TITLE: Symmetric matrix in $\mathbb{Z}_2$
QUESTION [1 upvotes]: Let $A\in (\mathbb{Z}_2)^{n\times n}$ be a symmetric matrix. Let $Row(A)$ denote its rowspace.
Prove that $$(a_{1,1}, a_{2,2},\cdots, a_{n,n}) \in Row(A)$$
I tried induction on $n$. If a row is completely empty we can remove the row and corresponding column. If $rank A=n$ we are also trivially done. Thus I tried doing some row operations and column operations of the form $r_i\rightarrow r_i-r_j, c_i\rightarrow c_i-c_j$ at the same time, which didn't work because I want to keep symmetry which also changes the rows.
I thought there might be a more combinatorial/graph theory solution.

REPLY [0 votes]: We proceed by induction on $n$. We use the inductive hypothesis to its fullest potential:
If I remove row $i$ and column $i$, then the statement is true by IH. Hence there exists $x_1, \cdots, x_{i-1}, x_{i+1}, \cdots, x_n$ such that $\sum x_ir_i = (a_{11}, a_{22}, \cdots, a_{i-1,i-1}, T, a_{i+1,i+1}, \cdots, a_{nn})$ where $r_i$ is the $i$th row vector of $A$.
If $T=a_{i,i}$ we are done. Otherwise, we get $V-e_i$ where $V=(a_{11}, \cdots, a_{nn})$ and $e_i=(\underbrace{0,\cdots,0}_{i-1}, 1, \underbrace{0,\cdots,0}_{n-i})$. We can do this to every $1\le i\le n$.
This means we can get $e_i+e_j$ and all vectors in the orthogonal component of $1_n$ (ie all vectors with an even number of 1's). Furthermore, if a row vector has an odd number of 1's, then the rank of $A$ is $n$ (ie all vectors are okay). If $V$ has an even number of 1's, I'm done. Otherwise, $V$ has an odd number of 1's and every row of $A$ must have an even number of $1$'s, which implies $A$ has an even number of $1$'s. But since $A$ is symmetric, the number of 1's in $A$ has the same parity as the number of 1's in $V$, absurd!


text:
\section{Concentration of measure inequalities}
\label{App: Concentration of Measure Inequalities}

In this final appendix, we present two well-known exponential concentration of measure inequalities that are used in this paper. The first of these results bounds the tail probability of the empirical average of a collection of independent bounded random variables.

\begin{lemma}[Hoeffding's Inequality {\cite[Theorems 1 and 2]{Hoeffding1963}}]
\label{Lemma: Hoeffding's Inequality}
Given independent random variables $X_1,\dots,X_n \in [a,b]$, for some constants $a < b$, we have for every $\varepsilon \geq 0$:
$$ \P\!\left(\frac{1}{n}\sum_{i = 1}^{n}{X_i - \E\!\left[X_i\right]} \geq \varepsilon\right) \leq \exp\!\left(-\frac{2 n \varepsilon^2}{(b-a)^2}\right) $$
where $\exp(\cdot)$ denotes the natural exponential function with base $e$ throughout this paper.
\end{lemma}

The second of these results provides a tighter bound on the tail probability of the empirical average of a collection of independent bounded random variables using information about the variances of the random variables.

\begin{lemma}[Bernstein's Inequality {\cite{Bernstein1946}}]
\label{Lemma: Bernstein's Inequality}
Given independent random variables $X_1,\dots,X_n$ such that for some constants $a,b > 0$, $\big|X_i - \E[X_i]\big| \leq a$ and $\VAR(X_i) \leq b$ for all $i \in [n]$, we have for every $\varepsilon \geq 0$:
$$ \P\!\left(\frac{1}{n}\sum_{i = 1}^{n}{X_i - \E\!\left[X_i\right]} \geq \varepsilon\right) \leq \exp\!\left(-\frac{n \varepsilon^2}{2 b + \frac{2}{3} a \varepsilon}\right) . $$
\end{lemma}

config: arxiv


TITLE: What happens when the universe runs out of fuel?
QUESTION [11 upvotes]: After some X billion years, one would think the stars in the entire universe will run out of hydrogen. What would happen next? Is there any way to get hydrogen out of heavy metals (extreme fission)? Just curious.

REPLY [8 votes]: Star formation will die out long before all the hydrogen runs out. Much of it will be trapped in very low mass stars and lots more will be in the very sparse intergalactic medium (where $\sim 50$% of it is now).
In terms of possible production mechanisms, almost all the hydrogen that exists now comes from the big bang. However, neutrons can be generated by spallation reactions between cosmic rays and heavy metal targets. The free neutrons then beta decay to yield hydrogen nuclei and electrons.
This isn't a very efficient production mechanism and is also likely to cease if as thought, supernova remnants are the sites of cosmic ray acceleration.


TITLE: Finding $\sup(a_n)$ and $\inf(a_n)$
QUESTION [0 upvotes]: I was trying to figure out $\inf(a_n)$ and $\sup(a_n)$ of the sequence defined by $a_n=\arctan(n^2-7n+13)$. Since $a_n$ is increasing for $n \geq 4$, and $\sup(a_n)$  cannot be $a_1, a_2, a_3$, I thought it was possible to apply the monotone theorem which states $\lim\limits_{n \rightarrow +\infty}a_n=\sup(a_n)$, so $\sup(a_n)$ should be $\pi/2$. Is it correct? What should I do to find $\inf(a_n)$?

REPLY [1 votes]: Yes, that is correct: $\sup\{a_n\mid n\in\Bbb N\}=\frac\pi2$ for that reason.
And $\inf\{a_n\mid n\in\Bbb N\}=\arctan(1)=\frac\pi4$, since $1$ is the smallest number of the form $n^2-7n+13$ ($n\in\Bbb N$). You can deduce this from the fact that$$n^2-7n+13=\left(n-\frac72\right)^2+\frac34$$and that $1=\left(\pm\frac12\right)^2+\frac34$.


text:
\begin{document}

  \maketitle

\begin{abstract}
We prove the necessity of the UMD condition, with a quantitative estimate of the UMD constant, for any inequality in a family of $L^p$ bounds between different partial derivatives $\partial^\beta u$ of $u\in C^\infty_c(\R^n,X)$. In particular, we show that the estimate $\|u_{xy}\|_p\leq K(\|u_{xx}\|_p+\|u_{yy}\|_p)$ characterizes the UMD property, and the best constant $K$ is equal to one half of the UMD constant. This precise value of $K$ seems to be new even for scalar-valued functions.
\end{abstract}

\section{Introduction} \label{sec:intro}

A priori estimates between different partial derivatives such as
\begin{equation}\label{eq:estim}
    \|\partial^\beta u\|_{L^{q}(\R^n,X)}
        \leq K \sum_{j=1}^N \|\partial^{\alpha^j} u\|_{L^{p_j}(\R^n,X)}, \quad u \in C_c^\infty(\R^n,X),
\end{equation}
play an important role in Analysis. Here $\partial^\beta=\partial_1^{\beta_1} \cdots \partial_n^{\beta_n}$ is the usual derivative associated to the multi-index
$\beta=(\beta_1, \dots, \beta_n)$. In the case of scalar-valued functions, $X=\C$ or $X=\R$, and exponents in the range $q,p_j\in(1,\infty)$, a complete characterization of admissible sets of $q,p_j$ and $\beta,\alpha^j$ is a classical result of O. Besov, V. Il$'$in and S. Nikol$'$ski{\u\i} \cite{BNS}. The second author showed in \cite{Hy4} that the same characterization remains valid for vector-valued functions, provided that the target space $X$ has the UMD property.

A Banach space $X$ has this property
if, for some (equivalently, for all) $1<p<\infty$, there exists a constant $C_p<\infty$, such that
\begin{equation}\label{eq:defUMD}
    \Big\| \sum_{\ell=1}^r \sigma_\ell d_\ell \Big\|_{L^p(\Omega,X)}
        \leq C_p \Big\| \sum_{\ell=1}^r d_\ell \Big\|_{L^p(\Omega,X)},
\end{equation}
for each $r \in \N$, every sequence of signs $(\sigma_\ell)_{\ell \in \N}$ in $\{-1,1\}$
and any martingale difference sequence
$(d_\ell)_{\ell \in \N}$ in $L^p(\Omega,X)$. Here
$L^p(\Omega,X)$ represents the space of all functions
that take values in the Banach space $X$ and their $X$ norm is
$p$-integrable on $\Omega$. We denote by $\beta_p(X)$ the smallest admissible constant $C_p$.
Main properties of UMD spaces can be found in the survey of J. L. Rubio de Francia \cite{Rub}.

Our present goals are two-fold. On the one hand, we show that the result of \cite{Hy4} is optimal in the sense that it establishes the most general class of Banach spaces  where the classical Besov--Il'in--Nikol'ski{\u\i} result \cite{BNS} can be generalized; namely, we exhibit instances of \eqref{eq:estim}, for which UMD is not only sufficient (as shown in \cite{Hy4}) but also necessary. This continues a tradition of diverse characterizations of the UMD property (see for instance \cite{Bou3,Bu2,CL,GMS,Gue,HTV,Hy3,Hy,KaWe,KuWe,Xu}, amongst others). On the other hand, as a byproduct, we obtain information on the size of the constant $K$ in \eqref{eq:estim}, which seems to be new already in the scalar case. Recall that the list of analytic inequalities, where the sharp constant can be determined, is somewhat restricted, whereas the UMD inequality \eqref{eq:defUMD} for $X=\C$ or $X=\R$ is one of the prominent positive examples with $\beta_p(\C)=\beta_p(\R)=(p^*-1)$; here $p^*=\!
 max\{p,p'\}$. This is a celebrated result of Burkholder; see \cite{Bu5}, \cite[p. 12]{Bu4} and \cite[Theorem 14]{Bu6}. Thus it is useful to relate other estimates to \eqref{eq:defUMD}.

A particular case of \eqref{eq:estim} that we have in mind is the basic bound
\begin{equation}\label{eq:mainExample}
  \|\partial_1\partial_2 u\|_{L^p(\R^2,X)}
  \leq K(\|\partial_1^2 u\|_{L^p(\R^2,X)}+\|\partial_2^2 u\|_{L^p(\R^2,X)}).
\end{equation}
Since $\partial_1\partial_2 u=R_1R_2\triangle u$, where $R_i$ is the Riesz transform in the $i$th direction, it follows at once that
\begin{equation}\label{eq:R12comput}
\begin{split}
  \|\partial_1\partial_2 u\|_{L^p(\R^2,X)}
  &\leq\|R_1R_2\|_{\mathcal{L}(L^p(\R^2,X))}\|\partial_1^2 u+\partial_2^2 u\|_{L^p(\R^2,X)} \\
   &\leq\|R_1R_2\|_{\mathcal{L}(L^p(\R^2,X))}(\|\partial_1^2 u\|_{L^p(\R^2,X)}+\|\partial_2^2 u\|_{L^p(\R^2,X)}),
\end{split}
\end{equation}
and $\|R_1R_2\|_{\mathcal{L}(L^p(\R^2,X))}=\beta_p(X)/2$ by \cite{GMS}, so that $K\leq \beta_p(X)/2$ in \eqref{eq:mainExample}. We prove that this constant, optimal for the first line in \eqref{eq:R12comput}, is also sharp for the seemingly weaker inequality \eqref{eq:mainExample}.

It is interesting that the best constant in such a simple-looking classical inequality appears to be previously unknown.
This constant is somewhat connected to the famous open problem of determining the $L^p$ norm of the Beurling transform, whose imaginary part is $2R_1R_2$; see \cite{GMS} for more details. It has been long conjectured \cite{Iwa} that the norm of the Beurling operator on $L^p(\R^2)$ is $\beta_p(\C)$, and a surprising contribution of \cite{GMS} was to show that this conjectured norm is already achieved by the imaginary part alone. Our result further amplifies this phenomenon.

