text:
\begin{document}

\maketitle

\begin{abstract}By proving a topological Paley-Wiener Theorem for Riemannian symmetric spaces of non-compact type, we show that a non-zero
invariant differential operator is a homeomorphism from the space of test functions
onto its image and hence surjective when extended to the space of
distributions. \end{abstract}
\vspace{0.7cm}

\section{Introduction}\label{sec:intro}
Let $\D(\R^n)$ be the space of compactly supported smooth functions on $\R^n$. Define the Euclidean Paley-Wiener space $\H^R(\C^n)$ to be the space of holomorphic maps $\f:\C^n\to\C$ such that
$$
|\f(\lambda)|\le C_N e^{R|\Ima\lambda|}(1+|\lambda|)^{-N}.
$$
This is topologized by the seminorms $\interleave \f\interleave_N$ which are the smallest constants $C_N$ for which the estimates hold. By the Paley-Wiener Theorem, the Euclidean Fourier transform
$$
\widetilde f(\xi):=\int_{\R^n}f(x)e^{-ix\cdot\xi}dx
$$
is a linear homeomorphism $\D(\R^n)\to \bigcup_{R>0}\H^R(\C^n)$ when the spaces are given the inductive limit topology.

Now let $D\neq 0$ be a constant coefficient differential operator on $\R^n$. It is well-known that conjugation of such a differential operator with the Fourier transform is just multiplication by a (non-zero) polynomial, $P_D$. This multiplication map turns out to be a homeomorphism onto its image, and consequently this holds also for $D$. An immediate consequence of this is that the differential operator is surjective on the space of distributions and hence always admit weak solutions (for more on this see \cite{55} Chapter VII).

In the article \cite{1012} it is stated that this result may be generalized to invariant differential operators on symmetric spaces of non-compact type. However the central arguments are either rather sketchy (\cite{1012}, Lemma 8) or non-existing (\cite{1012}, Theorem 7). In this article we remedy this by providing a different approach.

\section{Notation}
First we introduce some notation. Let $X$ be a Riemannian symmetric space of non-compact type, i.e. a quotient $G/K$ where $G$ is a semisimple non-compact Lie-group with finite center and $K$ is a maximal compact subgroup. The Lie algebra $\g$ has a Cartan decomposition $\g=\mathfrak k\oplus\mathfrak p$ and we pick in $\mathfrak p$ a maximally abelian subalgebra $\a$. Define $M:=Z_K(\a)$ and $B:=K/M$ and let $\a_\C^*$ denote the complexification of the dual of $\a$. Let $\Sigma\subseteq \a^*$ denote the set of restricted roots w.r.t. $(\g,\a)$ and let $\Sigma^+$ denote the set of positive roots relative to a fixed Weyl chamber. Put $\rho:=\frac12\sum_{\lambda\in\Sigma^+}\lambda$.

We let $\D(X)$ denote the set of test functions ($C^\infty$-functions with compact support) equipped with its usual inductive limit topology, and let $\D_R(X)$ denote the closed subspace of test functions whose support is contained in the closed $R$-ball $\overline{B_R(eK)}$ around $eK\in X$.

The Fourier transform $\widetilde f:\a_\C^*\times B\to \C$ of a function $f\in \D(X)$ is defined by (see e.g. \cite{47} Chapter III)
$$
\widetilde f(\lambda,b):=\int_X f(x)e^{(-i\lambda+\rho)A(x,b)}dx
$$
where $A(gK,kM):=A(k^{-1}g)$ with $A:G\to \a$ being the Iwasawa projection from the $NAK$-decomposition of $G$.

Let $\H^R(\a_\C^*\times B)$ denote the space of smooth functions $\p:\a_\C^*\times B\to \C$ which are holomorphic on $\a_\C^*$ and which satisfy the following growth condition
\begin{equation}\label{eq:growthcond1}
\forall N\in \Z_{>0}\:\exists C_N\:\forall\lambda,b:\ |\p(\lambda,b)|\le C_N e^{R|\Ima\lambda|}(1+|\lambda|)^{-N}.
\end{equation}

Furthermore, we denote by $\H^R(\a_\C^*\times B)_W$ the subset of functions satisfying the following Weyl invariance for each $s\in W$:
\begin{equation}\label{eq:Weylinv}
\forall x\in X\,\forall\lambda\in\a_\C^*:\quad \int_B e^{(is\cdot\lambda+\rho)A(x,b)}\p(s\cdot\lambda,b)db=\int_B e^{(i\lambda+\rho)A(x,b)}\p(\lambda,b)db.
\end{equation}

\section{A Topological Paley-Wiener Theorem}
First we need a topological Paley-Wiener Theorem and in order to do so, we have to topologize the space $\H^R(\a_\C^*\times B)_W$. For this, introduce the space $\H^R(\a_\C^*, L^2(B))$ to be the space consisting of holomorphic maps $\p:\a_\C^*\to L^2(B)$ satisfying
$$
\|\p(\lambda)\|_{L^2(B)}\le C_N e^{R|\Ima\lambda|}(1+|\lambda|)^{-N}
$$
for all $N$. We define $\interleave\p\interleave_N$ to be the smallest such constant $C_N$, and we topologize $\H^R(\a_\C^*, L^2(B))$ by this family of seminorms. This turns it into a Fréchet space. The Weyl invariance \eqref{eq:Weylinv} still makes sense in this generalized setting, and thus we define $\H^R(\a_\C^*, L^2(B))_W$ to be the subset of Weyl invariant elements. This is a closed subspace and hence a Fréchet space.

We have an obvious inclusion $\H^R(\a_\C^*\times B)_W\to \H^R(\a_\C^*, L^2(B))_W$ and this inclusion turns out to be surjective:

\begin{lem}
It holds that $\H^R(\a_\C^*, L^2(B))_W=\H^R(\a_\C^*\times B)_W$ as vector spaces.
\end{lem}
\begin{proof}
For $\p\in \H^R(\a_\C^*, L^2(B))_W$ define
$$
f(x):=\int_{\a^*\times B}\p(\lambda,b)e^{(i\lambda+\rho)A(x,b)}|c(\lambda)|^{-2}dbd\lambda=\int_{\a^*}\<\p(\lambda),e^{(-i\lambda+\rho)A(x,\cdot)}\>_{L^2(B)}|c(\lambda)|^{-2}d\lambda.
$$
Obviously, $f$ is a smooth function. By examining the proof of bijectivity of $\mathcal F:\D(X)\to \H(\a_\C^*\times B)_W$ (e.g. in \cite{47} p. 278--280), it is seen that $f$ is supported in the closed $R$-ball $B_R(eK)$ and that $\widetilde f-\p=0$ almost everywhere, and thus $f$ is a smooth representative of $\f$. Furthermore $\widetilde f$ satisfies the stronger growth condition \eqref{eq:growthcond1} and hence $f\in\H^R(\a_\C^*\times B)_W$.
\end{proof}

Now $\H^R(\a_\C^*\times B)_W$ inherits the topology from $\H^R(\a_\C^*, L^2(B))_W$ (given by the seminorms $\interleave\cdot\interleave_N$), and hence it becomes a Fréchet space. Furthermore we define
$$
\H(\a_\C^*\times B)_W:=\bigcup_{R\in \Z_{>0}}\H^R(\a_\C^*\times B)_W
$$
and give it the inductive limit topology.

\begin{thm}[\textbf{Topological Paley-Wiener}]
The Fourier transform $\mathcal F:\D(X)\to \H(\a_\C^*\times B)_W$ is a linear homeomorphism. Furthermore $\widetilde f\in \H^R(\a_\C^*\times B)_W$ if and only if $f\in \D_R(X)$.
\end{thm}
\begin{proof}
The bijectivity of $\mathcal F$ as well as the last claim is stated
and proved in \cite{Helg1} Theorem III.5.1.

Now we consider $\mathcal F:\D_R(X)\to \H^R(\a_\C^*\times B)_W$ for a given $R$. For $f\in \D_R(X)$ it is
straightforward to check the inequality for each $N$:
$$
\interleave\mathcal F f\interleave_N\le
C\sup_{\lambda\in\a_\C^*,b\in B}
e^{-R|\Ima\lambda|}\int_{\overline{B_R(eK)}}|Df(x)||e^{(-i\lambda+\rho)A(x,b)}|dx
$$
where $D$ is the invariant differential operator (of order $2N$) on $X$ corresponding to the invariant polynomial $(1+|\lambda|^2)^N$ (as in \eqref{eq:diag}) and where $C$ is a constant depending on $N$ and $R$. Since $x\in
\overline{B_R(eK)}$ we have by \cite{Helg2} p. 476 eq. (13) that
$|A(x,b)|\le R$ and hence we see that
$$
e^{-R|\Ima\lambda|}|e^{(-i\lambda+\rho)A(x,b)}|=e^{(\Ima\lambda+\rho)A(x,b)-R|\Ima\lambda|}\le
e^{\rho(A(x,b))}\le e^{R|\rho|}.
$$
Hence we get
$\interleave\mathcal Ff\interleave_N\le C\|f\|_{2N}$, where
$\|\cdot\|_{2N}$ is one of the standard seminorms on $\D_R(X)$, i.e.
$\mathcal F:\D_R(X)\to \H^R(\a_\C^*\times B)_W$ is continuous.

Thus the Fourier transform is a homeomorphism $\D_R(X)\iso\H^R(\a_\C^*\times B)_W$
since these spaces are Fréchet. Hence it is also a
homeomorphism when defined on $\D(X)$.
\end{proof}

\section{Consequences of the Paley-Wiener Theorem}

Now, let $D$ be a non-zero invariant differential operator. There exists a $W$-invariant polynomial $P_D\neq 0$ on $\a_\C^*$ (this is a consequence of \cite{45} Theorem II.4.6 and Lemma II.5.14 where we identify $\mathbb D(A)$ with $W$-invariant polynomials on $\a$) such that the following diagram commutes
\begin{equation}\label{eq:diag}
\xymatrix{\D_R(X)\ar[rr]^-\sim \ar[d]_D & &\H^R(\a_\C^*\times B)_W\ar[d]^{M_{P_{D}}}\\
\D_R(X)\ar[rr]^-\sim & & \H^R(\a_\C^*\times B)_W}
\end{equation}
where $M_{P_D}$ is multiplication by $P_D$. The first goal is to show that $M_{P_D}$ and hence $D$ are linear homeomorphisms onto their images. Injectivity of $M_{P_D}$ is clear by holomorphicity since $P_D\neq 0$.

Another payoff of considering $\H^R(\a_\C^*, L^2(B))$ rather than $\H^R(\a_\C^*\times B)$, is that
it admits a nice description as a tensor product. First, however, note that the spaces we defined in Section \ref{sec:intro} actually make sense for any finite-dimensional inner product space $V$ and its complexification $V_\C$. Thus we can define $\D(V)$, $\D_R(V)$, $\H^R(V_\C)$ and so on, as well as a Euclidean Fourier transform which will be a homeomorphism $\mathcal F:\D(V)\iso \H(V_\C)$. In the following we will take $V=\a^*$ with the Killing form as inner product.

Let $\H^R(\a_\C^*)\widehat\otimes L^2(B)$ denote the completion of the
algebraic tensor product in either the projective or injective
topology (they are both equal since $\H^R(\a_\C^*)\cong \D_R(\a^*)$ is
nuclear). Then

\begin{lem}
There exists a natural linear homeomorphism
$\H^R(\a_\C^*)\widehat\otimes L^2(B)\iso\H^R(\a_\C^*, L^2(B))$.
\end{lem}
\begin{proof}
Letting $\D_R(\a^*,L^2(B))$
denote the space of smooth $L^2(B)$-valued functions on $\a^*$ with
support in the $R$-ball, we define a Fourier transform
$$
\mathcal F:\D_R(\a^*,L^2(B))\to \H^R(\a_\C^*, L^2(B))
$$
by
$$
(\mathcal F f)(\lambda):=\int_{\a^*} f(x)e^{-i\<x,\lambda\>}dx
$$
($L^2(B)$-valued integration). For all $v\in L^2(B)$ it holds that $\<\mathcal
Ff(\lambda),v\>=\mathcal F(\<f,v\>)(\lambda)$. Since the function
$x\longmapsto\<f(x),v\>$ is an element of $\D_R(\a^*)$ we see that
$\lambda\longmapsto\<\mathcal Ff(\lambda),v\>$ is holomorphic, i.e.
$\mathcal Ff$ is weakly holomorphic, hence holomorphic (as it takes
values in a Hilbert space). Furthermore we note that $\mathcal F$ has
an inverse:
$$
(\mathcal
F^{-1}\p)(x)=\int_{\a^*}\p(\lambda)e^{i\<x,\lambda\>}d\lambda.
$$
Continuity of $\mathcal F$ is easily checked and since the spaces in question are both Fréchet, $\mathcal F$ is a linear homeomorphism. The lemma
now follows from the fact that
$\D_R(\a^*,L^2(B))\cong\D_R(\a^*)\widehat\otimes L^2(B)$ (which is a consequence of \cite{52} Theorem 44.1) and that
$\D_R(\a^*)\cong \H^R(\a^*_\C)$ by the Euclidean Fourier transform.
\end{proof}

Now returning to the commuting diagram \eqref{eq:diag} we see that under the
identification of $\H^R(\a_\C^*, L^2(B))$ with
$\H^R(\a_\C^*)\widehat\otimes L^2(B)$, the map
$M_{P_D}:\H^R(\a_\C^*, L^2(B))\to \H^R(\a_\C^*, L^2(B))$ is
replaced by $M_{P_D}\otimes \id_{L^2(B)}$. And from the Euclidean
theory we know that $M_{P_D}:\H^R(\a_\C^*)\to \H^R(\a_\C^*)$ is a
homeomorphism onto its image, hence the same holds for
$M_{P_D}\otimes \id_{L^2(B)}$ (cf. \cite{52} Proposition 43.7). By restriction to $\H^R(\a_\C^*\times
B)_W$ we get:

\begin{lem}
The multiplication map $M_{P_D}:\H^R(\a_\C^*\times B)_W\to
\H^R(\a_\C^*\times B)_W$ is a homeomorphism onto its image.
\end{lem}

Finally, we need to transfer this to the inductive limit. For this we
need the following lemma which is an immediate generalization of \cite{47} Lemma III.5.13 to functions with Hilbert space
values:

\begin{lem}\label{lem:pollemma}
Assume $P:\a_\C^*\to \C$ is a nonzero polynomial and that
$\p:\a_\C^*\to L^2(B)$ is a holomorphic function such that $P\p\in
\H^R(\a_\C^*, L^2(B))$, then $\p\in \H^R(\a_\C^*, L^2(B))$.
\end{lem}

Restricting to $\H^R(\a_\C^*, L^2(B))_W=\H^R(\a_\C^*\times B)_W$ and referring to the commutative diagram \eqref{eq:diag} we get that if $f\in\D(X)$ is such that $Df\in\D_R(X)$, then $f\in
\D_R(X)$.

