TITLE: Solving an equation that contains a logarithm
QUESTION [1 upvotes]: I have the follwing equation:
$$y=\frac 1 4x^2 -\frac 1 2 \ln{x}$$
How can $x$ be expressed in terms of $y$?

REPLY [1 votes]: We have
$$y=\frac14 x^2-\frac12 \log x\tag 1$$
Multiplying $(1)$ by $-4$ yields
$$\begin{align}
-4y&=- x^2+2 \log x\\\\
&=-x^2+\log x^2 \tag 2\\\\
\end{align}$$
Now, exponentiating both sides of $(2)$ and multiplying by $-1$ reveals that
$$-e^{-4y}=-x^2e^{-x^2} \tag 3$$
from which we find that
$$x=\sqrt{-W(-e^{-4y})}$$
where $W(z)$ is the Lambert W Function.  Note we rejected negative values of $x$ since in $(1)$ we have $\log x$.
We must have $W\le 0$ and $-e^{-4y}\ge -e^{-1}\implies y\ge \frac14$.

REPLY [1 votes]: As said in comments, the solution of equation $$\frac 1 4x^2 -\frac 1 2 \ln{x}-y=0$$ cannot be expressed in terms of elementary functions.
However, there are explicit solutions in terms of Lambert function. Chenging variable $x^2=t$, the equation write $$t-\log(t)-4y=0$$ and the solution is $$t=-W\left(-e^{-4 y}\right)$$
You must notice that $y'=\frac{x}{2}-\frac{1}{2 x}$ cancels at $x=1$ and, at this point $y=\frac 14$ and the second derivative is always positive. So, if $y<\frac 14$, there is no real root; if $y=\frac 14$, the only root is $x=1$ and if $y>\frac 14$, there are two real roots (one between $0$ and $1$, another larger than $1$).


TITLE: Total chromatic number and total clique number
QUESTION [0 upvotes]: Let $G=(V,E)$ be a finite, simple, unconnected graph. We define the total graph $T(G)$ of $G$ as follows:

$V(T(G)) = (V\times\{0\}) \cup (E\times\{1\})$,
$E(T(G)) = E_v \cup E_e \cup E_{v+e}$, where

$E_v = \big\{\{(v,0), (w,0)\}: \{v,w\}\in E\big\}$,
$E_e = \big\{\{(e,1), (f,1)\}: (e,f\in E) \land (e\neq f)\land (e\cap f \neq \emptyset\big)\}$, and
$E_{v+e} = \big\{\{(v,0), (e,1)\}: v\in e\big\}$.

I think that for the clique number we have $\omega(T(G)) = \Delta(G)+1$ (where $\Delta(G)$ is the maximum degree in $G$), except if $\Delta(G) = 1$, but I have to check this (it's probably an easy question).
Question: Is there a $G$ such that $\chi(T(G)) > \omega(T(G))$?

REPLY [3 votes]: Counterexamples are, for example, complete graphs with even number $2k$ of vertices. If we manage to color such a graph with $2k$ colors, then all vertices have different colors, and from each vertex of color, say, $s$, there is exactly one edge of each color except $s$. Therefore total number of edges of each color is $(2k-1)/2$, this is not possible.
Your observation on clique number is correct: each clique either contains at least two edges and at most one vertex, or it contains unique edge and at most 2 vertices, or it contains only vertices. All cases are clear.


TITLE: Is it possible to classify extensions $G$ of an abelian $A$ by an abelian $N$ such that the map $G\rightarrow A$ is abelianization?
QUESTION [3 upvotes]: Suppose we have a given action $\varphi : A\rightarrow\text{Aut}(N)$ with $A,N$ abelian groups. Is it possible describe the isomorphism classes of extensions $G$ of $A$ by $N$ realizing $\varphi$ such that the map $G\rightarrow A$ is abelianization (ie, such that $N = G'$)?
(Without the requirement $N = G'$ this is classified by the group cohomology $H^2(A,N)$)
Some remarks:
Nontrivial central extensions certainly give examples where classes in $H^2(A,N)$ do not necessarily satisfy the condition that $G' = N$.
I wonder if there is a condition on $\varphi$ which guarantees that every class in $H^2(A,N)$ represented by $G$ satisfies $G' = N$.

REPLY [4 votes]: When $A$ acts trivially on $N$, this $H^2(A,N)$ has a canonical map $\Phi$ into the group $\mathrm{Hom}(\Lambda^2A,N)$ ($\Lambda^2A$ being the second exterior power, quotient of $A\otimes_\mathbf{Z}A$ by the subgroup generated by elements of the form $x\otimes x$ when $x$ ranges over $A$), induced by the commutator map. Given a central extension with cocycle $c$, the resulting extension has $N$ as derived subgroup iff the corresponding element $\Phi(c)\in\mathrm{Hom}(\Lambda^2A,N)$ is surjective.
(Remark: there's a canonical inclusion $\Psi$ of $\mathrm{Hom}(\Lambda^2A,N)$ into $H^2(A,N)$ and $\Phi\circ\Psi=2\mathrm{Id}$.)
In general, let $N/M$ be the co-invariants of the $A$-action on $N$ (i.e., $M$ is generated as a group by the $\varphi(g)h-h$, when $h$ ranges over $N$ and $g$ over $A$). Then after modding out by $M$, every extension as given yields a cocycle in $H^2(A,N/M)$ and hence, taking $\Phi$ an element in $f\in\mathrm{Hom}(\Lambda^2A,N/M)$; then $N$ is the derived subgroup iff $f$ is surjective.


TITLE: Does the Law of Factor Sparsity apply to factorization?
QUESTION [1 upvotes]: Does the 80-20 Rule (also known as the Law of Factor Sparsity, the Law of the Vital Few, and the Pareto Principle) apply to factorization? That is, if x is a large positive integer, then are about 20% of the primes not exceeding x sufficient for factorizing about 80% of the positive integers not exceeding x? Has anyone done a statistical study of this?

REPLY [1 votes]: Yes. Hint: The set of primes less than 20% of $x$ constitute 20% of the set of primes less than $x$ for large $x$ (use the prime number theorem; using some explicit bounds you can show that the set constitutes at least 19% of the set of primes less than $x$ for all sufficiently large $x$). If any integer $n\leq x$ has a prime factor $p$ larger than 20% of $x$, then that integer $n$ must be one of $p,2p,3p,4p,5p$, so the number of such integers is less than 6 times the number of primes larger than 20% of $x$, and in particular, less than $6\pi(x)$. Thus since $\pi(x)/x\to0$ as $x\to\infty$, the set of primes less than 20% of $x$ are sufficient for factorizing almost all integers $\leq x$.


TITLE: How to find critical F-Value?
QUESTION [1 upvotes]: So for a problem in the book I'm using I need to find $F_{0.025,10,11}$ and $F_{0.025,11,10}$
So in my book I used the F-distribution table to help me find
$F_{0.025,10,11} = 3.53$
But how do I find
$F_{0.025,11,10} = \text{???}$
My book only has degrees of freedom in the numerator go from 1-10, then 12, 15, $20,$ etc. to infinity completely skipping 11. Other tables I find on google have the same issue. So how do I calculate $F_{0.025,11,10} = \text{???}$
Is there another way to do so without a table that doesn't involve using a programming language like R?
(I can't find the F-distribution tag, so I'm just tagging this under statistics)

REPLY [0 votes]: Suppose $F\sim F_{10,11}$ and $G\sim F_{11,10}.$
Then $\Pr(F>3.525672) = 1 - 0.025.$
The distribution of $1/G$ is the same as the distribution of $F.$ Suppose $\Pr(G> a) = 1 -0.025$ and you want to find $a.$ So $\Pr(1/G < 1/a) = 1-0.025.$ You can find $1/a$ in the table for $F_{10,11}$ and then takes its reciprocal.


TITLE: $R(T)$ and $N(T)$ are $T$-invariant subspaces
QUESTION [1 upvotes]: Let $T:V\to V$ be linear. Show that $R(T)$ and $N(T)$ are $T$-invariant.

I know that $\dim(V)=\dim(N(T))+\dim(R(T))$, but I'm confused on where to go from here.

REPLY [2 votes]: This is mostly a matter of stating definitions.  Let's go through the proof that $R(T)$ is $T$-invariant, for example:
Suppose that $y \in R(T)$.  By definition, this means that there is a vector $x$ such that $y = Tx$.  We note that, by this very definition, $Ty \in R(T)$.
Since $Ty \in R(T)$ for every $y \in R(T)$, we conclude that $R(T)$ is a $T$-invariant subspace

REPLY [1 votes]: $$ T(N(T))=\{ 0\}\subset N(T)$$ and $$T(R(T))\subset T(V)=R(T)$$


text:
\documentclass[11pt,a4paper]{article}
\usepackage[T1]{fontenc}
\usepackage{isabelle,isabellesym}

% this should be the last package used
\usepackage{pdfsetup}

% urls in roman style, theory text in math-similar italics
\urlstyle{rm}
\isabellestyle{it}

\begin{document}

\title{Cardinality of Set Partitions}
\author{Lukas Bulwahn}
\maketitle

\begin{abstract}

The theory's main theorem states that the cardinality of set partitions of
size $k$ on a carrier set of size $n$ is expressed by Stirling numbers of the
second kind.
In Isabelle, Stirling numbers of the second kind are defined in the AFP entry
`Discrete Summation'~\cite{Discrete_Summation-AFP} through their well-known
recurrence relation. The main theorem relates them to the alternative
definition as cardinality of set partitions.
The proof follows the simple and short explanation in Richard P. Stanley's
`Enumerative Combinatorics: Volume 1'~\cite{Stanley-2012} and
Wikipedia~\cite{Wikipedia-Stirling-Numbers-2015}, and unravels the full
details and implicit reasoning steps of these explanations.

\end{abstract}

\tableofcontents

% sane default for proof documents
\parindent 0pt\parskip 0.5ex

% generated text of all theories
\input{session}

\bibliographystyle{abbrv}
\bibliography{root}

\end{document}

%%% Local Variables:
%%% mode: latex
%%% TeX-master: t
%%% End:

subset_name: curated


TITLE: A primitive permutation group with abelian point stabilizars
QUESTION [1 upvotes]: Suppose that $G$ is a group acting primitively on a set $X$. Also suppose that all of its point stabilizers are abelian.
I want to prove that $G$ is either regular of prime degree or a Frobenius permutation group.
I have some ideas for proving this statement when $X$ is finite but in general, I don't have any useful ideas.
I will be so appreciate for any helpful answers and comments.

REPLY [1 votes]: If not, then there exists $1 \ne g \in G$ that fixes more than one point. But for each $\alpha$ in the fixed point set of $g$, $C_G(g)$ contains the stabilizer $G_\alpha$, because $G_\alpha$ is abelian.
Since $G$ is primitive, we cannot have  $G_\alpha = G_\beta$ for $\alpha \ne \beta$ (I'll leave that for you to prove). So $C_G(g)$ strictly contains $G_\alpha$ and hence $C_G(g)$ is not maximal in $G$. But $G$ primitive is equivalent to the stabilizers being maximal.
This argument does not use finiteness of $G$.


TITLE: No decay of current in $RLC$ circuits?
QUESTION [0 upvotes]: They are teaching alternating current in school. The book states that in an $RLC$ circuit, the instantaneous current is given as
$$i(t)=\frac{v_m}{\sqrt{R^2+(X_C-X_L)^2}}\sin(\omega t+\phi)$$
where $\tan\phi=\frac{X_C-X_L}R$, $v(t)=v_m\sin(\omega t)$. It's obvious from the equation that there is no damping of the current. The current is perfectly sinusoidal. This is also the case with RC circuit. Whereas, this is not the case in RL circuit, there is asymptotic growth in current.
Why is that?
To find more about the undergoing mathematical nuisance (the derivation in the book wasn't satisfactory enough for me, where we already assume that the current is sinusoidal), I tried Wolfram|Alpha. It seems like although an exponential factor exists, it has no effect asymptotically, except in the RL circuit where it leads to current asymptotically growing from $0$.
Then what's the problem?
What doesn't make sense to me is that we know for a fact that resistance dissipates energy, while pure inductor and pure capacitor don't. Then how it can be the case that the steady-state current is sinusoidal as the resistance will still be there in the circuit to dissipate energy. An "idea" I have for this is that maybe the inductor and capacitor work to store some amount of the energy (like in LC oscillations) and that portion, somehow, isn't affected by the resistance.
Please help me understand the undergoing mechanics. And if possible, how can we, without using Wolfram|Alpha, solve for the exact equations of the instantaneous current in RLC,LC circuits with alternating voltage? A reference to required material would be enough, as I want to do that on my own. I can solve it for the RL,RC circuits using integration factor method, but don't what to do for the others.
Please correct me if I got anything wrong.

REPLY [1 votes]: Whereas, this is not the case in RL circuit, there is asymptotic growth in current.
This is only true if the source is DC.
Your equation $i(t)=\dfrac{v_m}{\sqrt{R^2+(X_C-X_L)^2}}\sin(\omega t+\phi)$ is correct for all combination when there is a connection to an AC source, ie, $LCR,\,LC,\, LR, \, CR$ and the three circuit elements alone.
For those components which are not present you need to make their reactance/resistance equal to zero to have the correct equation for the current flowing in the circuit.
In the LR circuit with a DC supply the final (steady state) condition has the current reaching a maximum value and only the resistor dissipating electrical energy as heat.
With the AC source again the resistor is the only circuit element which dissipated electrical energy as heat whereas on average over a period of the AC the net electrical power dissipated is zero.
A steady state is reached after the transient current has decayed away.


TITLE: direct proof of an identity regarding certain symmetry of integer partitions?
QUESTION [5 upvotes]: Suppose $(a_1,\ldots, a_k)$ is an integer partition of $n$, and $(b_1,\ldots,b_k)$ is a rearrangement of the $a$-sequence. Prove the following identity (preferably combinatorially):
$$
\sum_{j_2+\cdots+j_k=l,\atop l\geq 0, \, a_t>j_t\geq 0} \quad\frac{(-1)^l l!}{(a_1+l+1)_{l+1}} {a_2\choose j_2}\cdots {a_k \choose j_k}=\sum_{j_2+\cdots+j_k=l,\atop l\geq 0, \, b_t>j_t\geq 0} \quad\frac{(-1)^l l!}{(b_1+l+1)_{l+1}} {b_2\choose j_2}\cdots {b_k \choose j_k},
$$
where $(x)_k$ is the falling factorial.

REPLY [4 votes]: I assume you meant $j_t\leq a_t$ (not $j_t<a_t$).
It's sufficient to prove that for any analytic function $f(x)$, the function
$$F(a_1,a_2) := \sum_{l\geq 0} \sum_{j=0}^{a_2} \frac{(-1)^ll!}{(a_1+l+1)_{l+1}} \binom{a_2}{j} [x^{l-j}]\ f(x)$$
is symmetric, i.e. $F(a_1,a_2) = F(a_2,a_1)$.
Using the property of beta function, we have
\begin{split}
F(a_1,a_2) &=\sum_{l\geq 0} \frac{(-1)^ll!}{(a_1+l+1)_{l+1}} [x^l]\ (1+x)^{a_2}f(x) \\
&=\sum_{l\geq 0} (-1)^l \int_0^1 t^l (1-t)^{a_1} {\rm d}t\ [x^l]\ (1+x)^{a_2}f(x)\\
&=\int_0^1 (1-t)^{a_1} (1-t)^{a_2} f(-t) {\rm d}t,
\end{split}
which is clearly symmetric.


TITLE: Difference between convergent in Measure and convergent a.e.
QUESTION [0 upvotes]: I’m thinking about two problems:
In $L^2([0,1],dx);$

If $f_n\rightarrow f$ In $L^2$, then $f_n\rightarrow f$ In measure.
If $f_n\rightarrow f$ In $L^2$, then $f_n\rightarrow f$ Almost everywhere.