We then turn to the precise formulation of our main results. We are only concerned with the case that $p_j\equiv q=:p$ in \eqref{eq:estim}, and all multi-indices $\beta,\alpha^j$ have equal length, which we denote by $|\beta|=\beta_1 + \dots +\beta_n$, when $\beta=(\beta_1, \dots, \beta_n) \in \N^n$. We also need to assume that the components of $\beta$ have sufficiently different parity from those of each $\alpha^j$.

\begin{Th}\label{Th:main}
Let $1<p<\infty$, $N \in \N$ and $\alpha^j$, $\beta \in \N^n$. Let $|\beta|=|\alpha^j|$ for all $j=1, \ldots, N$,
and suppose that there exists a set $F\subset\{1,\ldots,n\}$ such that $\sum_{\ell\in F}\alpha^j_\ell$ has the same parity (even or odd) for each $j=1,\ldots,N$, which is different from the parity of $\sum_{\ell\in F}\beta_{\ell}$. If the following estimate holds:
   \begin{equation}\label{eq:estim2}
    \|\partial^\beta u\|_{L^p(\R^n,X)}
        \leq K \sum_{j=1}^N \|\partial^{\alpha^j} u\|_{L^p(\R^n,X)}, \quad u \in C_c^\infty(\R^n,X),
    \end{equation}
then $X$ is UMD and $\beta_p(X) \leq KN$.
\end{Th}

\begin{Rm}
It is known (see e.g. \cite{Hy4}) that a necessary condition for \eqref{eq:estim2} to hold in any Banach space is that $\beta$ is a convex combination of the multi-indices $\alpha^j$. Although we will not explicitly use this fact, it is worth observing that the assumption \eqref{eq:estim2} is only meaningful for such sets of multi-indices.
\end{Rm}

The following corollaries demonstrate the somewhat technical conditions on the multi-indices assumed in the theorem:

\begin{Cor}
Let $X$ be a Banach space and $1<p<\infty$. Let $\beta\in\N^n$ be a multi-index such that $|\beta|$ is even, but $\beta_s$ is odd for at least one index $s\in\{1,\ldots,n\}$. Then the estimate
\begin{equation*}
  \|\partial^\beta u\|_{L^p(\R^n,X)}\leq K\sum_{j=1}^n\|\partial_j^{|\beta|}u\|_{L^p(\R^n,X)},\qquad u\in C^\infty_c(\R^n,X),
\end{equation*}
holds if and only if $X$ is a UMD space, and in this case $\beta_p(X)\leq Kn$.
\end{Cor}

\begin{proof}
The ``if'' part is well known; for instance, it is a special case of the results of \cite{Hy4}.
For the ``only if'' part, we apply Theorem~\ref{Th:main} with $\alpha^j=|\beta|e_j$ and $F=\{s\}$.
Then $\sum_{\ell\in F}\beta_\ell=\beta_s$ is odd, whereas $\sum_{\ell\in F}\alpha^j_\ell=|\beta|\delta_{js}$
is even for all $j$, as both $0$ and $|\beta|$ are even. Thus the Theorem implies that $X$ is UMD, with the
asserted estimate for the constant.
\end{proof}

\begin{Cor}\label{Cor:bestconst1}
    Let $X$ be a Banach space and $1<p<\infty$. Then \eqref{eq:mainExample} holds if and only if $X$ is a UMD space, and the best constant $K$ satisfies $K=\beta_p(X)/2$.
\end{Cor}

\begin{proof}
We already explained that $K\leq\beta_p(X)/2$ in \eqref{eq:R12comput}; this part of the result is not new.
The converse estimate is a special case of the previous Corollary with $n=2$ and $\beta=(1,1)$.
\end{proof}

There are many characterizations of the UMD property by estimates between $L^p$ norms of two different objects. A novel qualitative feature of the above characterizing conditions is that they involve a sum of several $L^p$ norms on the right hand side.

The proof of Theorem~\ref{Th:main} proceeds via the theory of Fourier multipliers.
We denote by $\widehat{f}$ the Fourier transform of the function $f$, defined as usual by
$$\widehat{f}(\xi)
    = \int_{\R^n} e^{-2 \pi i \xi \cdot x} f(x) dx, \quad \xi \in \R^n,$$
and by $\widecheck{f}$ its inverse, given as above but without the plus
sign in the complex exponential. The Fourier multiplier of a function $m$ is
defined by $T_m f = (m \widehat{f})^{\vee}$.

We write
\begin{equation}\label{eq:mmj}
    m(\xi)= \frac{\xi^\beta}{|\xi|^{|\beta|}}
        \quad \text{ and } \quad
    m_j(\xi)= \frac{\xi^{\alpha^j}}{|\xi|^{|\alpha^j|}}, \qquad \beta, \alpha^j \in \N^n, \ j=1, \dots, N,
\end{equation}
where $\xi^\beta=\xi_1^{\beta_1} \cdots \xi_n^{\beta_n}$.
By well-known properties of the Fourier transform we have that
$$\partial^\beta u = T_m (\Delta^{|\beta|/2}u),$$
where $\Delta=-\partial_1^2 - \dots - \partial_n^2$ is the positive Laplacian in $\R^n$
and $\Delta^{|\beta|/2}u=[(2\pi i|\xi|)^{|\beta|} \widehat{u}]^{\vee}$.
Hence, \eqref{eq:estim2} with $|\alpha^j|=|\beta|$,
$j=1, \dots N$, implies that
\begin{equation}\label{eq:estim3}
\|T_m v\|_{L^p(\R^n,X)}
        \leq K \sum_{j=1}^N \| T_{m_j}v\|_{L^p(\R^n,X)}, \quad v \in \mathcal{S}_0(\R^n,X),
\end{equation}
where $\mathcal{S}_0(\R^n,X)$ is the subspace of the Schwartz class constituted by functions having zero mean value.
Moreover, in order to prove that $X$ is UMD it is more convenient to have an estimate of type \eqref{eq:estim3},
but in terms of discrete Fourier multipliers (see Propositions~\ref{Prop:RnTn} and \ref{Prop:TnTrn} below).

This paper is organized as follows. In Section~\ref{sec:transf} we collect some transference results about
Fourier multipliers and in Section~\ref{sec:proofTh} we present the proof of Theorem~\ref{Th:main}.

\section{Transference results for Fourier multipliers} \label{sec:transf}

Throughout this section we assume that $m$ is a measurable bounded function verifying that
\begin{equation}\label{eq:mk}
    m(k)=\lim_{r \to 0} \frac{1}{|B(k,r)|} \int_{B(k,r)}m(\xi)d\xi,
\end{equation}
exists for every $k \in \Z^n$. The discrete Fourier multiplier
$T_{\widetilde{m}}f$, of a function $f$ defined in the $n$-dimensional torus $\T^n=[-1/2,1/2)^n$, is given by
\begin{equation}\label{eq:Tmdiscret}
    T_{\widetilde{m}}f (t)
        = \sum_{k \in \Z^n} m(k) \widehat{f}(k) e_k(t), \quad t \in \T^n,
\end{equation}
where $e_k(t)=e^{2\pi i t \cdot k}$ and $\widehat{f}(k)=\int_{\T^n} e_k(-t) f(t)dt$.

Next, we present some lemmas which will be very useful in the proof of
Proposition~\ref{Prop:RnTn}. We say that $f$ is a trigonometric polynomial in $X$ if
$$f(t)
    = \sum_{k \in \Z^n} a_k e_k(t), \quad t \in \T^n,$$
with finitely many nonzero coefficients $a_k \in X$. Note that $a_k=\widehat{f}(k)$.

\begin{Lem}\label{Lem:11.20}
    Let $\phi$, $\psi \in \mathcal{S}(\R^n)$ be radial functions satisfying
    $\int_{\R^n} \phi \psi d\xi=1 $.
    Then,
    $$\int_{\T^n}\langle g(t), T_{\widetilde{m}}f(t) \rangle dt
        = \lim_{\eps \to 0} \eps^n \int_{\R^n}\langle \psi(\eps x)g(x) , T_m(\phi(\eps \cdot )f)(x)\rangle dx,$$
    for every trigonometric polynomials $f:\T^n \longrightarrow X$, $g:\T^n \longrightarrow X^*$.
    Here $\langle \cdot , \cdot \rangle$ denotes the pairing between $X$ and its dual $X^*$.
\end{Lem}

\begin{proof}
    The key step in this proof is the identity
    \begin{equation}\label{eq:11.21}
        \lim_{\eps \to 0} \frac{1}{\eps^n}
            \int_{\R^n} m(\xi) \widehat{\phi}\Big(\frac{\xi - k}{\eps} \Big) \widecheck{\psi}\Big(\frac{\xi - \ell}{\eps} \Big) d\xi
            = \chi_{\{k=\ell\}} m(k), \quad k, \ell \in \Z^n,
    \end{equation}

    Assume first that $k=\ell$. Since the Fourier transform (and its inverse) of a radial function, is also radial
    (see for example \cite[p. 430]{Ste2}), we consider
    $\rho(|\xi|)=\widehat{\phi}(\xi) \widecheck{\psi}(\xi)$, which satisfies $\int_{\R^n}\rho(|\xi|)d\xi=\int_{\R^n}\widehat\phi\widecheck\psi\,d\xi=\int_{\R^n}\phi\psi\,dx=1$. We have that
    \begin{align*}
        & \frac{1}{\eps^n} \int_{\R^n} m(\xi) \widehat{\phi}\Big(\frac{\xi - k}{\eps} \Big) \widecheck{\psi}\Big(\frac{\xi - k}{\eps} \Big) d\xi
            = \int_{\R^n} m(\eps \eta + k) \rho(|\eta|) d\eta
            = - \int_{\R^n} m(\eps \eta + k) \int_{|\eta|}^\infty \rho'(r) dr d\eta \\
        & \qquad = - \int_0^\infty \rho'(r) \int_{B(0,r)} m(\eps \eta + k)  d\eta dr
            = - \int_0^\infty \rho'(r) \frac{|B(0,r)|}{|B(k,\eps r)|}\int_{B(k,\eps r)} m(\xi)  d\xi dr.
    \end{align*}
    By taking into account $\rho \in \mathcal{S}(\R)$, $m \in L^\infty(R^n)$ and \eqref{eq:mk}, we can apply
    dominated convergence and then integrate by parts to obtain
    \begin{align*}
        & - \lim_{\eps \to 0} \int_0^\infty \rho'(r) \frac{|B(0,r)|}{|B(k,\eps r)|}\int_{B(k,\eps r)} m(\xi)  d\xi dr
            = - m(k) |B(0,1)| \int_0^\infty  r^n \rho'(r)  dr \\
        & \qquad =  m(k) |B(0,1)| \int_0^\infty n r^{n-1} \rho(r) dr
            =  m(k)  \int_0^\infty \int_{\partial B(0,r)} \rho(r) d\sigma  dr
            =  m(k)  \int_{\R^n} \rho(|\xi|) d\xi
            = m(k),
    \end{align*}
    because $|\partial B(0,r)|=|B(0,1)| n r^{n-1}$.

    Suppose now that $k \neq \ell$. This time,
    $$\frac{1}{\eps^n} \int_{\R^n} m(\xi) \widehat{\phi}\Big(\frac{\xi - k}{\eps} \Big) \widecheck{\psi}\Big(\frac{\xi - \ell}{\eps} \Big) d\xi
        = \int_{\R^n} m(\eps \eta + k) \widehat{\phi}(\eta) \widecheck{\psi}\Big(\eta + \frac{k - \ell}{\eps} \Big) d\eta,$$
    where the integrand is bounded by $\|m\|_\infty \|\widecheck\psi\|_\infty |\widehat\phi(\eta)| \in L^1(d\eta)$,
    and converges pointwise to zero as $\eps \to 0$, since $\widecheck{\psi}(\eta + (k - \ell)/\eps)\to 0$.
    Applying again dominated convergence
    we readily see that the integral converges to zero as $\eps \to 0$.