We now arrive at our main theorem:

\begin{thm}
Let $D\neq 0$ be an invariant differential operator, then
$D:\D(X)\to \D(X)$ is a homeomorphism onto its image.
\end{thm}
\begin{proof}
Since $D:\D_R(X)\to \D_R(X)$ is a homeomorphism onto its image,
it is clear that $D$ is injective and continuous. We just need to
show that it's open. For this, let $U\subseteq \D(X)$ be open, i.e.
$U\cap D_R(X)$ is open for all $R$. We need to show that $D(U)\cap
\D_R(X)$ is open for all $R$. Claim: $D(U)\cap \D_R(X)=D(U\cap
\D_R(X))$. The inclusion ``$\supseteq$'' is clear, since $D$
decreases support. The inclusion ``$\subseteq$'' is a consequence of
Lemma \ref{lem:pollemma}. But the right hand side $D(U\cap \D_R(X))$
is open in $\D_R(X)$, and thus the result follows.
\end{proof}

Let $\D'(X)$ denote the set of distributions over $X$, i.e. the dual of $\D(X)$. A simple Hahn-Banach argument (see e.g. \cite{55} p. 236) yields
\begin{cor}
$D:\D'(X)\to \D'(X)$ is surjective.
\end{cor}


TITLE: Is the series $\sum_{n=1}^\infty\frac{n-1}{n+\ln n}$ convergent?
QUESTION [1 upvotes]: We have the series $\sum _{n=1}^{\infty}\frac{n-1}{n+\ln n}$. Does it converge? What test should I use, the comparison test?

REPLY [0 votes]: Observe that : $\ln n < n \Rightarrow \dfrac{n-1}{n+ \ln n} > \dfrac{n-1}{2n} > \dfrac{1}{4} \Rightarrow \displaystyle \sum_{n=1}^\infty \dfrac{n-1}{n+\ln n} \geq \displaystyle \sum_{n=1}^\infty \dfrac{1}{4} = \infty \Rightarrow \text{ the given series diverges to $\infty$}$.


TITLE: Expected value of a card game, with a fixed probability of playing it again.
QUESTION [1 upvotes]: A person has to select one of three cards. One card would get him a
reward of 1 dollar. Another one would not give him any return. The
third one would give him a return of 2 dollars and also compels him
to select one of the 3 cards once again- all similar to the first set
of 3 cards. The person has no way of identifying the cards beforehand
and is equally likely to take any of the 3 cards at any stage. Find
the expected value.

REPLY [1 votes]: If we had only the first two cards, the expected value would be
$$E[X] = \frac{1}{2}(1) + \frac{1}{2}(0) = 0.5$$
With the third card, it becomes:
$$E[X] = \frac{1}{3}(1) + \frac{1}{3}(0) + \frac{1}{3}\left(2 + E[X]\right).$$
Writing it like this allows you to avoid the geometric series that is lurking behind that third card -- provided that you know $E[X]$ exists. In this case it does, but if $E[X] = \infty$, this is invalid, and you need to expand this into a geometric series.
Edit: I fixed this, I was previously treating it as though the first card were worth 1 dollar, the second card 2 dollars, and the third worth a chance to pick another card, and to play again as well.


TITLE: essential range of a function
QUESTION [0 upvotes]: Suppose $\phi$ is an invertible function in $L^\infty(T)$ where $T$ is the unit circle such that the essential range of $\phi$, $R(\phi)$ is contained in the open right half-plane.  Show there exist an $\epsilon>0$ such that $\epsilon R(\phi)\subset \{z\in\mathbb{C}:|z-1|<1\}$.
This is part of a proof.  I cannot see why it is true.  I know $R(\phi)$ is compact and does not contain 0 as $\phi$ is invertible.
For this, we have $R(\phi)=\{\lambda\in\mathbb{C}:\mu(\{x\in X:|f(x)-\lambda|<\epsilon\})>0 \ \forall \epsilon>0\}$.  Here $\mu$ is the Lebesgue measure and $X$ is the unit circle.

REPLY [1 votes]: The essential range is compact, and the sets of the form $\varepsilon^{-1}\{z\in\mathbb{C}\colon\lvert z-1\rvert<1\}$ with $\varepsilon>0$ is an open cover (of the open right half plane, and hence of $R(\phi)$). So $R(\phi)$ is contained in a finite union of these sets, and therefore in one of them since these sets are nested.


TITLE: Show that $\frac{\log 2}{\log 3}$ is irrational.
QUESTION [1 upvotes]: I found this exercise in Beachy and Blair:Abstract algebra.
Show that
$$
\frac{\log 2}{\log 3}
$$
is not a rational number.
Here is what I did:
Assume that it is rational so
$$
\frac{\log 2}{\log 3}=\frac{a}{b}, \qquad a,b\in\mathbb{Z}, b\neq 0
$$
Then we have the following
$$
\frac{\log 2}{\log 3}=\frac{a}{b} \iff
$$
$$
b\cdot \log 2=a\cdot \log 3 \iff
$$
$$
\log 2^b=\log 3^a \iff
$$
$$
2^b=3^a
$$
which is trivially false and hence the result follows.
Now my problem is that all the previous exercises are a lot "harder" and this seemed really easy. Am I correct here or missing something? I can't see a problem in my solution.

REPLY [0 votes]: It is not the point of the question, but it is not difficult to prove that $\alpha=\frac{\log 2}{\log 3}$ is trascendental, once $\alpha\not\in\mathbb{Q}$ has been proved. If we assume that $\alpha$ is algebraic over $\mathbb{Q}$ from the Gelfond-Schneider theorem we get that $3^\alpha$ is trascendental. But $3^\alpha=2\in\mathbb{Q}$, hence $\alpha$ is trascendental.


TITLE: transforming vector potential with a coordinate rotation
QUESTION [0 upvotes]: In electrodynamics, given the vector potential $\vec{A}$, the magnetic field is defined as:
$\vec{B} = \nabla \times \vec{A}$
I'm having trouble figuring out how a coordinate transformation (a rotation) affects these vectors.
If in coordinate system number 1 we have $A_x=x,A_y=0,A_z=0$. Then $\vec{B}=0$.
Now change coordinate systems by rotating.
$x' = \cos(\theta) x - \sin(\theta) y$
$y' = \sin(\theta) x + \cos(\theta) y$
And the inverse is just
$x = \cos(\theta) x' + \sin(\theta) y'$
$y = -\sin(\theta) x' + \cos(\theta) y'$
So $A'_{x'} = \cos(\theta) x,\ A'_{y'} = - \sin(\theta) x,\ A'_{z'} = 0$
$A'_{x'} = \cos(\theta) (\cos(\theta)x' + \sin(\theta)y'),\ A'_{y'} = - \sin(\theta) (\cos(\theta)x' + \sin(\theta)y'),\ A'_{z'} = 0$
But now we have
$B'_{x'}=0,\ B'_{y'}=0,\ B'_{z'}=\frac{\partial}{\partial x'} A'_{y'} - \frac{\partial}{\partial y'} A'_{x'} \neq 0$
A magnetic field shoudn't appear from no-where if I merely rotate my coordinate system.  What am I doing wrong? What is the correct way to do these transformations?

REPLY [1 votes]: A magnetic field shoudn't appear from no-where if I merely rotate my coordinate system

That's right, it should not :-)

What am I doing wrong?

It seems to be a simple sign error (or you may have to look up the contravariant transformation law for vectors, e.g. on Wikipedia here). We would like to compute $A(x', y', z')$, so remember that we have to compute, for the first component,
$$
A_1(x', y', z') = \frac{\partial x'}{\partial x} * A_1(x, y, z) +  \frac{\partial x'}{\partial y} * A_2(x, y, z) + \frac{\partial x'}{\partial z} * A_3(x, y, z) = \frac{\partial x'}{\partial x} * x = x \cos(\theta)
$$
and inserting $x = \cos(θ) x'+ \sin(θ) y'$ we get
$$
A_1(x', y', z') = \cos(\theta) ( \cos(θ) x'+ \sin(θ) y')
$$
and for the second:
$$
A_2(x', y', z') = \frac{\partial y'}{\partial x} * A_1(x, y, z) +  \frac{\partial y'}{\partial y} * A_2(x, y, z) + \frac{\partial y'}{\partial z} * A_3(x, y, z) = \frac{\partial x'}{\partial x} * x = x \sin(\theta)
$$
Inserting again $x = \cos(θ) x'+ \sin(θ) y'$ we get
$$
A_2(x', y', z') = \sin(\theta)  ( \cos(θ) x'+ \sin(θ) y')
$$
For the z-component of $B$ we now get
$$
B_{z'} = \partial_{x'} A_2(x', y' ,z') - \partial_{y'} A_1(x', y' ,z') = \sin(\theta) \cos(\theta) - \sin(\theta) \cos(\theta) = 0
$$
I hope I did not screw up :-)


text:
\begin{document}
\begin{center}
{\bf  Van der Waerden Conjecture for Mixed Discriminants
 }

\vskip 4pt
{Leonid Gurvits }\\
\vskip 4pt
{\tt gurvits@lanl.gov}\\
\vskip 4pt
Los Alamos National Laboratory, Los Alamos , NM 87545 , USA.
\end{center}

\begin{abstract}
We prove that the mixed discriminant of doubly stochastic $n$-tuples of semidefinite hermitian $n \times n$  matrices is
bounded below by $\frac{n!}{n^{n}}$ and that this
bound is uniquely attained at the $n$-tuple $( \frac{1}{n} I,...,\frac{1}{n} I )$.
This result settles a conjecture posed by R. Bapat in 1989.
We consider various generalizations and applications
of this result.
\end{abstract}

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\def\Tps{TchebysDeto be the maximafine $A(n,d)$ l size of a code with distance
$d$hef polynomials}

\newcommand{\rarrow}{\rightarrow}

\newcommand{\larrow}{\leftarrow}

\newcommand{\grad}{\bigtriangledown}

\overfullrule=0pt
\def\setof#1{\lbrace #1 \rbrace}

\section{Introduction}

An $n \times n$ matrix $A$ is called doubly stochastic if it is nonnegative entry-wise
and its every column and row sum to one. The set of $n \times n$ doubly stochastic
matrices is denoted by $\Omega_{n}$.

Let $S_{n}$ be the symmetric group, i.e. the group
of all permutations of the set $\{1,2, \cdots, n\}$.
Recall that the permanent of a square matrix A is defined by
$$ per(A) = \sum_{\sigma \in S_{n}} \prod^{n}_{i=1} A(i,
\sigma(i)). $$
The famous Van der Waerden Conjecture \cite{M78} states that
$$ min_{A \in \Omega_{n}} D(A) = \frac{n!}{n^{n}}$$ and the minimum is
attained uniquely at the matrix $J_{n}$ in which every entry equals
$\frac{1}{n}$.  The ``modern'' attack at the conjecture began in
fifties (of 20th century), see \cite{M78} for some history, and
culminated with three papers \cite{fr}, \cite{Fa81}, \cite{Eg81}.  In
a very technical paper \cite{fr} S. Friedland got very close to the
desired lower bound by proving that $ min_{A \in \Omega_{n}} D(A) \geq
e^{-n}$. \\ In \cite{Fa81} D.I. Falikman proved the lower bound
$\frac{n!}{n^{n}}$ via ingenious and custom-made arguments.  Finally,
the full conjecture was proved by G.P. Egorychev in \cite{Eg81}.  The
paper \cite{Eg81} capitalized on the simple, but crucial, observation
that the permanent is a particular case of the mixed volume or the
mixed discriminant, which we will define below.  Having in mind this
connection, the main inequality in \cite{Fa81} is just a particular
case of the famous Alexandrov-Fenchel inequalities \cite{Al38}.

Let us consider an $n$-tuple
{ \bf A } = (A$_{1}$, A$_{2}$,... A$_{n}$), where A$_{i} = (A_{i}(k,l):
1 \leq k, l \leq n)$ is a complex $n \times n$ matrix $(1 \leq i
\leq n)$.  Then $\det (\sum t_{i}A_{i})$ is a homogeneous
polynomial of degree n in $t_{1},t_{2}, \cdots, t_{n}$.  The number
\begin{equation}
D({ \bf A }):=D(A_{1}, A_{2}, \cdots, A_{n}) =
\frac{\partial^{n}}{\partial t_{1} \cdots \partial t_{n}} \det
  (t_{1} A_{1} + \cdots + t_{n}A_{n})
\end{equation}
is called the mixed discriminant of $A_{1}, A_{2}, \cdots,
A_{n}$.\\

Mixed discriminants were introduced by A.D. Alexandrov
as a tool to derive mixed volumes of convex sets (\cite{Al38}, \cite{Sch93}). They are also a 3-dimensional
case of multidimemensional Pascal's determinants \cite{pasc}.

There exist many alternative ways to define mixed discriminants.  Let $S_{n}$ be the symmetric group, i.e. the group
of all permutations of the set $\{1,2, \cdots, n\}$.  Then the
following identities hold.
\begin{equation}
D(A_{1}, \cdots A_{n})= \sum_{\sigma, \tau  \in S_{n}}
(-1)^{sgn(\sigma \tau}) \prod^{n}_{i=1} A_{i} (\sigma (i), \tau (i)).
\end{equation}
\begin{equation}
D(A_{1}, \cdots A_{n}) = \sum_{\sigma \in S_{n}} \det(A_{\sigma}),
\end{equation}
where the $i$th column of $A_{\sigma}$ is the $i$th column of
$A_{\sigma(i)}$. \\
\begin{equation}
  D(A_{1}, \cdots A_{n}) = \sum_{\sigma \in
    S}(-1)^{sgn(\sigma)} per(B_{\sigma}),
\end{equation}
where $B_{\sigma}(k,l) = A_{l}(k, \sigma(k))$.  \\
\beqn
M(A_1,...,A_N) = < (A_1  \otimes...  \otimes A_N) V, V >
\eeqn
where the
$N^{N}$-dimensional vector $V = V(i_1,i_2,...,i_n): 1 \leq i_{k} \leq N,1 \leq k \leq N $ is defined as follows:

\noindent
$V(i_1,i_2,...,i_n)$ is equal to $(-1)^{sign(\tau)}$
if there exists a permutation $\tau \in S_N $,  and equal
to zero otherwise.

It follows from the definition of the permanent
that $per(A) = D(A_{1}, \cdots A_{n})$,
where $A_{j} = Diag(A(i,j): 1 \leq i \leq n), 1 \leq j \leq n$.  \\

In a 1989 paper \cite{BA89}  R.B. Bapat defined the set
$D_{n}$ of doubly stochastic $n$-tuples.  An $n$-tuple ${ \bf A }=(A_{1},
\cdots, A_{n})$ belongs to $D_{n}$ iff the following properties hold:\\

1. \ \  $A_{i} \succeq 0$, i.e. $A_{i}$ is a positive
semi-definite matrix, $1 \leq i \leq n$.\\

2. \ \ $tr A_{i} = 1$ for $1 \leq i \leq n$. \\

3. \ \ $\sum^{n}_{i=1} A_{i} = I$, where $I$, as usual,
stands for the identity matrix.  \\

One of the problems posed in \cite{BA89} is to determine the minimum
of mixed discriminants of doubly stochastic tuples
$$ min_{A \in D_{n}} D(A) = ?$$
Quite naturally, Bapat conjectured that
$$ min_{A \in D_{n}} D(A) = \frac{n!}{n^{n}}$$
and that it is attained uniquely at ${\bf J}_{n} =:(\frac{1}{n} I,...,
\frac{1}{n} I)$.

In \cite{BA89} this conjecture was formulated for real matrices. We
will prove it in this paper for the complex case, i.e. when matrices
$A_{i}$ above are complex positive semidefinite and, thus,
hermitian. (Recall that a square complex $n \times n$ matrix $A = \{
A(i,j) : 1 \leq i,j \leq n \}$ is called hermitian if $A = A^{*} = \{
\overline{A(j,i)} : 1 \leq i,j \leq n \}$. A square complex $n \times
n$ matrix $A$ is hermitian iff $<A x, x > = <x,A x >$ for all $x \in
C^{n}$.)  One of the main tools we will use below are necessary
conditions for a local minimum under semidefinite constraints. It is
very important,in this optimizational context, that the set of $n
\times n$ hermitian matrices can be viewed as an $n^2$-dimensional
real linear space with (real) inner product $<A,B> = : tr(AB) $.