So I know the definition of convergence in measure is:
$\lim_{n\rightarrow \infty}\mu(\{x\in X: \lvert f_n(x)-f(x)\lvert\geq\epsilon\})=0$, $\forall\epsilon>0.$
So can I prove the first one by
$\int\lvert f_n-f\lvert= \lvert \lvert f_n-f\lvert\lvert_1\leq\lvert\lvert f_n-f\lvert\lvert_2\cdot\lvert\lvert 1\lvert\lvert_2\rightarrow0$
Therefore $\lvert f_n-f\lvert\rightarrow0,$ (since it’s positive) this means it’s convergent in measure?
But what’s the difference of convergent almost everywhere? Thanks.

REPLY [1 votes]: Your proof is incorrect because $f_n>0$ and $\int f_n\to 0$ do not imply $f_n\to 0$ a.e.
Let $f_{n,k}$ be the characteristic function of $[(k-1)/n,k/n]$, $k=1,\ldots,n$ and $n\in\mathbb{Z}_+$. Rearrange it into a usual sequence:
$$g_1=f_{1,1},\ g_2=f_{2,1},\ g_3=f_{2,2},\ g_4=f_{3,1},\ g_5=f_{3,2},\ \ldots$$
It is easy to see that $g_n\to 0$ in $L^2$ but $g_n(x)\not\to 0$ for all $x$.
The first implication (convergence in norm implies convergence in measure) is correct, which follows easily from Chebyshev's inequality.


TITLE: $(x+1)^2y''+(x+1)y'+y=x^2+2\sin(\ln(x+1)), y(0)=\frac{1}{5},y'(0)=2$
QUESTION [0 upvotes]: $(x+1)^2y''+(x+1)y'+y=x^2+2\sin(\ln(x+1)), y(0)=\frac{1}{5},y'(0)=2$
My solution:
$y''+\frac{y'}{(x+1)}+\frac{y}{(x+1)^2}=\frac{x^2+2\sin(\ln(x+1))}{(x+1)^2}$
First of all , I found the equation solution of $y''+\frac{y'}{(x+1)}+\frac{y}{(x+1)^2}=0$
$y=c_1\cos(\ln(x+1))+c_2\sin(\ln(x+1))$
I try to solve this ode using the variation of parameters theorem
Get this system equation:
(I)$c'_1\cos(\ln(x+1))+c'_2\sin(\ln(x+1))=0$
(II)$-c'_1\sin(\ln(x+1))+c'_2\cos(\ln(x+1))=x^2+2\sin(\ln(x+1))$
Multiply (I) by $\sin(\ln(x+1))$ , (II) by $\frac{\cos(\ln(x+1))}{x+1}$.
By addtion i get:
$c'_2=\frac{\cos(\ln(x+1))[x^2+2\sin(\ln(x+1))]}{(x+1)}$
I do not know how I get $c_2$ by an integral ?
Help please
Thanks !

REPLY [0 votes]: $$(x+1)^2y''+(x+1)y'+y=x^2+2\sin(\ln(x+1))$$ $$y(0)=\frac{1}{5},y'(0)=2$$
Substitute $u=x+1$:
$$u^2y''+uy'+y=(u-1)^2+2\sin(\ln(u))$$ $$y(1)=\frac{1}{5},y'(1)=2$$
Substitute $u=e^t$:
$$y''(t)+y(y)=(e^t-1)^2+2\sin(t)$$
This is easy to solve with variation of constants method.
$$y''(t)+y(t)=2 \sin t$$
$$\implies y_p=At\cos t$$
$$y''(t)+y(t)=e^{2t}-2e^t+1$$
$$y_p=Ae^{2t}+Be^t+C$$

Edit:
Note that:
$$\dfrac {dy}{du}=\dfrac {dy}{dt}\dfrac{dt}{du}=e^{-t} \dfrac {dy}{dt}=\dfrac 1u\dfrac{dy}{dt}$$
Then $$u\dfrac {dy}{du}=\dfrac {dy}{dt}$$
Do the same for:
$$\dfrac {d^2y}{du^2}$$
Then the DE is simplified and you find that :
$$y''(t)+y(t)=(e^t-1)^2+2\sin(t)$$


TITLE: Compute in closed form $\int_0^1\frac{\arctan{ax}}{\sqrt{1-x^{2}}}dx$
QUESTION [3 upvotes]: I am trying to find closed form for this integral:
$$I(a)=\int_0^1\frac{\arctan{ax}}{\sqrt{1-x^{2}}}dx$$
Where $a>0$.
My try: Let: $$I(a)=\int_0^1\frac{\arctan{ax}}{\sqrt{1-x^{2}}}dx$$
Then:
$$\frac{dI(a)}{da}=\int_0^1\frac{x}{(1+(ax)^2)\sqrt {1-x^2}}dx$$

How I can complete this work or is there another way to
  approach it?

REPLY [3 votes]: Assuming $a$ to be real and positive, a CAS gives
$$\color{blue}{I(a)=\frac{\pi ^2}{4}-2 \sinh ^{-1}(a) \,\coth ^{-1}\left(a+\sqrt{a^2+1}\right)+\text{Li}_2\left(a-\sqrt{a^2+1}\right)-\text{Li}_2\left(\frac{1}{a+\sqrt{a^2+1}}
\right)}$$
$$I\left(\frac{1}{2}\right)=\frac{\pi ^2}{12}-\frac{1}{2} \sinh ^{-1}(2) \,\text{csch}^{-1}(2)$$
Using @clathratus's approach
$$I(a)=\sum_{n\geq0}\frac{(-1)^n a^{2n+1}}{2n+1}\int_0^1\frac{x^{2n+1}}{\sqrt{1-x^2}}dx$$ we have
$$\int_0^1\frac{x^{2n+1}}{\sqrt{1-x^2}}dx=\frac{\sqrt{\pi }\, \Gamma (n+1)}{2\, \Gamma \left(n+\frac{3}{2}\right)}$$ making
$$I(a)=a \, _3F_2\left(\frac{1}{2},1,1;\frac{3}{2},\frac{3}{2};-a^2\right)$$ the simplication of which giving the first result.


TITLE: Numerical Analysis: Using Forward Euler to approximate a system of Differential Equations
QUESTION [0 upvotes]: I'm giving the following system of ODE's:
  \begin{array}
d \begin{bmatrix}
x^{'}(t) \\
y^{'}(t)
\end{bmatrix} & = &
\begin{bmatrix}
7 & -1\\-1 & 2
\end{bmatrix}\cdot
\begin{bmatrix}
x(t)\\ y(t)
\end{bmatrix}
\end{array}
  with an arbitrary initial condition of $\vec{x}(0) = \vec{x}_0$.

We are asked to solve the linear system using the Forward Euler Method using MATLAB and then plotting our solution.

Now I am familiar with how to solve the closed form of this IVP using Fundamental Theorem of Linear Systems and Spectural Decomposition. However, we are asked to solve the problem numerically using Forward Euler.

The Forward Euler Method is derived using the forward difference approximation. In this case since we are dealing with matrices and vectors:

\begin{eqnarray}
\frac{\vec{x}_{n+1}-\vec{x}_n}{\Delta t}  \approx A\vec{x}_n
\end{eqnarray} which can be rewritten as
      \begin{eqnarray}
\vec{x}_{n+1} = \vec{x}_n + \Delta t  A\vec{x}_n
\end{eqnarray}

In order to advance time steps, the second equation stated above is recursively applied as

\begin{eqnarray}
\vec{x}_{1} &=& \vec{x}_0 + \Delta t  A\vec{x}_0\\
\vec{x}_{2} &=& \vec{x}_1 + \Delta t  A\vec{x}_1\\
\vec{x}_{3} &=& \vec{x}_2 + \Delta t  A\vec{x}_2\\
&\vdots&\\
\vec{x}_{n} &=& \vec{x}_{n-1}+ \Delta t  A\vec{x}_{n-1}\\
\end{eqnarray}

This is where I am stuck. I am not familiar with MATLAB so I am not sure how to go about coding the problem to do what I desire. Especially when I don't have a set initial condition. (I should add that I do know how to create matrices, vectors, and do some calculations in MATLAB, but nothing of this level... yet).
I am familiar with core basics of programming languages like C++ and Python and I am roughly familiar with the meta-language Mathematica; therefore, I am sure I will catch on right away. I can more a less guess that my algorithm will require defining my starting point, my $2\times 2 $ matrix $A$, the number of iterations, my time step, and my starting and ending value for $t$. (The professor didn't specify on the assignment so I am just going to say an interval from $0$ to $10$ should do). I would then apply a loop (most probably a for-loop) to recursively apply forward Euler for $n$th number of times. I would assume I would need to create a dynamical array to store all the values as I go along, that way I can eventually call them up when I decide to plot my data points.

I was wondering if someone could give me an idea and how to start the code.

Thank You for taking the time to read my post. I greatly appreciate any suggestions, comments, or feedback. Have a wonderful day.

REPLY [2 votes]: for i=1:n
    x(n+1) = x(n)+dt*A*x(n)
end

should do the trick, you need of course initialize the vectors and matrices.


TITLE: Prove that a splitting field is not an algebraic closure over $\mathbb{Z}_p$
QUESTION [0 upvotes]: Let $n > 0$ be some integer and let $G$ be a splitting field of the set of all polynomials of degree at most $n$ over a field $F$. I need to prove that $G$ is not an algebraic closure of $F$ in the case $F = \mathbb{Z}_{p}$
I've actually found a solution for this, but I would really appreciate it if someone could fill in some of the details because I am so confused:

There are only finitely many polynomials of degree $\leq n$ over a
  finite  field $F$, so $G/F$ will wind up having finite degree. Then
  $G^\times$ will  be a cyclic group of some finite order, and a
  generator $a$ will not have an  $m^{th}$ root for any nontrivial
  divisor d of |$G^\times$|, so $xd−a$ cannot  split in G

First off all, why is $G/F$ going to have finite degree?  Also what exactly does $G^\times$ denote? Just the cyclic group $G$?  Also, where did the fact that there is no $m^{th}$ root come from?

REPLY [2 votes]: First, $G/F$ has a finite degree since $G$ is obtained by adjoining roots of polynomials of $\deg\leq n$ over $F$, which are all algebraic and there are finitely many such elements. (There are finitely many polynomials of $\deg\leq n$ over finite fields.)
$G^{\times}$ stands for the multiplicative group of the field $G$, i.e. $G^{\times} = G\backslash\{0\}$. It is known that  $k^{\times}$ is cyclic for any given finite field $k$. (Here $G$ is finite since it is a finite extension of $F$.)
Now we choose generator $a$ of the cyclic group $G$. If $d\neq 1$ is a divisor of $|G^{\times}|$, then $x^{d}=a$ cannot have a solution - if it has, then it contradicts to the assumption that $a$ is a generator of $G^{\times}$. Hence $G$ is not algebraically closed.


TITLE: Endomorphisms of irreducible submodules of simple algebras
QUESTION [1 upvotes]: I am quite new to Representation Theory and I am trying to wrap my head around some basics. I have some difficulties with the following setup:
Suppose I start with a group ring $R=K[G]$, where $K$ is a field of characteristic $0$ (not necessarily algebraically closed) and $G$ is a finite group (not necessarily abelian).

I can decompose $R$ as a direct sum
$$ R = \bigoplus e_{\rho} R,$$
where each $e_{\rho}$ is a central idempotent and for each $\rho$, $e_{\rho}R$ is a simple algebra.
Suppose I fix some $\rho$ and take $A=e_{\rho}R$. Then $A$ can be decomposed further as direct sum of simple left-modules
$$ A= \bigoplus\limits_{j=1}^f M_j, $$
Consider now some $M_j$ as above and let $m \in M_j$. Then, as $M_j \subset A$, multiplication by $m$ seems to induce an $A$-endomorphism of $M_j$ and since $M_j$ is simple, this would imply by Schur's lemma that this map is an isomorphism.

Something seems off with the above reasoning, but I am unable to pinpoint where the error in the reasoning occurs. Any help or suggestions is greatly appreciated.

REPLY [2 votes]: Multiplication by an element of $A$ is typically not an endomorphism of an $A$-module, if $A$ is not commutative.  Indeed, suppose $M$ is a left $A$-module and $a\in A$.  For $f(x)=ax$ to be an endomorphism $M\to M$, we would need to have $f(bx)=bf(x)$ for all $b\in A$, which means $a(bx)=b(ax)$.  There is no reason for this to be true if $ab\neq ba$.
So, in your setup, left multiplication by $m$ is typically not an endomorphism of $M_j$.
On the other hand, though, right-multiplication by $m$ actually is an endomorphism of $M_j$ (it maps $M_j$ to itself since for any $x\in M_j$, $xm\in M_j$ because $m\in M_j$).  Your argument is then (almost) correct and says that right multiplication by $m$ is an isomorphism on $M_j$ unless it is $0$ (don't forget the latter possibility!).
This might seem surprising but you can verify it quite explicitly.  We can identify $A$ with a matrix ring $M_f(D)$ over some division ring $D$, with $M_j$ being the matrices whose only nonzero column is the $j$th column.  You can then verify that if $x,m\in M_j$ then $xm=xd$ where $d\in D$ is the $(j,j)$ entry of the matrix $m$ (none of the other entries matter since $x$ has only one nonzero column).  Since $D$ is a division ring, $d$ is either $0$ or a unit, and so $x\mapsto xd$ is either $0$ or a bijection on $M_j$.


TITLE: Compactness of the set in $\mathbb{R}$
QUESTION [0 upvotes]: Can anyone explain me is this set compact or not?
$$S:=\{x ∈ \mathbb{R} : x ∈ (2, 3] ∪ (4, 5]\text{ or } x=10\}$$
I already know that for instance $(2,3]$ and $(4,5]$ are not compact, but does it imply for the union?

REPLY [1 votes]: The open cover $\{U_n : n = 1,2,\cdots\}$ has no finite subcover, where
$$
U_n = S \cap \left(2+\frac{1}{n},\infty\right) .
$$


TITLE: A uniqueness proposition involving Erf, the error function
QUESTION [1 upvotes]: This is a MathOverflow cross-post (currently no answer there) and a generalization of a previous MathOverflow question, "Reducing system of equations involving Erf, Error Function".
Consider the system of equations:
$$1/2 + {\rm Erf}(x) - \alpha {\rm Erf}(\frac{x+y}{2})=0$$
$$-1/2 + {\rm Erf}(y) - \alpha {\rm Erf}(\frac{x+y}{2})=0,$$
where $x \le y$ and ${\rm Erf}$ is the Error Function. By ansatz it is clear that one solution is $(x,y)=(-{\rm Erf}^{-1}(1/2),{\rm Erf}^{-1}(1/2))$,  independent of $\alpha$. The question is to prove the uniqueness of this solution, for any $\alpha$.
(Update: Robert's visual below demonstrates how there are clearly multiple solutions for a robust interval of $\alpha \in (0.5, \approx 0.8)$, and so the general $\alpha$ formulation was not helpful. I am revising my question to specifically focus on uniqueness in the case of $\alpha =1/2$.)
I am particularly interested in a proof for $\alpha = 1/2$, but I hope that by phrasing the problem in terms of a general $\alpha$, someone may see an elegant elementary proof.
Observe that proving that $x+y=0$ is sufficient. My previous MO question includes a proof of uniqueness in this way for the case where $\alpha=1$, with two proofs (one by Noam Elkies, one by myself). Unfortunately these proofs do not appear to generalize to other values of $\alpha$ (Noam's proposition is false for $\alpha = 1/2$, my approach does not appear to generalize), even though I strongly suspect the general statement.
Numerics:
I have embarked on some numerical exploration. If one considers the equation
$${\rm Erf}(x) + {\rm Erf}(y) = 2\alpha {\rm Erf}(\frac{x+y}{2}),$$
For $\alpha=1$ we know that $y=-x$ and $y=x$ are solutions. Numerically, it appears as through $y=-x$ is always a solution, while the second solution is $\alpha$-dependent, and not always so nice. For $\alpha=1/2$ the second equation appears to be ${\rm Erf}(c_{1/2} y) - {\rm Erf}(c_{1/2}x) =1$ where $c_{1/2} \approx 0.57285884$. For other values of $\alpha$ the equation appears to be of the form ${\rm Erf}(c_\alpha y - d_\alpha) - {\rm Erf}(c_\alpha x +d_\alpha) =1$. Any insight on what the value of $c_{1/2}$ might be analytically or why $d_{1/2}$ might be zero would be much appreciated.