    Take the  trigonometric polynomials $f=\sum_{k \in \Z^n} a_k e_k$ and
    $g=\sum_{\ell \in \Z^n} b_\ell e_\ell$ with $a_k \in X$ and $b_\ell \in X^*$.
    An application of \eqref{eq:11.21} and making use of the usual properties of the Fourier transform, we deduce
    \begin{align*}
        & \int_{\T^n}\langle g(t), T_{\widetilde{m}}f(t) \rangle dt
            = \sum_{\ell, k \in \Z^n}  \langle b_\ell , a_k \rangle m(k) \int_{\T^n} e_\ell(t) e_k(t) dt
            = \sum_{\ell, k \in \Z^n}  \langle b_{-\ell} , a_k \rangle  \chi_{\{k=\ell\}} m(k) \\
        & \qquad = \lim_{\eps \to 0} \eps^n \int_{\R^n}
            \Big \langle \sum_{\ell\in \Z^n} b_{-\ell} \frac{1}{\eps^n} \widecheck{\psi}\Big(\frac{\xi-\ell}{\eps}\Big),
                m(\xi) \sum_{k \in \Z^n} a_k \frac{1}{\eps^n} \widehat{\phi}\Big(\frac{\xi-k}{\eps}\Big) \Big \rangle d\xi\\
        & \qquad = \lim_{\eps \to 0} \eps^n \int_{\R^n}
            \Big \langle \Big[ \psi(\eps \cdot) g \Big]^{\vee}(\xi),
                m(\xi) \Big[  \phi(\eps \cdot) f \Big]^{\wedge}(\xi) \Big \rangle d\xi
        = \lim_{\eps \to 0} \eps^n \int_{\R^n}\langle \psi(\eps x)g(x) , T_m(\phi(\eps \cdot )f)(x)\rangle dx.
    \end{align*}
\end{proof}

\begin{Lem}\label{Lem:11.21}
    Let $1<p<\infty$. Then, for every $\phi \in \mathcal{S}(\R^n)$ and $f \in L^p(\T^n,X)$, we have that
    $$\lim_{\eps \to 0} \eps^{n/p} \| \phi(\eps \cdot) f \|_{L^p(\R^n,X)}
        = \| \phi \|_{L^p(\R^n)} \| f \|_{L^p(\T^n,X)}.$$
\end{Lem}

\begin{proof}
    The periodicity of the function $f$ allow us to write
    \begin{align*}
        \eps^n \int_{\R^n} \| \phi(\eps x) f(x)\|_X^p dx
        & = \eps^n \sum_{k \in \Z^n} \int_{\T^n+k} |\phi(\eps x)|^p \|f(x)\|_X^p dx \\
        & = \int_{\T^n} \Big(\eps^n \sum_{k \in \Z^n} |\phi(\eps (x-k))|^p \Big) \|f(x)\|_X^p dx,
    \end{align*}
    where the quantity in parentheses is a Riemann sum of the function $|\phi(\cdot)|^p$,
    which is uniformly bounded in $x$ and $\eps$. Now, this lemma is a simple consequence of the
    dominated convergence theorem.
\end{proof}

\begin{Lem}\label{Lem:11.23}
    Let $1<p<\infty$ and $p'=p/(p-1)$. Then, for each $f \in \mathcal{S}(\R^n,X)$, we have that
    $$\lim_{\eps \to 0} \eps^{n/p'} \Big\| \sum_{k \in \Z^n} \widehat{f}(\eps k) e_k \Big\|_{L^p(\T^n,X)}
        =  \| f \|_{L^p(\R^n,X)}.$$
\end{Lem}

\begin{proof}
Recall that we identify $\T^n=[-1/2,1/2)^n$; whence the notation
$\eps^{-1}\T^n=[-1/(2\eps),1/(2\eps))^n$ is meaningful below.
    Define $f_\eps=\eps^{-n}f(\cdot/\eps)$. The Poisson summation formula (see \cite[Theorem 3.1.17]{Graf}),
    tells us that
    $$\sum_{k \in \Z^n} \widehat{f}(\eps k) e_k(t)
        = \sum_{k \in \Z^n} \widehat{f_\eps}(k) e_k(t)
        = \sum_{k \in \Z^n} f_\eps(t+k)
        = \frac{1}{\eps^n}\sum_{k \in \Z^n} f\Big(\frac{t}{\eps}+\frac{k}{\eps}\Big), \quad t \in \T^n.$$
    Then, making the change of variables $u=t/\eps$,
    \begin{align*}
        \eps^{n/p'} \Big\| \sum_{k \in \Z^n} \widehat{f}(\eps k) e_k \Big\|_{L^p(\T^n,X)}
            & = \eps^{-n/p} \Big\| \sum_{k \in \Z^n} f\Big(\frac{\cdot}{\eps}+\frac{k}{\eps}\Big) \Big\|_{L^p(\T^n,X)}
              = \Big\| \sum_{k \in \Z^n} f\Big(\cdot +\frac{k}{\eps}\Big) \Big\|_{L^p(\eps^{-1}\T^n,X)} \\
            & =\|f\|_{L^p(\eps^{-1}\T^n,X)}
                + \mathcal{O}\Big( \sum_{k \in \Z^n \setminus \{0\}} \Big\|  f\Big(\cdot +\frac{k}{\eps}\Big) \Big\|_{L^p(\eps^{-1}\T^n,X)} \Big).
    \end{align*}
    It is obvious that
    $$\|f\|_{L^p(\eps^{-1}\T^n,X)} \longrightarrow \| f \|_{L^p(\R^n,X)}, \quad  \text{as } \eps \to 0.$$
    We analyze the second term. Since $f \in \mathcal{S}(\R^n,X)$, for every $N \in \N$, we have that
    $$\Big\| f\Big(u +\frac{k}{\eps} \Big)\Big\|_X
        \leq C \Big(\frac{\eps}{|k|}\Big)^{N}, \quad u \in \eps^{-1}\T^n.$$
    Hence, it is enough to take $N>n$ to see that
    \begin{equation*}
\sum_{k \in \Z^n \setminus \{0\}} \Big\|  f\Big(\cdot +\frac{k}{\eps}\Big) \Big\|_{L^p(\eps^{-1}\T^n,X)}
        \leq C \eps^{N-n/p} \sum_{k \in \Z^n \setminus \{0\}} \frac{1}{|k|^N}
        \leq C \eps^{N-n/p} \longrightarrow 0, \quad \text{as } \eps \to 0.\qedhere
\end{equation*}
\end{proof}

We now establish our transference result between $\R^n$ and $\T^n$.
We say that $m$ is a homogeneous function (of order zero) when
$m(\lambda \xi)=m(\xi)$, $\xi \in \R^n$, $\lambda>0$. We denote
$$L^p_0(\T^n,X)=\Big\{f \in L^p(\T^n,X) : \int_{\T^n}f dt = 0\Big\}.$$

\begin{Prop}\label{Prop:RnTn}
    Let $1<p<\infty$, $N \in \N$ and $m$, $m_j \in C^\infty(\R^n \setminus \{0\})$, $j=1, \dots, N$,
    be homogeneous functions.  Then, the following assertions are equivalent:
    \begin{itemize}
        \item[$(i)$] $\displaystyle \|T_m f\|_{L^p(\R^n,X)}
                            \leq K \sum_{j=1}^N \| T_{m_j}f\|_{L^p(\R^n,X)}, \quad f \in L^p(\R^n,X),$
        \item[$(ii)$] $\displaystyle \|T_{\widetilde{m}} f\|_{L^p(\T^n,X)}
                            \leq K \sum_{j=1}^N \| T_{\widetilde{m}_j}f\|_{L^p(\T^n,X)}, \quad f \in L^p_0(\T^n,X).$
    \end{itemize}
\end{Prop}

\begin{proof}[Proof of Proposition~\ref{Prop:RnTn}, $(ii) \Rightarrow (i)$]
    Let $f \in \mathcal{S}(\R^n,X)$ be such that $\supp \widehat{f} \subset \R^n \setminus \{0\}$ is compact.
    Observe that this class of functions is dense in $L^p(\R^n,X)$. By the smoothness of the
    multiplier $m$, we also have that $T_m f \in \mathcal{S}(\R^n,X)$. Hence, we can apply Lemma~\ref{Lem:11.23}
    and get
    \begin{align*}
        \| T_m f \|_{L^p(\R^n,X)}
            = & \lim_{\eps \to 0} \eps^{n/p'} \Big\| \sum_{k \in \Z^n} \widehat{T_m f}(\eps k) e_k \Big\|_{L^p(\T^n,X)}
            =  \lim_{\eps \to 0} \eps^{n/p'} \Big\| \sum_{k \in \Z^n} m(k) \widehat{f}(\eps k) e_k \Big\|_{L^p(\T^n,X)} \\
            = & \lim_{\eps \to 0} \eps^{n/p'} \Big\| T_{\widetilde{m}}\Big(\sum_{k \in \Z^n} \widehat{f}(\eps k) e_k \Big) \Big\|_{L^p(\T^n,X)}.
    \end{align*}
    Notice that in the second equality we have applied that $m$ is a homogeneous function.
    Now, since the function between parentheses is in $L^p_0(\T^n,X)$, we use hypothesis $(ii)$
    and again Lemma~\ref{Lem:11.23} to conclude that
    \begin{align*}
        \| T_m f \|_{L^p(\R^n,X)}
            \leq & K  \lim_{\eps \to 0} \eps^{n/p'} \sum_{j=1}^N\Big\| T_{\widetilde{m}_j}\Big(\sum_{k \in \Z^n} \widehat{f}(\eps k) e_k \Big) \Big\|_{L^p(\T^n,X)} \\
            = & K  \lim_{\eps \to 0} \sum_{j=1}^N\eps^{n/p'} \Big\| \sum_{k \in \Z^n} \widehat{T_{m_j}f}(\eps k) e_k \Big\|_{L^p(\T^n,X)}
            =  K \sum_{j=1}^N \| T_{m_j}f \|_{L^p(\R^n,X)}.\qedhere
    \end{align*}
\end{proof}

\begin{proof}[Proof of Proposition~\ref{Prop:RnTn}, $(i) \Rightarrow (ii)$]
    Let $f$ be a trigonometric polynomial with $\widehat{f}(0)=0$. This family of functions is
    dense in $L^p_0(\T^n,X)$. We can write $f= \sum_{k \in \Z^n} \widehat{f}(k)e_k=\sum_{k\in I}\widehat{f}(k)e_k$, where
    $I=\{ k \in \Z^n : \widehat{f}(k) \neq 0\}$ is a finite set such that $0 \notin I$.

    We choose the auxiliary functions $\phi(x)=e^{-\pi|x|^2/p}$ and $\psi(x)=e^{-\pi|x|^2/p'}$, which
    clearly satisfy (see, for instance, \cite[Proposition 8.24]{Foll})
    $$1
        =\|\phi\|_{L^p(\R^n)}
        =\|\psi\|_{L^{p'}(\R^n)}
        =\int_{\R^n} \phi(x) \psi(x) dx
        =\int_{\R^n} \widehat{\phi}(\xi) \widecheck{\psi}(\xi) d\xi.$$
    Then, by Lemma~\ref{Lem:11.20}, Hölder's inequality,
    Lemma~\ref{Lem:11.21} and hypothesis $(i)$ we arrive at
    \begin{align*}
        \|T_{\widetilde{m}} f\|_{L^p(\T^n,X)}
            = & \sup_{g} \int_{\T^n}\langle g(t), T_{\widetilde{m}}f(t) \rangle dt
            =   \sup_{g} \lim_{\eps \to 0} \eps^n \int_{\R^n}\langle \psi(\eps x)g(x) , T_m(\phi(\eps \cdot )f)(x)\rangle dx \\
            \leq & \sup_{g} \lim_{\eps \to 0} \eps^{n/p'}  \|\psi(\eps \cdot)g \|_{L^{p'}(\R^n,X^*)} \eps^{n/p} \|T_m(\phi(\eps \cdot )f)\|_{L^{p}(\R^n,X)} \\
            \leq & K\lim_{\eps \to 0}   \sum_{j=1}^N \eps^{n/p} \|T_{m_j}(\phi(\eps \cdot )f)\|_{L^{p}(\R^n,X)},
    \end{align*}
    where the supremum is over all trigonometric polynomials $g : \T^n \to X^*$ such that
    $\|g\|_{L^{p'}(\T^n,X^*)} \leq 1$.

    To finish the reasoning we are going to prove that
    \begin{equation}\label{eq:goal1}
        \lim_{\eps \to 0} \eps^{n/p} \|T_{m_j}(\phi(\eps \cdot )f)\|_{L^{p}(\R^n,X)}
            \leq \|T_{\widetilde{m}_j}f\|_{L^{p}(\T^n,X)}, \quad j=1, \dots, N.
    \end{equation}
    For simplicity, from now on, we just write $m$ instead of $m_j$, $j=1, \dots, N$.

    Take $\theta_0 \in C^\infty_c(\R^n)$ such that $\supp \theta_0 \subset B(0,2)$,
    $\theta_0 \equiv 1$ in $B(0,1)$ and consider
    $$\theta_\ell (\xi)
        = \theta_0 \Big( \frac{\xi}{2^\ell} \Big) - \theta_0 \Big( \frac{\xi}{2^{\ell-1}} \Big), \quad \xi \in \R^n, \ \ell \geq 1.$$
    Note that the function $\theta_\ell$ is supported in the annulus $B(0,2^{\ell+1}) \setminus B(0,2^{\ell-1})$.
    Then, we have the partition of the unity
    $$1= \sum_{\ell \in \N} \theta_\ell (\xi), \quad \xi \in \R^n,$$
    where for each $\xi$ there exists at most two nonzero terms in the above sum
    (see \cite[p. 242]{Ste2} for details). Let also $\Theta_\ell =\theta_{\ell-1}+\theta_{\ell}+\theta_{\ell+1}$
    ($\theta_{-1}=0$), so that $\Theta_\ell$ is supported in $B(0,2^{\ell+2})$ for $\ell=0,1$,
    and in $B(0,2^{\ell+2})\setminus B(0,2^{\ell-2})$ for $\ell\geq 2$;
    and $\Theta_\ell=1$ on the support of $\theta_\ell$.

