instead that $W=1\pm\sigma$, each with probability $1/2$.
Then~\eqref{e.WlogW} becomes $|\sigma|<1$ and~\eqref{e.Da} transforms to
$$
2^{\Da}=\frac{2^\Db}{\frac 12 (1-\sigma)^\Db+\frac 12 (1+\sigma)^\Db}\,,
$$
or
$$
\Da = 1+\Db-\log_2\bigl((1-\sigma)^\Db+(1+\sigma)^\Db\bigr).
$$

\bigskip

We now proceed to prove Theorem~\ref{t.haus}.
Define
$$
\phi(s):=s-\log_2 \Eb{W^s}\,.
$$
Then~\eqref{e.Da} reads $\Da=\phi(\Db)$.
The following lemma implies the uniqueness of the $\Db$ satisfying~\eqref{e.Da}.

\begin{lemma}\label{l.psi}
The function $\phi$ is continuous, strictly monotone increasing in $[0,1]$
and maps $[0,1]$ onto $[0,1]$.
\end{lemma}

\proof
Set $\psi(s):= \Eb{(W/2)^s}$.
Continuity of $\psi$ follows from the dominated convergence theorem and convexity
of $\psi$ is immediate by the convexity of $(W/2)^s$ in $s$.
Since $\psi$ is convex and
$$
\psi'(1-)=\Eb{(W/2)\log(W/2)}=\frac 12\, \Eb{W\log W}-\frac 12\log 2<0,
$$
it is strictly monotone decreasing in $[0,1]$.
The lemma follows since $\phi=-\log_2 \psi$, $\phi(0)=0$ and $\phi(1)=1$.
\QED

The following simple lemma can serve to motivate Theorem~\ref{t.haus}, and
is also important in its proof.

\begin{lemma}\label{l.interval}
Let $x,y\in[0,1]$, and let $s\in(0,1]$.
Then
$$
\Eb{\rho(x,y)^s} \le
8\,|x-y|^{\phi(s)}.
$$
\end{lemma}
\proof
Let $[a,b]\in\II_n$.
Then by the construction of $\rho$ and the independence of the different
variables $W_I$, $I\in\II$,
$$
\Eb{\rho(a,b)^s}= 2^{-ns}\,\Eb{W^s}^n\,\Eb{\ell^s}
=|a-b|^{\phi(s)}\,\Eb{\ell^s}.
$$
Now, Jensen's inequality gives $\Eb{\ell^s}\le \Eb{\ell}^s\le 1$.
Note that if $|y-x|\in (2^{-n-1},2^{-n}]$, then
the interval joining $x$ and $y$ can be covered by
two consecutive intervals in $\II_n$, say $[a,b]$ and $[b,c]$.
Then
\begin{equation*}
\begin{aligned}
\Eb{\rho(x,y)^s}
&
\le \Eb{\bigl(\rho(a,b)+\rho(b,c)\bigr)^s}
\\&
\le \Eb{\bigl(2\,\rho(a,b)\bigr)^s+\bigl(2\,\rho(b,c)\bigr)^s}
\\&
= 2^{1+s}\,\Eb{\rho(a,b)^s}
\\&
\le 2^{1+s}\,|a-b|^{\phi(s)}
\le 2^{1+s+\phi(s)}\,|x-y|^{\phi(s)}\,.
\end{aligned}
\end{equation*}

The lemma follows, since $\phi(s)\le 1$ by Lemma~\ref{l.psi}.
\QED

\begin{theorem}
\label{t.upperbound}
Let $K$, $\Da$ and $\Db$ be as in Theorem~\ref{t.haus}.
Then a.s.  $\phi(\Db)\le \Da$.
\end{theorem}

It is worth pointing out that we are not assuming~\eqref{e.negmom} here.

\proof
Let $s\in[0,1]$ and assume that $t:=\phi(s)>\Da$.
We now show that $s \ge \Db$ a.s.
Let $\eps>0$.  Then there is a covering of
$K$ by at most countably many intervals $[x_i,y_i]$
such that $\sum_i |x_i-y_i|^t<\eps$.
By Lemma~\ref{l.interval}, we have
$$
\EB{\sum_i \rho(x_i,y_i)^s} \le
8\, \sum_i |x_i-y_i|^t \le 8\,\eps\,.
$$
By Markov's inequality, with probability at least $1-\sqrt\eps$
we have a covering of $K$ with balls whose radii in the $\rho$ metric
satisfy $\sum r_i^s \le 8\,\sqrt\eps$.
Thus $s\ge \Db$ a.s.
Hence $\Db\le\inf\phi^{-1}(\Da,1]$.
By Lemma~\ref{l.psi}, the theorem follows.
\QED

\begin{theorem}
\label{t.lowerbd}
Let $K$, $\Da$ and $\Db$ be as in Theorem~\ref{t.haus}.
Then a.s.
$$
\Db \ge \sup\bigl\{s\in(0,1): \phi(s)<\Da,\,
\Es{W^{-s}}<\infty\bigr\}.
$$