REPLY [1 votes]: The solution is not always unique.  For $1/2 < \alpha < .79654774210531568819$ approximately, there are two other solutions.  For example with $\alpha = 0.6$ the other solutions are approximately
$x = -1.910219091, y = .006118254868$ and $ x = -.006118254868, y =  1.910219091$.  This can  be seen graphically: here are the curves given by your two equations (the first in red, second in blue) for different values of $\alpha$ from $0.3$ to $1$.
The value near 0.8 is actually the value of $\exp{-(\text{erf}^{-1}(1/2))^2}$.  This is the value of $\alpha$ for which the two curves have the same slope at the point $(-\text{erf}^{-1}(1/2), \text{erf}^{-1}(1/2))$.


TITLE: Spectrum of a right shift operator.
QUESTION [5 upvotes]: I have some doubts on the following problem :
Let us consider $T : \ell^1(\mathbb N) \to \ell^1(\mathbb N) $by $(x_1,x_2..... ) \to (x_2, x_3 ........) $.
I want to find the eigen values and spectrum of T and also of $T' : \ell^\infty (\mathbb N)\to \ell^\infty(\mathbb N)$
let us consider $\lambda $ to be the eigen value , then $Tx=\lambda x$ for a $x \in \ell^1$  then we get $(x_2,x_3,......)=(\lambda x_1, \lambda x_2 ........)$ which holds equality if $x_1=x_2=.....=0$ , which means there is no eigen value for $T$ .
How do i find the spectrum of $T$ and $T'$ ?
Thank you for your help.

REPLY [2 votes]: The "iff" part is wrong. Consider: $(x_i)$ with $x_i=\lambda^i$. For what $\lambda$ is this in $\ell^1$? In $\ell^\infty$?
You can write out explicitly for $|\lambda|>1$ the inverse of $\lambda I -T$ by writing:
$$S_\lambda = (\lambda I - T)^{-1} = \lambda^{-1}(I-\lambda^{-1}T)^{-1} = \lambda^{-1}\sum_{k=0}^\infty \lambda^{-k} T^k$$
Writing $x=(x_i)$ and $(y_i)=S_\lambda x$, we get:
$$y_i = \sum_{k=0}^\infty \lambda^{-(k+1)} x_{i+k}$$
You need to show that if $x\in\ell^1$  (resp. $\ell^\infty$), then this series for $y_i$ coverges for all $i$, and $(y_i)\in\ell^1$ (resp. $\ell^\infty$.)


text:
\begin{document}

\maketitle

\begin{abstract}
The notions of $p$-convexity and $q$-concavity are mostly known because of their importance as a tool in the study of isomorphic properties of Banach lattices, but they also play a role in several results involving linear maps between Banach spaces and Banach lattices.
In this paper we introduce Lipschitz versions of these concepts, dealing with maps between metric spaces and Banach lattices, and start by proving nonlinear versions of two well-known factorization theorems through $L_p$ spaces due to Maurey/Nikishin and Krivine.
We also show that a Lipschitz map from a metric space into a Banach lattice is Lipschitz $p$-convex if and only if its linearization is $p$-convex.
Furthermore, we elucidate why there is such a close relationship between the linear and nonlinear concepts by proving characterizations of Lipschitz $p$-convex and Lipschitz $q$-concave maps in terms of factorizations through $p$-convex and $q$-concave Banach lattices, respectively, in the spirit of the work of Raynaud and Tradacete.
\end{abstract}

\section{Introduction}

The classical examples of Banach spaces of functions or sequences (say, $L_p[0,1]$ or $c_0$) come naturally endowed with an order structure that is compatible with the norm, and this is often a useful tool.
Banach spaces with such ``extra'' order structure are called Banach lattices, and, paraphrasing Lindenstrauss and Tzafriri \cite{Lindenstrauss-Tzafriri-II}, this additional ingredient makes
the theory of Banach lattices in some regards simpler, cleaner and more complete than the theory for general Banach spaces.
For a Banach-space flavored study of the fundamentals of the theory of Banach lattices, the reader is referred to \cite{Lindenstrauss-Tzafriri-II}.
Because of the extra structure present in Banach lattices, the morphisms of interest are not just bounded linear maps but in fact maps that take advantage of the order.
Two of the most important classes of such maps are the $p$-convex and $q$-concave ones.
These two notions play an important role in the study of isomorphic properties of lattices, for example uniform convexity in Banach lattices \cite[Sec. 1.f]{Lindenstrauss-Tzafriri-II} and the study of rearrangement invariant function spaces \cite[Sec. 2.e]{Lindenstrauss-Tzafriri-II}.
Let us recall their definitions. Consider $1 \le p \le \infty$.
A linear map $T : X \to E$ from a Banach space $X$ to a Banach lattice $E$ is called \emph{$p$-convex} if there exists a constant $M < \infty$ such that for all $v_1, \dotsc, v_n \in X$
$$
\n{ \Big( \sum_{j=1}^n |Tv_j|^p \Big)^{1/p} }_E \le M \Big( \sum_{j=1}^n \n{v_j}_X^p \Big)^{1/p}, \quad\text{ if $1\le p < \infty$ }
$$
or
$$
\n{ \bigvee_{j=1}^n |Tv_j| }_E \le M \max_{1 \le j \le n} \n{v_j}_X, \quad\text{ if $p =\infty$. }
$$
The smallest such constant $M$ is denoted $M^{(p)}(T)$.
On the other hand, a linear map $S : E \to Y$ from a Banach lattice $E$ to a Banach space $Y$ is called $q$-concave if there exists a constant $M < \infty$ such that for all $v_1, \dotsc, v_n \in E$,
$$
\Big( \sum_{j=1}^n \n{Sv_j}_Y^q \Big)^{1/q} \le M \n{ \Big( \sum_{j=1}^n |v_j|^q \Big)^{1/q} }_E , \quad\text{ if $1\le q < \infty$ }
$$
or
$$
\max_{1 \le j \le n} \n{Sv_j}_Y \le \n{ \bigvee_{j=1}^n |v_j| }_E, \quad\text{ if $q =\infty$. }
$$
The smallest such constant $M$ is denoted $M_{(q)}(S)$.
The constants $M^{(p)}(T)$ and $M_{(q)}(S)$ are called the \emph{$p$-convexity} constant of $T$, respectively \emph{$q$-concavity} constant of $S$.
When we say that a Banach lattice $E$ is $p$-convex (resp. $q$-concave) we mean that the identity map $I_E : E \to E$ is $p$-convex (resp. $q$-concave).

In this paper we develop nonlinear counterparts of these two concepts,
considering Lipschitz maps between a metric space and a Banach lattice,
and show how some of
the elementary results from the theory of $p$-convex and $q$-concave maps admit generalizations to the Lipschitz setting.
The rest of the paper is organized as follows.
In Section \ref{sec-preliminaries} we recall some preliminaries that will be needed in the sequence.
In Section \ref{sec-lip-p-convex} we prove a nonlinear version of the Maurey/Nikishin factorization theorem, which suggests our definition of Lipschitz $p$-convex maps.
We then prove that a map is Lipschitz $p$-convex if and only if its canonical linearization is $p$-convex.
In Section \ref{sec-lip-q-concave} we introduce the concept of Lipschitz $q$-concave maps, and prove that the composition of a Lipschitz $p$-convex map followed by a Lipschitz $p$-concave one factors through an $L_p$ space, a nonlinear version of a theorem due to Krivine.
Finally, Section \ref{sec-factorization-theorems} makes clear why there is a close parallel between the linear and nonlinear situations. This is done by proving characterizations of Lipschitz $p$-convex and Lipschitz $q$-concave maps in terms of factorizations through $p$-convex and $q$-concave Banach lattices, respectively.
This is a nonlinear generalization of the work of Raynaud and Tradacete \cite{Raynaud-Tradacete}.

\section{Notation and preliminaries}\label{sec-preliminaries}

For the basic background and notation on Banach lattices, the reader is directed to \cite{Lindenstrauss-Tzafriri-II}.
We use the convention of having \emph{pointed} metric spaces, i.e. with a designated special point always denoted by $0$.
As customary, if $E$ is a Banach space then $B_E$ denotes the closed unit ball of $E$ and $E^*$ its linear dual. If $X$ is a metric space,
$\Lip_0(X,E)$ denotes the Banach space of Lipschitz functions $T : X \to E$ such that $T(0)=0$, with addition defined pointwise and the Lipschitz constant $\Lip(T)$ as the norm of $T$.
We use the shorthand $X^\# := \Lip_0(X,\R)$.

Let us recall the definition and basic properties of the space of Arens and Eells \cite{Arens-Eells-56}.
We follow the presentation in \cite{Weaver}.
A \emph{molecule} on a metric space $X$ is a finitely supported function $m:X\rightarrow \R$ such that $\sum_{x\in X} m(x) =0$.
For $x, x' \in X$ we denote by $m_{xx'}$ the molecule $\chi_{\{x\}}-\chi_{\{x'\}}$.
The simplest molecules, i.e. those of the form $a m_{xx'}$ with $x,x'\in X$ and $a$ a real number are called
\emph{atoms}.
It is easy to show that every molecule can be expressed as a sum of atoms (for instance, by induction on the cardinality of the support of the molecule).
The Arens-Eells space of $X$, denoted $\fbs(X)$, is the completion of the space of molecules with the norm
\begin{equation}\label{eqn-def-arens-eells-norm}
\n{m}_{\fbs} := \inf \bigg\{ \sum_{j=1}^n |a_j| d(x_j,x'_j) \; : \; m = \sum_{j=1}^n a_j m_{x_jx_j'} \bigg\}.
\end{equation}
The fundamental properties of the Arens-Eells space are summarized in the following theorem \cite{Arens-Eells-56}, \cite[pp. 39-41]{Weaver}.

\begin{theorem}\label{thm-Arens-Eells}
\begin{enumerate}[(i)]
\item
$\n{\cdot}_{\fbs}$ is a norm on the vector space of molecules on $X$.
\item
The dual of $\fbs(X)$ is (canonically) isometrically isomorphic to $X^\#$. Moreover, on bounded subsets of $X^\#$ the weak$^*$ topology coincides with the topology of pointwise convergence.
\item\label{universal-property}
The map $\iota : x \mapsto m_{x0}$ is an isometric embedding of $X$ into $\fbs(X)$. Moreover,
for any Banach space $E$ and any Lipschitz map $T : X \to E$ with $T(0)=0$ there is a unique linear map $\hat{T} : \fbs(X) \to E$ such that $\hat{T} \circ \iota = T$. Furthermore, $\bign{\hat{T}} = \Lip(T)$.
\end{enumerate}
\end{theorem}

Because of the universal property \eqref{universal-property}, the space $\fbs(X)$ is also called the free Lipschitz space of $X$, or simply the free space of $X$.
These spaces have been recently used as tools in nonlinear Banach space theory, see \cite{Godefroy-Kalton-03}, \cite{Kalton-04} and the survey \cite{Godefroy-Lancien-Zizler}.

\section{Lipschitz $p$-convex maps}\label{sec-lip-p-convex}

The concept of Lipschitz $p$-convex map was inspired by our discovery of the following non-linear version of the Maurey/Nikishin factorization theorem.
The proof presented here follows very closely the presentation of Albiac and Kalton \cite[Thm. 7.1.2]{Albiac-Kalton}.

\begin{theorem}\label{thm-nonlinear-nikishin}
Let $X$ be a metric space and $\mu$ be a $\sigma$-finite measure on some measurable space $(\Omega, \Sigma)$ and $1 \le q < p <\infty$. Suppose that $T : X \to L_q(\mu)$ is a Lipschitz map and $C > 0$.
The following are equivalent:
\begin{enumerate}[(a)]
\item
There exists a density function $h$ on $\Omega$ such that
\begin{equation}\label{eqn-7-1}
\left[ \int_{\{h>0\}} \left|\frac{Tx-Tx'}{h^{1/q}}\right|^ph\,d\mu \right]^{1/p} \le C d(x,x'), \qquad x,x' \in X
\end{equation}
and
\begin{equation}\label{eqn-7-2}
\mu\{ |Tx-Tx'| >0, h=0 \} = 0 \qquad x, x' \in X.
\end{equation}
\item For every $x_1,\dots,x_n,x_1',\dots,x_n' \in X$ and $\lambda_1,\dots,\lambda_n \ge 0$,
\begin{equation}\label{eqn-7-3}
\n{ \bigg(\sum_{j=1}^n \lambda_j|Tx_j-Tx_j'|^p\bigg)^{1/p} }_{L_q(\mu)} \le C \bigg(\sum_{j=1}^n \lambda_j d(x_j,x_j')^p\bigg)^{1/p}
\end{equation}
\end{enumerate}
\end{theorem}

As in the linear case, condition $(a)$ is equivalent to the existence of a factorization diagram
\begin{equation}\label{eqn-factorization-maurey-nikishin-nonlinear}
	\xymatrix{
	X \ar[r]^{T}\ar[d]^S &L_{q}(\mu) \\
	L_p(hd\mu) \ar[r]^{i_{p,q}} & L_q(hd\mu) \ar[u]^{j}
	}
\end{equation}
where $S$ is a Lipschitz function with $\Lip(S)\le C$ and the isometry $j$ has, in fact, range $L_q(A,\mu)$ where $A = \{ h > 0\}$.
Also, if we consider $X$ as a pointed metric space with a designated point $0\in X$ and impose the condition $T(0)=0$, condition \eqref{eqn-7-2} can be replaced by the somewhat simpler one
$$
\mu\{ |Tx| > 0, h=0 \} = 0 \qquad x \in X\setminus\{0\}.
$$

\begin{proof}[Proof of Theorem \ref{thm-nonlinear-nikishin}]
Without loss of generality, via a first change of density, we may assume that $\mu$ is in fact a probability measure.

$(a) \Rightarrow (b)$
Let $x_1,\dots,x_n,x_1',\dots,x_n' \in X$ and $\lambda_1,\dots,\lambda_n \ge 0$. Since $\mu$ is a probability space and $q<p$, the $L_q(hd\mu)$ norm is smaller than the $L_p(hd\mu)$ norm and thus
\begin{align*}
\n{ \bigg(\sum_{j=1}^n \lambda_j|Tx_j-Tx_j'|^p\bigg)^{1/p} }_{L_q(\mu)} &= \n{ \bigg(\sum_{j=1}^n \lambda_j|Sx_j-Sx_j'|^p\bigg)^{1/p} }_{L_q(hd\mu)} \\
&\le \n{ \bigg(\sum_{j=1}^n \lambda_j|Sx_j-Sx_j'|^p\bigg)^{1/p} }_{L_p(hd\mu)} \\
&= \bigg( \sum_{j=1}^n \lambda_j\n{Sx_j-Sx_j'}_{L_p(hd\mu)}^p\bigg)^{1/p}\\
&\le C \bigg( \sum_{j=1}^n \lambda_jd(x_j,x_j')^p\bigg)^{1/p}.
\end{align*}

$(b) \Rightarrow (a)$
Assume $C$ is the best constant in \eqref{eqn-7-3}. Without loss of generality, we can assume $C=1$ (by considering $T/C$ instead of $T$).