The rest of the paper will provide a proof of Bapat's conjecture. \\
(The lower bound $\frac{n!}{n^{n}}$ for real symmetric doubly stochastic
$n$-tuples was proved in \cite{GS} .)

\section { Basic Facts about Mixed Discriminants}

\noindent
{\bf Fact 1.}
\begin{equation}
D(X \alpha_{1}A_{1} Y, \cdots, X \alpha_{i}A_{i}Y, \cdots, X
  \alpha_{n}A_{n}Y) = \det(X) \cdot \det(Y) \cdot \prod^{n}_{i=1}
\alpha_{i} \cdot D(A_{1}, \cdots A_{n}).
\end{equation}

\noindent
{\bf Fact 2.}
\begin{equation}
D(x_{i}y_{i}^{\ast}, \cdots, x_{n}y_{n}^{\ast})=\det
(\sum^{n}_{i=1} x_{i}y_{i}^{\ast}).
\end{equation}

Here $x_{i}y_{i}^{\ast}$ is an $n \times n$ complex matrix of rank one,
$x_{i}$ and $y_{i}$ are $n \times 1$ matrices (column-vectors), $y^{\ast}$ is an
adjoint matrix , i.e $y^{\ast} = \overline{y^{T}}$ .\\

\noindent
{\bf Fact 3.}
$$
D(A_{1},\cdots, A_{i-1}, \alpha A + \beta B, A_{(i+1)}, \cdots
A_{n}) = \\
$$
\begin{equation}
\alpha D(A_{1}, \cdots, A_{i-1}, A, A_{i+1}, \cdots
A_{n})\\ + \beta D(A_{1}, \cdots, A_{i-1}, B, A_{i+1}, \cdots, A_{n}).
\end{equation}

\noindent
{\bf Fact 4.}

\noindent
$D(A_{1}, \cdots, A_{n}) \geq 0$ if $A_{i} \succeq 0$
(positive semidefinite), $1 \leq i \leq n$.

This inequality follows, for instance, from the tensor product representation (5).

\noindent
{\bf Fact 5.}

Suppose that $A_{i} \succeq 0, 1 \leq i \leq n$.  Then
$D(A_{1}, \cdots, A_{n}) > 0$ iff for any $1 \leq i_{1} < i_{2}
\cdots < i_{k} \leq n$\\
the following inequality holds: $Rank (\sum^{k}_{j=1} A_{i_{j}}) \geq k$  \cite{Pa85}. \\
This fact is a rather direct corollary of the Rado theorem on the rank of intersection of
a matroid of transversals and a geometric matroid, which is a particular case
of the famous Edmonds' theorem on the rank of inrersection of two matroids \cite{gr:lo:sc}.

\noindent
{\bf Fact 6.} \\

$D(A_{1}, \cdots, A_{n}) >0$ if the $n$-tuple $(A_{1},
\cdots A_{n})$ is a doubly stochastic.
This fact follows from Fact 5.

\noindent

{\bf Fact 7.} \\
\beqn
D(A_{1}, \cdots A_{i-1}, X, A_{i+1}, \cdots A_{n}) = tr (X
\cdot Q_{i} ) .
\eeqn

where the matrix $Q_{i} =: (\frac{\partial D}{\partial
  A_{i}})^{T}$ ; it follows from the tensor product representation (5) that
if all the matrices $A_{i}$ are hermitian (i.e. $A_{i}
= A_{i}^{\ast})$ then the mixed discriminant $D(A_1,...,A_n)$ is a real number and  $Q_{i} = Q_{i}^{\ast}$ also
$(1 \leq i \leq n)$. \\

All previous facts $1-\cdots-7$ are well known
(see, e.g., \cite{BA89}).  The next, Fact 8, is quite simple, but
seems to be unknown (at least to the author ) .\\

{\bf Fact 8.(Eulerian matrix identity) }\\
\beqn
 \sum^{n}_{i=1} \langle Q_{i} \omega, A_{i}^{*} \omega \rangle =   \sum^{n}_{i=1} \langle A_{i} Q_{i} \omega,  \omega \rangle =
  D(A_{1}, \cdots , A_{n}) \cdot \langle \omega, \omega \rangle,
\eeqn

where $<\omega, \omega>$ stands for the standard inner
product in $C^{n}$.\\

\prf
Consider the following identity :
\begin{equation}
D(X A_{1} , \cdots, X A_{i}, \cdots, X A_{n}) = \det(X) \cdot D(A_{1}, \cdots ,A_{n}) ,
\end{equation}
where $X$ is real symmetric nonsingular matrix .
Differentiate its left ant right sides respect to this matrix $X$:
\begin{equation}
 \sum^{n}_{i=1} \hat{Q_{i}}^{T} A_{i}^{T} =  \det(X) X^{-1} \cdot D(A_{1}, \cdots ,A_{n}).
\end{equation}

Here $\hat{Q_{i}}^{T}=\frac{\partial D}{\partial \bullet_{i}}$ evaluated at $(X A_{1} , \cdots, X A_{i}, \cdots, X A_{n})$.\\
Puting $X=I$ we get that\\
\begin{equation}
 \sum^{n}_{i=1} Q_{i}^{T}A_{i}^{T} =  D(A_{1}, \cdots ,A_{n}) \cdot I = \sum^{n}_{i=1} A_{i} Q_{i}
\end{equation}
\eprf

\section{ Basic Facts About Minimizers}

\subsection{ Indecomposability}

Following \cite{GS},\cite{GS1}  we call n-tuple $A=(A_{1}, A_{2},
\cdots A_{n})$ consisting of positive semidefinite hermitian matrices
indecomposable if  $Rank (\sum^{k}_{i=1} A_{i_{j}}) > k$ for all
$1 \leq i_{1} < i_{2}< \cdots< i_{k} \leq n $, where $1 \leq k<n$.  \\

Also, as in \cite{GS},\cite{GS1} associate with a given $n$-tuple ${\bf
  A} = (A_{1},...,A_{n})$ a set $M({\bf A})$ of $n$ by $n$ matrices
$M({\bf A},W)$, where the indices $W$ run over the orthogonal group $O(n)$,
i.e. $WW^{*} =W^{*}W = I$ ,
and for any orthogonal matrix $W$ with columns $\omega_{1},
...,\omega_{n}$ the matrix $M({\bf A},W)$ has as its $(i,j)$ entry
the inner product $<A_{j} \omega_{i}, \omega_{i}>$.  Matrices
$M({\bf A},W)$ inherit many properties of ${\bf A}$, which means that
sometimes we can simplify things by dealing with matrices and not
$n$-tuples.\\
Let us write down a few of these shared properties in the
following proposition.

\pro
\begin{enumerate}
\item
The $n$-tuple ${\bf A}$ is nonnegative (i.e. consists of positive semidefinite hermitian
matrices ) iff for all $W \in O(n)$,
the matrix $M({\bf A}, W)$ has nonnegative entries.
\item
 The $n$-tuple ${\bf A}$ is indecomposable iff for all $W \in
O(n)$, the matrix $M({\bf A},W)$ is fully indecomposable in the sense of \cite{M78}.
\item
$\sum^{n}_{i=1} a_{i}=I$ iff for all $W \in O(n)$, the
matrix $M({\bf A},W)$ is row stochastic.
\item
$tr(A_{i})=1$, for $1 \leq i \leq n$, iff for all $W \in
O(n)$, the matrix $M({\bf A},W)$ is column stochastic.
\item
Therefore ${\bf A}$ is doubly stochastic iff for all $W \in
O(n)$, the matrix $M({\bf A},W)$ is doubly stochastic. \\
\end{enumerate}
\epro

The following fact \cite{GS},\cite{GS1}  states that doubly stochastic $n$-tuples can be
decomposed into indecomposable doubly stochastic tuples.

\noindent
{\bf Fact 9.}

Let ${\bf A}=(A_{i},\cdots,A_{n})$ be a doubly stochastic
n-tuple.  Then there exists a partition $C_1 \cup \cdots \cup C_{k}$ of
$\{1, \cdots, n\}$ such that\\

1. \ \ For all $1 \leq s \leq k$
$$ dim(Im(\sum_{i \in C_{s}} A_{i}))=: dim X_{s}= |C_{s}|=:c_{s}.$$

2. \ \  The linear subspaces $X_{s} (1 \leq s \leq k)$ are
pairwise orthogonal and $C^{n} = X_{1} \oplus X_{2} \oplus
\cdots \oplus X_{s}$.  \\

It follows from the definition that, in the notation of
Fact 9, the following identity holds : \\
$$D({\bf A}) = D({\bf A}_{1}) \cdots D({\bf A}_{s}) \cdots D({\bf A}_{k}),$$ \\
where ${\bf A}_{s}$ is a doubly stochastic $c_{s}$-tuple
formed of restrictions of matrices $A_{i}(i \in C_{s})$ on the
subspace $X_{s}= Im (\sum_{i \in C_{s}}A_{i}).$\\

\subsection {Fritz John's Optimality Conditions}

Recall that we are to find the minimum of D({\bf
  A}) on $D_{n}$.  The set of doubly stochastic $n$-tuples is
characterized by the following constraints
$$\sum^{n}_{i=1} A_{i} =I, tr A_{i} = 1, A_{i} \succeq 0.$$

Notice that our constrained optimization problem is defined on the
linear space $H=:H_{n} \oplus \cdots \oplus H_{n}$, where $H_{n}$ is
the linear space of $n \times n$ hermitian matrices. We view the
linear space $H_{n}$ as a $n^2$ dimensional real linear space with the
inner product $<A,B>=tr(AB)$. This inner product extends to tuples via
standard summation.  Though a rather straightforward application of John's
Theorem \cite{John48} (see \cite{GS} ) gives the next result, we
decided to include a proof to make this paper self-contained.

\dfn
Consider a doubly stochastic $n$-tuple ${\bf A}=(A_{1}, A_{2}, \cdots, A_{n})$.
Present positive semidefinite matrices $A_i \succeq 0$ in the following block form with respect
to the orthogonal decomposition $C^n = Im(A_i) \oplus Ker(A_i)$ :
\beqn
A_i = \left( \begin{array}{cc}
		  \tilde{A_{i}} & 0\\
		    0 & 0\end{array} \right), A_i \succ 0 ; 1 \leq i \leq n.
\eeqn
Define
a cone of admissible directions as follows:
$$
K_{0}=\{ (Z_{1}, Z_{2}, \cdots, Z_{n}): \mbox{ there exists } \epsilon > 0 \mbox{ such that the tuple }
$$
$$
 (A_{1}+ \epsilon Z_{1},A_{2}+ \epsilon Z_{2}, \cdots, A_{n}+ \epsilon Z_{n})  \mbox { is doubly stochastic.}\}
$$
I.e. $K_{0}$ is a minimal convex cone in the linear space of hermitian $n$-tuples $H=:H_{n} \oplus \cdots \oplus
H_{n}$,  which contains all $n$-tuples $\{{\bf B} -{\bf A} : {\bf B} \in D_{n} \}$.
We also define the following two convex cones $K_1$,$K_2$  and one linear subspace $K_3$ of $H=:H_{n} \oplus \cdots \oplus H_{n}$ : \\

$ K_{1} = \{ (B_{1}, B_{2}, \cdots, B_{n})$, where the matrices $B_i$ are hermitian and
$$
B_i = \left( \begin{array}{cc}
		  B_{i;1,1} & B_{i;1,2}\\
		   B_{i;2,1}   & B_{i;2,2}\end{array} \right) ; Im(B_{i;2,1}) \subset Im(B_{i;2,2}),B_{i;2,2} \succeq 0,  1 \leq i \leq n.
$$

$K_{2} = \{ (B_{1}, B_{2}, \cdots, B_{n})$, where the matrices $B_i$ are hermitian and
$$
B_i = \left( \begin{array}{cc}
		  B_{i;1,1} & B_{i;1,2}\\
		   B_{i;2,1}   & B_{i;2,2}\end{array} \right) ; B_{i;2,2} \succeq 0,  1 \leq i \leq n.
$$

$$
K_{3} = \{ (C_{1}, C_{2}, \cdots, C_{n}): C_i \in H_n, \  tr(C_{i})=0 ( 1 \leq i \leq n ) \mbox{ and } C_{1}+...+C_{n}=0 \}.
$$
\edfn

\pro
\begin{enumerate}
\item
$K_{0} = K_1 \cap K_3 $.
\item
The closure $\overline{K_1} = K_2$.
\item
The closure $\overline{K_0} = K_2 \cap K_3 $.
\end{enumerate}
\epro

\prf
\begin{enumerate}
\item
Recall that a hermitian block matrix with strictly positive definite block $D_{1,1} \succ 0$
$$
D = \left( \begin{array}{cc}
		  D_{1,1} & D_{1,2}\\
		   D_{2,1}   & D_{2,2}\end{array} \right)
$$
is positive semidefinite iff $D_{2,2} \succeq 0$ and
$D_{2,2} \succeq D_{2,1}D_{1,1}^{-1}D_{2,1}^{*}$. This proves that an $n$-tuple $(B_{1}, B_{2}, \cdots, B_{n}) \in K_1$
iff there exists $\epsilon > 0$ such that $A_{i} + \epsilon B_{i} \succeq 0, 1 \leq i \leq n $. Intersection with $K_3$
just enforces the linear constraints $tr(A_i) = 1,  1 \leq i \leq n $ and $\sum_{1 \leq i \leq n} A_i = I$.
\item
This item is obvious.
\item
Clearly, the closure $\overline{K_1 \cap K_3} \subset \bar{K_1} \cap \bar{K_3} = K_2 \cap K_3 $. We need to prove the reverse
inclusion $\bar{K_1} \cap \bar{K_3} = K_2 \cap K_3  \subset \overline{K_1 \cap K_3}$. Consider the following hermitian $n$-tuple
$$
\Delta_i = \frac{1}{n} I - A_i, \Delta_i = \left( \begin{array}{cc}
		  \frac{1}{n} I - \tilde{A_{i}} & 0\\
		    0 & \frac{1}{n} I\end{array} \right) ; 1 \leq i \leq n.
$$
Clearly, the tuple $(\Delta_1,...,\Delta_n)$ is admissible. The important thing is that the $(2,2)$ blocks of matrices $\Delta_i$
are strictly positive definite. Therefore if $(B_1,...,B_n) \in K_2 \cap K_3$ then for all $\epsilon > 0$ the
tuple $(B_1 +\epsilon \Delta_1,...,B_n +\epsilon \Delta_n) \in K_{0} = K_1 \cap K_3$. This proves that
$\bar{K_1} \cap \bar{K_3} = K_2 \cap K_3  \subset \overline{K_1 \cap K_3}$ and thus that $\bar{K_0} = K_2 \cap K_3 $.
\end{enumerate}
\eprf

\thm If a doubly stochastic $n$-tuple ${\bf
  A}=(A_{1}, A_{2}, \cdots, A_{n})$ is a (local) minimizer then
there exists a hermitian matrix R and scalars $\mu_{i} (1 \leq i
\leq n)$ such that
\beqn
(\frac{\partial D}{\partial A_{i}})^{T} =: Q_{i} = R+ \mu_{i}
I+P_{i},
\eeqn
where the matrices $P_{i}$ are positive semi-definite and
$A_{i}P_{i}= P_{i}A_{i} = 0, 1 \leq i \leq n $.
\ethm