\end{theorem}

\proof
Suppose that $s\in(0,1)$ satisfies $t:=\phi(s)<\Da$ and $\Es{W^{-s}}<\infty$.
We need to prove that $\Db\ge s$.
Since $\Es{W^s}$ is convex in $s$ and equals to $1$ at $s=0,1$,
we have $t\ge s\ge0$.

Since $\Da>t$, by Frostman's lemma~\cite[Chapter 8]{\Mattila} there is a Borel
probability measure $\nu_0$ supported on $K$
such that
\begin{equation}
\label{e.nuen}
\mathcal E_t(\nu_0):=\iint \frac{d\nu_0(x)\,d\nu_0(y)}{|x-y|^t}<\infty\,.
\end{equation}
Set $a:=\Es{W^s}$, $Z:= W^s/a$ and $Z_I:= W_I^s/a$.
Define
$$
f_n(x):= \prod_{j<n}Z_{I_j(x)},\qquad \nu_n:= f_n\,\nu_0\,.
$$

Since for every $a\in[0,1]$ the sequence $\nu_n[0,a]$ is
a non-negative martingale, it easily follows that
the weak limit $\nu:=\lim_{n\to\infty} \nu_n$ exists.
(See, e.g., the proof of Theorem~\ref{t.bg} in the appendix.)

Then the support of $\nu$ is contained in the
support of $\nu_0$ and therefore in $K$.