     For every $\ell \in \N$, we define $\phi_\ell \in \mathcal{S}(\R^n)$
    to be the function such that $\widehat{\phi}_\ell = \widehat{\phi} \ \theta_\ell$.
    Now, we can write
    \begin{align*}
        T_m(\phi(\eps \cdot )f)
            & = \sum_{k \in \Z^n} \widehat{f}(k) T_m(\phi(\eps \cdot )e_k)
            = \sum_{k \in \Z^n} \widehat{f}(k) \Big[ m(\xi) \frac{1}{\eps^n} \widehat{\phi} \Big( \frac{\xi-k}{\eps} \Big)  \Big]^{\vee} \\
            & = \sum_{\ell \in \N} \sum_{k \in \Z^n} \widehat{f}(k) \Big[ m(\xi) \frac{1}{\eps^n} \widehat{\phi}_\ell \Big( \frac{\xi-k}{\eps} \Big)  \Big]^{\vee}
            = \sum_{\ell \in \N} T_m(\phi_\ell(\eps \cdot )f).
    \end{align*}
    Since $\widehat{\phi}_\ell((\cdot - k)/\eps)$ is supported in $B(k,2^{\ell+1}\eps)$, it suggests that we may replace
    $m(\xi)$ by $m(k)$ with the advantage that
    $$ \sum_{\ell \in \N} \sum_{k \in \Z^n} \widehat{f}(k) \Big[ m(k) \widehat{\phi_\ell(\eps \cdot)e_k}(\xi)  \Big]^{\vee}
        = \sum_{\ell \in \N} \phi_\ell(\eps \cdot) T_{\widetilde{m}} f
        = \phi(\eps \cdot) T_{\widetilde{m}} f.$$
    With this motivation in mind, we write
    \begin{align*}
        \lim_{\eps \to 0} \eps^{n/p} \|T_{m}(\phi(\eps \cdot )f)\|_{L^{p}(\R^n,X)}
            \leq &  \lim_{\eps \to 0} \eps^{n/p} \|T_{m}(\phi(\eps \cdot )f) - \phi(\eps \cdot) T_{\widetilde{m}} f\|_{L^{p}(\R^n,X)} \\
            & +  \lim_{\eps \to 0} \eps^{n/p} \|\phi(\eps \cdot) T_{\widetilde{m}} f\|_{L^{p}(\R^n,X)}.
    \end{align*}
    Thus, by Lemma~\ref{Lem:11.21}, to establish \eqref{eq:goal1} we only need to see that
    the first term converges to zero. Even more, it is sufficient to show that
    \begin{equation}\label{eq:goal2}
        \lim_{\eps \to 0} \sum_{\ell \in \N} \eps^{n/p} \|T_{m(\cdot)-m(k)}(\phi_\ell(\eps \cdot )e_k)\|_{L^{p}(\R^n)} = 0, \quad k \in I.
    \end{equation}
    Notice that \eqref{eq:goal2} is a scalar property, so the Banach space $X$ does not play any role hereinafter.

We first observe an easy estimate:
\begin{equation*}
\begin{split}
  \eps^{n/p} \|T_{m(\cdot)-m(k)}(\phi_\ell(\eps \cdot )e_k)\|_{L^{p}(\R^n)}
 &\leq \eps^{n/p} \Big(\|T_m\|_{L^p(\R^n)\longrightarrow L^p(\R^n)}+\|m\|_{L^\infty(\R^n)}\Big)\|\phi_\ell(\eps \cdot )e_k\|_{L^{p}(\R^n)} \\
 &\leq C\|\phi_\ell\|_{L^{p}(\R^n)}, \\
\end{split}
\end{equation*}
where moreover we
    claim that
    \begin{equation}\label{eq:claim2}
        \|\phi_\ell\|_{L^p(\R^n)}
            \leq C 2^{-\rho \ell}, \quad \rho \in \N.
    \end{equation}
    Indeed, fixing some $\nu>n/(2p)$ and any (large) $\mu\in\N$, we have
    \begin{align*}
        \|\phi_\ell\|_{L^p(\R^n)}
            & \leq \|(1+|x|^2)^\nu \phi_\ell\|_{L^\infty(\R^n)} \| (1+|x|^2)^{-\nu} \|_{L^p(\R^n)} \\
            & \leq C \Big\|\Big[(1+ \Delta)^\nu \widehat{\phi_\ell}\Big]^{\vee}\Big\|_{L^\infty(\R^n)}
              \leq C \| (1+ \Delta)^\nu \widehat{\phi_\ell} \|_{L^1(\R^n)}
              \leq C \sum_{|\gamma| \leq 2 \nu } \| \partial^\gamma \widehat{\phi_\ell} \|_{L^1(\R^n)}\\
            &\leq C \sum_{|\gamma| \leq 2 \nu }\sum_{\delta\leq\gamma}\binom{\gamma}{\delta}
                \| \partial^{\delta} \widehat{\phi}\cdot \partial^{\gamma-\delta}\theta_{\ell} \|_{L^1(\R^n)}
            \leq C \sum_{|\delta| \leq 2 \nu }
                \| \partial^{\delta} \widehat{\phi}\|_{L^1(B(0,2^{\ell+1})\setminus B(0,2^{\ell-1}))} \\
            & \leq C 2^{(n-\mu) \ell} \sum_{|\delta| \leq 2 \nu } \| |\xi|^\mu \partial^\delta \widehat{\phi}(\xi) \|_{L^\infty(\R^n)}
             \leq C 2^{(n-\mu) \ell},
    \end{align*}
    since $\phi$ and thus $\widehat\phi$ belong to $\mathcal{S}(\R^n)$.

By dominated convergence, it suffices to show that each term in \eqref{eq:goal2}, for a fixed $\ell\in\N$, tends to zero as $\eps\to 0$.

    Fix $k \in I$, $\ell \in \N$ and $0<\eps<2^{-\ell-3}$.
    If we define $M_\ell^{\eps,k} = (m(\cdot)-m(k))\Theta_\ell((\cdot-k)/\eps)$, it is clear that
    $$T_{m(\cdot)-m(k)}(\phi_\ell(\eps \cdot )e_k)
        = T_{M_\ell^{\eps,k}}(\phi_\ell(\eps \cdot )e_k),$$
    and
    \begin{align}\label{eq:phiTM}
        \eps^{n/p} \|T_{M_\ell^{\eps,k}}(\phi_\ell(\eps \cdot )e_k)\|_{L^{p}(\R^n)}
            \leq & \| \phi_\ell \|_{L^{p}(\R^n)} \|T_{M_\ell^{\eps,k}}\|_{L^{p}(\R^n) \longrightarrow L^{p}(\R^n)}.
    \end{align}
We already estimate $\| \phi_\ell \|_{L^{p}(\R^n)}$ above, and we now turn to the multiplier norm.
     First of all, it is more convenient to consider the new multiplier
    $$\widetilde{M_\ell^{\eps,k}}(\eta)
        = M_\ell^{\eps,k}(2^\ell \eps \eta + k) = [m(2^\ell \eps \eta + k) - m(k)] \Theta_\ell (2^\ell \eta), \quad \eta \in \R^n,$$
    because the $L^p$-norm of Fourier multipliers is invariant under dilations and translations
    in the multiplier function (see for example \cite[(2.5.14) and (2.5.15)]{Graf}).
    Note that $\supp \Theta_\ell (2^\ell \cdot) \subset B(0,4)$, so that
    $|k+2^\ell\eps\eta|\geq|k|- |\eta| \cdot 2^\ell\eps \geq 1- 4 \cdot 1/8=1/2$,
    and thus $m$ is only evaluated in $B(0,1/2)^c$, where it is $C^\infty$. Also note that $\Theta_\ell(2^\ell\eta)=\Theta_2(2^2\eta)$ for all $\ell\geq 2$.

We first estimate $\widetilde{M_\ell^{\eps,k}}$ pointwise.
By applying the mean value theorem:
    \begin{align*}
        |\widetilde{M_\ell^{\eps,k}}(\eta)|
        &\leq \max_{z\in[k,k+2^\ell\eps\eta]} |\nabla m(z)|\cdot 2^\ell\eps|\eta|\cdot |\Theta_\ell(2^\ell\eta)| \\
            & \leq 2^{\ell+2} \eps \sup_{|z|\geq 1/2} |\nabla m(z)|
            \leq C 2^{\ell} \eps, \quad \eta \in \R^n,
    \end{align*}
    where the last bound follows from the homogeneity of $m$ (see for example \cite[p. 366]{Graf}). (Here
     $[x,y]$ stands for the segment connecting points $x$, $y \in \R^n$ and
    $C$ does not depend on $\eta$, $\ell$ or $\eps$.)

    Next, we estimate the size of the derivatives of $\widetilde{M_\ell^{\eps,k}}$. From the Leibniz rule,
    we get for every $\gamma \in \N^n \setminus \{0\}$, such that $|\gamma| \leq n+1$,
    \begin{equation*}
\begin{split}
         |\partial^\gamma \widetilde{M_\ell^{\eps,k}}(\eta)|
          &  \leq   \Big\{ |m(2^{\ell}\eps \eta + k) - m(k)|\cdot |(\partial^\gamma) (\Theta_\ell)(2^{\ell}\cdot))(\eta)| \\
                 &  \qquad\qquad+ \sum_{0 \neq \tau \leq \gamma} \binom{\gamma}{\tau} (2^\ell\eps)^{|\tau|} |(\partial^\tau m)(2^{\ell}\eps \eta + k)|
                  \cdot |(\partial^{\gamma - \tau})( \Theta_\ell) (2^{\ell}\cdot))(\eta)|\Big\} \\
           & \leq   C\Big\{ 2^{\ell} \eps+\sum_{0\neq\tau\leq\gamma}(2^\ell\eps)^{|\tau|}\Big\}\leq C 2^\ell\eps, \quad \eta \in \R^n,
\end{split}
\end{equation*}
    for certain constant $C$ independent of $\eta$, $\ell$ and $\eps$. The same bound holds for $|\eta|^{|\gamma|}|\partial^\gamma \widetilde{M_\ell^{\eps,k}}(\eta)|$, since $|\eta|\leq 4$ on the support of $ \widetilde{M_\ell^{\eps,k}}$. Hence, Mihlin's multiplier theorem
    (see for instance \cite[Theorem 5.2.7]{Graf}) implies that
    \begin{equation}\label{eq:Mihlin}
        \| T_{\widetilde{M_\ell^{\eps,k}}} \|_{L^{p}(\R^n) \longrightarrow L^{p}(\R^n)}
            \leq C 2^{\ell} \eps.
    \end{equation}

    Putting together \eqref{eq:claim2}, \eqref{eq:phiTM} and \eqref{eq:Mihlin} we conclude \eqref{eq:goal2}.
\end{proof}

To finish this section we present a transference result from $\T^n$ to $\T^{rn}$. Its idea when $n=1$ goes back to Bourgain \cite{Bou}.
Given an operator $S$ acting on $L^p_0(\T^n,X)$ we define its tensor extension
to $L^p_0(\T^n,L^p(\T^{(\ell-1)n},X)) \simeq L^p(\T^{(\ell-1)n},L^p_0(\T^{n},X))$, $\ell \in \N$, by
$$S_\ell = I_{L^p(\T^{(\ell-1)n})} \otimes S.$$

\begin{Prop}\label{Prop:TnTrn}
    Let $1<p<\infty$, $N,r \in \N$ and $m$, $m_j \in C^\infty(\R^n \setminus \{0\})$, $j=1, \dots, N$,
    be homogeneous functions.  If the following estimate holds
    $$\displaystyle \|T_{\widetilde{m}} f\|_{L^p(\T^n,X)}
                            \leq K \sum_{j=1}^N \| T_{\widetilde{m}_j}f\|_{L^p(\T^n,X)}, \quad f \in L^p_0(\T^n,X),$$
    then,
    $$\displaystyle \Big \| \sum_{\ell=1}^r (T_{\widetilde{m}})_{\ell} f_\ell \Big\|_{L^p(\T^{rn},X)}
                            \leq K \sum_{j=1}^N \Big\| \sum_{\ell=1}^r (T_{\widetilde{m}_j})_\ell f_\ell \Big \|_{L^p(\T^{rn},X)},
                            \quad f_\ell \in L^p_0(\T^n,L^p(\T^{(\ell-1)n},X)).$$
\end{Prop}

\begin{proof}
    By density arguments it is enough to consider trigonometric polynomials $(f_\ell)_{\ell=1}^r$.
    For each $\ell =1, \dots, r$, we write
    \begin{equation*}\label{eq:expansion}
        f_\ell(\bar{t}_{\ell-1},t_\ell)
            = \sum_{\substack{k \in \Z^n \setminus \{0\} \\ |k| \leq B}}
            \sum_{\substack{s \in \Z^{(\ell-1)n} \\ |s| \leq B}}
            a_{s,k}^{(\ell)} e_s(\bar{t}_{\ell-1}) e_k(t_\ell),
    \end{equation*}
    where $\bar{t}_{\ell-1}=(t_1, \dots, t_{\ell-1}) \in \T^{(\ell-1)n}$, $t_\ell \in \T^n$ and
    $a_{s,k}^{(\ell)} \in X$. Notice that it is posible to choose the same $B \in \N$ for all $f_\ell$.
    Then,
    \begin{align*}
        (T_{\widetilde{m}})_{\ell} f_\ell(\bar{t}_{\ell-1},t_\ell)
            & = \sum_{\substack{k \in \Z^n \setminus \{0\} \\ |k| \leq B}}
                \sum_{\substack{s \in \Z^{(\ell-1)n} \\ |s| \leq B}}
                a_{s,k}^{(\ell)} e_s(\bar{t}_{\ell-1}) T_{\widetilde{m}} \Big(e_k(t_\ell)\Big) \\
            & = \sum_{\substack{k \in \Z^n \setminus \{0\} \\ |k| \leq B}}
                \sum_{\substack{s \in \Z^{(\ell-1)n} \\ |s| \leq B}}
                a_{s,k}^{(\ell)} e_s(\bar{t}_{\ell-1}) m(k) e_k(t_\ell), \quad (\bar{t}_{\ell-1},t_\ell) \in \T^{\ell n}.
    \end{align*}