\prf

If a doubly stochastic $n$-tuple ${\bf A}$ is a (local) minimizer then $<Q_{1},Z_{1}>+ \cdots + <Q_{n},Z_{n}> \geq 0$
for all admissible tuples  $(Z_{1}, Z_{2}, \cdots, Z_{n}) \in K_{0} $.
In other words the hermitian $n$-tuple $(Q_{1}, Q_{2}, \cdots, Q_{n})$ belongs to a dual cone $K_{0}^{\prime}$.
It is well known and obvious that a dual cone $\bar{K_{0}}^{\prime}$ of a clossure is equal to $K_{0}^{\prime}$.
By Proposition 3.3 the closed cone $\bar{K_{0}} = K_2 \cap K_3 $, and the convex cones $K_2, K_3$ are closed.
Therefore, see, for instance, \cite{rok}, the closed convex dual cone $\bar{K_{0}}^{\prime} =  K_2 ^{\prime} + K_3^{\prime}$.
We get by a straigthforward inspection that
$$
K_2 ^{\prime} = \{(P_{1}, P_{2}, \cdots, P_{n}) : P_i = \left(
		  \begin{array}{cc} 0 & 0\\ 0 & \tilde{P_i}\end{array}
		  \right), \tilde{P_i} \succeq 0, 1 \leq i \leq n,
$$
$K_3 ^{\prime} = \{(R + \mu_{1}, R + \mu_{2}, \cdots, R + \mu_{n} ) $,
where the matrix $R$ is hermitian and $\mu_{i},1 \leq i \leq n $ are
real.\\ Therefore, we get that $ Q_{i}^{T} = R+ \mu_{i}I+P_{i} $, for
some hermitian matrix $R$, real $\mu_{i}$ and positive semidefinite
$P_{i} \succeq 0$ satisfying the equality $A_{i}P_{i}= P_{i}A_{i} = 0,
1 \leq i \leq n $.
\eprf

\cor
In notations of Theorem 3.4 the following identities
hold:
$$ D({\bf A}) = tr(A_{i} \cdot Q_{i}) = tr(A_{i}(R+\mu I);$$

$$<A_{i} \omega, (R+\mu_{i}I)\omega > =<A_{i} \omega, Q_{i} \omega>,
\omega \in C^{n} ;$$

$$ D({\bf A}) = \sum_{1 \leq i \leq n} <A_{i} \omega, (R+\mu_{i}I)\omega > , <\omega ,\omega> = 1 . $$
\ecor
\prf It follows directly from the identity
$A_{i}P_{i}=0$, Facts 7,8 and the hermiticity of all the matrices
involved here.
\eprf

The following simple Lemma will be used in the proof of
uniqueness.
\lem
Let us consider a doubly stochastic $n$-tuple
$$ {\bf A} = (A_{1}, A_{2}, \frac{1}{n} I \cdots,  \frac{1}{n} I),$$
where $A_{1}, A_{2} \geq 0, tr A_{1} = tr A_{2} = 1$ and $A_{1} +
A_{2} = \frac{2}{n} I$.  Then $D({\bf A}) = \frac {n!}{n^{n}} +
tr ((A_{1} - \frac{1}{n} I) \cdot (A_{1} - \frac{1}{n}
I)^{\ast}))\frac{(n-2)!}{n^{n-2}} $.\\
\elem

\prf First, notice that the matrices in this tuple commute.  Thus,
 $D({\bf A})=per(C_{ij}(1 \leq i,j \leq n)$, where the first column of
 matrix $C$ is equal to $\frac{1}{n} e +\tau$, second column to
 $\frac{1}{n} e -\tau$; all other columns are equal to $\frac{1}{n} e
 $.  Here, as usual, $e$ stands for vector of all ones; the vector
 $\tau$ consists of eigenvalues of $A_{1} - \frac{1}{n}I$. Notice
 that $\sum \tau_{i}=0$.  Using the linearity of the permanent in each
 column we get that
$$per(\alpha_{ij})= \frac {n!}{n^{n}} + Per (B),$$ where the first and
second columns of matrix $B$ are equal to $\tau$ and all others to
$\frac{1}{n} e $.

An easy computation gives that
$$ Per B = -2 (\sum_{i<j} \tau_{i}\tau_{j}) \frac
{(n-2)!}{n^{n-2}}.$$
But $0 = (\tau_{i} + \cdots +\tau_{n})^{2} = \tau_{1}^{2} + \cdots +
\tau_{n}^{2} + 2 \sum_{i<j} \tau_{i} \tau_{j}.$
Thus
$$
D({\bf A}) = per(C) = \frac{n!}{n^{n}} +
  \frac{(n-2)!}{n^{n-2}} \cdot (\sum_{i=1}^{n} \tau_{i}^{2}) =
    \frac{n!}{n^{n}} + \frac{(n-2)!}{n^{n-2}} tr ((A_{i} -
      \frac{1}{n}I)(A_{i} - \frac{1}{n}I)^{\ast})\;.
$$
\eprf

\section { Proof of Bapat's conjecture }

\thm
\begin{enumerate}
\item
$min_{{\bf A} \in D_{n}} D({\bf A}) = \frac {n!}{n^{n}}.$
\item
The minimum is uniquely attained at $J_{n} =
(\frac{1}{n}I, \frac{1}{n}I, \cdots \frac{1}{n}I).$
\end{enumerate}
\ethm

\prf
\begin{enumerate}
\item
 To make our proof a bit simpler, we will prove the first part of Theorem 4.1 by
induction.  Assume that the theorem is true for $m < n$.  (Case
$n=1$ is obvious).  Therefore, had a minimizing tuple {\bf A}
decomposed into two tuples $B_{1}$ and $B_{2}$, of dimensions
$m_{1}$ and $m_{2}$ respectively, this would imply, using Fact 9,
that $D(A)=D(B_{1})D(B_{2}) \geq \frac{m_{1}!}{m_{1}^{m_{1}}}
  \frac{m_{2}!}{m_{2}^{m_{2_{2}}}} > \frac {n!}{n^{n}}$.
The last inequality is clearly wrong as $D(A) \leq D (J_{n})=
\frac {n!}{n^{n}}$.  Thus we can assume that any minimizing tuple
is fully indecomposable.  Now apply  to a minimizing tuple
${\bf A}=(A_{1}, \cdots A_{n})$ Theorem 3.4:  there exists a
Hermitian matrix R and scalars $\mu_{1}, \cdots,\mu_{n}$ such
that $P_{i}=Q_{i}-R-\mu_{i}I \geq 0$ and $A_{i}P_{i}=0$.  From
Corollary 3.5 we get that $D(A)=tr(A_{i}(R+\mu_{i}I)$ and
$D(A)=\sum^{n}_{i=1}<A_{i} \omega, (R+
\mu_{i}I) \omega>$ for any normed vector $\omega \in C^{n}$.

Let $W^{\ast} RW=Diag (\theta_{i}, \cdots,\theta_{n})$ for some
  real $\theta_{i}$ and unitary W.  As in Proposition 3.1, define a doubly
  stochastic matrix $B=M(A,W)$.  i.e. $b_{ij} = <A_{i}
    \omega_{j}, \omega_{j}>.$ Here $\omega_{j}$ is a $j$th column
  of W ( or $j$th eigenvector of R).  Writing identities
  $D(A)=tr(A_{i}(R+\mu_{i}I))$ and $\sum^{n}_{i=1} <A_{i}
  \omega_{j}, (R+ \mu_{i}I)\omega_{j}>=D(A) (1 \leq j \leq n)$ in
  terms of the matrix $B$, we obtain the following  systems of linear equations
$$\mu_{i} + \sum^{n}_{i=1} b_{ij} \theta_{j} = D(A) (1 \leq i \leq n) ;$$
$$\theta_{j} + \sum^{n}_{i=1} B_{ij} \mu_{i} = D(A) (1 \leq j \leq n).$$
These equations are, of course, encountered also in the matrix
case, where they led to the crucial London's Lemma  \cite{lon} ,\cite{M78},
\cite{Fa81},\cite{Eg81}.  Proceeding
in exactly the same way, we easily deduce that $\mu = (\mu_{1},
\cdots,\mu_{n})$ is an eigenvector of $BB^{T}$ with eigenvalue 1;
$\theta$ is an eigenvector of $B^{T}B$ with eigenvalue 1.

It follows from Proposition 3.1 that $B$ is fully indecomposable,
which is equivalent to the fact that $1$ is a simple eigenvalue for
both $BB^{T}$ and $B^{T}B$ (see e.g. \cite{M78} ).
Therefore  both $\mu$ and $\theta$ are
proportional to the vector $e$ (all ones).  Thus $\mu_{i} =
\alpha$, $\theta_{i}= \beta$ and $\alpha+\beta = D({\bf A})$. \\
Returning to matrices $R+ \mu_{i}I$ we get that $R + \mu_{i} I =
D({\bf A}) \cdot I$.

Now comes the punch line,
a generalization of London's Lemma \cite{lon} to mixed
discriminants:

For a minimizing tuple ${\bf A}=(A_{i} \cdots
A_{n})$ the following inequality holds:
\begin{equation}
(\frac{D(A)}{\partial A_{i}})^{T} =: Q_{i} \succeq R+ \mu_{i}I=D({\bf
  A}) \cdot I.
\end{equation}
Indeed, by Theorem 3.4 $Q_{i} = R+ \mu_{i}I +P_i$ and $P_i \succeq 0$.
After the inequality (16) is established, we are back to
familiar grounds of the Van der Waerden's conjecture proofs \cite{Fa81},
\cite{Eg81}.  Let us introduce the following notations:
$$
{\bf A}^{i,j}=(A_1,...,A_i,...,A_i,...,A_n), f_{i,j}({\bf A})=\frac{{\bf A}^{i,j} + {\bf A}^{j,i}} {2}.
$$

I.e., we replace jth matrix in the tuple by the ith one to define ${\bf A}^{i,j}$,
replace ith and jth matrices by their arithmetic average to define $f_{i,j}({\bf A})$.

The celebrated Alexandrov-Fenchel inequalities state that

1. \ \ $D({\bf A})^{2} \geq D({\bf A}^{i,j}) \cdot D({\bf
  A}^{j,i})$. \\
2. \ \ If $A_{i} \succ 0 (1 \leq i \leq n)$ and $D({\bf A})^{2} = D({\bf A}^{i,j})
\cdot D({\bf A}^{i,j})$ \\
then $A_{i} = \tau A_{j}$ for some positive $\tau$.

 Suppose that ${\bf A}$ is a minimal doubly stochastic $n$-tuple, i.e. that $D({\bf A}) = min_{{\bf B} \in D_{n}} D({\bf B})$.
Then for $i \neq j$, $f_{i,j}({\bf A})$ is also minimal doubly
stochastic $n$-tuple.  Indeed, double stochasticity is obvious.

By the linearity of mixed discriminants in each matrix argument we get that
$D(f_{i,j} ({\bf A})) = \frac{1}{2} D({\bf A}) + \frac{1}{4}
D({\bf A}^{i,j}) + \frac{1}{4} D({\bf A}^{j,i})$.
Also,
\beqn D({\bf A}^{i,j}) = tr (A_{i} \cdot Q_{j}), D(A^{j,i}) = tr
(A_{j} \cdot Q_{i})\;. \eeqn  (We used here Fact 7).

As $Q_{k} \succeq D(A)\cdot I (1 \leq k \leq n)$ from inequality (16) and $tr A_{k}
\equiv 1$ then $D({\bf A}^{i,j}) \geq D(A)$ and $D(A^{j,i}) \geq D(A).$
Thus, we conclude based on Alexandrov-Fenchel's inequalities that
if {\bf A} is a minimal doubly stochastic n-tuple then $D({\bf
  A}^{i,j}) \equiv D({\bf A}) (i \neq j)$ and $D(f_{i,j} (A)) =
D({\bf A})$.

Additionally, if a minimal tuple ${\bf A} = (A_{1}, \cdots,
A_{n})$ consists of positive definite matrices then $A_{i}=A_{j}
(i \neq j)$.  Indeed $A_{i}=\tau A_{j}$ and $tr A_{i} =
tr A_{j}=1$, thus $ \tau= 1$.  As $A_{1} + \cdots + A_{n} = I$, we
conclude that the only ``positive'' minimal doubly stochastic
n-tuple is $J_{n} = (\frac{1}{n}I, \cdots, \frac{1}{n} I)$.  In
any case, as $D({\bf A}^{i,j}) \equiv D({\bf A})(i \neq j)$, then
$D(f_{i,j}({\bf A})) = D({\bf A})$ for minimal tuples {\bf A}. \\

Define the following iteration on n-tuples:
$${\bf A}_{o} = {\bf A},$$
$${\bf A}_{1} = f_{1,2}(A_{o})$$
$$ \cdots $$
$${\bf A}_{n-1} = f_{n-1,n} (A_{n-2})$$
$${\bf A}_{n} = f_{1,2} (A_{n-1})$$
$$ \cdots. $$\\
It is clear that $${\bf A}_{k} \rightarrow \left( \frac{A_{1} + \cdots
  A_{n}}{n}, \cdots, \frac{A_{1} + \cdots A_{n}}{n} \right).$$

As the initial tuple satisfies $A_{1} + \cdots + A_{n} = I$, then
$A_{K} \rightarrow (\frac{1}{n}I, \cdots \frac{1}{n}I)$.
(Indeed $ Pr_{1} = f_{1,2},...,Pr_{n-1} = f_{n-1,n}$ are orthogonal projectors in the linear finite-dimensional Hilbert
space of hermitian $n$-tuples. By a well known result,
$\lim_{k \rightarrow \infty} (Pr_{n-1}...Pr_{1})^{k} = Pr$,
where $Pr$ is an orthogonal projector on the linear subspace $L = Im(Pr_{1}) \cap Im(Pr_{2}) \cap...\cap Im(Pr_{n-1})$.
It is obvious that $L = \{(A_1,...,A_n) : A_1 = A_2 =... = A_n \}$. )
As the mixed discriminant is a continuous map from tuples to reals,
we conclude that
$J_{n} = (\frac{1}{n}I, \cdots \frac{1}{n}I)$ is a minimal tuple
and $min_{{\bf A} \in D_{n}} D({\bf A}) = D(J_{n}) =
\frac{n!}{n^{n}}$.\\

\item
Let us now prove the uniqueness.  We only have to prove that any
minimal doubly stochastic n-tuple consist of positive definite
matrices.  Suppose that not.  Then there exists an integer $k
\geq 0$ such that the tuple ${\bf A}_{k}$ has at least one
singular matrix and the next tuple ${\bf A}_{k+1}$ consists of
  positive definite matrices.\\

Assume without loss of generality that ${\bf A}_{k+1} = f_{1,2}
  ({\bf A}_{k})$ and ${\bf A}_{k} = (A_{1}, A_{2}, A_{3}, \cdots,
  A_{n})$.  \\

Then $\frac{A_{1} + A_{2}}{2} = A_{3} = \cdots = A_{n} =
\frac{1}{n} I.$  From Lemma 3.6 we conclude that \\
$$
D({\bf A}_{k}) = D(J_{n}) +  \frac{(n-2)!}{n^{n-2}}\cdot tr ((A_{2} -\frac{1}{n}I) \cdot (A_{2} - \frac{1}{n}I)^{\ast}).
$$