Let
$$
W_0 = \left\{ f : \Omega \to \R \,:\, 0 \le f \le  \left(\sum_{j=1}^n \lambda_j|Tx_j-Tx_j'|^p\right)^{q/p}, \;\; \sum_{j=1}^n \lambda_j d(x_j,x_j')^p \le 1 \right\},
$$
and let $W$ be the closure of $W_0$ in $L_1(\mu)$.
Since $1$ is the best constant in \eqref{eqn-7-3},
\begin{equation}\label{eqn-sup}
\sup \left\{ \int_\Omega f \,d\mu : f\in W_0 \right\} = \sup \left\{ \int_\Omega f \,d\mu : f\in W \right\}  =1.
\end{equation}

CLAIM 1: $W^{p/q}$ is a convex set.

It suffices to show that $W_0^{p/q}$ is a convex set. Let $f,g \in W_0$ and $a,b \ge 0$ with $a+b=1$.
From the definition of $W_0$, there exist $x_1,\dots,x_n,x_1',\dots,x_n' \in X$ and $\lambda_1,\dots,\lambda_n \ge 0$ such that
$$
0 \le f \le \left(\sum_{j=1}^n \lambda_j|Tx_j-Tx_j'|^p\right)^{q/p} \quad \text{and} \quad \sum_{j=1}^n \lambda_j d(x_j,x_j')^p \le 1,
$$
and there also exist
$y_1,\dots,y_m,y_1',\dots,y_m' \in X$ and $\sigma_1,\dots,\sigma_m \ge 0$ such that
$$
0 \le g \le \left(\sum_{k=1}^m \sigma_k|Ty_k-Ty_k'|^p\right)^{q/p} \quad \text{and} \quad \sum_{k=1}^m \sigma_k d(y_k,y_k')^p \le 1.
$$
Now
\begin{align*}
0 \le \big( af^{p/q}+b^{p/q} \big)^{q/p} &\le \left( a\sum_{j=1}^n \lambda_j|Tx_j-Tx_j'|^p + b\sum_{k=1}^m \sigma_k|Ty_k-Ty_k'|^p \right)^{q/p} \\
&\le \left( \sum_{j=1}^n a\lambda_j|Tx_j-Tx_j'|^p + \sum_{k=1}^m b\sigma_k|Ty_k-Ty_k'|^p \right)^{q/p},
\end{align*}
and since
$$
\sum_{j=1}^n a\lambda_jd(x_j,x_j')^p + \sum_{k=1}^m b\sigma_kd(y_k,y_k')^p \le a + b =1,
$$
we conclude that $\big( af^{p/q}+b^{p/q} \big)^{q/p} \in W_0$ and therefore $W_0^{p/q}$ is a convex set.

CLAIM 2: There exists $h \in W$ such that $\int hd\mu = 1$.

Since $\mu$ is a probability measure, the map $f \mapsto \int fd\mu$ is a continuous linear functional and therefore it will suffice to show that $W$ is a weakly compact set in $L_1(\mu)$.
By definition, $W$ is norm closed. Moreover, it is convex
so $W$ is weakly closed.

In order to show that $W$ is weakly compact, all that is left to check is equi-integrability. Suppose that $W$ is not equi-integrable. Then there exist $\delta>0$, a sequence $(E_n)_{n=1}^\infty$ of disjoint subsets of $\Omega $ and a sequence $(f_n)_{n=1}^\infty$ in $W$ such that for all $n\in\N$,
$$
\int_{E_n} f_n\,d\mu >\delta.
$$
Thus given any $N\in \N$, since the sets $(E_n)$ are disjoint,
\begin{align*}
N\delta &\le \sum_{n=1}^N \int_{E_n} f_n\,d\mu \le \int \max\{f_1,\dots,f_n\}\,d\mu \\
&\le N^{q/p} \int \left( \sum_{n=1}^N \frac{1}{N} f_n^{p/q} \right)^{q/p}\,d\mu.
\end{align*}
By Claim 1, this last integral is at most $1$, so $\delta \le N^{q/p-1}$. Since $q/p<1$, this is a contradiction for large enough $N$.

Now, let $f \in W$ and $\tau>0$. By Claim 1,
$$
\frac{1}{1+\tau}\big( h^{p/q} + \tau f^{p/q} \big) \in W^{p/q},
$$
so from \eqref{eqn-sup}
\begin{equation}\label{eqn-variation}
(1+\tau)^{q/p} \ge \int \big( h^{p/q} + \tau f^{p/q} \big) ^{q/p} \,d\mu.
\end{equation}
But
\begin{multline*}
\int \big( h^{p/q} + \tau f^{p/q} \big) ^{q/p} \,d\mu. \ge \int_{\{h>0\}} h\,d\mu + \tau^{q/p}\int_{\{h=0\}} f\,d\mu
= 1 + \tau^{q/p} \int_{\{h=0\}} f\,d\mu.
\end{multline*}
so, since $0<q/p<1$,
$$
0 \le \int_{\{h=0\}} f\,d\mu \le \frac{(1+\tau)^{q/p}-1}{\tau^{q/p}} \underset{\tau\rightarrow 0^+}{\longrightarrow} 0,
$$
from where we get \eqref{eqn-7-2}.
by considering $f$ of the form $|Tx-Tx'|/d(x,x')$.

From \eqref{eqn-variation},
\begin{equation}\label{eqn-fatou}
\frac{(1+\tau)^{q/p}-1}{\tau} \ge \int_{\{h>0\}} \left[ \frac{ \big( 1+\tau(f/h)^{p/q} \big)^{q/p} -1 }{\tau} \right] hd\mu.
\end{equation}
Letting $\tau\rightarrow 0^+$, the left-hand side of \eqref{eqn-fatou} converges to $q/p$.
By Fatou's lemma, the right-hand side is at least
$$
\frac{q}{p} \int_{\{h>0\}} \left( \frac{f}{h} \right)^{p/q} hd\mu.
$$
By considering once more $f$ of the form $|Tx-Tx'|/d(x,x')$ we get \eqref{eqn-7-1}.
\end{proof}

Note that the equivalent conditions in Theorem \ref{thm-nonlinear-nikishin} are in fact equivalent to the $p$-convexity of the linear extension $\hat{T} : \fbs(X) \to L_q(\mu)$ of $T : X \to L_q(\mu)$. If we have a Lipschitz factorization as in \eqref{eqn-factorization-maurey-nikishin-nonlinear}, that immediately gives a linear factorization
$$
\xymatrix{
\fbs(X) \ar[r]^{\hat{T}}\ar[d]^{\hat{S}} &L_{q}(\mu) \\
L_p(hd\mu) \ar[r]^{i_{p,q}} & L_q(hd\mu) \ar[u]^{j}
}
$$
and vice versa; such a linear factorization of $\hat{T}$ is equivalent to the $p$-convexity of $\hat{T}$ by the classical Maurey/Nikishin theorem.
Hence, the following definition is a natural one:

\begin{definition}
Let $1 \le p \le \infty$.
Let $X$ be a metric space and $E$ a Banach lattice. A Lipschitz map $T : X \to E$ is called \emph{Lipschitz $p$-convex} if there exists a constant $C\ge 0$
for any $x_j,x_j' \in X$ and $\lambda_j \ge 0$,
$$
\n{ \bigg(\sum_{j=1}^n \lambda_j|Tx_j-Tx_j'|^p\bigg)^{1/p} }_E \le C \bigg(\sum_{j=1}^n  \lambda_jd(x_j,x_j')^p\bigg)^{1/p},
$$
(with the obvious adjustment if $p=\infty$).
The smallest such constant $C$ is called the \emph{Lipschitz $p$-convexity constant} of $T$ and is denoted by $M^{(p)}_{\Lip}(T)$.
\end{definition}

Note that this is a generalization of the linear concept: a linear map $T : X \to E$ from a Banach space $X$ to a Banach lattice $E$ is $p$-convex if and only if it is Lipschitz $p$-convex, and with the same constant.

One could be tempted to follow the footsteps of \cite{Farmer-Johnson-09} and ``get rid of the constants'' in the Lipschitz $p$-convexity condition; that is, checking that it suffices to have the condition with all $\lambda_j$ being equal to 1.
For that to be true we need extra continuity conditions on the lattice, so we do not take such simplification as the definition of Lipschitz $p$-convexity.

The situation of Theorem \ref{thm-nonlinear-nikishin}, where a Lipschitz map turned out to be Lipschitz $p$-convex if and only if its linearization is $p$-convex, is in fact the general case as demonstrated below.

\begin{theorem}\label{theorem-lipschitz-convexity}
Let $X$ be a metric space and $E$ a Banach lattice. A Lipschitz map $T : X \to E$ is Lipschitz $p$-convex if and only if $\hat{T} : \fbs(X) \to E$ is $p$-convex. Moreover, in this case the $p$-convexity constants are the same.
\end{theorem}

\begin{proof}
The ``if'' part is trivial: $p$-convexity of $\hat{T}$ clearly implies Lipschitz $p$-convexity of $T$ with no increment in the constant, since $\n{m_{xx'}}_{\fbs(X)} = d(x,x')$ and $\hat{T}m_{xx'} = Tx-Tx'$.

Now suppose that $T$ is Lipschitz $p$-convex.
Let $\varphi_j^* \in E^*$ be arbitrary.
For any $x_j,x_j' \in X$ with $x_j \not=x'_j$ we obviously have
$$
\bigg( \sum_j \Big| \frac{\pair{\varphi_j^*}{Tx_j-Tx'_j}}{d(x_j,x'_j)} \Big|^{p'} \bigg)^{1/p'}
= \sup_{ \sum_j |\alpha_j|^p \le 1} \sum_j \alpha_j \frac{\pair{\varphi_j^*}{Tx_j-Tx'_j}}{d(x_j,x'_j)}.
$$
Using \cite[Prop. 1.d.2.(iii)]{Lindenstrauss-Tzafriri-II}, the latter is bounded by
\begin{multline*}
\sup_{ \sum_j |\alpha_j|^p \le 1} \bigg( \bigg( \sum_j |\varphi_j^*|^{p'} \bigg)^{1/p'} \bigg) \bigg( \bigg( \sum_j |\alpha_j|^p \frac{\big|Tx_j-Tx'_j\big|^p}{d(x_j,x'_j)^p} \bigg)^{1/p} \bigg) \\
\le \n{ \bigg( \sum_j |\varphi_j^*|^{p'} \bigg)^{1/p'} }_{L^*} \sup_{ \sum_j |\alpha_j|^p \le 1} \n{ \bigg( \sum_j |\alpha_j|^p \frac{\big|Tx_j-Tx'_j\big|^p}{d(x_j,x'_j)^p} \bigg)^{1/p} }_E
\end{multline*}
The Lipschitz $p$-convexity of $T$ allows us to bound this by
\begin{multline*}
\n{ \bigg( \sum_j |\varphi_j^*|^{p'} \bigg)^{1/p'} }_{E^*} M^{(p)}_{\Lip}(T)\sup_{ \sum_j |\alpha_j|^p \le 1} \bigg( \sum_j |\alpha_j|^p \frac{d(x_j,x'_j)^p}{d(x_j,x'_j)^p} \bigg)^{1/p} \\
= M^{(p)}_{\Lip}(T) \n{ \bigg( \sum_j |\varphi_j^*|^{p'} \bigg)^{1/p'} }_{E^*}.
\end{multline*}
Therefore,
$$
\bigg( \sum_j \Big| \frac{(\hat{T}^*\varphi_j^*)(x_j)-(\hat{T}^*\varphi_j^*)(x'_j)}{d(x_j,x'_j)} \Big|^{p'} \bigg)^{1/p'}  \le  M^{(p)}_{\Lip}(T) \n{ \bigg( \sum_j |\varphi_j^*|^{p'} \bigg)^{1/p'} }_{E^*},
$$
so taking the supremum over all pairs $x_j,x'_j \in X$ with $x_j \not=x'_j$ we conclude
$$
\bigg( \sum_j \n{ \hat{T}^*\varphi_j^* }_{X^\#}^{p'} \bigg)^{1/p'}  \le  M^{(p)}_{\Lip}(T) \n{ \bigg( \sum_j |\varphi_j^*|^{p'} \bigg)^{1/p'} }_{E^*}.
$$
Since the $\varphi_j^* \in L^*$ were arbitrary, this means that $\hat{T}^* : L^* \to X^\#$ is $p'$-concave with $M_{(p')}(\hat{T}^*) \le M^{(p)}_{\Lip}(T)$, and by duality \cite[Prop. 1.d.4]{Lindenstrauss-Tzafriri-II}
 $\hat{T} : \fbs(X) \to L$ is $p$-convex with $M^{(p)}(\hat{T}) \le M^{(p)}_{\Lip}(T)$.
\end{proof}

Let us note that the argument in the previous result is based on the duality between $p$-convexity and $p'$-concavity, so it seems unlikely that it could be used to prove a similar result for other classes of maps obtained by replacing the expression $ \Big(\sum_j |x_j|^p\Big)^{1/p}$ by other homogeneous functions given by the Krivine functional calculus for Banach lattices.

\begin{remark}
	It is worth pointing out that Theorem \ref{thm-nonlinear-nikishin} follows immediately from Theorem \ref{theorem-lipschitz-convexity} and the classical Maurey/Nikishin factorization theorem \cite[Thm. 7.1.2]{Albiac-Kalton}.
\end{remark}

\section{Lipschitz $q$-concave maps}\label{sec-lip-q-concave}

Following up on the previous work on $p$-convexity we now point our attention to the natural companion concept, that of Lipschitz $q$-concavity.

\begin{definition}
Let $X$ be a metric space and $E$ a Banach lattice. A map $T : E \to X$ is called \emph{Lipschitz $q$-concave} if there exists a constant $C\ge 0$ such that for any $v_j,v'_j \in E$ and any $\lambda_j \ge 0$,
$$
\bigg(\sum_{j=1}^n  \lambda_j d(Tv_j,Tv'_j)^q\bigg)^{1/q} \le C  \n{ \bigg(\sum_{j=1}^n \lambda_j|v_j-v'_j|^q\bigg)^{1/q} }_E.
$$
The smallest such constant $C$ is the \emph{Lipschitz $p$-concavity constant} of $T$ and is denoted by $M_{(p)}^{\Lip}(T)$.
\end{definition}

This time we can apply the idea from \cite{Farmer-Johnson-09} and note that it suffices to check the condition with all the $\lambda_j$ being equal to one. From that it one can deduce the same inequality for integer values of $\lambda_j$, hence rational values, and for arbitrary values of $\lambda_j$ we approximate with increasing sequences of rationals.
We will primarily be interested in the case when $X$ is a Banach space.
Note that when $X$ is a Banach space and $T : E \to X$ is linear, clearly $T$ is $p$-concave if and only if it is Lipschitz $p$-concave (and with the same constant).

Unlike in the Lipschitz $p$-convexity situation, there is no natural candidate for a linearized map that is $q$-concave if and only if $T$ is Lipschitz $q$-concave, because now the metric space is not the domain and thus there is no canonical linearization.

