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The answer is no. Indeed, consider the case when $n=2$ and $$f(x,y)=\max(| x|,2 | y|)$$ for all $(x,y)\in\mathbb R^2$. Then for $(x,y)$ near $(\pm1,0)$ we have $f(x,y)=|x|$ and hence $\partial f(\pm1, …

Of course, your "if and only if" statement is incorrect. E.g., let $$f(x,y):=x^2/2+y^2/4.$$ Then your conditions $(-f_{xx})(-f_{yy})>0$ and $f_{xx}>0$ obviously hold. Further, here $f_x(x,y)=y$ mean …

$\newcommand{\R}{\mathbb R}\newcommand{\la}{\lambda}$Let $\R_-:=(-\infty,0]$ and $\R_+:=[0,\infty)$. The desired statement is equivalent to the following: Suppose that \begin{equation*} \forall x …

$\newcommand{\ep}{\varepsilon}$The "convex" part of this conjecture is not true in general. Indeed, suppose it is true. Then (letting $x:=\phi$, $t:=\theta_1$, and $\theta_2\downarrow\theta_1=t$) we s …

Yes: Just take $u(x,y):=v(x)$, which will be assumed in what follows. Indeed, one can use Green's formula to show this, as is done in Christian Remling's answer. More generally, the result holds for a …

$\newcommand\ep\varepsilon$The answer is yes, and we only need the uniform convergence of $f_k$ to $f$. Moreover, then $co(f_k)\to co(f)$ uniformly as well. Indeed, for each real $x$, $$co(f)(x)=\sup\ …

Actually, your desired conclusion does "follow just from $D$ having full measure". Indeed, without loss of generality, $U=(-1,1)^2$. Let $$X:=\{x\in(-1,1)\colon|D_x|=2\},$$ where $D_x:=\{y\in(-1,1)\co …

The left-hand side, $\int_{0}^t f^2(x)\, dx$, of your inequality does not contain $\mu$. Since you wanted "to find lowerbound on $\int_{0}^t f^2(x) d\mu(x)$", it appears that your desired inequality i …

Suppose the contrary, so that $g(s)<0$ for some $s\in(a,b)$. Replacing now $a$ and $b$ by $\max\{t\in[a,s)\colon g(t)\ge0\}$ and $\min\{t\in(s,b]\colon g(t)\ge0\}$, respectively, we see that without l …

By the equivalence of conditions (ii) and (iv) in Theorem A.3 on page 14, the condition $y=Ax$ for $y=[y,\dots,y_n]^T$ and $x=[x,\dots,x_n]^T$ is equivalent to the condition that $\sum_1^n g(y_i)\ge\s …

The answer is no. E.g., take any real $b>1$ and let $a:=2e^b$. Let $f(x):=a|x|-b$ for all real $x$. Then $\int_{\mathbb R}e^{-f(x)} dx=1$ but $$\int_{\mathbb R} f(x) e^{-f(x)} dx =1-b<0.$$

It is not true in general that the function $p$ is concave. For instance, let $\Omega$ be the conical hull of the ball of radius $1$ centered at the point $(2,0,\dots,0)\in\mathbb R^n$. Then $p(x)=c_n …

$\newcommand\vpi\varphi$The answer, which is a modification of the previous answer, is still no. E.g., suppose that $X=Y=\mathbb R$, $c=0$, and $\vpi_a(x)=\min(1,\max(0,x-a))$ for $a\ge0$ and all real …

$\newcommand{\ep}{\varepsilon}$This conjecture is not true in general. Indeed, suppose the "convex" part of your conjecture is true. Then (letting $x:=\phi$, $t:=\theta_1$, and $\theta_2\downarrow\the …

$\newcommand{\ep}{\varepsilon}$This conjecture is not true in general. Indeed, suppose it is true. Then (letting $x:=\phi$, $t:=\theta_1$, and $\theta_2\downarrow\theta_1=t$) we see that for any stric …