    Fix some $\bar{t}_\ell=(\bar{t}_{\ell-1},t_\ell)=(t_1, \dots, t_\ell) \in \T^{\ell n}$ and take
    $\bar{M}_\ell = (\bar{M}_{\ell-1},M_\ell)=(M_1, \dots, M_\ell) \in (\N \setminus\{0\})^\ell$
    to be chosen below. We introduce some operations between vectors of different lengths,
    $$\bar{M}_\ell \otimes t
        = (\bar{M}_{\ell-1} \otimes t, M_\ell t)= (M_1 t , \dots, M_\ell t) \in (\T^{n})^\ell,
        \quad t \in \T^n,$$
    $$\bar{M}_{\ell-1} \odot s
        = M_1 s_1 + \dots + M_{\ell-1}s_{\ell-1} \in \Z^n, \quad s=(s_1, \dots, s_{\ell-1}) \in (\Z^n)^{\ell-1}.$$
    It is verified that
    $s \cdot (\bar{M}_{\ell-1} \otimes t) = (\bar{M}_{\ell-1} \odot s) \cdot t$,
    and hence $e_s(\bar{M}_{\ell-1} \otimes t)=e_{\bar{M}_{\ell-1} \odot s}(t)$.

    Now, we consider Bourgain's transformation of $f_\ell$, which is given by
    \begin{align*}\label{eq:11.22}
        \widetilde{f}_\ell(t)
            = f_\ell(\bar{t}_{\ell} + \bar{M}_\ell \otimes t)
            = \sum_{\substack{k \in \Z^n \setminus \{0\} \\ |k| \leq B}}
            \sum_{\substack{s \in \Z^{(\ell-1)n} \\ |s| \leq B}}
            a_{s,k}^{(\ell)} e_s(\bar{t}_{\ell-1}) e_k(t_\ell)e_{\bar{M}_{\ell-1} \odot s + M_\ell k}(t),
            \quad t \in \T^n.
    \end{align*}
    The function $\widetilde{(T_{\widetilde{m}})_\ell f_\ell}$ is defined analogously.

    We want to compare $\widetilde{(T_{\widetilde{m}})_\ell f_\ell}$ with $T_{\widetilde{m}} \widetilde{f}_\ell$.
    Observe that both of them are multipliers transforms of $\widetilde{f}_\ell$, where in the first case
    the exponential factor $e_{\bar{M}_{\ell-1} \odot s + M_\ell k}$ is multiplied by $m(k)$; and in the second
    one by $m(\bar{M}_{\ell-1} \odot s + M_\ell k)$. Using the homogeneity of $m$ and the mean value theorem,
    we arrive at
    \begin{align*}
        |m(k) - m(\bar{M}_{\ell-1} \odot s + M_\ell k)|
            & = \Big| m\Big( \frac{k}{|k|} \Big) - m\Big( \frac{\bar{M}_{\ell-1} \odot s}{M_\ell|k|} + \frac{k}{|k|}\Big) \Big| \\
            & \leq \Big| \frac{\bar{M}_{\ell-1} \odot s}{M_\ell|k|} \Big| \sup_z |\nabla m(z)|
            \leq  \frac{|\bar{M}_{\ell-1}|}{|M_\ell|}  \sup_z |\nabla m(z)|,
    \end{align*}
    where the supremum is taken over certain compact set away from the origin. This supremum is
    finite because $m \in C^\infty(\R^n \setminus \{0\})$. If we choose the sequence
    $M_1 < M_2 < \dots$ to be sufficiently rapidly increasing, we can make above difference smaller
    than any preassigned $\eps>0$.

    In conclusion, denoting by $\|g\|_A$ the sum of the $X$-norms of the
    Fourier coefficients of a trigonometric polynomial $g$ (on a torus of any dimension), we have
    deduced that
    $$\Big\|\widetilde{(T_{\widetilde{m}})_\ell f_\ell} - T_{\widetilde{m}} \widetilde{f}_\ell \Big\|_{L^p(\T^n,X)}
        \leq \Big\|\widetilde{(T_{\widetilde{m}})_\ell f_\ell} - T_{\widetilde{m}} \widetilde{f}_\ell \Big\|_{A}
        \leq \eps \|f_\ell\|_A, \quad \ell=1, \dots r,$$
    provided that $M_1 < M_2 < \dots$ is sufficiently rapidly increasing.

    Now, we apply the hypothesis in order to get
    \begin{align*}
        \Big\| \sum_{\ell=1}^r \widetilde{(T_{\widetilde{m}})_\ell f_\ell} \Big\|_{L^p(\T^n,X)}
        & \leq \Big\| T_{\widetilde{m}} \Big( \sum_{\ell=1}^r \widetilde{f}_\ell \Big) \Big\|_{L^p(\T^n,X)}
            + \eps \sum_{\ell=1}^r \|f_\ell\|_A \\
        & \leq K \sum_{j=1}^N \Big \| T_{\widetilde{m}_j} \Big( \sum_{\ell=1}^r \widetilde{f}_\ell \Big) \Big\|_{L^p(\T^n,X)}
            + \eps \sum_{\ell=1}^r \|f_\ell\|_A.
    \end{align*}
    So far, we have taken $L^p$-norms with respect to the variable $t \in \T^n$.
    It is time to take $L^p$-norms with respect to the fixed variables
    $\bar{t}_r=(t_1, \dots, t_r) \in \T^{rn}$,
    \begin{align*}
        & \Big( \int_{\T^{rn}} \int_{\T^n} \Big\| \sum_{\ell=1}^r
            (T_{\widetilde{m}})_\ell f_\ell(\bar{t}_\ell + \bar{M}_\ell \otimes t) \Big\|^p_X dt d\bar{t_r} \Big)^{1/p} \\
        & \qquad \leq K \sum_{j=1}^N \Big( \int_{\T^{rn}} \int_{\T^n} \Big\| \sum_{\ell=1}^r
             T_{\widetilde{m}_j} \widetilde{f}_\ell (\bar{t}_\ell + \bar{M}_\ell \otimes t) \Big\|^p_X dt d\bar{t_r} \Big)^{1/p}
            + \eps \sum_{\ell=1}^r \|f_\ell\|_A.
    \end{align*}
    Finally, exchanging the order of integration we notice that the dependence in $t$ and $\bar{M}_\ell$
    disappears, resulting in
    $$\Big \| \sum_{\ell=1}^r (T_{\widetilde{m}})_{\ell} f_\ell \Big\|_{L^p(\T^{rn},X)}
        \leq K \sum_{j=1}^N \Big\| \sum_{\ell=1}^r (T_{\widetilde{m}_j})_\ell f_\ell \Big \|_{L^p(\T^{rn},X)}
            + \eps \sum_{\ell=1}^r \|f_\ell\|_A,$$
    which finishes the proof of this lemma, once we take $\eps \to 0$.
\end{proof}

\section{Proof of Theorem~\ref{Th:main}} \label{sec:proofTh}

We start this section with an alternative, but equivalent, definition of a UMD Banach space.
Let $(\Omega,d\mu)$ be a probability space. We call $(\eps_\ell d_\ell(\eps_1, \dots, \eps_{\ell-1}))_{\ell \in \N}$
a Paley--Walsh martingale difference sequence when $d_\ell : \R^{\ell-1} \longrightarrow X$, $\ell \geq 2$; $d_1$ is
a constant map in $X$ and $(\eps_\ell)_{\ell \in \N}$ are independent Bernoulli random variables, that is,
$\mu(\{\omega \in \Omega : \eps_\ell(\omega)=\pm 1\})=1/2$, $\ell \in \N$.

The Banach space $X$ has UMD if
for some (equivalently, for every) $1<p<\infty$, there exists a constant $c_p$ such that
\begin{equation}\label{eq:defUMD2}
    \Big\| \sum_{\ell=1}^r \sigma_\ell \eps_\ell d_\ell(\eps_1, \dots, \eps_{\ell-1}) \Big\|_{L^p(\Omega,X)}
        \leq c_p \Big\| \sum_{\ell=1}^r \eps_\ell d_\ell(\eps_1, \dots, \eps_{\ell-1}) \Big\|_{L^p(\Omega,X)},
\end{equation}
for each $r \in \N$, all Walsh-Paley martingale difference sequence $(\eps_\ell d_\ell)_{\ell=1}^r$ and
every $(\sigma_\ell)_{\ell =1}^r \in \{-1,1\}^r$.
It is well-known that $\beta_p(X)= \inf c_p$ (see \cite[p. 12]{Bu4} and \cite{Mau}).

The next lemma reduces checking the condition \eqref{eq:defUMD2} to the following inequality, more convenient  for
our purposes.

\begin{Lem}\label{Lem:reduction}
    Let $1<p<\infty$. Suppose that there exists $C_p>0$ such that the following holds:
    For every $r\in\N$ and a sequence $(\sigma_\ell)_{\ell=1}^r$ in $\{-1,+1\}$, there is a sequence
    $(b_\ell)_{\ell=1}^r$ in $\{-1,+1\}^n$
    such that
    \begin{equation}\label{eq:reduction}
        \Big\| \sum_{\ell=1}^r \sigma_\ell a(b_\ell \cdot t_\ell) \Phi_\ell(t_1, \dots, t_{\ell-1}) \Big\|_{L^p(\T^{rn},X)}
            \leq C_p \Big\| \sum_{\ell=1}^r a(b_\ell \cdot t_\ell) \Phi_\ell(t_1, \dots, t_{\ell-1}) \Big\|_{L^p(\T^{rn},X)},
    \end{equation}
    for every $1$-dimensional trigonometric polynomial $a$ with zero mean value and every trigonometric polynomial
    $\Phi_\ell:\T^{(\ell-1)n}\longrightarrow X$.
    Then $\beta_p(X) \leq C_p$.
\end{Lem}

\begin{proof}
    Let $r \in \N$, let $(\eps_\ell d_\ell)_{\ell=1}^r$ be a given Paley--Walsh martingale difference sequence  and
    $(\sigma_\ell)_{\ell =1}^r \in \{-1,1\}^r$.
    Take also $b_\ell \in \{-1,1\}^n$, $\ell=1, \dots, r$, satisfying \eqref{eq:reduction}.

    It is obvious that the function $\sgn(\theta)$, $\theta \in \T$, is a Bernoulli random variable.
    Moreover, for every $b \in \{-1,1\}^n$, the  function $\sgn( b \cdot t)$, $t \in \T^n$, is also a
    Bernoulli random variable.
    Since $(\eps_1(\omega), \dots, \eps_r(\omega))$ and $(\sgn(b_1 \cdot t_1), \dots, \sgn(b_r \cdot t_r))$ have the same
    distribution,
    we can write
\begin{equation}\label{eq:equid}
\begin{split}
     &\Big\| \sum_{\ell=1}^r \sigma_\ell' \eps_\ell d_\ell\Big(\eps_1, \dots, \eps_{\ell-1})\Big) \Big\|_{L^p(\Omega,X)}  \\
     &= \Big\| \sum_{\ell=1}^r \sigma_\ell' \sgn(b_\ell \cdot t_\ell) d_\ell\Big(\sgn(b_1 \cdot t_1), \dots, \sgn(b_{\ell-1} \cdot t_{\ell-1})\Big) \Big\|_{L^p(\T^{rn},X)},
\end{split}
\end{equation}
for both $\sigma_\ell'=\sigma_\ell$ and $\sigma_\ell'=1$.