The following factorization theorem and its proof are inspired by the corresponding result for linear maps proven by Krivine \cite{Krivine-factorisation-reticules}.
The presentation follows that of \cite[Thm. 1.d.11]{Lindenstrauss-Tzafriri-II}.

\begin{theorem}\label{theorem-lipschitz-concavity}
Let $X$, $Y$ be metric spaces with $Y$ complete and $E$ a Banach lattice.
Suppose that $T : X \to E$ is Lipschitz $p$-convex and $S : E \to Y$ is Lipschitz $p$-concave.
Then the map $ST$ can be factored through an $L_p(\mu)$ space.
Moreover, we may arrange to have $ST = S_1T_1$ with $T_1 : X \to L_p(\mu)$, $S_1 : L_p(\mu)\to Y$, $\Lip(T_1) \le M^{(p)}_{\Lip}(T)$ and $\Lip(S_1) \le M^{\Lip}_{(p)}(S)$.
\end{theorem}

\begin{proof}
Let $I_T$ be the (in general non-closed) ideal of $E$ generated by the range of $T$.
We define new operations on $I_T$ as in the usual $p$-concavification procedure, that is, for $x, y \in I_T$ and real $\alpha$ put
$$
x \oplus y := (x^p+y^p)^{1/p}, \qquad \qquad \alpha \odot x := \alpha^{1/p} x,
$$
and let $\check{I}_T$ denote the vector lattice obtained when $I_T$ is endowed with the original order and the operations $\oplus, \odot$. Set
\begin{multline*}
 F_1 := \conv \Big\{ x \in \check{I}_T : |x| \le \lambda|Tv-Tv'| \\ \text{ for some } v,v' \in X, \lambda>0  \text{ with } \lambda d(v,v') < 1/M^{(p)}_{\Lip}(T) \Big\}
\end{multline*}
and
\begin{multline*}
 F_2 := \conv \Big\{ x \in \check{I}_T : x>0 \text{ and } \eta d(Sy,Sy') \ge M^{\Lip}_{(p)}(S) \\ \text{ for some } y,y' \in L, \eta>0 \text{ with } \eta|y-y'| \le x \Big\}.
\end{multline*}
where both convex hulls are taken in the sense of $\check{I}_T$, i.e. using the operations $\oplus, \odot$.

If $x$ belongs to $F_1$, then it can be written the form $\bigoplus_j \alpha_j \odot x_j$ where $\alpha_j \ge 0$, $\sum_j \alpha_j =1$ and $|x_j| \le \lambda_j |Tv_j -Tv'_j|$ with $\lambda_j d(v_j,v'_j) < 1/ M^{(p)}_{\Lip}(T)$. Therefore,
\begin{align*}
\n{x} &= \bigg\| \Big( \sum_j |\alpha_j^{1/p} x_j|^p  \Big)^{1/p} \bigg\|  \le \bigg\| \Big( \sum_j \alpha_j \lambda_j^p |Tv_j-Tv'_j|^p  \Big)^{1/p} \bigg\| \\
&\le M^{(p)}_{\Lip}(T) \Big( \sum_j \alpha_j \lambda_j^p d(v_j,v'_j)^p \Big)^{1/p} < 1.
\end{align*}

On the other hand, if $x$ belongs to $F_2$ then it can be written as $\bigoplus_j \beta_j \odot x_j$ where $\beta_j \ge 0$, $\sum_j \beta_j =1$ and $x_j \ge \eta_j |y_j-y'_j|$ with $\eta_j \ge 0$ and $\eta_j d(Sy_j,Sy'_j) \ge M^{\Lip}_{(p)}(S)$.
Therefore
\begin{align*}
\n{x} &= \bigg\| \Big( \sum_j |\beta_j^{1/p} x_j|^p  \Big)^{1/p} \bigg\| \ge \bigg\| \Big( \sum_j \beta_j \eta_j^p|y_j-y'_j|^p  \Big)^{1/p} \bigg\| \\
&\ge \frac{1}{M^{\Lip}_{(p)}(S)}  \Big( \sum_j \beta_j \eta_j^p d(Sy_j,Sy'_j)^p \Big) \ge 1.
\end{align*}

Hence, $F_1 \cap F_2 = \emptyset$ and since $0$ is an internal point of $F_1$
it follows from the separation theorem that there exists a linear functional $\varphi$ on $\check{I}_T$ such that $\varphi(x) \le 1$ for all $x \in F_1$ and $\varphi(x) \ge 1$ for all $x \in F_2$.
Note that from the definition of $F_2$, for any positive real $\alpha$, any positive $x$ in $\check{I}_T$ and any $x_0 \in F_2$ we have that $\alpha \odot x \oplus x_0$ belongs to $F_2$.
It follows that $\varphi(x) \ge 0$ whenever $0 < x \in \check{I}_T$ and, thus, we can define a seminorm on $I_T$ by putting
$$
\n{x}_0 := \varphi(|x|)^{1/p}, \qquad x \in I_T.
$$
Let $\alpha$ be a real number and $x \in I_T$. Then
$$
\n{\alpha x}_0 = \varphi(|\alpha||x|)^{1/p} = \varphi(|\alpha|^p \odot |x|)^{1/p} = \big[|\alpha|^p \varphi(|x|)  \big]^{1/p} = |\alpha| \n{x}_0.
$$
Let $x,y \in I_T$. Note that $|x|+|y| = \big(|x|^{1/p}\oplus|y|^{1/p}\big)^{p}$.
On the other hand, from the lattice functional calculus and H\"older's inequality, whenever $\alpha$ and $\beta$ are positive reals with $\alpha^{p'}+\beta^{p'}=1$ we have
$$
\big(|x|^{1/p}\oplus|y|^{1/p}\big)^{p} \le \alpha^{-p} \odot |x| + \beta^{-p} \odot |y|.
$$
Hence
\begin{align*}
\n{x+y}_0^p &= \varphi\big(|x+y|\big) \le \varphi\big(|x|+|y|\big) = \varphi\big( \big(|x|^{1/p}\oplus|y|^{1/p}\big)^{p} \big) \\
&\le  \varphi\big( \alpha^{-p} \odot |x| + \beta^{-p} \odot |y| \big) = \alpha^{-p}\varphi(|x|) + \beta^{-p}\varphi(|y|) \\
&= \alpha^{-p} \n{x}_0^p + \beta^{-p}\n{y}_0^p.
\end{align*}
Therefore, setting
$$
\alpha := \frac{\n{x}_0^{1/p'}}{(\n{x}_0+\n{y}_0)^{1/p'}} \qquad\text{and}\qquad \beta := \frac{\n{y}_0^{1/p'}}{(\n{x}_0+\n{y}_0)^{1/p'}}
$$
we satisfy the condition $\alpha^{p'}+\beta^{p'}=1$, while
$$
\alpha^{-p} = \frac{(\n{x}_0+\n{y}_0)^{p-1}}{\n{x}_0^{p-1}} \qquad\text{and}\qquad  \beta^{-p} = \frac{(\n{x}_0+\n{y}_0)^{p-1}}{\n{y}_0^{p-1}}
$$
so we conclude
$$
\n{x+y}_0^p \le \frac{(\n{x}_0+\n{y}_0)^{p-1}}{\n{x}_0^{p-1}}\n{x}_0^p + \frac{(\n{x}_0+\n{y}_0)^{p-1}}{\n{y}_0^{p-1}}\n{y}_0^p  = \big( \n{x}_0 + \n{y}_0 \big)^p,
$$
and thus $\n{x+y}_0 \le \n{x}_0 + \n{y}_0$.

Observe now that for any $x,y \in I_T$ we have
$$
|x|+|y| \ge \big( |x|^p + |y|^p \big)^{1/p} \ge |x| \vee |y|
$$
since these inequalities are valid for reals. By the fact that $\varphi$ is non-negative, we get that
\begin{align*}
\big\| |x|+|y| \big\|_0^p &= \varphi\big(|x|+|y|\big) \ge \varphi\big((|x|^p+|y|^p)^{1/p}\big) = \varphi(|x|\oplus|y|) = \varphi(|x|) + \varphi(|y|) \\
&= \n{x}_0^p + \n{y}_0^p \ge \varphi\big( |x|\vee|y| \big) = \big\| |x|\vee|y| \big\|_0^p.
\end{align*}
This inequality concerning $\n{\cdot}_0$ clearly remains valid in the completion $Z$ of $I_T$ modulo the ideal of all $x \in I_T$ for which $\n{x}_0 = 0$.
Therefore, if $|x| \wedge |y| = 0$ for some $x$ and $y$ in the lattice $Z$ then (recalling that $|x|\wedge|y| = |x|+|y|-|x|\vee|y|$) we obtain
$$
\big\| |x|+|y| \big\|_0^p = \n{x}_0^p + \n{y}_0^p,
$$
i.e. $Z$ is an abstract $L_p$ space. It follows from the $L_p$ version of Kakutani's representation theorem that $Z$ is order isometric to an $L_p(\mu)$ space for a suitable measure $\mu$.

Let $T_1 : X \to Z$ be defined by $T_1v = Tv$, $v \in X$, i.e. the same as $T$ but considered as an map into $Z$.
For $v,v' \in X$ and $\lambda>0$, if $\lambda d(v,v') < 1/M^{(p)}_{\Lip}(T)$ then $\lambda(T_1v - T_1v') \in F_1$, which implies that $\varphi(\lambda(T_1v-T_1v')) \le 1$ and thus $\n{\lambda(T_1v-T_1v')}_0 \le 1$, from where it follows that
$\n{T_1v-T_1v'}_0 \le M^{(p)}_{\Lip}(T) d(v,v')$, i.e. $\Lip(T_1) \le M^{(p)}_{\Lip}(T)$.

Let $S_1 : I_T/\ker(\n{\cdot}_0) \to Y$ be defined by $S_1x = Sx$, $x \in I_T$.
Note that this is well defined: if $Sx\not= Sx'$, then $\tfrac{M^{\Lip}_{(p)}(S)}{d(Sx,Sx')}|x-x'|$ belongs to $F_2$, so $\varphi\Big(\tfrac{M^{\Lip}_{(p)}(S)}{d(Sx,Sx')}|x-x'|\Big) \ge 1$ and in particular $\n{x-x'}_0\not= 0$.
By an argument similar to the one for $T_1$, this defines a Lipschitz map from $I_T/\ker(\n{\cdot}_0)$ to $Y$ with Lipschitz constant at most $M^{\Lip}_{(p)}(S)$.
Since $I_T/\ker(\n{\cdot}_0)$ is dense in $Z$ and $Y$ is complete, this can be extended to a Lipschitz map $S_1 : Z \to Y$ with the same Lipschitz constant, giving the desired factorization.
\end{proof}

\section{Factorization theorems}\label{sec-factorization-theorems}

In this section we prove some results that make very clear why our Lipschitz versions of the  Maurey/Nikishin and Krivine  factorization Theorems hold.
They are nonlinear generalizations of the beautiful factorization theorems of Raynaud and Tradacete \cite{Raynaud-Tradacete}, that characterize $p$-convex and $q$-concave linear maps in terms of factorizations through $p$-convex and $q$-concave Banach lattices.

\subsection{Factorization of Lipschitz $p$-convex maps}

We start by proving the following characterization of Lipschitz $p$-convex maps, compare to \cite[Thm. 3]{Raynaud-Tradacete}.

	\begin{theorem}\label{thm-factorization-lipschitz-p-convex}
		Let $E$ be a Banach lattice, $X$ a metric space and $1 \le p \le \infty$. A
		Lipschitz map $T : X \to E$ is Lipschitz $p$-convex if and only if there exist a $p$-convex Banach
		lattice $W$, a positive operator (in fact, an injective interval preserving lattice
		homomorphism) $\psi : W \to E$ and another Lipschitz map $R : E \to W$
such that $T = \psi \circ R$.
	$$
	\xymatrix{
	X \ar[rr]^{T}\ar[dr]_{R} & &E  \\
	 &W \ar[ru]_{\psi} &
	}
	$$
	Moreover, $M^{(p)}_{\Lip}(T) = \inf \Lip(R) \cdot M^{(p)}(I_W) \cdot \n{\psi}$ where the infimum is taken over all such factorizations.
	\end{theorem}

\begin{remark}
	Theorem \ref{thm-factorization-lipschitz-p-convex} in fact follows as an immediate corollary from Theorem \ref{theorem-lipschitz-convexity} and \cite[Thm. 3]{Raynaud-Tradacete}, but we have chosen to show a direct proof.
In this way, Theorem \ref{thm-factorization-lipschitz-p-convex} (together with \cite[Thm. 3]{Raynaud-Tradacete}) can be used to provide an alternative proof of Theorem \ref{theorem-lipschitz-convexity} that does not rely on duality arguments.
\end{remark}

\begin{proof}[Proof of Theorem \ref{thm-factorization-lipschitz-p-convex}]
We will assume that $1\le p<\infty$, the proof of the case $p=\infty$ can be obtained via the usual changes.