As we already know that $J_{n}$ is a minimal tuple and ${\bf
  A}_{k}$ is also a minimal tuple, thus $A_{1}=A_{2} =
\frac{1}{n}I$.  But at least  one of $A_{1}, A_{2}$
suppose to be singular. We got the desired contradiction.\\
\end{enumerate}
\eprf

\cor
 Let us define the set $D_{n,P}$ of hermitian $n$-tuples as follows
$$
 D_{n,P}= \{ {\bf A}=(A_{1}, A_{2}, \cdots, A_{n}) :  A_{i} \mbox{ is positive semi-definite }, tr(A_{i}) \equiv 1 \mbox{ and }
 \sum^{n}_{i=1} A_{i} = P.
$$
Notice that $tr(P)=n$ necessarily.  If the matrix $P$ is sufficiently
close to the identity matrix $I$ then $ min_{A \in D_{n,P}} D(A) =
\frac{n!}{n^{n}}\det(P)$. \ecor

\prf
It follows from the uniqueness part of Theorem 4.1 that if $P$ is sufficiently close to the identity matrix $I$
then there exists a minimal $n$-tuple ${\bf A}=(A_{1}, A_{2}, \cdots, A_{n}) \in D_{n,P}$ consisting of positive definite
matrices. Very similarly to Theorem 3.4 ,
it follows that $Q_{i}=R + \mu_i I (1 \leq i \leq n ), \mu_i \mbox{ is real and } R \mbox{ is hermitian }$.
It is straightforward to prove that under this condition $D(f_{i,j} (A)) =D({\bf A})$.
 Also, $f_{i,j} (A) \in D_{n,P}$.
Indeed,
\beqn D({\bf A}^{i,j}) = tr(A_{i} \cdot Q_{j})=tr(A_{i} \cdot (R + \mu_j I) \mbox{ and } D({\bf A}) = tr(A_{i} \cdot (R + \mu_i I)\;.\eeqn
Thus, $D({\bf A}^{i,j})=tr(A_{i})(\mu_j-\mu_i)+D({\bf A})=D({\bf A})+(\mu_j-\mu_i).$\\
Therefore,
$$
D(f_{i,j} ({\bf A})) = \frac{1}{2} D({\bf A}) + \frac{1}{4}D({\bf A}^{i,j}) + \frac{1}{4} D({\bf A}^{j,i})= D({\bf A}).
$$
As in the proof of Theorem 4.1 , the last equality leads to the
minimality of the tuple $( \frac{P}{n}, \cdots, \frac{P}{n})$.
\eprf

\section{ Motivations and Connections }

The author came across Bapat's conjecture because of the following
theorem \cite{GS},\cite{GS1}.
\thm
Let us consider an indecomposable $n$-tuple $ {\bf A} =(A_1,...,A_n)$ consisting of positive semi-definite
matrices. Then

1. There exist a unique vector $\alpha$ of positive scalars $\alpha_i; 1 \leq i \leq n $ with product
equal to $1$ and a positive definite matrix $S$, such that the
 $n$-tuple $ {\bf B} =(B_1,...,B_n)$, defined by
$B_i=\alpha_i S A_i S $, is doubly stochastic.

2. The vector $\alpha$ above is the unique minimum of $det (\sum
t_{i}A_{i})$ on the set of positive vectors with product equal to $1$
and $\min_{x_i > 0, \prod_{i=1}^N x_i=1} \det (\sum x_i A_i) =
(\det(S))^{-2}$.  \ethm

Let us define the following important quantity, the capacity of ${\bf A}$:
$$
Cap ({\bf A}) = \inf_{x_i > 0, \prod_{i=1}^N x_i=1} \det (\sum x_i
A_i). $$

Using Theorem 4.1 and Theorem 5.1 we get the following inequality \cite{GS}, \cite{GS1}:

\begin{equation}
        1 \leq \frac {Cap({\bf A})} {D({\bf A})} \leq \frac {n^{n}}{n!}.
\end{equation}

This last inequality played the most important role
in \cite{GS}, \cite{GS1}.  With many other technical details it led to a
deterministic poly-time algorithm to approximate mixed volumes of
ellipsoids within a simply exponential factor.  In the following
subsection we will use it to obtain a rather unusual extension of
the Alexandrov-Fenchel inequality.

\subsection{Generalized Alexandrov-Fenchel inequalities }

Define $L_n$ as a set of all integer vectors $\alpha = (\alpha_1, \ldots,
\alpha_n)$ such
that $\alpha_i \geq 0$ and $\sum_{i=1}^n \alpha_i=n$.\\

For an integer vector
$$
\alpha = (\alpha_1, \alpha_2, \ldots, \alpha_n): \alpha_i \geq 0, \ \sum
\alpha_i = N \
 $$
we define a matrix tuple
$$
{\bf A}^{(\alpha)} = (\underbrace{A_1, \ldots, A_1}_{\alpha_1}, \ldots,
\underbrace{A_k,
\ldots, A_k}_{\alpha_k}, \ldots, \underbrace{A_n, \ldots, A_n}_{\alpha_n}) \,
$$
i.e., matrix $A_i$ has $\alpha_i$ copies in ${\bf A}^{(\alpha)}$.  For a vector
$x=(x_1,\ldots, x_n)$, we define a monomial $x^{(\alpha)} = x_1^{\alpha_1} \ldots
x_n^{\alpha_n}$.

We will use the notations $M^{(\alpha)}$ for the mixed discriminant $D({\bf A}^{(\alpha)})$ and
$Cap^{(\alpha)}$ for the capacity $Cap({\bf A}^{(\alpha)})$.

\thm
Consider a tuple $A=(A_1, \ldots, A_n)$ of semidefinite hermitian $n \times n$
matrices.  If vectors
$\alpha$, $\alpha^1, \ldots, \alpha^m$ belong to $L_n$ and
$$
\alpha = \sum_{i=1}^n \gamma_i \alpha^i; \quad \gamma_i \geq 0 \,, \quad
\sum \gamma_i =
1$$
then the following hold inequalities hold:
\beqn
\log ( Cap^{(\alpha)} ) \geq \sum \gamma_i \log ( Cap^{(\alpha^i)} ),
\eeqn
\begin{equation}
\log ( M^{(\alpha)} ) \geq \sum \gamma_i \log ( M^{(\alpha^i)} ) -\log (\frac {n^{n}}{n!}).
\end{equation}
\ethm

\prf
Notice that the inequality (21) follows from (20)  via a direct application
of the inequality (19). It remains to prove (20).\\

First, using the arithmetic/geometric means inequality, we get that
 $Cap^{(\alpha)}\geq d $ iff
$$
\log ( \det (\sum_{i=1}^n A_i \alpha_i e^{x_i}) ) \geq \sum
\alpha_i x_i + \log d
$$
for all real vectors $x= (x_1, \ldots, x_n)$.

Now, suppose that
$$
\alpha, \alpha^1, \ldots, \alpha^m \in L_n $$
and
$$
\alpha = \sum_{i=1}^m \gamma_i \alpha^i, \quad \gamma_i \geq 0, \quad
\sum_{i=1}^m
\gamma_i = 1.
$$
Then
$$
\log ( \det ( \sum_{i=1}^n A_i \alpha_i^j e^{x_i}) ) \geq
\langle
\alpha^j, x \rangle + \log ( Cap^{(\alpha^j)} ).$$
Multiplying each of the inequalities above by the corresponding $\gamma_i$
and adding
afterwards we get that
$$
\sum_{j=1}^m \gamma_j \log (\det ( \sum_{i=1}^n A_i \alpha_i^j
e^{x_i} ) )
\geq \langle \alpha, X \rangle + \sum_{j=1}^m \gamma_i \log ( Cap^{(\alpha^j)} ).
 $$
As $\log (\det(X))$ is concave for $X \succ 0$, we eventually get the inequality
$$
\log( Cap^{(\alpha)} ) \geq \sum_{j=1}^m \gamma_j \log ( Cap^{(\alpha^j)} ).
$$
\eprf

The Alexandrov-Fenchel inequalities  can be written as
$$
\log ( M^{(\alpha)} ) \geq   \frac{\log ( M^{(\alpha^{1})} ) +  \log ( M^{(\alpha^{2})} )}{2}.
$$
Here the vector $\alpha=(1,1,...,1); \alpha^{1}=(2,0,1,...,1) ; \alpha^{2}=(0,2,1,...,1)$.
Perhaps the extra $-\log (\frac {n^{n}}{n!})$ in Theorem 5.2 is just an artifact of our proof?
We will show below that an extra factor is needed indeed.

Define $AF(n)$ as the smallest( possibly infimum ) constant one has to subsract from $\sum \gamma_i \log ( M^{(\alpha^i)} )$
in the right
side of (21) in order to get the inequality. Then by Theorem 5.2 , $AF(n) \leq \log(\frac {n^{n}}{n!}) \cong n$.
We will prove below that $AF(n) \geq n \log(\sqrt{2})$ even for tuples consisting of diagonal
matrices. In this diagonal case the mixed discriminant coincides with the permanent.

Consider the following $N \times N$ matrix with nonnegative entries:
$$
B = \left( \begin{array}{ccccc}
1 & 1\ddots & \ddots & 0 \\[-0.2cm]
0&1 \ddots & \ddots & \ddots \\[-0.2cm]
0 & \ddots & \ddots \\[-0.2cm]
1 & & & \ddots & 1 \end{array} \right) \, \mbox{ i.e. } B = I+ J \,
$$
where $J$ is a cyclic shift.  Assume  without a ``big" loss of
generality that $N =
2k$.  Define
$$
\alpha^1 = (\underbrace{2,2,2,...,2}_k,0,0,...,0), \quad \alpha^2 =
(\underbrace{0,0,0,...,0}_k, 2,2,
\ldots, 2) \; $$
Then,$e =\frac{\alpha^1 + \alpha^2}{2} , e = (1,1,\ldots, 1) $ and
$per(B^{(e)}) = 2 $,
$per( B^{(\alpha^1)}) = per( B^{(\alpha^2)}) = 2^k =
2^{\frac{N}{2}}.$
(Here the notation $B^{(\alpha)}$ stands for the matrix having $\alpha_{i}$ copies of $i$th column
of the matrix $B$.)
Thus,
$$
\frac{per(B^{(e)})}{\sqrt{per( B^{(\alpha^1)} \cdot per( B^{(\alpha^2)}}}
= \frac{2}{2^{\frac{N}{2}}} \cong \sqrt{2} ^{-N}. $$
\cnj
$$
\lim_{n \rightarrow \infty} \frac{AF(n)}{n} = 1.$$
\ecnj

\section{Further analogs of van der Waerden conjecture }
\subsection{ $4$-dimensional Pascal's determinants }
The following open question has been motivated by the author's
study of the quantum entanglement \cite{stoc}.

Consider a block matrix

\beqn
\rho = \left(\begin{array}{cccc}
		  A_{1,1} & A_{1,2} & \dots & A_{1,n}\\
		  A_{2,1} & A_{2,2} & \dots & A_{2,n}\\
		  \dots &\dots & \dots & \dots \\
		  A_{n,1} & A_{n,2} & \dots & A_{n,n}\end{array} \right),
\eeqn
where each block is a $n \times n$ complex matrix. Define

\beqn
QP(\rho) =:  \sum_{\sigma \in S_n} (-1)^{sign(\sigma)}D(A_{1,\sigma(1)},...,A_{n,\sigma(n)}),
\eeqn
where $D(A_1,...,A_n)$ is the mixed discriminant. If instead of the block form (22), to present
$\rho = \{\rho(i_1,1_2,i_3,i_4) : 1 \leq i_1,1_2,i_3,i_4 \leq n \}$, e.g. as a $4$-dimensional tensor,
then
\begin{eqnarray}
QP(\rho) & = & \frac{1}{N!}\sum_{\tau_{1}, \tau_{2}, \tau_{3}, \tau_{4} \in S_N}(-1)^{sign(\tau_{1}\tau_{2}\tau_{3}\tau_{4})} \nonumber \\
   & &\prod_{i=1}^N \rho(\tau_{1}(i),\tau_{2}(i),\tau_{3}(i), \tau_{4}(i)).
\end{eqnarray}

In other words, $QP(\rho)$ is, up to $\frac{1}{N!}$ factor, equal to the  $4$-dimensional Pascal's determinant \cite{pasc}.

Call such a block matrix $\rho$ doubly stochastic if the following conditions hold:
\begin{enumerate}
\item
$\rho \succeq 0$, e.g. the $n^2 \times n^2$ matrix $\rho$ is positive semidefinite.
\item
$\sum_{1 \leq i \leq n} A_{i,i} = I $.
\item
The matrix of traces $\{ tr(A_{i,j} ) : 1 \leq i,j \leq n \} = I$.
\end{enumerate}
A positive semidefinite block matrix $\rho$ is called separable if
$\rho = \sum_{1 \leq i \leq k < \infty} P_i \otimes Q_i$, where the
matrices $P_i \succeq 0, Q_i \succeq 0 : 1 \leq i \leq k$ ; nonseparable
positive semidefinite block matrices are called entangled.  Notice
that in the block-diagonal case $QP(\rho) = D(A_{1,1},...,A_{n,n})$
and our definition of double stochasticity coincides with double
stochasticity of $n$-tuples from \cite{BA89}. Let us denote the closed
convex set of doubly stochastic $n \times n$ block matrices as
$BlD_{n}$, a closed convex set of separable doubly stochastic $n
\times n$ block matrices as $SeD_{n}$.  It was shown in
\cite{stoc} that $\min_{\rho \in BlD_{n}} QP(\rho) = 0$ for $n \geq
3$ and , on the other hand , $\min_{\rho \in SeD_{n}} QP(\rho) > 0$ for $n
\geq 1$.  \cnj $\min_{\rho \in SeD_{n}} QP(\rho) = \frac{n!}{n^n}.$
\ecnj

It is easy to prove this conjecture for $n=2$ , moreover the following equalities hold :
$$
\min_{\rho \in BlD_{2}} QP(\rho) = \min_{\rho \in SeD_{2}} QP(\rho) = \frac{2!}{2^2} = \frac{1}{2} .
$$

\subsection{Hyperbolic polynomials}
The following concept  of hyperbolic polynomials originated in the theory of partial differential equations \cite{gar}.
\dfn
A homogeneous polynomial $p(x), x \in R^m$ of degree $n$ in $m$ real varibles is called hyperbolic in the direction $e \in R^m$
(or $e$- hyperbolic) if for any $x \in R^m$ the polynomial
 $p(x - \lambda e)$  in the one variable $\lambda$ has exactly $n$ real roots counting their multiplicities. We assume in this
paper that $p(e) > 0$.

Denote the ordered vector of roots of $p(x - \lambda e)$ as
$\lambda(x) = (\lambda_{1}(x) \geq \lambda_{2}(x)
\geq... \lambda_{n}(x)) $. It is well known that the product of roots
is equal to $p(x)$. Call $x \in R^m$ $e$-positive ($e$-nonnegative) if
$\lambda_{n}(x) > 0$ ($\lambda_{n}(x) \geq 0$).  Define $tr_{e}(x) =
\sum_{1 \leq i \leq n} \lambda_{i}(x)$.