Define
$$
\rho_n(x,y):=\rho(x,y)\vee \mu\bigl(I_n(x)\bigr)\vee\mu\bigl(I_n(y)\bigr).
$$
In order to estimate the expectation of
\begin{equation}
\label{e.nunun}
\mathcal E_s(\nu_n;\rho_n):=\iint \frac{d\nu_n(x)\,d\nu_n(y)}{\rho_n(x,y)^s}
\end{equation}
we fix $x,y\in [0,1]$, and estimate
$$
\Eb{f_n(x)\,f_n(y)\,\rho_n(x,y)^{-s}}.
$$
Let $k$ be the smallest integer such that $[x,y]$ contains
some interval of $\II_k$. Then
\begin{equation}
\label{e.xyd}
  |x-y|< 4\cdot 2^{-k}.
\end{equation}
Let $J\in \II_k$ satisfy $J\subset [x,y]$, and let
$J'\in \II_{k-1}$ satisfy $J'\supset J$. Then $x\in J'$ or $y\in J'$.
By symmetry, assume $x\in J'$. Let $\ev G$ denote the $\sigma$-field
$\langle W_{I_j(x)}, W_{I_j(y)}:j< n\rangle$.
Assume first that $k\le n$.
Then
$$
\Eb{\rho(x,y)^{-s}\md \ev G} \le
\Eb{\mu(J)^{-s}\md \ev G} = 2^{ks}\,\Eb{\ell^{-s}}\prod_{j=0}^{k-1} W_{I_j(x)}^{-s}\,.
$$
By Lemma~\ref{l.lowertail} in the appendix and our assumption that
$\Eb{W^{-s}}<\infty$, we have $\Eb{\ell^{-s}}<\infty$. Using~\eqref{e.xyd},
we therefore obtain
$$
\Eb{\rho(x,y)^{-s}\md \ev G} \le O(1)\,
|x-y|^{-s}\,\prod_{j=0}^{k-1} W_{I_j(x)}^{-s}\,,
$$
where the implied constant may depend on $s$ and the law of $W$.
Now,
$$
f_n(x)=a^{-n}\prod_{j=0}^{n-1} W_{I_j(x)}^s,
$$
and we have a similar expression for $f_n(y)$.
Thus,
$$
\Eb{f_n(x)\,f_n(y)\,\rho(x,y)^{-s}\md \ev G} \le
O(1)\,a^{-2n}\,|x-y|^{-s}\prod_{j=k}^{n-1} W_{I_j(x)}^s\prod_{j=0}^{n-1} W_{I_j(y)}^s\,.
$$
Note that for $j\ge k$ we have $I_j(x)\ne I_j(y)$.
Taking expectations and using the definition of $a$ yields
\begin{multline*}
\Eb{f_n(x)\,f_n(y)\,\rho_n(x,y)^{-s}}\le
\Eb{f_n(x)\,f_n(y)\,\rho(x,y)^{-s}}
\\
\le O(1)\,|x-y|^{-s}\,a^{-k}
\overset{\eqref{e.xyd}}{\le} O(1)\,|x-y|^{-s+\log_2 a}
=
O(1)\,|x-y|^{-t}.
\end{multline*}
Now, if $k>n$, we have instead
$$
\Eb{\rho_n(x,y)^{-s}\md \ev G} \le O(1)\,\Eb{\mu\bigl(I_n(x)\bigr)^{-s}\md\ev G} \le
O(1)\,2^{sn}\prod_{j=0}^{n-1} W_{I_j(x)}^{-s}\,.
$$
The above argument therefore gives in this case,
$$
\Eb{f_n(x)\,f_n(y)\,\rho_n(x,y)^{-s}}
\le O(1)\, 2^{ns}\,a^{-n}\le O(1)\,|x-y|^{-t}.
$$
Thus, we have
$
\Eb{f_n(x)\,f_n(y)\,\rho_n(x,y)^{-s}} \le O(1)\,|x-y|^{-t}
$
for every $x,y\in[0,1]$. Integrating this with respect to $d\nu_0(x)\times d\nu_0(y)$
and applying Fubini, one obtains
\begin{equation}
\label{e.nun}
\Eb{\mathcal E_s(\nu_n;\rho_n)} \le O(1)\,\mathcal E_t(\nu_0)\,.
\end{equation}
Since $\rho_n(x,y)\le \ell$ holds for $x,y\in[0,1]$, this estimate gives
$$
\Eb{(\nu_n[0,1])^2\,\ell^{-s}} \le O(1)\,\mathcal E_t(\nu_0)\,.
$$
Now H\"older's inequality comes into play:
$$
\Eb{(\nu_n[0,1])^{2/(1+s)}} \le
\Eb{(\nu_n[0,1])^2\,\ell^{-s}}^{1/(1+s)} \,\Eb{\ell}^{s/(1+s)}
\le O(1) \,\ev E_t(\nu_0)^{1/(1+s)}.
$$
Thus, the martingale sequence $\nu_n[0,1]$ is uniformly bounded in
$L^p$ with $p=2/(1+s)>1$. It follows by the corresponding martingale
convergence theorem that $\Eb{\nu[0,1]}=\nu_0[0,1]=1$,
and in particular, $\nu[0,1]>0$ with positive probability.
The event $\nu[0,1]>0$ is clearly independent of $\sigma$-field
generated by any finite number of the random variables
$W_{I}$, and therefore has probability $0$ or $1$, and in this case,
$\Pb{\nu[0,1]>0}=1$.

Since a.s.\ $\rho$ is continuous, $\rho_n\to\rho$ uniformly as $n\to\infty$ and $\nu_n\to\nu$
weakly, we have a.s.
$$
\ev E_s(\nu;\rho) \le \liminf_{n\to\infty} \ev E_s(\nu_n;\rho_n)\overset{\eqref{e.nun}}<\infty\,.
$$
The proof is now completed by appealing to
Frostman's criterion~\cite[Chapter 8]{\Mattila},
since $\nu[0,1]>0$ a.s.
\QED

\proofof{Theorem~\ref{t.haus}}
The theorem follows immediately from Lemma~\ref{l.psi} and Theorems~\ref{t.upperbound}
and~\ref{t.lowerbd}.
\QED

\appendix
\section{Some multiplicative cascades background}

\begin{lemma}
\label{l.xlogx}
Our standing assumption $\Eb{W\log_2 W}<1$ implies that $\ell>0$ a.s.
\end{lemma}

\proof
Set $a:=\Eb{W\log_2W}<1$.
We first prove that $\ell>0$ with positive probability.
Since $\ell_n:=\mu_n[0,1]$ is a positive martingale, we have a.s.\
convergence $\ell_n\to \ell$.

The proof will come out of a recurrence relation for
the sequence $b_n:=\Eb{\ell_n\log_2\ell_n}$.

