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$\newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\varepsilon}$ Take any probability measures $P_0,P_1$ absolutely continuous with respect (w.r.) to $Q$. We shall prove the followi …

It should oftentimes be the case that, analyzing a "thoughtless" Lagrange multiplier solution, one finds a more elegant, "clever" solution. At least, this is the case here. Analyzing the previous Lagr …

$\newcommand{\1}{\mathbf 1}\newcommand{\ep}{\varepsilon}\newcommand{\tr}{\operatorname{tr}}$The min-max value is $\sqrt n$. Indeed, take any real $n\times n$ matrix $H$ with $|\det H|=1$. By the singu …

We need to show that $$\sum x_j^2 y_j^2\le1/2+\Big(\sum x_j y_j\Big)^2\tag{0}\label{0}$$ given that $\sum x_j^2=1$ and $\sum y_j^2=1$. Consider the maximization of $\sum x_j^2 y_j^2$ for a fixed value …

$\newcommand{\R}{\mathbb{R}}$ The constant $c(n)$ can be improved from $\frac{n-1}2$ to the optimal value \begin{equation} c_*(n):=\frac{3n(n-1)}{2(2n-1)} \end{equation} for $n\ge2$. Indeed, for $ …

$\newcommand\la\lambda$The answer is no. E.g., suppose that $X=\ell^\infty$, $z_1=e_1$, $z_2=-e_1+e_2$, and $z_k=e_2/2+e_k/k$ for $k\ge3$, where $(e_1,e_2,\dots)$ is the standard basis of $\ell^\infty …

Rewrite the constraint as $$x_n \le f_n(\rho_1,\dots,\rho_n):=\ln\Big(B\log_2\Big(1+\frac{e^{\rho_n} g_n^2}{\sum_{i=1}^{n-1} e^{\rho_i} g_i^2+\sigma^2}\Big)\Big).$$ The problem is then to prove the co …

$\newcommand\de\delta$If the function $f$ were convex, then a convolution of $f$ with (say) an even positive mollifier would do. However, $f$ is not convex. Yet, it would be quite easy to construct ju …

This is a special case (with $f=1_S$) of the duality $$s=i,\tag{1}$$ where $$s:=\sup\Big\{\int f\,d\mu\colon\mu\text{ is a measure, }\int g_j\,d\mu=c_j\ \;\forall j\in J\Big\},$$ $$i:=\inf\Big\{\sum b …

We have \begin{equation} f_2(x):=f''(x)\Big/\frac{2^{x-3}}{125 \left(2^x+1\right)^2}= -x \ln2-2 \left(2^x+1-500 \ln2\right) \end{equation} and \begin{equation} f''_2(x)=-2^{1 + x} \ln^2 2<0, \en …

We shall prove the more general result: for $n\ge2$ distinct points and any positive real $a,b$, we have $D(R):=d_1^2+\cdots+d_n^2\le 2a^2+2b^2$, where the points $P_j$ are now in an $a\times b$ recta …

For $a\ge0$ and $u\ge0$, let $$q(u):=\ln Q(a+\sqrt u). $$ Then $$q_2(t):=q''(u)\frac{8 \sqrt{2 \pi } t^3 e^{\frac{1}{2} (a+t)^2} Q(a+t)^2}{a t+t^2+1} =2 Q(a+t)-\frac{\sqrt{\frac{2}{\pi }} t e^{-\fr …

First, a few simplifications. Note that $g(-t)\ge g(t)$ and $|-t-1|\ge|t-1|$ if $t\ge0$. So, without loss of generality (wlog) $t\ge0$ and $$g(t)=\sqrt{(t-1)^2 + a^2} + bt. \tag{1}\label{1}$$ Since $g …

The answer is no. E.g., if $d=2$ and $$f(x,y)=f_c(x,y):=(x - 1)^2 + (y - 1)^2 - 2 c (x - 1) (y - 1) \tag{1}\label{1}$$ for some real $c\in(-1,1)$ and all real $x,y$, then the function $f$ is convex. H …

$\newcommand\R{\mathbb R}$The answer is no. E.g., let $A=B=\R$, $B_a=[1-a^2,2]$ for $a\in A$, and $\ell(b)=b^3-3b$ for $b\in B$. Then $\ell(b)$ and $B_a$ are perfectly smooth, but $L$ is not different …