A $k$-tuple of vectors
$(x_1,...x_k)$ is called $e$-positive ($e$-nonnegative) if $x_i, 1
\leq i \leq k$ are $e$-positive ($e$-nonnegative).  Let us fix $n$
real vectors $x_i \in R^m, 1 \leq i \leq n$ and define the following
homogeneous polynomial: \beqn P_{x_1,..,x_n}(\alpha_1,...,\alpha_n) =
p(\sum_{1 \leq i \leq n} \alpha_i x_i) \eeqn Following \cite{khov}, we
define the $p$-mixed value of an $n$-vector tuple ${\bf X} =
(x_1,..,x_n)$ as
\beqn M_{p}({\bf X}) = : M_{p}(x_1,..,x_n) =
\frac{\partial^n}{\partial \alpha_1...\partial \alpha_n} p(\sum_{1
\leq i \leq n} \alpha_i x_i)
\eeqn
\noindent
Finally, call an $n$-tuple of real
$m$-dimensional vectors $(x_1,...x_n)$ $e$-doubly stochastic if it is
$e$-nonnegative. $tr_{e} x_i = 1 (1 \leq i \leq n)$ and $\sum_{1
\leq i \leq n} x_i = e $.  Denote the closed convex set of $e$-doubly
stochastic $n$-tuples as $HD_{e,n}$.  \edfn

\xam
Consider the following  homogeneous polynomial $p(\alpha_1,...,\alpha_n) = \det(\sum_{1 \leq i \leq n} \alpha_i A_i)$.
If $A_i \succeq 0 : 1 \leq i \leq n $ and $\sum_{1 \leq i \leq n} A_i \succ 0$ then $p(.)$ is hyperbolic
in  the direction $e$, where $e$ is a vector of all ones. If $\sum_{1 \leq i \leq n} A_i = I$ then $e$-double stochasticity
of $n$-tuple ${\bf X}= (e_1,e_2,...,e_n)$ of $n$-dimensional canonical axis vectors is the same as double stochasticity of
$n$-tuple of matrices ${\bf A} = (A_1,...,A_n)$. Moreover $M_{p}({\bf X}) = D({\bf A})$.
\exam

It was proved in a very recent paper \cite{hyp} that $\min_{{\bf X} \in HD_{e,n}} M_{p}({\bf X}) > 0 $.

\cnj
$\min_{{\bf X} \in HD_{e,n}} M_{p}({\bf X}) = p(e) \frac{n!}{n^n} $
\ecnj

\section{Acknowledgements}

It is my great pleasure to thank my friend and colleague Alex
Samorodnitsky. Without our many discussions on the subject this paper
would have been impossible.  I got the first ideas for the proof of Bapat's
conjecture during my Lady Davis professorship at Technion (1998-1999).  \
Many thanks to that great academic institution.

config: arxiv


TITLE: A Car Pushes a Truck up to Speed on the Highway
QUESTION [2 upvotes]: I have a disagreement with my teacher, we recently reviewed a problem where a car was pushing a disabled truck up to speed on the highway, and then pushes the truck at a constant speed, and asked what the forces exerted on the truck by the car were and vice versa.

I argued that when the car was pushing the truck up to speed, the car must be exerting more force on the truck than the truck on the car, because in order to have an acceleration there must be net force.
My teacher argued that due to Newtons 3rd law that the forces must be the same.
We both agreed that during the second part when there was constant speed the forces were equal and opposite.

Who is right?

REPLY [1 votes]: If $F$ is the force exerted by the engine of the car, then let the contact forces between the car and the truck be $N$. So if the car was pushing the truck up to speed and both were accelerating at an acceleration of $a$. Then according to the free body diagram of the car,
$$F-N=M_ca$$
Then according to the free body of the truck
$$N=M_Ta$$
Hence $$F=(M_c+M_T)a$$
Herein, as the car and the truck were speeding up the highway, the forces exerted were equal and opposite, $N$.
When they are up to speed and maintaining a constant speed, $a=0$, hence $N=0$ and also $F=0$. The forces acting on each other are always $N$ (equal and opposite) according to Newton's 3rd Law.

REPLY [1 votes]: Your teacher is right that the force exerted on the truck and the car are equal, according to Newton's 3rd law.
However, you are not wrong in thinking that, in order for there to be an acceleration, there needs to be a net force. When the truck pushes back on the car, that does not detract from the force that the truck feels; in other words, when the truck exerts a force on the car, that force is NOT included in the truck's free-body diagram.
Following this, the truck is happily accelerating because it has a net force on it. The car is also accelerating though, right? It has a force being applied to it in the opposite direction, so there must be some other force to offset that. That force is from the wheels on the pavement, pushing the car forward like you are used to.
Picture:
[Car][Truck]--Accel.-->
Free-body diagrams:
[Truck]--1-->
<--1--[Car]----2---->
Note that the force exerted by the car on the truck (--1-->) and the force exerted by the truck on the car (<--1--) are equal and opposite, following Newton's 3rd law. Force 2 is the car pushing against the ground.
Note: In this case, we do not need the net forces on the car and truck to be equal to each other. In fact, since they are accelerating at the same rate, and the car has a smaller mass, the net force the car feels will in face be smaller than the net force the truck feels.
$$
a_t=\frac{F_t}{m_t}
$$
$$
a_c=\frac{F_c}{m_c}
$$
and
$$
a_c=a_t,
$$
so we get
$$
\frac{F_c}{m_c}=\frac{F_t}{m_t}
$$


TITLE: Eigenvectors and generalized Eigenvectors not same!!
QUESTION [1 upvotes]: Let $A = \begin{bmatrix}
0 & 1 & 0\\
0 & 0 & 1\\
4 & 0 & -3\\
\end{bmatrix}$. One can see that the corresponding eigenvalues are $\{-2,-2,1 \}$. Finding the eigenvectors we see that we get one one eigen vector corr to $-2$ which is $\begin{bmatrix}
1\\
-2\\
4\\
\end{bmatrix}$ and the eigenvector corr to $1$ is $\begin{bmatrix}
1\\
1\\
1\\
\end{bmatrix}$. We have eigenvalue $-2$ of multiplicity $2$ but got only one independent eigen vector so the matrix $A$ is defective. So we use method to fing generalized eigen vector.
But when I calculate $(A+2I)^2(x)=0$ to find eigen vectors I get a completely different set of eigen vectors $\{\begin{bmatrix}
1\\
0\\
-4\\
\end{bmatrix} , \begin{bmatrix}
0\\
1\\
-4\\
\end{bmatrix} \}.$
But how is this this possible??!!

Having a great confusion. Please Help!

REPLY [2 votes]: There is no problem. So when you solve $(A+2I)^2X=0$, you find two generalized eigenvectors $(1,0,-4)$ and $(0,1,-4)$. Now you don't know that one of these has to be an eigenvector, but you do know that there is an eigenvector in the space spanned by these two vectors. And indeed, $(1,-2,4)=(1,0,-4)-2(0,1,-4)$.
Generalized eigenvectors generalize eigenvectors in the sence that the space spanned by eigenvectors belonging to some eigenvalue is contained in the space spannen by the corresponding generalized eigenvectors.


TITLE: Need Help With Proof Regarding Prime Numbers
QUESTION [1 upvotes]: some friends and I are stuck on a proof regarding primes.
It goes as follows:

Let $p_1, p_2$ be primes with $p_1 < p_2$. Show that there is a $n \in
> \mathbb{N}$ such that $p_1 + n \cdot (p_2 - p_1)$ is prime, but $p_2 +$
  $n \cdot (p_2 - p_1)$ is not. $Hint:$ Chose a small $n$.

We tried several things but didn't come close to a solution.
I'd appreciate any help, approaches or solutions.
Thanks.

REPLY [4 votes]: First let's try $n=1$.  Of course $p_2=p_1+1\times (p_2-p_1)$ is prime.  If $2p_2-p_1$ is composite we are done, so assume it is prime.
Accordingly, try $n=2$.  Under the assumption  that $2p_2-p_1$ is prime, we now ask if $3p_2-2p_1$ is composite.  If it is composite, we are done, so assume it is prime.
Continue in this manner.  At each point we ask whether $np_2-(n-1)p_1$ is composite.  The first composite of this form would be an example of what you want (though a priori they might all be prime).
To see that they can not all be prime, note that the sequence of test cases $\{2p_2-p_1,3p_2-2p_1,\cdots\}$ is just the sequence $2p_2-p_1+k(p_2-p_1)$  It is easy to see that no arithmetic progression can contain only primes (take $k=2p_2-p_1$ for example).  Thus this sequence contains infinitely many composites, so we are done.  (the first composite in the list gives you your example).
Worth remarking that this argument only shows existence.  Well, it shows that there has to be an example with $n≤2p_2-p_1$ but other than that it does not actually produce an example.


TITLE: How to prove that solution have at most $m$ fractional coordinates
QUESTION [1 upvotes]: I started to study a bit of mixed linear programming, and I am facing the following exercise that after quite some time I don't know how to approach:

Let $A\in\mathbb{R}^{m,n}$, $b\in\mathbb{R}^{m}$, $c\in\mathbb{R}^{n}$
and $u\in\mathbb{Z}^{n}$. Prove that if the problem
$\max\{c^{T}x|Ax\leq b,\ 0\leq x\leq u\}$ is feasible and bounded,
then you have a solution with at most $m$ fractional coordinates.

Could you help me? Greetings!

REPLY [0 votes]: Assuming you are familiar with the concept of basic feasible solution (BFS), convince yourself that an optimal solution exists, express it as a BFS, and count the basic and nonbasic variables.


text:
\begin{document}

\begin{center}
{\Large
	{\sc  Scale matrix estimation under data-based loss in high and low dimensions}
}
\bigskip

 Mohamed Anis Haddouche $^{1}$,\, Dominique Fourdrinier $^{2}$ \& Fatiha Mezoued $^{3}$
\bigskip

{\it
$^{1}$ Université de Normandie, INSA Rouen, UNIROUEN, UNIHAVRE, LITIS, avenue de l’Université, BP 8, 76801 Saint-Étienne-du-Rouvray, France.    mohamed.haddouche@insa-rouen.fr et  \'Ecole Nationale Sup\'erieure de Statistique et d\textquoteright
\'Economie Appliqu\'ee (ENSSEA), LAMOPS, Tipaza, Algeria.

$^{2}$ Université de Normandie, UNIROUEN, UNIHAVRE, INSA Rouen, LITIS, avenue de l’Université, BP 12, 76801 Saint-Étienne-du-Rouvray, France.  dominique.fourdrinier@univ-rouen.fr

$^{3}$ \'Ecole Nationale Sup\'erieure de Statistique et d\textquoteright
\'Economie Appliqu\'ee (ENSSEA), LAMOPS, Tipaza, Algeria.\\
famezoued@yahoo.fr
}
\end{center}
\bigskip

{\bf R\'esum\'e.}

Nous considérons  le problème d'estimation de la matrice d'échelle $\Sigma$ du modèle additif $Y_{p\times n} = M + \mathcal{ E}$, du point de vue de la théorie de la décision. Ici, $p$ représente  le nombre de variables, $n$ le nombre d'observations, $M$ une matrice de paramètres inconnus de rang $q <p$ et  $\mathcal{ E}$  un bruit aléatoire  de distribution à symétrie  elliptique, de matrice de covariance proportionnelle à $I_n \otimes \Sigma$. Ce problème d'estimation est abordé sous une représentation canonique  où la matrice d'observation $Y$ est décomposée en deux matrices, à savoir, $Z_{q\times p}$ qui résume l'information contenue dans $M$ et une matrice $U_{m\times p}$, où $m=n-q$, qui résume  l'information suffisante pour l'estimation de $\Sigma$.

Comme les estimateurs naturels de la forme $  {\hat \Sigma} _a = a \, S $ (où $  S=U^{\T}\,U $ et $ a $ est une constante positive) se comportent mal lorsque  $p > m$ ($S$ n'est pas inversible), nous proposons des estimateurs alternatifs de la forme
$ {\hat {\Sigma}} _ {a, G} = a \big (  S +  S \,  {S ^ {+} G (Z,S)} \big) $ où ${S^{+}} $ est l'inverse de Moore-Penrose de $ S$ (qui coïncide avec l'inverse $S^{-1}$ lorsque $S$ est inversible). Nous fournissons  des conditions sur la matrice de correction $SS^{+} {G (Z,S)} $ telles que $  {\hat {\Sigma}} _ {a,  {G}} $ améliore  $  {\hat {\Sigma}} _a $ sous le coût basé sur les données $L _S( \Sigma, \hat { \Sigma})
= \tr \big ( S^{+}\Sigma\,({\hat{\Sigma}} \,  {\Sigma} ^ {- 1} -  {I}_ {p} )^{2}\big) $. Nous adoptons une approche unifiée des deux cas où $  S $ est inversible  ($p \leq m$) et $  S $ est non inversible ($p > m$).

{\bf Mots-cl\'es.} Distribution à symétrie elliptique, coût basé sur les données, identité de type Stein-Haff , matrice de covariance, matrice d'échelle.
\vspace{0.5cm}

\medskip

{\bf Abstract.}

We consider  the problem of estimating the scale matrix $\Sigma$ of the additif model  $Y_{p\times n} = M + \mathcal{E}$, under a theoretical decision point of view. Here, $ p $ is the number of variables, $ n$ is the number of observations, $ M $ is a matrix of unknown parameters with rank $q<p$ and   $ \mathcal {E}$ is  a random noise, whose distribution is elliptically symmetric  with covariance matrix  proportional to $ I_n \otimes \Sigma $\,.

We deal  with a canonical form of this model where $Y$ is decomposed in two matrices, namely,  $Z_{q\times p}$ which summarizes the information contained in $ M $, and $ U_{m\times p}$, where $m=n-q$, which summarizes  the sufficient information  to estimate $ \Sigma $.

As the natural estimators of the form ${\hat {\Sigma}}_a=a\,  S$ (where $ S=U^{\T}\,U$ and  $a$ is a positive constant) perform poorly when $p >m$ (S non-invertible), we propose estimators of the form
${\hat{\Sigma}}_{a, G} = a\big(  S+  S \,   {S^{+}\,G(Z,S)}\big)$ where ${S^{+}}$ is the Moore-Penrose inverse of $ S$ (which coincides  with $S^{-1}$ when $S$ is invertible).  We provide conditions on the correction matrix $SS^{+}{G(Z,S)}$ such that ${\hat {\Sigma}}_{a,  G}$ improves over ${\hat {\Sigma}}_a$ under the data-based loss $L _S( \Sigma, \hat { \Sigma}) =\tr \big ( S^{+}\Sigma\,({\hat{\Sigma}} \,  {\Sigma} ^ {- 1} -  {I}_ {p} )^ {2}\big) $.

We adopt a unified approach of the two cases where $ S$  is invertible ($p \leq m$) and $ S$ is non-invertible ($p>m$).

{\bf Keywords.} Eliptically symmetric distributions, data-based loss, Stein-Haff type identity, covariance matrix, scale matrix.

\bigskip\bigskip

\section{Introduction}

Consider  the following additive model

\begin{align}\label{additive.model}
{Y}={M}+  {\mathcal{E}}  ,\hspace{1.cm} {\mathcal{E}}  \sim  {ES}  ({0}_{np},I_n\otimes \Sigma),
\end{align}

where ${Y}$ is an observed $n\times p$ matrix, ${M}$ denotes an  $n\times p$ matrix of  unknown parameters and ${\mathcal{E}}$ is an $n\times p$ elliptically symmetric distributed noise with unknown covariance matrix proportional to $I_n\otimes \Sigma$, where $\Sigma $ is an unknown $p\times p$   invertible scale matrix and $I_n$ is the $n$-dimensional identity matrix.
Note that, the class of  elliptically symmetric distributions encompasses a large number of important distributions such as Gaussian, Cauchy, exponential, Student and   Weibull distributions. Our main assumption is that $ M$ is of low-rank, that is,

\begin{align}\label{low-rank}
{\rm rank}( M)= q < p
\end{align}

Note that Model (\ref{additive.model}) is a common alternative representation of the multivariate low-rank regression model $Y = X\,{\beta} + \mathcal{E}$, where $X$ is an $n\times q$ matrix of known constants of rank $q < p$ and ${\beta}$ is an $q\times p$ matrix of unknown parameters.
In the Gaussian setting, Model (\ref{additive.model}) arises in many fields that require to estimate $M$ as in signal processing, image processing, collaborative filtering. Thus, it has been considered by various authors such as  Cand\`es and Recht (2009), Ji et al. (2010) and Cand\`es et al. (2013).
Recently,  Canu and Fourdrinier (2017) introduced the extended elliptical setting in Model (\ref{additive.model}).
It is worth noting that many estimation procedures of $M$ rely on an accurate estimation of the scale matrix $\Sigma$, which is the aim of this paper.