Let $\ell_n'$ and $\ell_n''$ have the law of
$\ell_n$ and be independent and independent
from $W$. Then the law of $\ell_{n+1}$ is the same
as the law of $W\,(\ell_n'+\ell_n'')/2$.
Thus,
$$
\begin{aligned}
&b_{n+1}

=
\Eb{(W/2)\,(\ell_n'+\ell_n'')\log_2 W}+
{}
\\& \qquad\qquad
{}+
\Eb{(W/2)\,(\ell_n'+\ell_n'')\log_2 (\ell_n'+\ell_n'')}+
\Eb{(W/2)\,(\ell_n'+\ell_n'')\log_2 (1/2)}.
\end{aligned}
$$
We now use independence, $\Eb{\ell_n'}=\Eb{W}=1$, the symmetry
between $\ell_n'$ and $\ell_n''$, and the definition of $a$,
and simplify the above to
$$
b_{n+1} = a+ \Eb{\ell_n'\log_2 (\ell_n'+\ell_n'')}-1\,.
$$
Thus,
$$
\begin{aligned}
b_{n+1}-b_n
&
=
a-1+
\Eb{\ell_n'\log_2 (\ell_n'+\ell_n'')}-
\Eb{\ell_n'\log_2 \ell_n'}
\\&
=
a-1+\Eb{\ell_n'\log_2\bigl((\ell_n'+\ell_n'')/\ell_n'\bigr)}.
\end{aligned}
$$
Since $\log_2(1+x/\ell_n')$ is concave as a function of $x$,
and since $\ell_n''$ is independent from $\ell_n'$, Jensen's inequality
applied to the above gives
\begin{multline}
\label{e.bb}

b_{n+1}-b_n\le a-1+\Eb{\ell_n'\log_2(1+\Es{\ell_n''}/\ell_n')}
\\
 =a-1+\Eb{\ell_n\log_2(1+1/\ell_n)}.

\end{multline}
Because $\inf\{b_n:n\in\N\} \ge \inf \{x\log_2 x:x>0\}>-\infty$ and
$a<1$, the set $S:=\{n\in\N: b_{n+1}-b_n>(a-1)/2\}$
is infinite.
For $n\in S$,~\eqref{e.bb} implies
$$
\Eb{g(\ell_n)}>(1-a)/2\,,\qquad \text{where}\qquad
g(x):=x\log_2(1+1/x)\,.
$$
Note that $c:= \sup \{g(x):x>0\}<\infty$.
For all $\eps>0$ and $n\in S$, we have
$$
(1-a)/2 <\Eb{g(\ell_n)} \le
c\,\Pb{\ell_n\ge \eps} + \sup\bigl\{g(x):0<x<\eps\bigr\}\,.
$$
By taking $\eps$ sufficiently small, we can make sure that the last summand
is at most $(1-a)/4$. Then, $ \Pb{\ell_n>\eps}> (1-a)/(4\,c) $.
This proves that $\ell_n$ does not tend to zero in probability, and
hence does not tend to zero a.s.
Thus $\Pb{\ell>0}>0$.

Set $\ell':=2\,\mu[0,1/2]/W_{[0,1]}$ and $\ell'':=2\,\mu[1/2,1]/W_{[0,1]}$.
Then $\ell',\ell''$ and $W_{[0,1]}$ are independent, and each of $\ell'$ and
$\ell''$ has the law of $\ell=(W_{[0,1]}/2)\,(\ell'+\ell'')$.
 But $\ell=0$ if and only if $\ell'=0=\ell''$, since $W>0$ a.s.
Thus,
$$
\Pb{\ell=0}=\Pb{\ell'=0,\,\ell''=0}=\Pb{\ell=0}^2.
$$
 Since $\Pb{\ell>0}>0$, we conclude that $\ell>0$ a.s., which completes the proof.
\QED

\proofof{Theorem~\ref{t.bg}}
Since $\Eb{\mu_n[0,1]}=1$, the sequence of measures $\mu_n$ is tight
in the space of Borel measures on $[0,1]$ with the topology of
weak convergence. Thus, some subsequence
converges to a limit $\mu$.
For every rational $r\in[0,1]\cap\Q$ the sequence $\mu_n[0,r]$ is a positive martingale,
and hence the limit $f(r):=\lim_{n\to\infty} \mu_n[0,r]$ exists
for all $r\in\Q\cap[0,1]$ a.s.
It is immediate to verify that $\mu[0,x)=\sup\bigl\{f(r):r\in\Q\cap[0,x)\bigr\}$
and $\mu[0,1]=f(1)$. This implies that the subsequential limit $\mu$  is
unique, and hence is the weak limit of the entire sequence $\mu_n$.
Thus, the theorem follows from Lemma~\ref{l.xlogx} and the next lemma.
\QED