     For a fixed $\delta>0$, we choose a
    $1$-dimensional trigonometric polynomial $a$ with zero mean value, verifying that
    $$\|\sgn(\cdot) - a \|_{L^p(\T)} < \delta,$$
where $\sgn$ is the $1$-periodic extension of the sign on $[-1/2,1/2)$.
Then it easily follows that, for every $\ell=1, \dots, r$,
    $$\|\sgn(b_\ell \cdot t) - a(b_\ell \cdot t) \|_{L^p(\T^n)} < \delta,$$
 and so in particular
 \begin{equation*}
  \|a(b_\ell \cdot t) \|_{L^p(\T^n)} \leq \|\sgn(b_\ell \cdot t)\|_{L^p(\T^n)} +\delta=1+\delta.
\end{equation*}

We also choose trigonometric polynomials $\Phi_\ell:\T^{(\ell-1)n}\longrightarrow X$ such that
\begin{equation*}
   \Big\|\Phi_\ell(t_1,\ldots,t_{\ell-1})-d_\ell\Big(\sgn(b_1 \cdot t_1), \dots, \sgn(b_{\ell-1} \cdot t_{\ell-1})\Big) \Big\|_{L^p(\T^{(\ell-1)n},X)}<\delta,
   \quad \ell=1, \dots, r.
\end{equation*}
It follows that
\begin{equation}\label{eq:approx}
\begin{split}
  &\Big\|a(b_\ell\cdot t_\ell)\Phi_\ell(t_1,\ldots,t_{\ell-1})
  -\sgn(b_\ell\cdot t_{\ell})d_\ell\Big(\sgn(b_1 \cdot t_1), \dots, \sgn(b_{\ell-1} \cdot t_{\ell-1})\Big) \Big\|_{L^p(\T^{\ell n},X)} \\
  &\leq\|a(b_\ell\cdot t_{\ell})\|_{L^p(\T^n)}\Big\|\Phi_\ell(t_1,\ldots,t_{\ell-1})
  -d_\ell\Big(\sgn(b_1 \cdot t_1), \dots, \sgn(b_{\ell-1} \cdot t_{\ell-1})\Big) \Big\|_{L^p(\T^{(\ell-1) n},X)} \\
  &\quad+\|a(b_\ell\cdot t_\ell)-\sgn(b_\ell\cdot t_\ell)\|_{L^p(\T^n)}
  \Big\|d_\ell\Big(\sgn(b_1 \cdot t_1), \dots, \sgn(b_{\ell-1} \cdot t_{\ell-1})\Big) \Big\|_{L^p(\T^{(\ell-1) n},X)} \\
  &\leq (1+\delta)\cdot\delta+\delta\cdot C \leq C\delta, \quad \ell=1, \dots, r,
\end{split}
\end{equation}
where the first $C$ is the maximum of the $L^p$ norms of the given functions $d_\ell(\eps_1,\ldots,\eps_{\ell-1})$.

A combination of \eqref{eq:equid} and \eqref{eq:approx} with the assumption \eqref{eq:reduction} now shows that
\begin{equation*}
\begin{split}
  &\Big\| \sum_{\ell=1}^r \sigma_\ell \eps_\ell d_\ell(\eps_1, \dots, \eps_{\ell-1}) \Big\|_{L^p(\Omega,X)}
  \leq \Big\| \sum_{\ell=1}^r \sigma_\ell a(b_\ell \cdot t_\ell) \Phi_\ell(t_1,\ldots,t_{\ell-1}) \Big\|_{L^p(\T^{rn},X)} +Cr\delta \\
  &\leq C_p \Big\| \sum_{\ell=1}^r  a(b_\ell \cdot t_\ell) \Phi_\ell(t_1,\ldots,t_{\ell-1}) \Big\|_{L^p(\T^{rn},X)} +Cr\delta \\
  &\leq C_p\Big\| \sum_{\ell=1}^r  \eps_\ell d_\ell(\eps_1,\ldots,\eps_{\ell-1}) \Big\|_{L^p(\Omega,X)}+(C_p+1)Cr\delta.
\end{split}
\end{equation*}
    Since $\delta>0$ is arbitrarily small, we conclude that \eqref{eq:defUMD2} holds and $\beta_p(X) \leq C_p$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{Th:main}]
Let us first make an additional useful reduction. Note that the set $F$ can be neither $\varnothing$ or $\{1,\ldots,n\}$, because for these sets we have $\sum_{\ell\in F}\beta_\ell=\sum_{\ell\in F}\alpha^j_\ell$ (using $|\beta|=|\alpha^j|$ if $F=\{1,\ldots,n\}$), contradicting the fact that these have different parity. Thus, there exists an index $s\in F$ as well as $t\in\{1,\ldots,n\}\setminus F$. By applying the assumption \eqref{eq:estim2} to $\partial_k u$ in place of $u$, we see that \eqref{eq:estim2}  implies a similar estimate with $\beta+e_k$ in place of $\beta$ and $\alpha^j+e_k$ in place of $\alpha^j$. Thus, if $\sum_{\ell\in F}\beta_\ell$ is not already odd, we can make it odd by adding $e_s$ to both $\beta$ and $\alpha^j$; note that this preserves the other assumptions. After, if the new $|\beta|$ is not even, we can make it even by adding $e_t$ to both $\beta$ and $\alpha^j$, which again preserves the other assumptions, including the value of $\sum_{\ell\in F}\b!
 eta_\ell$, since $t\notin F$. Thus, if the assumptions of Theorem~\ref{Th:main} hold for some $\beta,\alpha^j$, they also hold for some (possibly different) $\beta,\alpha^j$ which satisfy in addition that $|\beta|=|\alpha^j|$ is even and $\sum_{\ell\in F}\beta_{\ell}$ is odd (so that $\sum_{\ell\in F}\alpha^j_\ell$ is again even). We henceforth assume this, and proceed to prove the claim that $X$ is UMD with $\beta_p(X)\leq KN$.

So, let $\alpha^j$, $\beta$ and $F$ be as in the
statement of the theorem, with the additional assumptions just made, and $m$, $m_j$ be the multipliers defined in \eqref{eq:mmj}.
The evenness of $|\beta|=|\alpha^j|$ gives the advantage that the multipliers
$m(\xi)=\xi^\beta/|\xi|^{|\beta|}$ and $m_j(\xi)=\xi^{\alpha^j}/|\xi|^{|\alpha^j|}$ are not only homogeneous, but also even, so that $m(\lambda\xi)=m(\xi)$ holds for all $\lambda\in\R\setminus\{0\}$, not just $\lambda>0$.

Fix a 1-dimensional trigonometric polynomial
$$a(\theta)
    = \sum_{\ell \in \Z} \widehat{a}(\ell) e_\ell(\theta), \quad \theta \in \T,$$
with $\widehat{a}(0)=0$, and consider the function $a_b(t)=a(b \cdot t)$, $t \in \T^n$, for certain $b \in \{-1,1\}^n$.
Also, if $J \subset \{1, \dots, n\}$ we define $b_J \in \{-1,1\}^n$ to be the vector
whose $\ell$-th component is equal to $-1$, if $\ell \in J$; or $1$ otherwise. With this notation,
take $a^+=a_{(1, \dots, 1)}$ and $a^-=a_{b_{F}}$.

Since $m(\ell b)=m(b)$ for all $\ell\in\Z\setminus\{0\}$, we find that
\begin{equation}\label{eq:claim}
    T_{\widetilde{m}} a_b(t)
    =T_{\widetilde{m}}\sum_{\ell\in\Z}\widehat{a}(\ell)e_{\ell b}(t)
    =\sum_{\ell\in\Z}\widehat{a}(\ell)m(\ell b)e_{\ell b}(t)
        = m(b)a_b(t), \quad b \in \{-1,1\}^n,
\end{equation}
where $T_{\widetilde{m}}$ the discrete Fourier multiplier defined in \eqref{eq:Tmdiscret}.
Now, taking in mind the assumptions imposed over $F$, we deduce
as a direct consequence of \eqref{eq:claim} that
\begin{equation}\label{eq:pm}
    T_{\widetilde{m}} a^{\pm} = \pm n^{-|\beta|/2} a^{\pm}
    \quad \text{and} \quad
    T_{\widetilde{m}_j} a^{\pm} = n^{-|\beta|/2} a^{\pm}, \ j=1, \dots, N.
\end{equation}

Let $r \in \N$, $\sigma_\ell \in \{-1,1\}$ and
$\Phi_\ell:\T^{(\ell-1)n}\longrightarrow X$ be trigonometric polynomials for $\ell=1, \dots, r$.
Moreover, we define
$\zeta_\ell = a^+$, if $\sigma_\ell=1$; and $\zeta_\ell = a^-$, when $\sigma_\ell=-1$.
As it was commented in the introduction, \eqref{eq:estim2} implies \eqref{eq:estim3}. Then,
Propositions~\ref{Prop:RnTn} and \ref{Prop:TnTrn} together with \eqref{eq:pm}, allows us to write
\begin{align*}
     \Big\| \sum_{\ell=1}^r \sigma_\ell \zeta_\ell \Phi_\ell \Big\|_{L^p(\T^{rn},X)}
     & = n^{|\beta|/2} \Big\| \sum_{\ell=1}^r T_{\widetilde{m}}\zeta_\ell \Phi_\ell \Big\|_{L^p(\T^{rn},X)}
      \leq  K n^{|\beta|/2} \sum_{j=1}^N \Big\| \sum_{\ell=1}^r T_{\widetilde{m}_j}\zeta_\ell \Phi_\ell \Big\|_{L^p(\T^{rn},X)} \\
     & = K N \Big\| \sum_{\ell=1}^r \zeta_\ell \Phi_\ell \Big\|_{L^p(\T^{rn},X)}.
\end{align*}
Finally, an application of Lemma~\ref{Lem:reduction} implies that $\beta_p(X) \leq KN$, and hence the proof of
this theorem is completed.
\end{proof}

\def\cprime{$'$}


TITLE: Pull back of a vector representing a 2-form in $\mathbb R^3$
QUESTION [1 upvotes]: Let $\omega$ be a $2$-form defined in $\mathbb R^3$. I know that it can be represented by a vector field $\xi$ in such a way:
$$
\omega_x (v,w) = \xi(x)\cdot (v\times w)
$$
($x,v,w \in \mathbb R^3$ and $\times$ is the vector product). And I know that the differential becomes the divergence:
$$
d \omega = (\nabla\cdot \xi)\ dx_1\wedge dx_2\wedge dx_3.
$$
Now if $T\colon \mathbb R^3 \to \mathbb R^3$ is a change of variables, I know that
$$
d(T^*\omega) = T^* (d\omega).
$$
I would like to prove the corresponding equality for vector fields (without passing to forms). I was able to make the corresponding computation for $1$-forms (and it wasn't trivial) but I'm stuck with $2$-forms.
What I've found is that if $\omega$ corresponds to $\xi$ as above, then
$$
T^* \omega(v,w) = \omega(DTv,DTw) = \xi \cdot(DTv\times DTw) = (\det DT) (DT)^{-1}\xi\cdot(v\times w)
$$
hence the vector-field representing $T^*\omega$ is
$$
\xi^* = (\det DT)(DT)^{-1}\xi = \mathrm{adj}(DT) \xi.
$$
Now I would like to prove that
$$
\nabla \cdot \xi^* = (\det DT)\  \nabla \cdot \xi
$$
which should be the correct change of variables. But I have no clue on how to work on this...

REPLY [2 votes]: Divergence of a vector field $Y_p$ shows by how much the volume form $\mu$ (or infinitesimal volume elements) changes along the flow:
$$
(\operatorname{div} Y)\mu=\mathcal{L}_Y(\mu)=d(Y\lrcorner\mu),
$$
 where I also used the Cartan's magic formula. The interior product is the $(n-1)$-form $$[Y\lrcorner\mu](Y_1,Y_2,\ldots,Y_{n-1})=\mu(Y, Y_1,Y_2,\ldots,Y_{n-1}).
$$
The two expressions for divergence form will help us pull it back along a given map, or coordinate change, $\phi$. We now see the vector (field) Y as a result of pushforward by the map derivative $Y_{\phi(p)}=\phi_*(X_p)$. The pullback of the volume form shows the expected determinant of the Jacobian matrix, $f=det(\phi_*)$,
$$
\phi^*\mu=f\nu
$$
-- this is another factor that may cause the volume form $\nu$ to change.
The pull back of the interior product form is
$$
\phi^*(Y\lrcorner\mu)=X\lrcorner \phi^*\mu=X\lrcorner (f\nu).
$$
The pullback commutes with exterior derivative, and so
$$
\phi^*d(Y\lrcorner\mu)=d(\phi^*(Y\lrcorner\mu))=d(X\lrcorner f\nu)=\mathcal{L}_X(f\nu),
$$
where we've again used the Cartan's formula.
We will now use properties of Lie derivative and the atiderivation of the inner product to show that
$$
\mathcal{L}_X(f\nu)=(\mathcal{L}_X\;f)\nu +f\mathcal{L}_X\nu=\\
(X\lrcorner df)\nu+\left(\mathcal{L}_{fX}(\nu)-df\wedge(X\lrcorner\nu) \right)=\\
\mathcal{L}_{fX}(\nu)+\left((X\lrcorner df)\nu -df\wedge(X\lrcorner\nu) \right)=\\
=\mathcal{L}_{fX}(\nu)+X\lrcorner(df\wedge\nu)=\\
\mathcal{L}_{fX}(\nu)+0=\operatorname{div}(fX)\nu.
$$
Therefore,
$$\phi^*((\operatorname{div} Y)\mu)=\operatorname{div}(fX)\nu,$$
which suggests that the original volume form $\nu$ may change due to both divergence of the field along which we observe it and the mapping with the determinant scalar field $f$.