Thanks to the low-rank assumption in  (\ref{low-rank}), there exists a $n\times n$ orthogonal matrix $Q=(Q_1 Q_2)$, with $Q_2^{\T}\,M = 0$, so that  the canonical form of Model (\ref{additive.model}) is given by

\begin{align}\label{canonical.model}
Q^{\T}\,Y=
\begin{pmatrix}
{Q_1^{\T}} \\ {Q_2^{\T}}
\end{pmatrix}
\,Y
=
\begin{pmatrix}
{Z} \\ {U}
\end{pmatrix}
=
\begin{pmatrix}
{\theta} \\ {0}
\end{pmatrix}
+ Q^{T }\,{\mathcal{E}},
\end{align}

where $ Z$ and $ U$ are respectively $q\times p$ and $m\times p$ matrices (\textit{cf.} Fourdrinier and Canu~(2017)  for more details). Note that, the canonical form $(\ref{canonical.model})$ separates information about the mean structure $Z$ and the information concerning the scale $U$, since $ S=U^{T}\,U$   summarizes  the   information to estimate $ \Sigma $.
Now, we restrict our attention to the setting where   the joint density of ${Z}$ and ${U}$ is of the form

\begin{align}\label{density}
({z},{u})
&\mapsto

|\Sigma|^{-n/2} \,f \! \left[\,\tr \{({z}
- {\theta})\Sigma^{-1}({z}-
{\theta})^{\T }\} + \tr \{\Sigma^{-1}\,{u}^{\!\T }
{u}\}\,\right]\,,
\end{align}

for some function $f$.

In the following, $\Exp{}$ will denote the expectation with respect to the density  (\ref{density}) and $E^{\ast}_{\theta,\Sigma}$ the expectation with respect to the density

\begin{align*}
({z},{u}) \mapsto \frac{1}{K^{*}}|\Sigma|^{-n/2}\,F^{*}
\! \left[\,\tr\{\,({z} -
{\theta})\Sigma^{-1}({z}-{\theta})^{\top}\} + \tr\{\Sigma^{-1}\,u^{\T}\,u\}\,\right]\,,
\end{align*}

where

${F^{*}(t) =\frac{1}{2} \, \int^{\infty}_{t} f(\nu) \,d\nu}$  and the normalizing constant $K^{*}$ is assumed to be finite. Note that, in the setting of a multivariate normal distribution, since $F^{*}=f$,
these two expectations coincide.

As mentioned by James and Stein (1961),  the natural estimators of the form  ${\hat {\Sigma}}_a=a\,  S$ (where  $a$ is a positive constant) perform poorly. Therefore, we consider  alternative estimators   of the form $\hat{\Sigma}_{{a},G}= a\,(S + SS^{+}G(Z,S))$ and we derive dominance results under the data-based loss function

\begin{align}\label{loss}
L_{S}(\hat{\Sigma},\Sigma) = \tr\big(S^{+}\Sigma\,\big({\hat\Sigma}\Sigma^{-1}- I_{p}\big)^{2}\big),
\end{align}

and its associated risk

\begin{align}\label{risk}
R(\hat{\Sigma},\Sigma)
=\Exp{} [L_S(\hat{\Sigma},\Sigma)]\,,
\end{align}

where $\hat{\Sigma}$ is an estimator of $\Sigma$ and $S^{+}$ is the Moore-Penrose inverse of $S$. It is worth  noticing that this type of loss function, called data-based since it involves $ S ^ {+} $, was introduced by Efron and Morris (1976). Since then, it was considered by various authors, in a Gaussian setting by Kubokawa and Srivastava (2008) and Tsukuma and Kubokawa (2015), and in a spherical setting by Fourdrinier and Strawderman (2015).

The two main features of our approach is that we consider the general elliptically symmetric distribution context and we unify the two cases where $S$ is non-invertible ($p>m$) and $S$ is invertible ($p \leq m$). The primary decision-theoretic results are presented in Section \ref{Improved estimators}. More precisely, we derive a sufficient condition on the correction matrix function $SS^{+}G(Z,S)$ for which  $\hat{\Sigma}_{{a},G}$ improves on $\hat{ \Sigma}_{a}$ under the data-based loss in  (\ref{loss}). In Section \ref{Simulation}, we provide  numerical results through simulations.

\section{Main result}\label{Improved estimators}
Among the  usual estimators $\hat{\Sigma}_a=a\,S$, there exists $a_o >0$ such that $\hat{\Sigma}_{a_o}$ is optimal (that is, the risk of $\hat{\Sigma}_{a_o}$ is less than or equal to the risk of $\hat{\Sigma}_{a}$, for any $a>0$); this is

\begin{align*}
a_o = \frac{1}{K^{*}(p \vee m)}\,,
\end{align*}

where $p\vee m = \max\{p,m\}$ (\textit{cf.} Haddouche (2019) for a proof).

The improvement over the class of $a\,S$'s will be shown through the improvement of

\begin{align}\label{alternative.esti}
\hat{\Sigma}_{{a_o},G}= a_o\,(S + SS^{+}G(Z,S))\,,
\end{align}

over $\hat{ \Sigma}_{a_o}=a_o\,S$,
where $$G(Z,S)=\frac{t}{\tr(S^{+})}\,SS^{+}$$ and $t$ is a positive constant. Note that the choice of this specific form of $G(Z,S)$ is motivated by the estimator considered by Konno (2009) in the normal case.

We give sufficient conditions on the corrected factor $SS^{+}G({Z},S)$, that is on the constant $t$, such that the risk difference

\begin{align*}
\Delta(G) = R(\hat{\Sigma}_{{a_o},G}, \Sigma) - R(\hat{\Sigma}_{a_o} , \Sigma)
\end{align*}

between $\hat{\Sigma}_{{a_o},G}$ and $\hat{\Sigma}_{a_o}$  is non-positive. Of course, $\Delta(G) \leq 0$ makes only sense if and only if $R(\hat{\Sigma}_{{a_o},G}, \Sigma) < \infty$. It is shown in Haddouche (2019) that this occurs as soon as the expectations $\Exp{}\left[\|S^{+}\,G \|^{2}_F\right]
$, $\Exp{}\left[\|\Sigma^{-1}SS^{+}G \|^{2}_F\right]
$, $\Exp{} \left[\tr (\Sigma\,S^{+})\right]$ and  $\Exp{} \left[\tr (\Sigma^{-1}\,S)\right]$ are finite. In that case,

\begin{align}\label{Delta.SG.1.S}
\Delta(G)
&=
a_{o}^2\,K^{*}\,\Exp{}\big[\tr\left(\Sigma^{-1}SS^{+}G\left\{I_p+ S^{+}G + SS^{+}	\right\}\right)\big]
-2\,a_{o}\,\Exp{}\big[\tr\left(S^{+}\,G\right)\big]\,.
\end{align}

The dependence of the risk difference in (\ref{Delta.SG.1.S}) on the unknown parameter $\Sigma^{-1}$ is
problematic. As a remedy, we apply the  Stein-Haff type identity in the framework of elliptically symmetric distribution given in Fourdrinier Haddouche and Mezoued~(2019).

\begin{lemma}
	Let $G(z,s)$ be a $p \times p$ matrix function such that, for any fixed $z$, $G(z,s)$ is weakly differentiable with respect to $s$. Assume that $\Exp{} \left[\big| \tr \left(\Sigma^{-1} S \,S^{+}\,G\right) \big| \right] < \infty$. Then we have

	\begin{align*}
	\Exp{}\left[\tr\left(\,\Sigma^{-1}\,SS^{+}\,G\right)\right]
	=
	K^{*}\,\Exp{*}\left[\tr \left(
	2\,SS^{+}\,{\cal D}_s
	\{SS^{+}G\}^{\T}\, + (m-(p\wedge m)-1)\,S^{+}\,G\right)\,\right].
	\end{align*}

\end{lemma}

Thanks to this identity, sufficient conditions for improvement of
$\hat{\Sigma}_{a_o,G}$ over $\hat{\Sigma}_{a_o}$, are given in the following
theorem  (\textit{cf.} Haddouche (2019) for a proof) through an upper bound of the risk difference in (\ref{Delta.SG.1.S}).

\begin{theorem}
	Consider a density of the form (\ref {density}). Let

	\begin{align}\label{Est.Konno.New}
	\hat{\Sigma}_{a_o,G} =  a_o\left( S + \frac{t}{\tr(S^{+})}\, SS^{+} \right)
	\end{align}

	where $t$ is a positive constant. Then $\hat{\Sigma}_{a_o,G}$ improves over  $\hat{\Sigma}_{a_o}$ as soon as

	\begin{align*}
	0 \leq t
	\leq \frac{2\,((p\wedge m) -1 ) }{(p\vee m) - (p \wedge m) + 1}\,.
	\end{align*}
where $p \wedge m = \min\{p,m\}$\,.
\end{theorem}

\section{Numerical study}\label{Simulation}
We deal here with the non-invertible case ($p>m$) for  a Gaussian distribution ($K^{*}=1$) where the scale matrix have an autoregressive structure of the form $(\Sigma)_{ij} = 0.9^{|i-j|}$. Note that simulation on the Student distributions are under study. We evaluate numerically the performance of the alternative estimator $\hat{\Sigma}_{a_o,G}$ in (\ref{Est.Konno.New}) where ${a_o}=1/p$ and $t= 2\,(m-1)/(p-m+1)$, through  the  percentage relative improvement in average loss PRIAL of $\hat{\Sigma}_{a_o,G}$ over $\hat{\Sigma}_{a_o}$ defined as
\begin{align*}
\text{PRIAL}(\hat{\Sigma}_{a_o,G})= \frac{\text {average loss of }\,\,\hat{ \Sigma}_{a_o}- \text {average loss of}\,\, \hat{\Sigma}_{a_o,G} }{\text {average loss of} \,\,\hat{ \Sigma}_{a_o}}\, \times 100\,,
\end{align*}
which is reported in the following table.
\vspace{.5cm}

	\begin{tabular*}
		{1\textwidth}{@{\extracolsep{\fill}}  c  c  c  }
		\hline $p$& $m$ &PRIAL\,(\%)   \\
		\hline
		20 & 4 	 &15.00  \\
		20 & 8 	  &18.56  \\
		20 & 12   &25.56  \\
		20 & 16   &47.034  \\

		100 & 20    &3.39  \\
		100 & 40    &4.19  \\
		100 & 60    &5.76  \\
		100 & 80    &10.42  \\
		\hline
	\end{tabular*}
		 \begin{center}
Results of 1000 Monte Carlo simulation for $(\Sigma)_{ij} = 0.9^{|i-j|}$.
			\end{center}
For $p=20$ and $p=100$, the PRIAL increases with the values of $m$. Note that, when $p=20$ and $m=16$ the PRIAL is close de $50\,\%$. Note that the data-based Loss is much more discriminant then the usual quadratic loss for which the PRIAL is lower.

\newpage
\section*{Bibliographie}

\noindent Cand{\`{e}}s, E. J. and  Sing{-}Long, C. A. and  Trzasko, J. D. (2013).  Unbiased Risk Estimates for Singular Value Thresholding and Spectral Estimators.\textit{{IEEE} Transactions on signal processing} 61:4643-4657

\noindent Cand{\`e}s, E. J. and Recht, B. (2009). Exact Matrix Completion via Convex Optimization, \textit{Foundations of Computational Mathematics}, 9(6): 717.

\noindent Efron, B. and Morris, C. (1976). Multivariate empirical Bayes and estimation of covariance matrices. \textit{Annals of Statistics.} 4(1):22-32

\noindent Fourdrinier, D. and Canu, S. (2017). Unbiased risk estimates for matrix estimation in the elliptical case, \textit{Journal of Multivariate Analysis}, 158, pp. 60-72

\noindent Foudrinier, D. Haddouche, M. A. and Mezoued, F. (2019).  Scale matrix estimation of an elliptically symmetric distribution in high and low dimensions, \textit{ Université de Rouen Normandie et ENSSEA Tipaza,Technical report}

\noindent Fourdrinier, D. and Strawderman, W.E. (2015). Robust minimax {S}tein estimation under invariant data-based loss for spherically and elliptically symmetric distributions. \textit{Metrika.} (4)78:461-484

\noindent Haddouche, M.A. (2019). Scale matrix estimation. \textit{Ph.D disertation, Normandie Universit{\'e}}. Chapter 4

\noindent James, W. and Stein, C. (1961). Estimation with Quadratic Loss. \textit{Proceedings of the Fourth Berkeley Symposium on Mathematical Statistics and Probability, Volume 1: Contributions to the Theory of Statistics.} 361--379

\noindent  Ji, H. and Liu, C. and Shen, Z. and Xu, Y. (2010). Robust video denoising using low rank matrix completion  \textit{2010 IEEE Computer Society Conference on Computer Vision and Pattern Recognition.} 100:2237-2253

\noindent Kubokawa, T. and Srivastava, M.S. (2008). Estimation of the precision matrix of a singular {W}ishart distribution and its application in high-dimensional data. \textit{Journal of Multivariate Analysis.} (9)99:1906-1928

\noindent Konno, Y. (2009). Shrinkage estimators for large covariance matrices in multivariate real and complex normal distributions under an invariant quadratic los. \textit{Journal of Multivariate Analysis} 100:2237-2253

\noindent Tsukuma, H. and Kubokawa, T. (2015). A unified approach to estimating a normal mean matrix in high and low dimensions. \textit{Journal of Multivariate Analysis} 139:312-328

\end{document}


TITLE: Showing the inverse map of a ring homomorphism of a prime ideal is again a prime ideal
QUESTION [4 upvotes]: Let $\phi : A \rightarrow B$ be a ring homomorphism and $I$ be a prime ideal of $B$.
  (i) Show that $\phi^{-1}(I)$ is a prime ideal of $A$, and
  (ii) find an example of $A$, $B$ and $I$ so that $I$ is a maximal ideal, but $\phi^{-1}(I)$ is not (a maximal ideal).

(i)
Let $x,y \in A$ with $xy \in \phi^{-1}(I)$
$\Rightarrow \exists z \in I: \phi(x) \phi(y) = \phi(xy) = z \in I$
$\Leftrightarrow x \in \phi^{-1}(I)$ or $y \in \phi^{-1}(I) \Rightarrow \phi^{-1}(I)$ is a prime ideal of A.
Can you please check my answer, is it (formally) correct?
(ii)
I have no clue how the example could look like, can you please help me with this?