\begin{lemma}\label{l.noatom}
A.s.\ $\mu$ has no atoms.
\end{lemma}
\proof
Let $Z$ have the distribution of the $\mu$-measure of the
largest atom in $[0,1]$.  Clearly, $\Eb{Z}\le 1$.
Let $Z_1$ and $Z_2$ have the law of
$Z$ with $W,Z_1$ and $Z_2$ independent. Then $W\max(Z_1,Z_2)/2$
has the law of $Z$.  Therefore,
$$
\Eb{Z_1+Z_2}/2 = \Eb{Z}= \Eb{W\max(Z_1,Z_2)/2}=
\Eb{\max(Z_1,Z_2)}/2\,.
$$
Since $Z_1+Z_2\ge \max(Z_1,Z_2)$, and the expectations are the same,
it follows that they are equal a.s.
Thus, on the event $Z_1>0$, we have $Z_2=0$ a.s.
Since $Z_1$ and $Z_2$ are independent, we get $Z_1=0$ a.s.,
as required.
\QED

\begin{lemma}\label{l.lowertail}
Suppose that
\begin{equation}
\label{e.Wr}
\Eb{W^{-r}}<\infty
\end{equation}
for some constant $r>0$.
Then also
$$
\Eb{\ell^{-r}}<\infty\,.
$$
\end{lemma}

A very slightly weaker form of this lemma can be found
in~\cite{MR1847093}, where the assumption~\eqref{e.Wr} is the same,
but the conclusion is that $\Eb{\ell^{-s}}<\infty$ for all $s\in[0,r)$.
However, the setup in~\cite{MR1847093} is more general.

\proof
Let $\ell_1:= 2\,\mu[0,1/2]/W$ and $\ell_2:= 2\,\mu[1/2,1]/W$.
By construction $\ell_1,\ell_2$ and $W$ are independent and each of
$\ell_1$ and $\ell_2$ has
the law of $\ell$.
Moreover
\begin{equation}
\label{e.recursion}
\ell=W\,(\ell_1+\ell_2)/2\,.
\end{equation}
Assume, for a moment, that there is some $\delta>0$
such that
\begin{equation}
\label{e.del}
  \Eb{\ell^{-\delta}}<\infty\,.
\end{equation}
By the means inequality and~\eqref{e.recursion}, we have
\begin{equation}
\label{e.recm}
\ell \ge W \sqrt{\ell_1\,\ell_2}\,.
\end{equation}
Now independence gives for every $s>0$
\begin{equation}
\label{e.boot}
\Eb{\ell^{-s}}\le \Eb{W^{-s}}\,\Eb{\ell_1^{-s/2}}\,\Eb{\ell_1^{-s/2}}
=
\Eb{W^{-s}}\Eb{\ell^{-s/2}}^2.
\end{equation}
Let $S$ denote the set of $s\in[0,\infty)$ such that $\Eb{\ell^{-s}}<\infty$.
By~\eqref{e.del}, we have $\delta\in S$.
Since $\Eb{W^{-s}}$ is a convex function of $s$, the set
$S$ must be an interval. Moreover, $[0,\delta]\subset S$.
Similarly, $\Eb{W^{-s}}<\infty$ for every $s\in[0,r]$.
Now,~\eqref{e.boot} shows that $[0,r]\cap (2 \,S)\subset S$, where
$2\,S=\{2\,s:s\in S\}$.  This implies that $[0,r]\subset S$, as needed.

It remains to prove~\eqref{e.del}. From~\eqref{e.Wr} we know that
\begin{equation}
\label{e.Wtail}
\lim_{x\searrow 0}\, x^{-r}\,\Pb{W<x}=0\,.
\end{equation}
By~\eqref{e.recursion}, for every $b,x>0$, we have
that if $\ell<b\,x/2$ then $W<x$ or $\ell_1+\ell_2<b$.
Thus,
\begin{equation}\label{e.rec1}
\begin{aligned}
\Pb{\ell<b\,x/2}
&
\le
\Pb{W<x}+\Pb{\ell_1+\ell_2<b}
\\&
\le
\Pb{W<x}+\Pb{\ell_1<b,\,\ell_2<b}
\\&
=
\Pb{W<x}+\Pb{\ell<b}^2.
\end{aligned}
\end{equation}
Set $\eps_j:=2^{-2^j-1}$ for $j\in\N$.
Then $\eps_{j+1}=2\,\eps_j^2$.
Let $b_0>0$ satisfy
$$
\Pb{\ell<b_0}\le \eps_0
$$
(Theorem~\ref{t.bg} implies the existence of such a $b_0$).
Set
$$
x_j:=\sup\Bigl\{x:\Pb{W<x}\le \eps_j^2\Bigr\}\,.
$$
Then our assumption~\eqref{e.Wtail} implies
that $x_j\ge \eps_j^{2/r}$ for all
but finitely many $j$.
Inductively define
$$
b_{j+1}=b_j\,x_j/2\,.
$$
Then using induction, the relation $\eps_{j+1}=2\,\eps_j^2$ and
the estimate~\eqref{e.rec1} give
\begin{equation}
\label{e.jj}
\Pb{\ell<b_j}\le \eps_j\,.
\end{equation}
The definition of $b_j$ gives
$$
b_j=2^{-j}\,b_0\,\prod_{k=0}^{j-1} x_k \ge C\, 2^{-j}\prod_{k=0}^{j-1}\eps_k^{2/r}\,,
$$
where $C>0$ is the product of $b_0$ and the finitely many $x_k$
that satisfy $x_k< \eps_k^{2/r}$.
Taking into account the definition of $\eps_j$, we get
$$
b_j\ge C\, 2^{-j-2(j-1+2^j)/r}.
$$