TITLE: Exercise 1.7 from Silverman's AEC
QUESTION [3 upvotes]: I'm working through Silverman's Arithmetic of Elliptic Curves Exercise 1.7 (3 parts) and I have a question about the last part.

(a) In $\mathbb{P}^1$ at least one of $S$ or $T$ is non-zero, so the map is well-defined at all $[S,T] \in \mathbb{P}^1$. Therefore, $\phi$ is a morphism.
(b) You can easily check that $\psi: [X,Y,Z]\mapsto [Y,X]$ works.
(c) The only problematic point on $\psi$ is $[0,0,1]$. I tried to do a bit of the usual manipulation, but was not able to make $\psi$ well-defined on $[0,0,1]$. For instance:
$$ [Y,X] = [YX^2,X^3] = [YX^2,Y^2Z] = [X^2,YZ] $$
This leads me to believe that $\psi$ is not actually a morphism. However, for $\phi$ to be an isomorphism there just has to exist an inverse morphism $\phi^{-1}$ to $\phi$, and this does not necessarily have to be $\psi$ as I have defined it.
I'm wondering how can I (formally) show that $\phi$ is not an isomorphism?
EDIT: I realize that $Y^2Z=X^3$ has a cusp at $[0,0,1]$. So I guess can someone show me how to prove that an isomorphism of projective varieties must preserve singularities?

REPLY [1 votes]: If $i: X\rightarrow \mathbb{P}^n_k$ is a projective variety over a field $k$ and if $x\in X$ is a point, you may define the local ring $\mathcal{O}_{X,x}$ at $x$. By definition
L1. $\mathcal{O}_{X,x} \cong lim_{x \in U}\mathcal{O}_{X}(U)$,
The local ring in L1 is invariant wrto choice of embedding $i$, it only depends on the structure sheaf $\mathcal{O}_X$ (the structure sheaf is an "intrinsic" object that does not depend on $i$). The maximal ideal $\mathfrak{m}(x) \subseteq \mathcal{O}_{X,x}$ is also independent of choice of embedding. Hence the notion "non singular" does not depend on choice of embedding or isomorphism.


TITLE: How to find $F(x) = \int_x^{x^2} (2+\sqrt t )\, dt$ ?
QUESTION [0 upvotes]: I have this problem:
$$
F(x) = \int_x^{x^2} (2+\sqrt t )\, dt
$$
I have to solve the integral.
I got $2x^2+\frac{2x^3}{3}-2x-\frac{2x^{3/2}}{3}$
However, I don't think that it correct.

REPLY [2 votes]: It's correct:
$$F(x) = \int_x^{x^2} (2+\sqrt t )\, dt = \left[2t + \frac{2t^{3/2}}{3}\right]_{x}^{x^2}
= 2x^2 + \frac{2x^3}{3} - 2x - \frac{2x^{3/2}}{3}$$
My guess is that you inputted it incorrectly into your calculator.


TITLE: Integrating non uniform grid data from an accelerometer
QUESTION [1 upvotes]: Is there a better method than the trapezoidal rule as outlined here?
P.S. not familiar with the site, sorry if this has the wrong tag.

REPLY [3 votes]: I'd have to understand the nature of the data and how you expect that data to behave, but in principle, you could use a modified Simpson's Rule.  The difference here is that you have to perform a quadratic interpolation for every triplet of points.  That is, given a triplet of data $(x_{2 i-1},y_{2 i-1}), (x_{2 i},y_{2 i}), (x_{2  i+1},y_{2 i+1})$, where $i \in \{1,2,\ldots,N/2\}$ and $N$ is the number of data points, set
$$f_i(x) = a_i + b_i (x-x_{2 i}) + c_i (x-x_{2 i})^2$$
Plug in the triplet of data and solve for $a_i$, $b_i$, and $c_i$.  Then integrate the resulting expression for $f_i(x)$ over $[x_{2 i-1},x_{2 i+1}]$:
$$\int_{x_{2 i-1}}^{x_{2 i+1}} dx \: f_i(x) = \frac{x_{2 i-1}-x_{2 i+1}}{6
   (x_{2 i}-x_{2 i-1}) (x_{2 i}-x_{2 i+1})}
\left[\left (-3 x_{2 i}^2 (y_{2 i-1}+y_{2 i+1})-2 x_{2 i-1}
   (x_{2 i+1} (y_{2 i}+y_{2 i-1}+y_{2 i+1})-x_{2 i} (y_{2 i-1}+2 y_{2 i+1}))\right) \\ \left (+2 x_{2 i} x_{2 i+1} (2 y_{2 i-1}+y_{2 i+1})+x_{2 i-1}^2
   (y_{2 i}-y_{2 i+1})+x_{2 i+1}^2 (y_{2 i}-y_{2 i-1})\right)\right]$$
This is for one triplet; you will run this over all (nonoverlapping) triplets of data to get your integral. No doubt that this is more complicated and expensive to use than the trapezoidal expression, but if your data represent a piecewise smooth function, this representation should be more accurate.


TITLE: Can someone tell me if my proof about irrational ^ irrational = rational is valid?
QUESTION [1 upvotes]: So I'm trying to solve the problem irrational ^ irrational = rational. Here is my proof
    Let $i_{1},i_{2}$ be two irrational numbers and r be a rational number such that $$i_{1}^{i_{2}} = r$$
    So we can rewrite this as $$i_{1}^{i_{2}} = \frac{p}{q}$$
    Then by applying ln() to both sides we get $$i_2\ln(i_1) = \ln(p)-\ln(q)$$
    which can be rewritten using the difference of squares as $$ i_2\ln(i_1) = \left(\sqrt{\ln(p)}-\sqrt{\ln(q)}\right)\left(\sqrt{\ln(p)}+\sqrt{\ln(q)}\right)$$
    so now we have $$i_1 = e^{\sqrt{\ln(p)}+\sqrt{\ln(q)}}$$
$$i_2 = \sqrt{\ln(p)}-\sqrt{\ln(q)}$$
    because I've found an explicit formula for $i_1$ and $i_2$ we are done.
So I'm new to proofs and I'm not sure if this is a valid argument. Can someone help me out?

REPLY [2 votes]: When writing a proof - any proof - the first and most important step is to decide what you are proving. If the thing you end up with is not exactly the thing you were proving, you're not done with the proof.
In this case, it's not clear to me what you're trying to prove - are you trying to prove that $i_1^{i_2}$ is always rational for every irrational $i_1,i_2$? Hopefully not, because that isn't true - but even if it were, your proof only supplies one example. Are you trying to show that for every rational $r$ there exists irrationals $i_1,i_2$ so that $i_1^{i_2} = r$? If so, then you have a problem - in your first sentence you assumed the thing you were trying to prove!
Your reasoning itself is good, assuming you're shooting for the second conclusion, but this looks more like the work you should use to get the proof, not the proof itself. If I were you, I'd have the proof look sort of like this:
Let $r$ be any rational. Then $r = \frac{p}{q}$ for some integers $p,q$. Let $i_1 = e^{\sqrt{ln(p)} + \sqrt{\ln(q)}}$ and $i_2 = \sqrt{\ln(p)} - \sqrt{\ln(q)}$. Then... (insert calculations here)... so $i_1^{i_2} = r$. Therefore, for any $r$, there exist $i_1$ and $i_2$ so that $i_1^{i_2} = r$.
Now, there's still a problem - how do you know that your $i_1$ and $i_2$ are irrational? For example, if $r = 1$, then you end up with $i_1 = 1$ and $i_2 = 0$ no matter which $p$ and $q$ you choose. You may be able to do some sort of clever argument here, but I'd recommend trying a different technique.


TITLE: Magnetic field lines are not perfectly straight between magnetic poles
QUESTION [1 upvotes]: Why is fringing between magnetic poles inevitable?
The diagram given below is the incorrect representation of magnetic field lines between two poles:

REPLY [1 votes]: There can't be such straight field lines between magnetic poles as the magnetic field can always form closed loops. However the tangential vectors  through each loops point in the same direction which is illustrated in the figure. Remember that the magnetic field is a pseudo-vector. Consider a solenoid for example.

We know that the magnetic field forms closed loops around a current carrying wire. (Can you possibly say the direction of a closed loop?) More the current more will be the field line density. But that's not the reason we draw straight lines between two opposite poles of a magnet. You can only specify the direction of field at a point. That we do by drawing a tangent through the point on the loop. So if you take a region inside a solenoid, or in between the north pole and south pole, you can draw the direction of field at that point by drawing tangent on the loops at that point which will be parallel pointing in the same direction
What that has been illustrated in your picture is just the direction in which the magnetic field acts, not the field lines.


TITLE: Dual form of $\frac{1}{2}\|y-x\|^2_2+\lambda\|z\|_1$
QUESTION [0 upvotes]: Can you help me to find the dual of Dual form of
$$
\min_{x,z} \frac{1}{2}\|y-x\|^2_2+\lambda\|z\|_1 \\
\text{subject to }Dx=z
$$
My attempt has been to find the Lagrangian first:
$$
L(x,z,\nu)=\frac{1}{2}\|y-x\|^2_2+\lambda\|z\|_1+\nu^T(Dx-z)
$$
And now to find the dual we need to minimize $L(x,z,\nu)$ w.r.t. $x,z$. That is, take the derivative of the Lagrangian and equal to zero. But I can't find how to take derivatives of $L$. We can express $\|y-x\|^2_2$ as $(y-x)^T(y-x)$ but how to deal with $\|z\|_1$?
Your help will be kindly appreciated! Thank you for your time.

REPLY [0 votes]: you can minimize the lagrangian first with repect to $z$, then with respect to $x$. you can find the minimun value with respect to z without the derivative, intuitively.
you can write the lagrangian as $\frac{1}{2}||y-x||^2 + v^TDx + \lambda||z||_1-v^Tz$. to minimize with repect to $z$, you just need to find minimum of $\lambda||z||_1-v^Tz$. you can find that intuitively by using various cases. I did three cases. $\lambda = 0, \lambda> 0, \lambda< 0$. if $\lambda = 0$, $-v^Tz$ is affine, so its minimum is $0$ if $v = 0$ or $-\infty$ otherwise. if $\lambda > 0$, if $\lambda \succeq v$, the minimum possible value is $0$, otherwise it is $-\infty$. also, same condition when $\lambda > 0$. so, generally when $\lambda \succeq v$, the minimum is $0$. and it is $-\infty$ otherwise. after minimizing with respect to $z$ , you have to minimize $min \frac{1}{2}||y-x||^2 + v^TDx$ with respect to $x$ which can be easily done with the gradient as you said. then, after getting the final result, don't forget to write $\lambda \succeq v$ as the constraint for the dual problem. NOTE: I used $\lambda \succeq v$ to mean $\lambda$ is greater than every component of the vector $v$


TITLE: Point and a normal vector question
QUESTION [1 upvotes]: I'm trying to calculate a normal vector given three points in the plane.  My results do not agree with what I expect, and I would appreciate guidance.  I'm using the following Wiki definition:
$n = (p_2 - p_1) \times (p_3 - p_1)$
Points:
$p_1 = (1, 1, 1), p_2 = (1, 0, 0), p_3 = (0, 1, 0)$
$p_2 - p_1 = (0, -1, -1)$
$p_3 - p_1 = (-1, 0, -1)$
It's my understanding that $p_2 - p_1$ and $p_3 - p_1$ represent two perpendicular vectors specifying the plane through the origin.  We then take the cross product to get the norm and the determinant can be used to calculate the cross product.
$\begin{vmatrix}
i & j & k \\
0 & -1 & -1 \\
-1 & 0 & -1 \\
\end{vmatrix}$
$i \begin{vmatrix}
   -1 & -1 \\
   0 & -1 \\
   \end{vmatrix}
 + j \begin{vmatrix}
   0 & -1 \\
   -1 & -1 \\
   \end{vmatrix}
 - k \begin{vmatrix}
   0 & -1 \\
   -1 & 0 \\
   \end{vmatrix}
$
$i(-1*-1 - 0) + j(-1*-1 - 0) - k(-1*-1 - 0)$
$i + j - k$
This vector does not appear normal to the plane when I look at a graph.  Does anyone know if I made an error?