REPLY [3 votes]: (ii): Take any integral domain $D$ which is not a field and let $k$ be it's field of fractions. Consider the natual inclusion $i :D[t] \to k[t].$ The ideal $M = tk[t]$ is a maximal ideal in $k[t].$ But $i^{-1}(M) = tD[t]$ is not a maximal ideal in $D[t].$


TITLE: The Laplace transform $\mathcal L$ is a self-adjoint operator on $L^2(\mathbb R_+)$
QUESTION [4 upvotes]: Let $\mathbb R_+ = [0, \infty)$ and consider the kernel $K(x,y) = e^{-xy}$ on $\mathbb R_+ \times \mathbb R_+$. The associated integral operator on $L^2(\mathbb R_+)$ is called the Laplace transform $\mathcal L$, i.e.
$$(\mathcal L f)(x) = \int_0^\infty K(x,y) f(y)\, dy$$
Show that the Laplace transform $\mathcal L$ is a self-adjoint operator on $L^2(\mathbb R_+)$.

This question is similar to another question that I'd asked a few days ago, but the same approach doesn't work since the kernel is not square-integrable - so Fubini's theorem may not be applicable.

My work:
We want to show $\langle \mathcal Lf, g\rangle = \langle f, \mathcal Lg\rangle$ for all $f,g\in L^2(\mathbb R_+)$. We have
$$\begin{align*}
\langle \mathcal Lf,g\rangle &= \int_0^\infty \mathcal Lf(x) \overline{g(x)} \, dx\\
&= \int_0^\infty \int_0^\infty e^{-xy} f(y) \overline{g(x)}\, dy\, dx
\end{align*}$$
and
$$\begin{align*}
\langle f, \mathcal L g\rangle &= \int_0^\infty f(x) \overline{\mathcal Lg(x)}\, dx\\ &= \int_0^\infty \int_0^\infty e^{-xy} f(x) \overline{g(y)} \, dy\, dx\\&=
\int_0^\infty \int_0^\infty e^{-xy} f(y) \overline{g(x)} \, dx\, dy
\end{align*}$$
So, we must show $$\int_0^\infty \int_0^\infty e^{-xy} f(y) \overline{g(x)}\, dy\, dx =  \int_0^\infty \int_0^\infty e^{-xy} f(y) \overline{g(x)} \, dx\, dy $$
This can be done easily if Fubini's theorem is applicable. For Fubini's theorem, we require $$\int_0^\infty \int_0^\infty |e^{-xy}| |f(y)| |g(x)| \, dx\, dy < \infty$$
By applying Cauchy-Schwarz inequality twice (as in the linked post's answer), I was able to get $$\int_0^\infty \int_0^\infty |e^{-xy}| |f(y)| |g(x)| \, dx\, dy \le \|f\|_2 \|g\|_2 \left(\int_0^\infty \int_0^\infty e^{-2xy}\, dx\, dy \right)^{1/2}$$
Unfortunately, $$\int_0^\infty \int_0^\infty e^{-2xy}\, dx\, dy$$ diverges, so this bound is useless.
Two possibilities arise: (i) either Fubini's theorem is applicable, and we must find a stronger bound to show $\int_0^\infty \int_0^\infty |e^{-xy}| |f(y)| |g(x)| \, dx\, dy < \infty$, or (ii) we must show that the two integrals are equal using some technique other than Fubini's theorem. To me, (ii) seems less likely.
I'd appreciate any help in completing the proof. Thank you!

Note:

$\mathcal L$ is a bounded bijective operator on $L^2(\mathbb R_+)$, in fact, $\| \mathcal L\| = \sqrt\pi$. The explicit calculations can be found in Rajendra Bhatia's Notes on Functional Analysis, Pg. $26-27$, and also in Setterqvist E., Unitary Equivalence: A New Approach to the Laplace transform and the Hardy operator. Master’s Thesis, Department of Mathematics Lule ̊a University of Technology, 2005:329 CIV here. Further, see Post 1, Post 2 and Post 3.

REPLY [3 votes]: Use an approximation argument.  Since you know that $\mathcal{L}$ is bounded, it suffices to show $\langle \mathcal{L}f, g \rangle = \langle f, \mathcal{L}g \rangle$ for $f,g$ in a dense subspace of $L^2$.  For instance, you can assume that $f,g$ are bounded and have compact support, in which case the hypothesis of Fubini's theorem is obviously satisfied.


text:
\section{The hypersurface case}
\label{sec:hypersurfaces}

Fix a toric Fano manifold $Y$, a line bundle $L$, and a pair of distinct sets $S$ and $S' \subset \{1,\ldots, R\}$ -- using the notation introduced in \S\ref{sec:introduction} -- such that
\[
L = \sum_{i \in S}{D_i} = \sum_{j \in S'}{D_j}.
\]
Given a ray generator $v_i$ of $\Sigma$, we let $X_i$ denote the variable in the total homogeneous co-ordinate ring of $Y$ corresponding to $e_i \in \ZZ^R$; recalling that $v_i$ is equal to $\rho(e_i)$. Crucially, we have that both $\prod_{i \in S}{X_i}$ and $\prod_{j \in S'}{X_j}$ are elements of $\Gamma(Y,L)$. In other words, the monomial
\[
\frac{\prod_{i \in S}{X_i}}{\prod_{j \in S'}{X_j}}
\]
is a rational function on $Y$, regular on $\TT_N$. Since $M = \Hom(N,\ZZ)$ is the character lattice of $\TT_N$, there is a linear function $w \colon N \to \ZZ$ such that
\begin{equation}
\label{eq:weight_vector}
w(v_i) =
\begin{cases}
1 & \text{if $i \in S\setminus S'$}\\
-1 & \text{if $i \in S' \setminus S$} \\
0 & \text{otherwise.}
\end{cases}
\end{equation}

\begin{rem}
	\label{rem:weight_vectors}
	We note that the construction of $w$ is not restricted to the hypersurface case: given $(Y;L_1,\ldots, L_c)$, and a pair of nef partitions $\{S_i : 1 \leq i \leq c+1\}$ and $\{S'_i : 1 \leq i\leq c+1\}$, there is a unique $w_i$ for each $i \in \{1,\ldots c\}$ such that $w_i(v_j) = 1$ if $j \in S_i\setminus S'_i$, $w_i(v_j) = -1$ if $j \in S'_i \setminus S_i$, and such that $w_i$ evaluates to zero on all other ray generators.
\end{rem}

The second geometric ingredient which will play a key role in the proof of Theorem~\ref{thm:main_result} are certain \emph{integral reflections}.

\begin{dfn}
	Given a linear function $w \colon N \to \ZZ$, fix a pair of vectors $u^{\pm} \in w^{-1}(\pm1)$ respectively. We let $T_{u^+,u^-}$ denote the unique $\GL_n(\ZZ)$ transformation which fixes $w^{-1}(0)$ and sends $u^\pm$ to $u^\mp$. We refer to such lattice isomorphisms as integral reflections.
\end{dfn}

The isomorphism $T_{u^+,u^-}\colon N \to N$ defines an isomorphism $\CC[N] \to \CC[N]$, which we also denote by $T_{u^+,u^-}$. We next recall the notion of \emph{(algebraic) mutation}, as described by Akhtar--Coates--Galkin--Kasprzyk~\cite{ACGK}. These maps were introduced by Galkin--Usnich~\cite{Galkin--Usnich} as \emph{mutations of potential}.

\begin{dfn}
	\label{dfn:mutation}
	Fix a lattice vector $w \in M \cong \ZZ^n$, and a Laurent polynomial $F \in \CC[w^\bot]$. We let $\mu_{w,F} \colon \TT_N \dashrightarrow \TT_N$ be the birational map defined by setting
	\[
	\mu^\star_{w,F}(z^v) = z^vF^{-\langle w,v\rangle}
	\]
	for each $v \in N$. Given a Laurent polynomial $f \in \CC[N]$ such that $\mu^\star_{w,F}f \in \CC[N]$ we say that $\mu_{w,F}$ defines a mutation of $f$.
\end{dfn}

We recall that, for any $w$ and $F$ as in Definition~\ref{dfn:mutation}, $\mu^\star_{w,F}(\omega_n) = \omega_n$; that is, algebraic mutations are volume preserving birational maps, or -- in the terminology used by Katzarkov--Przyjalkowski in~\cite{Katzarkov--Przyjalkowski} -- \emph{symplectomorphisms of cluster type}. Following \cite{ACGK}, we refer to $w$ as the \emph{weight vector} and $F$ the \emph{factor} of the mutation $\mu_{w,F}$.

\begin{pro}
	\label{pro:hypersurface_case}
	Theorem~\ref{thm:main_result} holds in the case $c=1$, setting $L_1 = L$.
\end{pro}
\begin{proof}
	We first let $w \in M$ be the vector given in \eqref{eq:weight_vector}, and choose elements $u^\pm$ in $w^{-1}(\pm 1)$. Consider the mutation $\mu_{w,F}$ of $W = z^{v_1} + \cdots + z^{v_R}$ where $F$ is given by
	\[
	F := \frac{1}{z^{u^+}}\Big(\sum_{i \in S\setminus S'}{z^{v_i}}\Big).
	\]
	We compute that
	\[
	W_1 := \mu^\star_{w,F}W = z^{u^+} + \frac{1}{z^{u^+}}\sum_{i \in S\setminus S'}{z^{v_i}}\sum_{j \in S'\setminus S}{z^{v_j}} + \sum_{i \notin S \ominus S'}{z^{v_i}}.
	\]
	Next consider the mutation $\mu_{-w,F'}$ of $W_1$ where
	\[
	F' := \frac{1}{z^{u^-}}\Big(\sum_{i \in S'\setminus S}{z^{v_i}}\Big).
	\]
	Setting $W_2 := \mu^\star_{-w,F'}W_1$, we compute that
	\[
	W_2 = \frac{z^{u^+}}{z^{u^-}}\sum_{j \in S'\setminus S}{z^{v_j}} + \frac{z^{u^-}}{z^{u^+}}\sum_{i \in S\setminus S'}{z^{v_i}} + \sum_{i \notin S \ominus S'}{z^{v_i}}.
	\]
	Moreover, we compute that
	\begin{align*}
		\mu_{-w,F'}^\star\circ \mu^\star_{w,F}\Big(\sum_{i \in S\setminus S'}{z^{v_i}}\Big) &= \mu_{-w,F'}^\star z^{u^+}\\
		&= \frac{z^{u^+}}{z^{u^-}}\Big(\sum_{i \in S'\setminus S}{z^{v_i}}\Big).
	\end{align*}
	Finally, since $w(v_i) = 1$ for any $i \in S \setminus S'$, we have that $T_{u^+,u^-}(v_i) = v_i - u^+ + u^-$; that is, $T_{u^+,u^-}(z^{v_i}) = \frac{z^{u^-}}{z^{u^+}}z^{v_i}$. Similarly, for any $i \in S' \setminus S$, $T_{u^+,u^-}(v_i) = v_i - u^- + u^+$. Thus the composition $\varphi := T_{u^+,u^-} \circ \mu^\star_{-w,F'} \circ \mu^\star_{w,F}$ satisfies the conditions required by the statement of Theorem~\ref{thm:main_result}.
\end{proof}

In order to exploit the symmetry of the situation, the proof of the general case of Theorem~\ref{thm:main_result} is a generalisation of a rephrased version of the above proof. Indeed, let $G := \sum_{i \in S\setminus S'}{z^{v_i}}$ and let $G' = \sum_{i \in S'\setminus S}{z^{v_i}}$. Applying $\mu^\star_{w,F}$ to $W$ and $G$ we obtain $W_1$ and $z^{u^+}$ respectively. However, applying $\mu^\star_{-w,F'}$ to $W$ and $G'$ we obtain
\[
\mu^\star_{-w,F'}W = z^{u^-} + \frac{1}{z^{u^-}}\sum_{i \in S\setminus S'}{z^{v_i}}\sum_{j \in S'\setminus S}{z^{v_j}} + \sum_{i \notin S \ominus S'}{z^{v_i}}
\] and $\mu^\star_{-w,F'}G' = z^{u^-}$ respectively. We note that $T_{u^+,u^-}$ exchanges $z^{u^+}$ and $z^{u^-}$, while every term in either
\[
\Big(\sum_{i \in S\setminus S'}{z^{v_i}}\sum_{j \in S'\setminus S}{z^{v_j}}\Big) \text{, or } \sum_{i \notin S \ominus S'}{z^{v_i}}
\]
is fixed by $T_{u^+,u^-}$. Hence we can describe $\varphi$ via the following diagram:
\begin{equation}
\label{eq:hypersurface_diagram}
\xymatrix{
 & \TT_N \ar^{T_{u^+,u^-}}[rr] \ar@{-->}_{\mu_{w,F}}[dl] & & \TT_N \ar@{-->}^{\mu_{-w,F'}}[dr] & \\
\TT_N  & & & & \TT_N \ar@{-->}_{\varphi}[llll]
}
\end{equation}

\begin{rem}
	\label{rem:degenerate_cases}
	We observe that the proof of Proposition~\ref{pro:hypersurface_case} breaks down if $S=S'$, and this case -- despite its triviality -- has to be considered separately. In particular, the vector $w$ becomes equal to zero and there are no possible choices of vectors $u^\pm$. The degeneration of the argument at this case generalises in higher codimension, see Definition~\ref{dfn:communicating}.
\end{rem}


TITLE: Continuity Question- Suppose that f and g are continuous at a point c E R and g(c) = h(c). Define
QUESTION [0 upvotes]: Let $g$ and $h$ be functions defined on all of $\mathbb R$. Suppose that $f, g$ are continuous at a
point $c$ in $\mathbb R$ and $g(c) = h(c)$. Define
$f(x)$ =
      $ g(x)$, if $x  \leq c $ ;
$f(x)= h(x)$, if $x \geq c $.
Prove that $f$ is also continuous at $c$.

REPLY [0 votes]: $\lim_{x\to c^-}f(x)=\lim_{x\to c^-}g(x)=g(c)=f(c)$
$\lim_{x\to c^+}f(x)=\lim_{x\to c^+}h(x)=h(c)=f(c)$
Since $g(c)=h(c)$, $\lim_{x\to c^-}f(x)=lim_{x\to c^+}f(x)=f(c)$.
so $f$ is continuous.


TITLE: When $x$ is a real number and $x>1$, why is $x^x>(x+1)^{x-1}$?
QUESTION [8 upvotes]: When $x$ is a real number and $x>1$, why is the following true?
$x^x>(x+1)^{x-1}$
I tried finding the minimum of $x^x-(x+1)^{x-1}$ with my limited calculus knowledge, but it shortly appeared out of my range.
It's good when I can understand a good answer, but I'd still be happy to come back years later when I'm better at math, so please don't hesitate to share your knowledge.

REPLY [2 votes]: Let $f(x)=x\ln x-(x-1)\ln(x+1),\;$ so $\color{red}{f(1)=0}$
and $\displaystyle f^{\prime}(x)=x\left(\frac{1}{x}\right)+\ln x-(x-1)\cdot\frac{1}{x+1}-\ln(x+1)=\frac{2}{x+1}-(\ln(x+1)-\ln x)$.
Since $\displaystyle \ln(x+1)-\ln x<\frac{1}{x}\;\;$ (by considering the area under $y=\frac{1}{x}$ from $x$ to $x+1$),
$\displaystyle \color{red}{f^{\prime}(x)>\frac{2}{x+1}-\frac{1}{x}=\frac{x-1}{x(x+1)}>0}\;$ for $x>1$;
so  for $x>1,\;\;$ $\color{red}{f(x)>0}\implies x\ln x>(x-1)\ln(x+1)\implies x^x>(x+1)^{x-1}$