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Yes. If $K \subset \Omega$ is compact, let $\text{dist}(K, \Omega^c) > r > 0$. Then for any $p \in K$ and convex function $f$, $f(p)$ is bounded above by the average of $f$ over the ball of radius $ …

If your function was strictly convex and $C^3$ (let's say on a compact interval $I$) and you had bounds on the third derivative there, it might be possible to prove its convexity on $I$ by evaluating …

Try $d = 2$, $f(x_1,x_2) = (x_1+1)^2 + (x_1+1)(x_2 + t) + (x_2 + t)^2$ where $t$ is a parameter. The minimum is at $(-1,-t)$. The sign of the second component of the gradient changes at $t = -1/2$, …

$$f(x) - f(x') \le \sum_i \left(f(x_i) - f(x')\right)$$ iff the function $g(x) = f(x + x') - f(x')$ is subadditive.

Assume wlog $\|c\|=1$. $C = \{0\}$ if $\alpha > 1$, while it is the ray $(-\infty, 0] c$ if $\alpha = 1$. So let's suppose $0 < \alpha < 1$. If $y = t c + w$ where $t \in \mathbb R$ and $\langle c, …

The last paragraph is wrong. Consider the case $m=1$. $\tilde{M}$ is the intersection of $K$ with one hyperplane. In general this will not contain any extreme point of $K$, so its extreme points wi …

If $z_k = x_k y_k$, the quantity you're looking at is $$ y^T D Q D y = \left(\sum_k z_k\right)^2 - \sum_k z_k^2$$ where $Q$ is the $n \times n$ symmetric matrix with diagonal terms $0$ and off-diagon …

I'll assume $m, n \ge 2$. I claim that $$\Lambda = \left\{\lambda \in \mathbb R^m: \; \sum_{i=1}^m \lambda_i = 1,\; 0 \le \lambda_i \le \frac{1}{2} \ \text{for all}\ i\right \}$$ The necessity o …

It's not true. Consider the $2 \times 2$ matrix $$F(X) = \pmatrix{f(X) & 0\cr 0 & f(X-2)\cr}$$ where $f$ is an even function, everywhere $> 0$, and decreasing on $[0,\infty)$. Take $a = (1,1)^T$. T …

The standard example of a non-convex but midpoint-convex function is additive: given a basis $B$ of $\mathbb R$ over the rationals $\mathbb Q$, choose one member $\alpha$ of $B$ and if $x = \sum_{\bet …

This can occur even in linear programming, in the presence of degeneracy. At an optimal basic solution, the slack variable for some binding constraint may be basic (but with value $0$ since it is bin …

In fact $\ln(x^2 + x^\beta)$ is concave for $x > 0$ iff $3-2\sqrt{2} \le \beta \le 3 + 2 \sqrt{2}$. This comes from writing $$ \dfrac{d^2}{dx^2} \ln(x^2 + x^\beta) = \dfrac{x^2}{(x^2 + x^\beta)^2} \l …