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I have since found a few results in the literature, which are somewhat similar to what I had in mind, but not quite the same. Of course, if the functions $f$ and $f_n$ are assumed to be real-valued, …

Suppose that there is an open subset $U$ of $E$ such that the Lebesgue measure of $E\setminus U$ is $0$. Since $E$ is compact, the function $E^n\ni(x_1,\dots,x_n)\mapsto|y-x_i|^2$ is $L$-Lipschitz for …

Of course, nothing definite can be said here. E.g., let $b=c=1$ and $g(x,y)=x+y-x^2-y^2+axy$ for some real $a$ and all real $x,y$, so that $f(x,y)=-x^2-y^2+axy$. Then, if $|a|<2$, then $(0,0)$ is the …

$\newcommand{\R}{\mathbb R}\newcommand{\la}{\lambda}\newcommand{\al}{\alpha}\newcommand{\be}{\beta}\newcommand{\tal}{\tilde\al}\newcommand{\ga}{\gamma}\newcommand{\K}{\mathfrak K}$The answer is yes -- …

$\newcommand\R{\mathbb R}$Since $f$ is nondecreasing, concave, and continuous on $[0,\infty)$, it has a nonnegative nonincreasing right derivative $g=f'_+$ on $[0,\infty)$, which is of course right co …

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 …

Replacing $x_k^*$ by $x_k$ and noting that $|x_k^*|=|x_k|$, we see that the problem is to show that $$|s|^2-\sum_{k=1}^n|x_k|^2|y_k|^2\ge-1/2$$ for $x,y\in\mathbb{C}^n$ with $\|x\|_2=\|y\|_2=1$, where …

First, let us answer question 2. Write $R=Q'Q$, where $Q=[q_1,\dots,q_n]$, a square matrix with columns $q_1,\dots,q_n$. Let $v_i$ and $r_i$ denote the coordinates of the vectors $v$ and $r$. Then the …

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 …

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 …

The answer is yes. Indeed, let $X:=\boldsymbol{X}$, $H:=\boldsymbol{H}$, and $a:=\boldsymbol{\alpha}$. By approximation and compactness, without loss of generality (wlog), the matrix $H$ is nonsingul …

$$\|r_k\|=\|A(x_*-x_k)\|\le\|A^{1/2}\|\,\|A^{1/2}(x_*-x_k)\| \\ =\|A\|^{1/2}\|x_*-x_k\|_A \le 2\|A\|^{1/2} \left(\frac{\sqrt{\kappa}-1}{\sqrt{\kappa}+1}\right)^k \|x_0-x_*\|_A.$$

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 …

$\newcommand\R{\mathbb R}$It does not make sense to define a convex function on the nonconvex set $[-\infty,\infty]^m$. Even if you replace $[-\infty,\infty]^m$ by $\mathbb R^m$ and even you just ta …

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 …