If we do have such a factorization,
consider $x_j,x'_j \in X$. By the positivity of $\psi$, using \cite[Prop. 1.d.9]{Lindenstrauss-Tzafriri-II} and the $q$-convexity of $W$,
\begin{multline*}
	\n{ \bigg(\sum_{j=1}^n |Tx_j-Tx'_j|^p\bigg)^{1/p} }_E
	= \n{ \bigg(\sum_{j=1}^n |\psi (Rx_j-Rx'_j)|^p\bigg)^{1/p} }_E \\
\le \n{\psi} \n{ \bigg(\sum_{j=1}^n |\psi (Rx_j-Rx'_j)|^p\bigg)^{1/p} }_W \\
\le \n{\psi} M^{(p)}(I_W) \bigg(\sum_{j=1}^n  \n{Rx_j - Rx'_j}_W^p\bigg)^{1/p} \\
\le \n{\psi} M^{(p)}(I_W) \Lip(R) \bigg(\sum_{j=1}^n  d(x_j,x'_j)^p\bigg)^{1/p},
\end{multline*}
showing that $T$ is Lipschitz $p$-convex with
$M^{(p)}_{\Lip}(T) \le \n{\psi} M^{(p)}(I_W) \Lip(R)$.
\\
\\
Now let $T : X \to E$ be Lipschitz $p$-convex.
Consider the set
\begin{multline*}
A := \bigg\{ u \in E \; : \; |u| \le \bigg( \sum_{i=1}^k \lambda_i|Tx_i-Tx_i'|^p \bigg)^{1/p}, \\
\text{ where } \lambda_i\ge 0 \text{ and } \sum_{i=1}^k\lambda_id(x_i,x_i')^p \le 1 \bigg\}.
\end{multline*}
We can consider the Minkowski functional defined by its closure $\bar{A}$ in $E$,
$$
\n{z}_W = \inf\{ \mu>0 \; : \; z \in \mu \bar{A} \}.
$$
Clearly $\bar{A}$ is solid, and since $T$ is Lipschitz $p$-convex, it is also a bounded subset of $E$.
Let us consider the space $W = \{ z \in L \; : \; \n{z}_W < \infty \}$.
We claim that for any $z_1 , \dotsc, z_n$ in $W$ it follows that $\Big(\sum_{i=1}^n|z_i|^p\Big)^{1/p}$ is in $W$ and moreover
$$
\n{\bigg(\sum_{i=1}^n|z_i|^p\bigg)^{1/p}}_W \le \bigg(\sum_{i=1}^n\n{z_i}_W^p\bigg)^{1/p}.
$$
Given $\varepsilon>0$, for each $i = 1, \dotsc, n$ there exist $\mu_i$ with $z_i \in \mu_i\bar{A}$ such that $\mu_i^p \le \n{z_i}_W + \frac{\varepsilon^p}{n}$.
Thus, for every $\delta>0$ there exist $\{z_i^\delta\}_{i=1}^n$ in $E$ such that $\n{z_i-z_i^\delta}_E \le \delta$ and
$$
|z_i^\delta| \le \bigg( \sum_{j=1}^{m_{i,\delta}} \lambda^{i,\delta}_j|Tx^{i,\delta}_j-Ty^{i,\delta}_j|^p \bigg)^{1/p}
$$
where the nonnegative numbers $\{ \lambda^{i,\delta}_j \}_{j=1}^{m_{i,\delta}}$    and the points $\{x^{i,\delta}_j, y^{i,\delta}_j\}_{j=1}^{m_{i,\delta}}$ in $X$ satisfy
$$
\bigg( \sum_{j=1}^{m_{i,\delta}} \lambda^{i,\delta}_j d(x^{i,\delta}_j,y^{i,\delta}_j)^p \bigg)^{1/p} \le \mu_i
$$
for each $i = 1, \dotsc, n$ and each $\delta>0$.
For each $\delta>0$, define
$$
w_\delta = \bigg(\sum_{i=1}^n|z^\delta_i|^p\bigg)^{1/p}.
$$
Then
\begin{multline*}
	\n{ \bigg(\sum_{i=1}^n|z_i|^p\bigg)^{1/p} - w_\delta }_E \le
	\n{ \bigg(\sum_{i=1}^n|z_i - z_i^\delta|^p\bigg)^{1/p} }_E  \\
	\le \bigg(\sum_{i=1}^n \n{z_i - z_i^\delta}_E^p\bigg)^{1/p} \le n\delta.
\end{multline*}
Moreover, we will show that for every $\delta>0$, $w_\delta$ belongs to $\big(\sum_{i=1}^n \mu_i^p\big)^{1/p} A$.
Indeed,
$$
|w_\delta| = \bigg(\sum_{i=1}^n|z^\delta_i|^p\bigg)^{1/p} \le
\bigg(\sum_{i=1}^n \sum_{j=1}^{m_{i,\delta}} \lambda^{i,\delta}_j|Tx^{i,\delta}_j-Ty^{i,\delta}_j|^p \bigg)^{1/p}
$$
and
$$
\bigg(\sum_{i=1}^n \sum_{j=1}^{m_{i,\delta}} \lambda^{i,\delta}_j d(x^{i,\delta}_j,y^{i,\delta}_j)^p \bigg)^{1/p} \le \bigg(\sum_{i=1}^n\mu_i^p\bigg)^{1/p}.
$$
Hence, $\big(\sum_{i=1}^n|z_i|^p\big)^{1/p}$ belongs to $\big(\sum_{i=1}^n\mu_i^p\big)^{1/p}\bar{A}$. Therefore, from the definition of $\n{\cdot}_W$ and the choice of $\mu_i$,
\begin{multline*}
	\n{\bigg(\sum_{i=1}^n|z_i|^p\bigg)^{1/p}}_W \le \bigg(\sum_{i=1}^n\mu_i^p\bigg)^{1/p} \\ \le \bigg(\sum_{i=1}^n (\n{z_i}_W^p +\varepsilon^p/n ) \bigg)^{1/p}
	\le  \bigg(\sum_{i=1}^n \n{z_i}_W^p \bigg)^{1/p} + \varepsilon.
\end{multline*}
Letting $\varepsilon$ go to 0, we conclude
$$
\n{\bigg(\sum_{i=1}^n|z_i|^p\bigg)^{1/p}}_W \le \bigg(\sum_{i=1}^n \n{z_i}_W^p \bigg)^{1/p}.
$$
It follows that the Minkowski functional $\n{\cdot}_W$ is in fact a norm on $W$.
Since $\bar{A}$ is bounded, $\n{z}_W = 0$ implies that $z = 0$.
Moreover if $u,v \in W$ are not zero, set
$$
\tilde{u} = \frac{|u|}{\n{u}_W}, \quad, \tilde{v} = \frac{|v|}{\n{v}_W},
\alpha = \frac{\n{u}_W}{\n{u}_W + \n{v}_W}, \quad \beta = \frac{\n{v}_W}{\n{u}_W + \n{v}_W}.
$$
Since $\n{\tilde{u}}_W = \n{\tilde{v}}_W = 1$, $\alpha,\beta \ge 0$ and $\alpha + \beta = 1$ we have
\begin{align*}
	\n{u+v}_W &\le \n{ |u|+|v| }_W
	= \big( \n{u}_W + \n{v}_W \big) \n{\alpha \tilde{u} + \beta \tilde{v}}_W \\
	&\le \big( \n{u}_W + \n{v}_W \big) \n{(\alpha \tilde{u}^p + \beta \tilde{v}^p)^{1/p}}_W \\
	&\le \big( \n{u}_W + \n{v}_W \big) \big( \alpha \n{\tilde{u}}_W^p + \beta \n{\tilde{v}}_W^p \big) \\
	&= \n{u}_W + \n{v}_W.
\end{align*}
Thus, $(W, \n{\cdot}_W)$ is a $p$-convex normed lattice with constant 1.
Let $(w_i)_{i=1}^\infty$ be a Cauchy sequence in $W$.
Since for every $z \in E$ it holds that $\n{z}_E \le M_{\Lip}^{(p)}(T) \n{z}_W$,
it follows that  $(w_i)_{i=1}^\infty$ is also a Cauchy sequence in $E$ and this has a limit $w$ in $E$.
Notice that since the $w_i$ are bounded in $W$, there exists some finite $\mu$ such that $w_i \in \mu \bar{S}$ for every $i \in \N$ and since $\bar{S}$ is closed in $E$, we must have $w \in \mu \bar{S}$.
Thus, $w$ belongs to $W$, and we will show that $(w_i)_{i=1}^\infty$ converges to $w$ also in $W$.
To this end, let $\varepsilon>0$. Since $(w_i)_{i=1}^\infty$ is a Cauchy sequence, there exists $N\in\N$ such that $w_i-w_j \in \varepsilon \bar{S}$ whenever $i, j \ge N$.
Thus, if $i \ge N$ we can write
$$
w - w_i = (w - w_j) + (w_j - w_i)
$$
for every $j \ge N$, and letting $j \to \infty$ we obtain that $w - w_i \in \varepsilon\bar{S}$.
This shows that $(w_i)_{i=1}^\infty$ converges to $w$ in $W$, hence $W$ is complete and therefore a Banach lattice.

Let us observe that from the definition of $W$ we clearly have $\n{Tx-Tx'}_W \le d(x,x')$ for every $x,x' \in X$, so the map $R : X \to W$ given by $Rx = Tx$ is Lipschitz with $\Lip(R) \le 1$.
Moreover, as noticed above it also holds that $\n{z}_E \le M_{\Lip}^{(p)}(T) \n{z}_W$ for each $z \in E$, hence the formal inclusion $\psi : W \to E$ is clearly and injective interval preserving lattice homomorphism with norm at most $M_{\Lip}^{(p)}(T)$, and $T = \psi \circ R$.
\end{proof}

Similarly to the notation in \cite{Raynaud-Tradacete}, since the lattice $W$ constructed in the proof of Theorem \ref{thm-factorization-lipschitz-p-convex} depends on $T$ and $p$ we will denote it by $W_{T,p}$.
As in \cite[Remark 4]{Raynaud-Tradacete}, note that whenever $1 \le p' \le p$ and $T : X \to E$ is Lipschitz $p$-convex, it always holds that the inclusion $W_{T,p'} \to W_{T,p}$ has norm at most one.
Moreover, as in \cite[Remark 5]{Raynaud-Tradacete}, note that when $E$ is a $p$-convex Banach lattice and $T$ is the identity map on $E$, then $W_{T,p}$ is a renorming of $E$ with $p$-convexity constant equal to one.
The Banach lattice $W_{T,p}$ also satisfies a certain minimality property that follows from \cite[Prop. 4]{Raynaud-Tradacete}, see that paper for the details.

\subsection{Factorization of Lipschitz $q$-concave maps}

The following characterization of Lipschitz $q$-concave maps is a nonlinear generalization of \cite[Thm. 1]{Raynaud-Tradacete}.

\begin{theorem}\label{thm-factorization-lipschitz-q-concave}
	Let $E$ be a Banach lattice, $X$ a complete metric space and $1 \le q \le \infty$. A
	Lipschitz map $T : E \to X$ is Lipschitz $q$-concave if and only if there exist a $q$-concave Banach
	lattice $V$, a positive operator $\phi : E \to V$ (in fact, a lattice homomorphism with dense image), and another Lipschitz map $S : V \to X$ such that $T = S \circ \phi$.
	$$
	\xymatrix{
	E \ar[rr]^{T}\ar[dr]_{\phi} & &X  \\
	 &V \ar[ru]_{S} &
	}
	$$
	Moreover, $M_{(q)}^{\Lip}(T) = \inf \n{\phi} \cdot M_{(q)}(I_V) \cdot \Lip(S)$.
\end{theorem}

\begin{remark}
	The proof in our case is inspired by the linear one, but as far as we can tell the Lipschitz case does not follow from the linear one.
	Unlike in the case of Theorem \ref{theorem-lipschitz-convexity}, it does not seem to be possible to fully reduce Lipschitz $q$-concave maps to (linear) $q$-concave maps.
	However, comparing Theorem \ref{thm-factorization-lipschitz-q-concave} with  \cite[Thm. 1]{Raynaud-Tradacete} shows that Lipschitz $q$-concave maps are in fact (linear) $q$-concave maps followed by Lipschitz maps.
\end{remark}

\begin{proof}[Proof if Theorem \ref{thm-factorization-lipschitz-q-concave}]
	The case $q=\infty$ is trivial because every Banach lattice is $\infty$-concave, so let us assume that $1 \le q < \infty$.
	The specific construction carried out below, however, has an analogue in the case $q=\infty$.

	First, let us assume that such a factorization exists.
	Consider $u_j,u'_j \in E$. By the positivity of $\phi$, using \cite[Prop. 1.d.9]{Lindenstrauss-Tzafriri-II} and the $q$-concavity of $V$,
	\begin{multline*}
		\bigg(\sum_{j=1}^n  d(Tu_j,Tu'_j)^q\bigg)^{1/q}
		= \bigg(\sum_{j=1}^n  d(S\phi u_j,S\phi u'_j)^q\bigg)^{1/q} \\
		\le \Lip(S) \bigg(\sum_{j=1}^n  \n{\phi u_j - \phi u'_j}_V^q\bigg)^{1/q} \\
		\le \Lip(S) M_{(q)}(I_V) \n{ \bigg(\sum_{j=1}^n |\phi(v_j-v'_j)|^q\bigg)^{1/q} }_V \\
		\le \Lip(S) M_{(q)}(I_V) \n{\phi} \n{ \bigg(\sum_{j=1}^n |v_j-v'_j|^q\bigg)^{1/q} }_E,
	\end{multline*}
	showing that $T$ is Lipschitz $q$-concave with
	$M_{(q)}^{\Lip}(T) \le \n{\phi} \cdot M_{(q)}(I_V) \cdot \Lip(S)$.

	Now suppose that $T$ is Lipschitz $q$-concave.
	Given $u \in E$, define
	$$
	\rho(u) = \sup \left\{ \bigg(\sum_{j=1}^n  \lambda_j d(Tu_j,Tu'_j)^q\bigg)^{1/q}
	\; : \; \lambda_j\ge0, \; \bigg(\sum_{j=1}^n \lambda_j |u_j-u'_j|^q\bigg)^{1/q} \le |u| \right\}.
	$$
	Note that since
	$$
	\bigg(\sum_{j=1}^n  \lambda_j d(Tu_j,Tu'_j)^q\bigg)^{1/q} \le M_{(q)}^{\Lip}(T)  \n{\bigg(\sum_{j=1}^n \lambda_j |u_j-u'_j|^q\bigg)^{1/q} }_E,
	$$
	we have for all $u,v \in E$
	$$
	d\big(Tu,Tv\big) \le \rho(u-v) \le M_{(q)}^{\Lip}(T) \n{u-v}.
	$$
	Moreover, we claim that $\rho$ is a lattice seminorm on $E$.
	First notice that for any $u,v \in E$ and $\lambda \ge 0$, it is clear that $\rho(\lambda u) = \lambda \rho(u)$ and that $|v| \le |u|$ implies $\rho(v) \le \rho(u)$.
	To prove the triangle inequality, let $u, v \in E$ and $w = |u| + |v|$, and let $I_w \subset E$ be the ideal generated by $w$ in $E$, which is identified with a space $C(K)$ in which $w$ corresponds to the function identically equal to 1 \cite[II.7]{Schaefer}.
	Given $\varepsilon>0$, find $w_1, w_1', \dotsc, w_n,w_n' \in E$ and $\lambda_1, \dotsc, \lambda_n \ge 0$ such that
	$$
	\bigg(\sum_{j=1}^n \lambda_j |w_j-w'_j|^q\bigg)^{1/q} \le |w| \quad\text{and}\quad
	\rho(w) \le \bigg(\sum_{j=1}^n  \lambda_j d(Tw_j,Tw'_j)^q\bigg)^{1/q} + \varepsilon.
	$$
	Since $u,v \in I_w$, they correspond to functions $f, g \in C(K)$ such that $|f(t)| + |g(t)| = 1$ for all $t \in K$.
	Similarly, each $w_j-w_j'$ corresponds to $h_j \in C(K)$ with $(\sum_{j=1}^n \lambda_j|h_j(t)|^q)^{1/q} \le 1$ for all $t \in K$.
	Let us now consider
	$$
	f_j(t) = h_j(t) f(t), \qquad g_j(t) = h_j(t) g(t)
	$$
	which belong to $C(K)$ and satisfy
	$$
	\bigg(\sum_{j=1}^n \lambda_j |f_j(t)|^q\bigg)^{1/q} \le |f(t)|
	\quad\text{and}\quad
	\bigg(\sum_{j=1}^n \lambda_j |g_j(t)|^q\bigg)^{1/q} \le |g(t)|.
	$$
	This means there are $u_j, v_j \in I_w \subset E$ with $u_j + v_j = w_j-w_j'$ for each $1\le j \le n$ and satisfying
	$$
	\bigg(\sum_{j=1}^n \lambda_j |u_j|^q\bigg)^{1/q} \le |u|
	\quad\text{and}\quad
	\bigg(\sum_{j=1}^n \lambda_j |v_j|^q\bigg)^{1/q} \le |v|.
	$$
	Notice that $w_j-u_j = v_j+w_j'$, and hence
	$$
		d(Tw_j,Tw'_j) \le d\big(Tw_j, T(w_j-u_j)\big) + d\big(T(v_j+w'_j),Tw'_j\big).
	$$
	Then,
	\begin{multline*}
		\rho(u+v) \le \rho(w) \le \bigg(\sum_{j=1}^n  \lambda_j d(Tw_j,Tw'_j)^q\bigg)^{1/q} + \varepsilon \\
		\le \bigg(\sum_{j=1}^n  \lambda_j d\big(Tw_j, T(w_j-u_j)\big)^q\bigg)^{1/q}
		 + \bigg(\sum_{j=1}^n  \lambda_j d\big(T(v_j+w'_j),Tw'_j\big)^q\bigg)^{1/q} + \varepsilon.
	\end{multline*}
	Since $w_j - (w_j-u_j) = u_j$ and $(v_j+w'_j) - w'_j = v_j$, it follows from the definition of $\rho$ that
	\begin{multline*}
		\bigg(\sum_{j=1}^n  \lambda_j d\big(Tw_j, T(w_j-u_j)\big)^q\bigg)^{1/q} \le \rho(u) \qquad \text{and} \\
		\bigg(\sum_{j=1}^n  \lambda_j d\big(T(v_j+w'_j),Tw'_j\big)^q\bigg)^{1/q} \le \rho(v)
	\end{multline*}

	and thus $\rho(u+v) \le \rho(u) + \rho(v) + \varepsilon$.
	Letting $\varepsilon$ go to 0, we have proved that $\rho$ satisfies the triangle inequality.

	Let $V$ denote the Banach lattice obtained by completing $E/\rho^{-1}(0)$ with the norm induced by $\rho$, and let $\varphi$ be the quotient map $E \to E/\rho^{-1}(0)$ seen as a map to $V$.
	For $u \in E$ let us define $S(\phi(u)) = T(u)$.
	Since $d\big(Tu,Tv\big) \le \rho(u-v)$ for any $u, v \in E$ the map $S$ is well defined on $E/\rho^{-1}(0)$. Moreover, since $X$ is complete we can extend $S$ to a Lipschitz map $S : V \to X$ such that $\Lip(S) \le 1$ and $T = S \varphi$.