REPLY [1 votes]: Does anyone know if I made an error?

It should be $i(-1\times-1 - 0) - j(0 - (-1\times-1)) + k(0 - (-1\times-1))= i \color{red}+ j - k$.


text:
subsection \<open>Determined games\<close>

theory GaleStewartDeterminedGames
  imports GaleStewartDefensiveStrategies
begin

locale closed_GSgame = GSgame +
  assumes closed:"e \<in> A \<Longrightarrow> \<exists> p. lprefix (llist_of p) e \<and> (\<forall> e'. lprefix (llist_of p) e' \<longrightarrow> llength e' = 2*N \<longrightarrow> e' \<in> A)"

(* Perhaps a misnomer, GSgames are supposed to be infinite, but this is very much the GS variation
   of finite games, as definitions still use coinductive lists rather than the more common
   inductive (and finite) ones. *)
locale finite_GSgame = GSgame +
  assumes fin:"N \<noteq> \<infinity>"
begin

text \<open> Finite games are closed games. As a corollary to the GS theorem, this lets us conclude that finite games are determined. \<close>
sublocale closed_GSgame
proof
  fix e assume eA:"e \<in> A"
  let ?p = "list_of e"
  from eA have len:"llength e = 2*N" using length by blast
  with fin have p:"llist_of ?p = e"
    by (metis llist_of_list_of mult_2 not_lfinite_llength plus_eq_infty_iff_enat)
  hence pfx:"lprefix (llist_of ?p) e" by auto
  { fix e'
    assume "lprefix (llist_of ?p) e'" "llength e' = 2 * N"
    hence "e' = e" using len by (metis lprefix_llength_eq_imp_eq p)
    with eA have "e' \<in> A" by simp
  }
  with pfx show "\<exists>p. lprefix (llist_of p) e \<and> (\<forall>e'. lprefix (llist_of p) e' \<longrightarrow> llength e' = 2 * N \<longrightarrow> e' \<in> A)"
    by blast
qed
end

context closed_GSgame begin
lemma never_winning_is_losing_even:
  assumes "position p" "\<forall> n. \<not> winning_position_Even (((augment_list \<sigma>) ^^ n) p)"
  shows "induced_play \<sigma> p \<notin> A"
proof
  assume "induced_play \<sigma> p \<in> A"
  from closed[OF this] obtain p' where
    p':"lprefix (llist_of p') (induced_play \<sigma> p)"
    "\<And> e. lprefix (llist_of p') e \<Longrightarrow> llength e = 2 * N \<Longrightarrow> e \<in> A" by blast
  from lprefix_llength_le[OF p'(1)] have lp':"llength (llist_of p') \<le> 2 * N" by auto
  show False proof (cases "length p' \<le> length p")
    case True
    hence "llength (llist_of p') \<le> llength (llist_of p)" by auto
    from lprefix_llength_lprefix[OF p'(1) _ this]
      induced_play_is_lprefix[OF assms(1)]
      lprefix_trans
    have pref:"lprefix (llist_of p') (induced_play strat p)" for strat by blast
    from assms(2)[rule_format,of 0] assms(1) have "\<not> strategy_winning_by_Even \<sigma> p" for \<sigma> by auto
    from this[unfolded strategy_winning_by_Even_def] obtain strat where
      strat:"induced_play strat p \<notin> A" by auto
    from strat p'(2)[OF pref] show False by simp
  next
    case False
    let ?n = "length p' - length p"
    let ?pos = "(augment_list \<sigma> ^^ ?n) p"
    from False have "length p' \<ge> length p" by auto
    hence [simp]:"length ?pos = length p'"
      by (auto simp:length_augment_list)
    hence pos[intro]:"position ?pos"
      using False lp'(1) unfolding position_def by auto
    have "llist_of p' = llist_of ?pos"
      using p'(1)
      by(intro lprefix_antisym[OF lprefix_llength_lprefix lprefix_llength_lprefix],auto)
    hence p'_pos:"p' = ?pos" by simp
    from assms(2)[rule_format,of ?n] assms(1) have "\<not> strategy_winning_by_Even \<sigma> ?pos" for \<sigma> by auto
    from this[unfolded strategy_winning_by_Even_def] obtain strat where
      strat:"induced_play strat ?pos \<notin> A" by auto
    from p'_pos induced_play_is_lprefix[OF pos, of strat]
    have pref:"lprefix (llist_of p') (induced_play strat ?pos)" by simp
    with p'(2)[OF pref] strat show False by simp
  qed
qed

text \<open> By proving that every position is determined, this proves that every game is determined
       (since a game is determined if its initial position [] is) \<close>
lemma every_position_is_determined:
  assumes "position p"
  shows "winning_position_Even p \<or> winning_position_Odd p" (is "?we \<or> ?wo")
proof(rule impI[of "\<not> ?we \<longrightarrow> \<not> ?wo \<longrightarrow> False",rule_format],force)
  assume "\<not> ?we"
  from defensive_strategy_Odd[OF this] never_winning_is_losing_even[OF assms]
  have js_no:"induced_play
         (joint_strategy s defensive_strategy_Odd) p \<notin> A" for s
    by auto
  assume "\<not> ?wo"
  from this[unfolded strategy_winning_by_Odd_def] assms
  have "\<exists> s. induced_play
         (joint_strategy s defensive_strategy_Odd) p \<in> A" by simp
  thus False using js_no by simp
qed
lemma empty_position: "position []" using zero_enat_def position_def by auto
lemmas every_game_is_determined = every_position_is_determined[OF empty_position]

text \<open> We expect that this theorem can be easier to apply without the 'position p' requirement,
       so we present that theorem as well. \<close>
lemma every_position_has_winning_strategy:
  shows "(\<exists> \<sigma>. strategy_winning_by_Even \<sigma> p) \<or> (\<exists> \<sigma>. strategy_winning_by_Odd \<sigma> p)" (is "?we \<or> ?wo")
proof(cases "position p")
  case True
  then show ?thesis using every_position_is_determined by blast
next
  case False
  hence "2 * N \<le> enat (length p)" unfolding position_def by force
  from induced_play_lprefix_non_positions[OF this]
  show ?thesis unfolding strategy_winning_by_Even_def strategy_winning_by_Odd_def by simp
qed

end

end

subset_name: curated


TITLE: $UL$ factorization from $LU$ factorization with permutation matrix
QUESTION [2 upvotes]: The answer at A = UL factorization shows that, if a matrix $A$ has an $LU$ factorization, then one can also get a $UL$ factorization. But it does so by starting with an $LU$ factorization not of $A$ but of the matrix $B = PAP$, where $P$ is the permutation matrix \begin{equation*}\begin{pmatrix}0 & \dots & 0 & 1 \\ 0 & \dots & 1 & 0 \\ \vdots & \ddots & \vdots & \vdots \\ 1 & \dots & 0 & 0\end{pmatrix}\end{equation*} My question is how do we know that an $LU$ factorization of $B$ actually exists? It does not seem to easily follow just from the fact that $A$ has an $LU$ factorization. Thanks in advance for assistance.

REPLY [1 votes]: UL factorization from LU factorization with permutation matrix
According to the preprint Necessary And Sufficient Conditions For Existence of the LU Factorization of an Arbitrary Matrix, a both necessary and sufficient condition for the existence of an LU factorization of a square $n\times n$ matrix $A$ is given by:
$$(1)\,\text{$A$ has LU factorization} \iff \forall k\colon\, \operatorname{rk}(A[\colon k,\, \colon k]) \ge \operatorname{rk}(A[\colon k,\, \colon n]) + \operatorname{rk}(A[\colon n,\, \colon k]) -k$$
Where $A[\colon i, \colon j]$ is the matrix consisting of the first $i$ rows and $j$ columns of $A$. We shall see that:
$$(2)\,\text{$A$ has UL factorization} \iff \forall k\colon\, \operatorname{rk}(A[k\colon,\, k\colon]) \ge \operatorname{rk}(A[k\colon,\, n\colon]) + \operatorname{rk}(A[n\colon,\, k\colon]) -k$$
Where $A[i\colon, j\colon]$ is the matrix consisting of the last $i$ rows and $j$ columns of $A$.
Now, without further notice, given any $k\times l$ matrix  $X$ let $\overline X = P_k^T X P_l$, be the flip of $X$, where $(P_n)_{ij}=\delta_{i, n-i}$ is the exchange matrix of size $n$. We will show:
$$(3)\qquad \text{$A$ has UL factorization} \iff \text{$\overline A$ has LU factorization}$$
So the existence of an LU factorization of $B=\overline A$ is in fact equivalent to $A$ having a UL factorization.

Proof of (2): Note that the way the preprint proves (1) is by splitting $A$ into a block matrix at index $k$ for every $k=1 \ldots n$
$$
A =\left[\begin{array}{cc}
A_{11} & A_{12} \\
A_{21} & A_{22}
\end{array}\right]
=\left[\begin{array}{ll}
L_{11}U_{11} & L_{11}U_{12} \\
L_{21}U_{12} & L_{21}U_{12} + L_{22} U_{22}
\end{array}\right]
=\left[\begin{array}{cc}
L_{11} & 0 \\
L_{21} & L_{22}
\end{array}\right]\left[\begin{array}{cc}
U_{11} & U_{12} \\
0 & U_{22}
\end{array}\right]
=LU
$$
and then apply rank inequalities of the type $\operatorname{rk}(XY) \ge \operatorname{rk}(X) + \operatorname{rk}(Y) -k$ to the blocks.
Note that $A=LU$ is equivalent to $P^TAP = (P^T L P)(P^T U P)$, i.e. $\overline A = \overline L \cdot \overline U$, hence
$$
\overline A =\left[\begin{array}{cc}
\overline A_{22} & \overline A_{21} \\
\overline A_{12} & \overline A_{11}
\end{array}\right]
=\left[\begin{array}{ll}
\overline L_{21}\overline U_{12} + \overline L_{22} \overline U_{22} & \overline L_{21}\overline U_{12} \\
\overline L_{11}\overline U_{12} & \overline L_{11}\overline U_{11}
\end{array}\right]
=\left[\begin{array}{cc}
\overline L_{22} & \overline L_{21} \\
0 & \overline L_{11}
\end{array}\right]\left[\begin{array}{cc}
\overline U_{22} & 0 \\
\overline U_{12} & \overline U_{11}
\end{array}\right]
=\overline L \overline U
$$
which is a block UL factorization! So we can in principle apply the exact same argumentation of the proof of (1) given in the preprint "module flipping" to conclude (2).

Proof of (3): This now follows immediately because conditions (1) and (2) are equivalent modulo flipping, since for any matrix $X$ holds $\operatorname{rk}(X) = \operatorname{rk}(\overline X)$, and the submatrices in the block decomposition of $\overline A$ are precisely the flips of the submatrices of the block decomposition of $A$ since
$$\begin{aligned}
\forall ij:\; (P_n^T AP_n)[\colon i,\, \colon j] = P_{i}^TA[n-i\colon,\, n-j\colon] P_{j}
\end{aligned}$$
I.e. $\overline{A}[\colon i,\, \colon j] = \overline{A[n-i\colon,\, n-j\colon]}$, which allows us to conclude
$$\begin{aligned}
   &\text{$A$ has LU factorization}
\\ &\iff \forall k\colon\, \operatorname{rk}(A[\colon k,\, \colon k]) \ge \operatorname{rk}(A[\colon k,\, \colon n]) + \operatorname{rk}(A[\colon n,\, \colon k]) -k
\\&\iff \forall k\colon\, \operatorname{rk}(\overline{A[\colon k,\, \colon k]}) \ge \operatorname{rk}(\overline{A[\colon k,\, \colon n]}) + \operatorname{rk}(\overline{A[\colon n,\, \colon k]}) -k
\\&\iff \forall k\colon\, \operatorname{rk}(\overline{A}[n-k\colon,\, n-k\colon ]) \ge \operatorname{rk}(\overline{A}[n-k\colon,\, n\colon]) + \operatorname{rk}(\overline{A}[n\colon,\, n-k\colon]) -k
\\&\iff \forall k\colon\, \operatorname{rk}(\overline{A}[k\colon,\, k\colon]) \ge \operatorname{rk}(\overline{A}[k\colon,\, n\colon]) + \operatorname{rk}(\overline{A}[n\colon,\, k\colon]) -k
\\ &\text{$\overline A$ has UL factorization}
\end{aligned}$$