Now,
$$
\Eb{\ell^{-\delta}}
\le
b_0^{-\delta} +
\sum_{j=0}^\infty b_{j+1}^{-\delta}\,\Pb{b_{j+1}\le \ell< b_j}.
$$
For all but finitely many $j$, the above estimate on $b_j$ gives
$b_{j+1}^{-\delta}\le 2^{5 \delta 2^j/r}$, while
$\Pb{b_{j+1}\le \ell< b_j} \le \Pb{\ell< b_j} \overset{\eqref{e.jj}}{\le}
\eps_j =2^{-2^j-1}$.
Thus, we have~\eqref{e.del} for every $\delta\in (0,r/5)$,
and the proof is thus complete.
\QED

\bibliographystyle{halpha}
\bibliography{mr,prep,notmr}

\end{document}

config: arxiv


TITLE: converting ordinary Differential Equations to Polar Coordinate system
QUESTION [0 upvotes]: I want to convert this system into Polar coordinates
$$\dfrac{dx}{dt} = y,\\
\dfrac{dy}{dt} = -\mu(x^2 + y^2 - 1)y - x$$
I know in order to convert this ODE system in polar coordinates, you would use these formulas, $$x \,dx/dt+y\,dy/dt=r\,dr/dt,\\ \,ydx/dt-x\,dy/dt=-r^2\,dθ/dt.$$  Here is what I have so far,
$$xy+y(-μ(r^2-1)y-x)=r\,dr/dt,\\ y^2-x(-μ(r^2-1))*y-x.$$
Can someone please tell me how I convert the system of $dy/dt$ and $dx/dt$ into polar coordinates?

REPLY [1 votes]: You can get a polar coordinates relationship linked to stability by adding
$$
\cases{
x x'= x y\\
y y'=-x y - \mu y^2(x^2+y^2-1)
}\Rightarrow \frac 12(x^2+y^2)'=-\mu y^2(x^2+y^2-1)
$$
and now using the polar coordinates
$$
\cases{
x=r\cos\theta\\
y=r\sin\theta
}
$$
we get
$$
\frac 12(r^2)' = -\mu r^2\sin^2\theta(r^2-1)
$$
NOTE
$$
\left(\begin{array}{c}\dot x\\ \dot y \end{array}\right) = \left(\begin{array}{cc}\cos\theta & -r\sin\theta\\ \sin\theta & r\cos\theta \end{array}\right)\left(\begin{array}{c}\dot r\\ \dot \theta \end{array}\right) = \left(\begin{array}{c}r\sin\theta\\ -\mu(r^2-1)r\sin\theta-r\cos\theta\end{array}\right)
$$
and finally
$$
\cases{
\dot r = -\mu  r \left(r^2-1\right) \sin ^2\theta\\
\dot\theta = -\mu  \left(r^2-1\right) \sin\theta\cos\theta-1
}
$$


TITLE: Transpose applications and matrices properties
QUESTION [0 upvotes]: Good Afternnon
If $f$ is a linear transformation mapping then if the transformation matrix of linear transformation mapping has been defined and its transpose properties are known as well.
Why, what I have written below, is called a definition rather than a proposition?? Is it possible to deduce it from the matrices?

$E,F$ two finite vector-spaces.
$\psi\in \mathcal{L}(E,F)$. The transpose of $\psi$ is the linear map :
$\begin{array}l ^t\psi :&F^*&\to E^*\\&\phi&\to \phi\circ\psi\end{array}$

REPLY [2 votes]: Because the transpose of a matrix is one thing and the tranpose of a linear map is a distinct concept.