	Now consider $\{u_i\}_{i=1}^n$ in $E$, and let $\varepsilon>0$.
	For each $i =1 , \dotsc, n$ there exist $\{v_j^i,w_j^i\}_{j=1}^{k_i}$ in $E$ and nonnegative numbers $\{\lambda_j^i\}_{j=1}^{k_i}$ such that
\begin{multline*}
	\bigg(\sum_{j=1}^{k_i} \lambda_j^i |v_j^i-w^i_j|^q\bigg)^{1/q} \le |u_i|  \qquad \text{and} \\
		\rho(u_i)^q \le \sum_{j=1}^{k_i}  \lambda_j^i d(Tv_j^i,Tw_j^i)^q + \frac{\varepsilon^q}{n}.
\end{multline*}
Adding up, we have that
	$$
	\bigg(\sum_{i=1}^{n} \rho(u_i)^q\bigg)^{1/q} \le
	\rho\bigg( \bigg( \sum_{i=1}^{n} |u_i|^q\bigg)^{1/q} \bigg) + \varepsilon.
	$$
	Letting $\varepsilon$ go to 0 we conclude that the normed lattice $E/\rho^{-1}(0)$ is $q$-concave with constant 1, and thus so is its completion $V$.

	Finally, note that
	$$
	\n{\phi} \cdot M_{(q)}(I_V) \cdot \Lip(S) \le M_{(q)}^{\Lip}(T) \cdot 1 \cdot 1 = M_{(q)}^{\Lip}(T).
	$$
\end{proof}

Since the lattice $V$ constructed in the previous proof depends on the map $T : E \to X$ and $q$, we will denote it by $V_{T,q}$ whenever needed.
As pointed out in the proof, $V_{T,q}$ has $q$-concavity constant one. In particular if $E$ is $q$-concave and $T$ is the identity on $E$, then $V_{T,q}$ is the usual lattice renorming of E with $q$-concavity constant one.

The factorization given by the proof of Theorem \ref{theorem-lipschitz-concavity} is maximal in a certain sense, as the following proposition shows. We omit the proof, since it is an easy combination of that of \cite[Prop. 2]{Raynaud-Tradacete} and the ideas we used to prove Theorem \ref{thm-factorization-lipschitz-q-concave}.

\begin{proposition}
	Let $E$ be a Banach lattice, $X$ a complete metric space, $1 \le q \le \infty$ and
	 $T : E \to X$ be a Lipschitz $q$-concave map. Suppose that $T$ factors through a $q$-concave Banach lattice $\tilde{V}$ with factors $A : E \to \tilde{V}$ and $B : \tilde{V} \to X$, such that $A$ is a lattice homomorphism whose image is dense in $\tilde{V}$, $B$ is a Lipschitz map, and $T = B \circ A$.
Then there is a lattice homomorphism $\varphi : \tilde{V} \to V_{T,q}$ such that
$\phi = \varphi \circ A$ and $S \circ \varphi = B$.

$$
\xymatrix{
E \ar[rr]^{T}\ar[ddr]_{\phi} \ar[dr]^{A} & &X  \\
 &\tilde{V} \ar@{-->}[d]^{\varphi}
 \ar[ru]^{B} & \\
 &V_{T,q} \ar[ruu]_{S}&
}
$$
\end{proposition}

\begin{remark}
	In \cite[Sec. 3]{Raynaud-Tradacete}, duality relations are established between the lattices constructed in the factorization theorems for $p$-convex and $q$-concave operators. We do not try to do something similar here, since the Lipschitz dual $L^\#$ of a Banach lattice is not necessarily a Banach lattice.
\end{remark}

\subsection{Factorization theorems \`a la Krivine}\label{sec-factorization-theorems-a-la-Krivine}

Once again taking \cite{Raynaud-Tradacete} as a model, the nonlinear Krivine factorization theorem (Theorem \ref{theorem-lipschitz-concavity}) follows easily from the factorization Theorems \ref{thm-factorization-lipschitz-p-convex} and \ref{thm-factorization-lipschitz-q-concave}.

\begin{proof}[Alternative proof of Theorem \ref{theorem-lipschitz-concavity}]
Let $X$, $Y$ be metric spaces with $Y$ complete and $E$ a Banach lattice.
Suppose that $T : X \to E$ is Lipschitz $p$-convex and $S : E \to Y$ is Lipschitz $p$-concave.
Apply Theorems \ref{thm-factorization-lipschitz-p-convex} and \ref{thm-factorization-lipschitz-q-concave} (more specifically, their proofs) to obtain factorizations
$$
\xymatrix{
X \ar[rr]^{T}\ar[dr]_{\tau} & &E \ar[rr]^{S}\ar[dr]_{\phi} & &Y  \\
 &W \ar[ru]_{\psi}  &  &V \ar[ru]_{\sigma} & \\
}
$$
where $W$ (resp. $V$) is a $p$-convex (resp. $p$-concave) Banach lattice with constant 1,
$\tau$ and $\sigma$ are Lipschitz maps with constant at most $1$, and
$\psi$ (resp. $\phi$) is a lattice
homomorphism of norm at most $M_{\Lip}^{(p)}(T)$ (resp. $M^{\Lip}_{(p)}(S)$ ).
From \cite[Lemma 17]{Raynaud-Tradacete}, $\phi \circ \psi$ factors through an $L_p(\mu)$ space with the factors being lattice homomorphisms whose norms have product at most
 $M_{\Lip}^{(p)}(T) \cdot M^{\Lip}_{(p)}(S)$.
The conclusion of Theorem \ref{theorem-lipschitz-concavity} is now clear.
\end{proof}

Similarly, Theorems \ref{thm-factorization-lipschitz-p-convex} and \ref{thm-factorization-lipschitz-q-concave} allow us to reduce the following two Lipschitz results to their linear counterparts (namely Remark 10, Prop. 18 and Cor. 19 in \cite{Raynaud-Tradacete}).

\begin{proposition}\label{prop-generalized-krivine}
	Let $X$, $Y$ be metric spaces with $Y$ complete and $E$ a Banach lattice.
	Suppose that $T : X \to E$ is Lipschitz $p$-convex and $S : E \to Y$ is Lipschitz $q$-concave, with $1 \le q < p < \infty$. Then $ST$ factors through a canonical inclusion $i_{p,q} : L_p(\mu) \to L_q(\mu)$. In fact, there is a factorization
	$$
	\xymatrix{
	X \ar[r]^{T}\ar[d]_{T_1} &E \ar[r]^{S} &Y  \\
	L_p(\mu) \ar[rr]^{i_{p,q}} &   & L_q(\mu) \ar[u]_{S_1} & \\
	}
	$$
	with $\Lip(T_1) \le M_{\Lip}^{(p)}(T)$ and $\Lip(S_1) \le M^{\Lip}_{(q)}(S)$.
\end{proposition}

\begin{proposition}
	Let $X$, $Y$ be metric spaces with $Y$ complete and $E$ a Banach lattice, and $1 \le p,q \le \infty$.
	Suppose that $T : X \to E$ is Lipschitz $p$-convex and $S : E \to Y$ is Lipschitz $q$-concave. Then for every $\theta\in(0,1)$,  $ST$ factors through a Banach lattice $U_\theta$ that is $\tfrac{p}{p(1-\theta)+\theta}$-convex and $\frac{q}{1-\theta}$-concave.
\end{proposition}

\begin{corollary}
	Let $E$ be a Banach lattice, $1 \le p,q \le \infty$, and assume that $T : E \to E$ is both Lipschitz $p$-convex and Lipschitz $q$-concave. Then for each $\theta \in (0,1)$, $T^2$ factors through a $\tfrac{p}{p(1-\theta)+\theta}$-convex and $\frac{q}{1-\theta}$-concave Banach lattice. In particular, if $p>1$ and $q< \infty$ then $T$ factors through a super reflexive Banach lattice.
\end{corollary}

A natural question in this context is: if a linear map $T : X \to Y$ between Banach spaces can be factored as a Lipschitz $p$-convex map followed by a Lipschitz $q$-convex one, is there a factorization where the factor maps are in addition linear?
Under certain conditions, the answer is yes.
The following result (and its proof) are in the spirit of
\cite[Thm. 1]{Johnson-Maurey-Schechtman}, \cite[Thm. 2]{Farmer-Johnson-09}, \cite[Thm. 2.1]{Chen-Zheng-12}.

\begin{theorem}\label{thm-equivalence-linear-and-nonlinear}
	Let $T : X \to Y$ be a linear map between a Banach space $X$ and a dual Banach space $Y$, and assume that $T$ admits a factorization $T= T_2 T_1$ where $T_1$ is Lipschitz $p$-convex and $T_2$ is Lipschitz $q$-concave, with $1 \le q < p < \infty$. Then there is also a factorization $T = \tau_2 \tau_1$ where $\tau_1$ is $p$-convex and $\tau_2$ is $q$-concave, and moreover
	$M^{(p)}(\tau_1) \le M_{\Lip}^{(p)}(T_1)$ and $M_{(q)}(\tau_2)\le M^{\Lip}_{(q)}(T_2)$.
\end{theorem}

\begin{proof}
	By Proposition \ref{prop-generalized-krivine}, $T$ can be factored as
	$$
	\xymatrix{
	X \ar[rr]^{T}\ar[d]_{B} &  &Y  \\
	L_p(\mu) \ar[rr]^{i_{p,q}} &   & L_q(\mu) \ar[u]_{A} & \\
	}
	$$
	with $\Lip(B) \le M_{\Lip}^{(p)}(T_1)$ and $\Lip(A) \le M^{\Lip}_{(q)}(T_2)$.

	Let's assume first that $X$ is separable. Since $L_p(\mu)$ is reflexive (hence has the Radon-Nikod\'ym property), it follows from \cite[Thm. 6.42]{Benyamini-Lindenstrauss} that $B$ has a point of G\^ateaux differentiability. By translations we can assume that $B$ is G\^ateaux differentiable at 0, and also that $B(0) = 0$. Define $B_n : X \to L_p(\mu)$ and $A_n : L_q(\mu) \to Y$ by
	$B_n(x) = nB(x/n)$, $A_n(z) = nA(z/n)$.
	Since $A \circ i_{p,q} \circ B = T$ is linear, we have that
	$A_n \circ i_{p,q} \circ B_n = T$.
	By the G\^ateaux differentiability condition, there is a linear map $\beta : X \to L_p(\mu)$ such that $\n{ B_n(x) - \beta(x) } \to 0$ for all $x \in X$.
	For each $z \in L_q(\mu)$, let $\tilde{A}(z)$ be the weak$^*$ limit of $(A_n(x))_{n=1}^\infty$ along a fixed free ultrafilter of natural numbers.
	Using the norm convergence of $B_n(x)$ to $\beta(x)$, we deduce that $\tilde{A} \circ i_{p,q} \circ \beta = T$.
	Note that $\n{\beta} \le \Lip(B)$ and $\Lip(\tilde{A}) \le \Lip(A)$.
	Since $\tilde{A}$ is linear on $i_{p,q} \circ \beta(X)$, by \cite[Thm. 6.42]{Benyamini-Lindenstrauss} there is a linear operator $\alpha : L_q(\mu) \to Y$ that coincides with $\tilde{A}$ on $i_{p,q} \circ \beta(X)$ and satisfies $\n{\alpha} \le \Lip(\tilde{A})$.
	The maps $\tau_1 = i_{p,q} \circ \beta$ and $\tau_2 = \alpha$ give the desired factorization.

	For a nonseparable Banach space $X$, we can apply the previous argument to the restrictions of $T$ to finite-dimensional subspaces of $X$. Taking an ultraproduct over an ultrafilter on the family of those subspaces, we obtain a factorization
	$$
	\xymatrix{
	X \ar[rr]^{(T_j)_{\mathcal{U}}}\ar[d]_{(\beta_j)_\mathcal{U}} &  &Y_\mathcal{U}  \\
	\big(L_p(\mu_j))_\mathcal{U} \ar[rr]^{(i^j_{p,q})_\mathcal{U}} &   & \big(L_q(\mu_j)\big)_\mathcal{U} \ar[u]_{(\alpha_j)_\mathcal{U}} & \\
	}
	$$
	By taking weak$^*$ limits over the ultrafilter, we can replace $Y_\mathcal{U}$ by $Y$ in the above diagram to get a linear factorization of $T : X \to Y$ through an ultraproduct of the inclusion maps $i^j_{p,q}$.
	By Kakutani's representation theorem, $\big(L_q(\mu_j)\big)_\mathcal{U}$ is lattice isometrically isomorphic to $L_q(\nu)$ for some measure $\nu$. The map $(i^j_{p,q})_\mathcal{U}$ is positive and hence $p$-convex, because $\big(L_p(\mu_j)\big)_\mathcal{U}$is $p$-convex, so by the Maurey/Nikishin factorization theorem $(i^j_{p,q})_\mathcal{U}$ factors through an inclusion $i_{p,q} : L_p(\tilde{\nu}) \to L_q(\tilde{\nu})$, giving the desired result.
\end{proof}

\begin{remark}
When $q=2$ Theorem \ref{thm-equivalence-linear-and-nonlinear} holds for a general Banach space $Y$, because in a Hilbert space all subspaces are 1-complemented.
\end{remark}

\subsection{Factorization for maps that are both Lipschitz $p$-convex and Lipschitz $q$-concave}\label{sec-factorization-theorems-for-both-convex-and-concave}

In  \cite[Sec. 5]{Raynaud-Tradacete} the natural question of factorizations for operators that are both $p$-convex and $q$-concave is studied, and in this section we consider the Lipschitz counterpart.

The motivation is simple: if we have a Lipschitz map $T : E \to F$ between Banach lattices $E$ and $F$ that is both Lipschitz $p$-convex and Lipschitz $q$-concave, we know from Theorems \ref{thm-factorization-lipschitz-p-convex} and \ref{thm-factorization-lipschitz-q-concave} that $T$ can be factored through a $p$-convex Banch lattice and also through a $q$-convex Banach lattice.
Could we have both conditions at the same time?
Note that from the same Theorems, if the map $T$ has a factorization of the form
$$
\xymatrix{
E \ar[r]^{T}\ar[d]_{\phi} &F  \\
E_1 \ar[r]_{R}  &F_1 \ar[u]_{\psi}
}
$$
where $\phi$ and $\psi$ are positive linear maps, $E_1$ is $q$-concave, $F_1$ is $p$-convex and $R$ is a Lipschitz map, then $T$ is both Lipschitz $p$-convex and Lipschitz $q$-concave.
The question is then: if the map $T : E \to F$ is Lipschitz $p$-convex and Lipschitz $q$-concave, do there exist maps $T_1$ and $T_2$, such that $T = T_1 T_2$, where $T_1$ is Lipschitz $p$-convex and $T_2$ is Lipschitz $q$-concave?

The answer to the corresponding linear question is negative, as shown in \cite[Sec. 5]{Raynaud-Tradacete}, and it turns out that allowing for Lipschitz maps does not change that. Thanks to the following Theorem, the linear examples will also work in the Lipschitz context.

\begin{theorem}\label{thm-nonlinear-implies-linear}
	Let $T : E \to F$ be a linear map between Banach lattices $E$ and $F$, and assume that $T$ admits a factorization $T= T_2 T_1$ where $T_1$ is Lipschitz $q$-concave and $T_2$ is Lipschitz $p$-convex. Then there is also a factorization $T = \tau_2 \tau_1$ where $\tau_1$ is $q$-concave and $\tau_2$ is $p$-convex, and moreover
	$M^{(p)}(\tau_2) \le M_{\Lip}^{(p)}(T_2)$ and $M_{(q)}(\tau_1)\le M^{\Lip}_{(q)}(T_1)$.
\end{theorem}

\begin{proof}
Apply the factorization Theorems \ref{thm-factorization-lipschitz-p-convex} and \ref{thm-factorization-lipschitz-q-concave} to obtain
$$
\xymatrix{
E \ar[rr]^{T_1}\ar[dr]_{\phi} & &G \ar[rr]^{T_2}\ar[dr]_{R} & &F  \\
 &V \ar[ru]_{S}  &  &W \ar[ru]_{\psi} & \\
}
$$
where $G$, $W$ and $V$ are Banach lattices with $W$ (resp. $V$) being $p$-convex (resp. $p$-concave) with constant 1;
$R$ and $S$ are Lipschitz maps with constant at most $1$, and
$\psi$ (resp. $\phi$) is a lattice
homomorphism of norm at most $M_{\Lip}^{(p)}(T_2)$ (resp. $M^{\Lip}_{(q)}(T_1)$ ) that is injective (resp. has dense range).
Let $J = RS$. We claim that $J$ is a linear map.
Given $u,v \in V$, choose sequences $(u_n)_{n=1}^\infty, (v_n)_{n=1}^\infty$ in $V$ such that $\phi(u_n) \to u$ and $\phi(v_n)\to v$.
Since $\phi$ is linear and continuous, $\phi(u_n+v_n) \to u+v$.
$T$ is also linear, so $\psi J \phi(u_n+v_n) = \psi J \phi(u_n) + \psi J \phi(v_n) = \psi(J\phi(u_n) + J\phi(v_n))$.
We conclude that $\psi J(u+v) = \psi(J(u) + J(v))$, and by the injectivity of $\psi$, $J$ is linear.
Choosing $\tau_1 = J \phi$ and $\tau_2 = \psi$, we get that
$\tau_1$ is $q$-concave and $\tau_2$ is $p$-convex, and in addition
	$M^{(p)}(\tau_2) \le M_{\Lip}^{(p)}(T_2)$ and $M_{(q)}(\tau_1)\le M^{\Lip}_{(q)}(T_1)$.
\end{proof}

Theorem \ref{thm-nonlinear-implies-linear} and the results from \cite[Sec. 5]{Raynaud-Tradacete} immediately imply the following Corollary.

\begin{corollary}
	\begin{enumerate}[(a)]
		\item Let $T : E \to F$ be a linear map from an $\infty$-convex Banach lattice (an AM-space) $E$ to a $q$-concave Banach lattice $F$ ($q < \infty$). If $T$ can be factored as $T = SR$, with $R$ Lipschitz $q$-concave and $S$ Lipschitz $\infty$-convex, then $T$ is compact.
		\item The formal inclusion $T : C(0,1) \to L_q(0,1$) is Lipschitz $q$-concave and Lipschitz $\infty$-convex, but it does not factor as $T = T1 \circ T2$, with $T_1$ Lipschitz $\infty$-convex, and $T_2$ Lipschitz $q$-concave.
		\item Let $T : E \to F$ be a linear map from a $p$-convex Banach lattice $E$ to a $1$-concave Banach lattice (an AL-space) $F$. If $T$ can be factored as $T = SR$, with $R$ Lipschitz 1-concave and $S$ Lipschitz $p$-convex, then $T$ is compact.
		\item If a lattice homomorphism $T : L_p(0,1) \to Lq(0,1)$ ($1 < q < p < \infty$) can be factored as $T = T_2 \circ T_1$ with $T_1$ Lipschitz $q$-concave and $T_2$ Lipschitz $p$-convex, then $T$ is AM-compact (recall that a linear map $T : E \to X$ between a Banach lattice $E$ and a Banach space $X$ is called \emph{AM-compact} if $T[-x,x]$ is relatively compact for every positive $x \in E$.).
		\item 	For $1 < q < p < \infty$, the formal inclusion $i_{p,q} : L_p(0,1)\to L_q(0,1)$ cannot be factored as $i_{p,q} = T_2 \circ T_1$ with $T_1$ Lipschitz $q$-concave and $T_2$ Lipschitz $p$-convex.
	\end{enumerate}
\end{corollary}

Even though the na\"ive factorization scheme for linear maps that are both $p$-convex and $q$-concave did not work out, Raynaud and Tradacete were able to prove that if one ``gives up'' a little on the exponents of convexity and concavity involved, one still gets such a factorization \cite[Thm. 15]{Raynaud-Tradacete}.
The following would be a Lipschitz version of that result.

\begin{question}
	Suppose that a Lipschitz map $T: E \to F$ between Banach lattices
	is Lipschitz $p$-convex and Lipschitz $q$-concave, with $1 < p \le \infty$ and $1 \le q < \infty$.
	Can we find $1 < p_0 < p$ and $q < q_0 < \infty$ so that there is a factorization of $T$ as
	$$
	\xymatrix{
	E \ar[r]^{T}\ar[d]_{\phi} &F  \\
	E_0 \ar[r]_{R}  &F_0 \ar[u]_{\varphi}
	}
	$$
	where $\phi$ and $\varphi$ are positive linear maps, $E_0$ is $p_0$-convex, $F_0$ is $q_0$-concave and $R$ is a Lipschitz map?
	Moreover: given $\theta\in(0,1)$, can we have $p_0 = \frac{p}{\theta + (1-\theta)p}$ and $q_0 = \frac{q}{1-\theta}$?
\end{question}

\begin{remark}
	The proof from \cite[Thm. 15]{Raynaud-Tradacete} cannot be easily adapted to the Lipschitz context.
	The arguments in that proof are heavily based on complex interpolation because that method works very well for lattices, as pointed out by Calder\'on \cite{Calderon}.
	Complex interpolation, however, is not well suited to work with Lipschitz maps.
	The results available, e.g. the main one from \cite{Bergh}, require strong extra assumptions due to the fact that a Lipschitz function generally does not preserve analyticity.
\end{remark}

\section*{Acknowledgments}
The author thanks Professor W. B. Johnson for many helpful discussions and suggestions.

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config: arxiv


TITLE: modified Dominated convergence theorem
QUESTION [1 upvotes]: Dominated convergence theorem:
If for some random variable $Z$, we have that $|X_{n}|\leq |Z|$ for all $n$ and $\mathbb{E}[Z]<\infty$, then $X_{n}\stackrel{d}{\to}X$ implies that $\mathbb{E}[X_{n}] \to \mathbb{E}[X]$. This is well known.
I got stuck with the following "modification": assume, that instead of $|X_{n}|\leq |Z|$ we have the following:
$|X_{n}|\leq |Z_{n}|$, where $Z_{n}$ is a sequence s. t. $Z_{n}\stackrel{d}{\to}Z$, $\mathbb{E}[Z_{n}]\to \mathbb{E}[Z]$ and $\mathbb{E}[Z_{n}]=a<\infty$, THE SAME! for all $n$.
Is the statement about convergence of means true in this case?
I am thinking to introduce a new variable $Z_{max} = \max\{Z_{1}, Z_{2}, \dots, Z_{n}, \dots\}$, but I doubt that $\mathbb{E}[Z_{max}]<\infty$ in this case...

REPLY [0 votes]: I propose the following solution:
Let $Z_{n}>0$.
Since $Z_{n}$ is a sequence s. t. $Z_{n}\stackrel{d}{\to}Z$, $\mathbb{E}[Z_{n}]\to \mathbb{E}[Z]$ and $\mathbb{E}[Z_{n}]=a<\infty$, then $Z_{n}$ is asymptotically uniformly integrable.
Next, we can show, that, since $X_{n}\leq Z_{n}$, and, let's say that $X_{n}\geq 0$, then $X_{n}$ is also asymptotically uniformly integrable.
$$
\mathbb{E}[X_{n}\mathbb{I}\{X_{n}>M\}]\leq \mathbb{E}[Z_{n}\mathbb{I}\{X_{n}>M\}]\leq\mathbb{E}[Z_{n}\mathbb{I}\{Z_{n}>M\}],
$$
so that
$$
\limsup \mathbb{E}[X_{n}\mathbb{I}\{X_{n}>M\}] \leq \limsup\mathbb{E}[Z_{n}\mathbb{I}\{Z_{n}>M\}]\to 0,
$$
as $n\to\infty$.
Then, we have done.


TITLE: Why is this a non-holonomic constraint?
QUESTION [6 upvotes]: Wikipedia states:

holonomic constraints are relations between the position variables (and possibly time1) which can be expressed in the following form:
  $$f(q_{1},q_{2},q_{3},\ldots ,q_{n},t)=0$$
where $\{q_{1},q_{2},q_{3},\ldots ,q_{n}\}$ are the $n$ coordinates which describe the system. For example, the motion of a particle constrained to lie on the surface of a sphere is subject to a holonomic constraint, but if the particle is able to fall off the sphere under the influence of gravity, the constraint becomes non-holonomic. [...] the second non-holonomic case may be given by:
  $$r^{2}-a^{2}\geq 0.$$

Is this really a non-holonomic constraint? Consider the following function $f(r)=\min(r^2-a^2,0)$. Then we have
$$r^{2}-a^{2}\geq 0\quad\text{iff}\quad f(r)=0.$$
Doesn't this mean that the constraint is in fact holonomic?

REPLY [5 votes]: No. A system with $n$ degrees of freedom is said to be under a holonomic constraint if imposition of the constraint reduces the number of degrees of freedom by $1$ , to $n-1$. Every holonomic constraint can be captured by some relation of the form $f(q_1...q_n,t)=0$, but not every constraint that can be captured in this form is necessarily holonomic. You have provided an example.
When $(x^2+y^2+z^2) = a^2$ is the constraint satisfied, transforming to spherical coordinates (where one d.o.f is explicitly $r= \sqrt{x^2+y^2+z^2}$ ) allows us to describe the motion constrained on the sphere using $\theta$ and $\phi$, ie, reduction of the number of d.o.f from $3$ to $2$. Sometimes, it may not be straightforward to find this coordinate transform ; in such cases, we automate the process of describing the resultant $n-1$ dimensional motion by using the Lagrange Multiplier method.
So, calling a constraint on a system with $n$ d.o.f holonomic implies that imposing this constraint turns the system into one with n-1 d.o.f. This is by definition (see comments below)
What you have constructed indeed has the functional form described above, but it does not describe a situation with $1$ d.o.f less. The motion is very clearly still $3$-dimensional.


TITLE: Why does hair dry faster when I have less hair?
QUESTION [2 upvotes]: I recently had a haircut, and have noticed that water in my air dries up faster when I short hair. I assume that hair loses water due to evaporation. We have been taught that evaporation is faster when the surface area is increase (which is the reason why behind drinking tea off a saucer)
But, when we have longer hair, the surface area of the hair is supposed to be more. So, why does hair dry faster when it is shorter and has lower surface area? Just genuinely curious to know this.

REPLY [1 votes]: You are confusing the total rate of evaporation with the time it takes for the water to completely evaporate. The rate of evaporation increases with the surface, true. But this don't tell which system will have all the water evaporated first. The ocean has much large surface area than the puddle left by the last rain. But the ocean is still there after the puddle is long gone. Isn't it? Besides, there are other factors that affect the rate of evaporation like the circulation of air which removes the vapors from the surface allowing more evaporation. If you take a wet cloth and roll it it will dry much slower than if it is all exposed to the air. So, short hair has less water to evaporate, a better circulation of air and less overlaping of threads of hair. You should consider that the surface expsoed to air in case of long, wet hair is not the sum of all surfaces of individual hairs due to clumping toghether in thicker threads.


TITLE: Is CLIQUE PROBLEM polynomial on a class of graphs $\mathbb{G}$, if $\mathbb{G}$ has one graph.
QUESTION [0 upvotes]: By CLIQUE PROBLEM I mean whether $\omega(G) \geq k$ for $k \geq 3$.
If $\mathbb{G}$ has just one graph, is the CLIQUE PROBLEM polynomial when restricted to $\mathbb{G}$?
My Attempt: I can check whether there is a clique of size $k$ for $k=1,2,\ldots,n$. So it should be polynomial.

REPLY [1 votes]: [Comment promoted to answer at request of OP]
The question of whether a problem can be solved in polynomial time is only asked when there are infinitely many instances of the problem, not when there's just one (or any finite number). Any single problem can always be solved in time polynomial in the input size by just taking a big enough polynomial!


TITLE: What are the minimal conditions for the exactness of a 1-form on an open connected subset?
QUESTION [0 upvotes]: When considering a continuous $\omega\in\Lambda^1(U)$, where $U\subset\mathbb{R}^n$ is some open connected subset, $d\omega=0$ is insufficient to guarantee exactness since Poincare's Lemma holds for contractible manifolds, and $U$ may not be one.
(Note in below discussion I assume all paths/loops are $\mathcal{C}^1$ singular 1-cells) If every loop $\gamma$ in $U$ satisfies $\int_\gamma\omega=0$, then we know that for any two points $\textbf{a}, \textbf{b}\in U$ any two paths $\gamma_1,\gamma_2\in U$ from $\textbf{a}$ to $\textbf{b}$ can be concatenated such that $\gamma_1-\gamma_2$ forms a loop so we have path independence for every path in $U$ since $\int_{\gamma_1-\gamma_2}\omega =0$.
Complete path independence in $U$ results in exactness since it ensures that $\forall\textbf{a},\textbf{x}\in U$, $f(\textbf{x})=\int_{[\textbf{a},\textbf{x}]}\omega$ is well-defined, which we can use to show $d f=\omega$ in an open neighborhood of every $\textbf{x}$ by looking at the difference of $\lim_{h\rightarrow 0}f(\textbf{x}+h{\boldsymbol\delta}_i)-f(\textbf{x})$ for $1\le i \le n$.
However, this question's answer by @Seub (see "Lemma 3") seems to imply a sufficient condition for exactness is path independence solely about the 2 paths of the oriented cell $\Delta(\textbf{a},\textbf{x})$ (in $U\subset\mathbb{R}^2$). How is this the case? It make sense, looking at how the proof for exactness works given that $\omega$ is completely path independent, but the first statement in that proof ($f$ is well-defined) is now in question.
If, as @Seub mentions, we restrict all possible paths $[\textbf{a},\textbf{x}]$ to just ones on $\partial \Delta(\textbf{a},\textbf{x})$, then I agree $f$ is well defined, and $df=\omega$ on the boundary of $\Delta(\textbf{a},\textbf{x})$. Why can we then "extend" its exactness to an open interval around $\textbf{b}$? Does it have something to do with continuity?

REPLY [1 votes]: No, well-definedness isn't in question. As you yourself say in the last paragraph, you define $f$ by integrating on paths that are (locally) parallel to the coordinate axes. You end up with a smooth function whose partial derivatives are computed by taking exactly such paths, and hence you've verified that $df=\omega$. Your sentence "$df=\omega$ on the boundary of $\Delta(\mathbf a,\mathbf x)$" is not right; this equality holds everywhere on our domain, as you verify that it holds at every $\mathbf a\in U$.


TITLE: Constructing a set with exactly three limit points
QUESTION [5 upvotes]: Construct a set of real numbers having exactly three limit points.

Progress:
I know the set $\{ 1/n   + k\}$ has one limit point $k$, but I am unable to justify why.
Please be a bit elaborate in your explanation because I need to understand this question if I face such a question in my exam.

REPLY [1 votes]: You can prove no other element in the set is a limit point because you can let 1/n be any of the elements and show that there is an r that gives you a neighborhood that doesn't have any of the other the elements in it. Because to be a limit point you have to have another element in the set in the neighborhood. for example, if you have 1/5 then you can make a neighborhood where r is really small then 1/4, and 1/6 wont be in the set and there is no other numbers in the neighborhood. So you would do 1/n-1 and 1/n and 1/n+1 and show that for a very small r like half the distance between 1/n-1 and 1/n then there wont be any elements. It must be for EVERY neighborhood.
Correct me if I'm wrong.


TITLE: Interior of a set than sounds easy
QUESTION [0 upvotes]: Let $\Omega\subseteq\mathbb{R}^N$ an open set (it can be cosidered bounded if needed), and
$\phi:\Omega\to\mathbb{R}$ be a $C^1$ function with $\phi^{-1}(0)\neq\emptyset$ and $\nabla\phi(x)\neq 0,\ \forall\ x\in\phi^{-1}(0)$. Is it true that:
$$\mathrm{int}\overline{\phi^{-1}([0,\infty))}=\phi^{-1}((0,\infty))$$
?
Remark that one inclusion is obvious. I cannot find a counterexample.

REPLY [1 votes]: A counterexample is given by the function
$$\phi : \mathbb{R} \to \mathbb{R} : x \mapsto (x - 1)^2 - 1$$