TITLE: Is it true that a processor with a higher temperature uses inevitably more power in comparison with a processor with a lower temperature?
QUESTION [1 upvotes]: If a processor has a higher temperature, can you be sure it it consumes more power than a processor with a lower temperature?
And is the reverse relationship true?
A processorchip with a higher mean power consumption will always work at a higher temperature than a processer with a lower mean power consumption?

REPLY [3 votes]: No.  Power and temperature are related, but not directly.  Thermal environment and surface area also matter a great deal.
A tiny chip has less area to dissipate heat than a larger one.  So for the same power draw, it may easily produce a higher temperature than a larger chip.
Imagine a 40W incandescent light bulb.  Both the filament and the bulb as a whole are dissipating 40W of power.  But the tiny filament can do so only by climbing to a very high temperature of a few thousand $K$.  The larger bulb surface dissipates the same amount at a few hundred $K$.
Design affect things as well.  If you have a chip that happens to be quite robust at high temperature, you might allow it to reach high temperatures in your product.  But a more powerful one might need aggressive cooling, which keeps it at a lower temperature.


TITLE: Find $A,B\in\Bbb K^{2\times 2}$ such that $AB\neq BA$ but $e^{A+B}=e^Ae^B$
QUESTION [6 upvotes]: Find $A,B\in\Bbb K^{2\times 2}$ such that $AB\neq BA$ but $e^{A+B}=e^Ae^B$. Hint: $e^{2k\pi i}=1$ for all $k\in\Bbb Z$.

This is the exercise 10 in page 146 of Analysis II of Amann and Escher. I dont know exactly what to do here more than just try things blindly (trying to guess what is the hint about).
I know that
$$A:=\begin{bmatrix}a&b\\0&c\end{bmatrix}\implies e^A=\begin{cases}\begin{bmatrix}e^{a}&\frac{b}{c-a}(e^{c}-e^{a})\\0&e^{c}\end{bmatrix},& c\neq a\\\begin{bmatrix}e^{a}&be^{a}\\0&e^{a}\end{bmatrix},& c= a\end{cases}\tag1$$
$$A:=\begin{bmatrix}0&-\omega\\\omega&0\end{bmatrix}\implies e^{A}=\begin{bmatrix}\cos \omega&-\sin\omega \\\sin\omega &\cos\omega \end{bmatrix}\tag2$$
What I thought about the hint is setup some matrices as in $(1)$ such that $i(A+B)=i2\pi I$, by example
$$A:=\begin{bmatrix}a&b\\0&c\end{bmatrix},\quad B:=\begin{bmatrix}2\pi-a&-b\\0&2\pi-c\end{bmatrix}\implies e^{i(A+B)}=I$$
However we have that $AB=BA$, then I must find other way. Some help will be appreciated, thank you.

REPLY [3 votes]: E.g.
$$
A=\pmatrix{0&2\pi\\ -8\pi&0},
\ B=\pmatrix{0&8\pi\\ -2\pi&0}.
$$
The key is that, since $e^{2k\pi i}=1$, it suffices to generate two matrices $A$ and $B$ so that $A,B$ and $A+B$ are similar to integer multiples of $J=\pmatrix{0&2\pi\\ -2\pi&0}$. If $A$ and $B$ are chosen at random, they usually do not commute.
For instance, if we pick
$$
A=\pmatrix{0&2a\pi\\ -2b\pi&0},
\ B=\pmatrix{0&2c\pi\\ -2d\pi&0},
$$
for some $a,b,c,d\in\mathbb N$, then $A$ and $B$ are similar to $\sqrt{ab}J$ and $\sqrt{cd}J$ respectively, and hence $e^A=e^B=I$ when $ab$ and $cd$ are perfect squares. Similarly, if $(a+c)(b+d)$ is also a perfect square, then $e^{A+B}=I=e^Ae^B$. Finally, if $(a,b)$ and $(c,d)$ are linearly independent, then $AB\ne BA$. So, we can generate a bunch of desired examples using a simple Octave/Matlab script:
for a=1:10,for b=1:10,for c=1:10,for d=1:10,
  if det([a b;c d])~=0
    && abs(sqrt(a*b)-floor(sqrt(a*b)))<1e-6
    && abs(sqrt(c*d)-floor(sqrt(c*d)))<1e-6
    && abs(sqrt((a+c)*(b+d))-floor(sqrt((a+c)*(b+d))))<1e-6,
    sprintf("%d %d %d %d", a,b,c,d),
  end;
end;end;end;end


TITLE: Arrangement Question
QUESTION [3 upvotes]: I'm trying to represent the number of ways to arrange n blue and 4 red balls whose colors are indistinguishable
I've come up with, $${{n+4} \choose 4}$$
but I'm not sure how it works... How come I choose 4 but not n, and how come i am not permuting anything, since I'm dealing with arrangements?

REPLY [4 votes]: First, note that $\dbinom{n+4}{4}=\dbinom{n+4}{n}$, so "choosing $n$" would give the same answer.
In general, $\dbinom{m}{k}=\dbinom{m}{m-k}$. Rows of Pascal's Triangle read the same forwards as backwards. If we wish, we can verify this by using the formula for binomial coefficients in terms of factorials.
But there is a better way. There are exactly as many ways of picking $k$ "winners" from $m$ people as there are of picking $m-k$ "losers."
Second, yes, we are counting permutations, but binomial coefficients, which are useful in counting certain kinds of combinations, have many other uses.
To justify the answer $\dbinom{n+4}{4}$, imagine that we have $n+4$ small boxes in a row. There are $\dbinom{n+4}{4}$ ways to choose the boxes that the red marbles will occupy.  Probably that you had an argument like this in mind.
To put it another way, suppose we want to make an $(n+4)$-letter "word," using $4$ R's  and $n$ B's. There are $\dbinom{n+4}{4}$ ways to choose the position of the R's. Once we have chosen where the R's will go, the whole word is determined.

REPLY [2 votes]: The symmetry of the binomial coefficients tells you that ${{n+4} \choose 4}={{n+4} \choose n}$ so you can choose $n$ if you like.  Your argument is one way to see this symmetry.  In fact ${{n+4} \choose 4}=\frac {(n+4)!}{n!4!}$ which makes the symmetry evident.


TITLE: What is the value of: $\lim_{n\rightarrow \infty}\sum_{r=1}^{n-1}\frac{\cot^2(r\pi/n)}{n^2}$
QUESTION [0 upvotes]: $$\lim_{n\rightarrow \infty}\sum_{r=1}^{n-1}\frac{\cot^2(r\pi/n)}{n^2}$$

How can I calculate the value of this trigonometric function where limits tends to infinity?  I have thought and tried various of methods like using:

Tan(x)/x property
Converting lim to
Integration
Sandwich theorem
Trigonometric series

But none worked out for me.
So please tell me a good and easy approach for this question :)

REPLY [2 votes]: Here I came up with a solution involving only elementary calculation, I hope it's appreciable...
(Sorry as it's in image form)


TITLE: The work-energy theorem
QUESTION [2 upvotes]: Well here's the question.
From some previous excercises we know that from
\begin{align}
A&=\int F\;ds,\\
 &=\int ma\;ds,   &&(F=ma)\\
 &=\int m \frac{dv}{dt}\;ds, &&(a=dv/dt)\\
 &=m \int_{v_1}^{v_2}v\; dv,\\
 &=m  \frac{v_2^2}{2}-m  \frac{v_1^2}{2},\\
 &=W_2-W_1,   &&(W_i=\frac12mv_i^2)\\
 &=\Delta W.
\end{align}
Meanwhile for potential energy we have the shown figure
\begin{align}
A&= \int m a\;ds,\\
 &= \int m \frac{dv}{dt}\;ds,
\end{align}
Here the professor did something like:
$$ds \times \cos \alpha =-dh$$
and then the equation goes
\begin{align}
A&=- \int m \frac{dv}{dt}\;dh,\\
 &=- \int m v \;dv,\\
 &=-m \int v \text{ }dv
\end{align}
and up to
$$A=-\Delta W_p$$
Now what I'd like to understand from you is one logic explanation for
$$ds \times \cos \alpha=-dh$$
I'd be very grateful!

REPLY [2 votes]: The height $h$ is probably the vertical displacement pointing downwards. Therefore:
$$
h = \left(-\mathbf{\hat j}\right)\cdot\mathbf s = -|\mathbf{\hat j}||\mathbf s|\cos\alpha = -s\cos\alpha
$$
Now we can derive:
$$
\frac{dh}{ds} = -\frac{d}{ds}\left(s\cos\alpha\right) = -\cos\alpha
\quad\Longrightarrow\quad \frac{dh}{ds} = -\cos\alpha
$$
Therefore, multiplying both sides by $ds$, we get:
$$
dh = -\cos\alpha ds
$$