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For all $x$ and $y$ in $[0,1]^2$ $$f(x,y)= \left\{ \begin{aligned} \int_x^y f_y(x,z)\,dz&\text{ if }x\le y, \\ -\int_y^x f_y(x,z)\,dz&\text{ if }x\ge y, \end{aligned} \right. $$ where $f_y(x,z):=\fra …

$\newcommand{\R}{\mathbb R}\newcommand{\ep}{\varepsilon}$Re-define, if needed, the function $u$ on a set of Lebesgue measure zero so that \begin{equation} u(y)-u(x):=\int_x^y dt\, u'(t) \end{equat …

For $k$ to make sense, we should assume that $\|u\|_p\ne0$. Let $l(x)$ and $r(x)$ denote the left- and right-hand sides of your displayed inequality. Then, by Tonelli's theorem, for any real $p>0$ $$ …

No. E.g., let $n=2$ and $g(x,y)=(y,0)$ for all $(x,y)$ in the unit disk $D$. Then $\nabla f=g$ a.e. would imply $f(x,y)=h(x)$ for some function $h$ and almost all $(x,y)\in D$ and hence $\nabla_{x,y}h …

Assume that $\Omega$ is open and connected (if $\Omega$ is not connected, then the desired conclusion is clearly false). By convolution with a mollifier and approximation, we may assume that $w_1$ and …

$\newcommand{\R}{\mathbb R}$ The square root of the multiplication operator does not exist unless the function $f$ is constant (and, obviously, the square root exists if $f$ is constant). Indeed, s …

$\newcommand\R{\mathbb R}\newcommand\ep\epsilon\newcommand{\de}{\delta} $For $(s,t)\in\R^2$, let \begin{equation} u(s,t):=\sum c_k g\Big(\frac{R-r_k}{h_k}\Big), \end{equation} where $g(z):=\max(0, …

Indeed, disjoint tiny smooth spikes, of small heights and even much-much smaller widths, will do. Let $K\in C^\infty(\mathbb R)$ be such that $K\ge0$, $K(x)=0$ if $|x|>1/2$, and $a:=K(1/3)-K(0)\ne0 …

$\newcommand{\R}{\mathbb R}$ The answer is: $A_f^*$ is a multiplication operator iff the function $f$ is constant. Indeed, recall that for all $x$ and $y$ in $H^1:=H^1(\mathbb R^n)$ \begin{equation* …

We shall assume that $l\in(0,\infty)$, so that for any $h\in H_0^1$ we have $$\int_0^l|h|^2=\int_0^l dx\,\Big|\int_0^x h'\Big|^2 \le\int_0^l dx\,\Big(\int_0^l|h'|\Big)^2 \\ \le\int_0^l dx\,l\,\int_0 …

The answer is no. Indeed, let $t:=\alpha\in(0,1)$ and $c:=\|f\|_1:=\|f\|_{L^1(0,1)}\in(0,\infty)$. Suppose that $f\in C^k(0,1)$ and $f=0$ on $(a,b)$, with $0\le a<b\le1$. Suppose that the inequality i …

Let $f:=\varphi$, $a:=\|f\|_2$, $b:=\|f'\|_2$, so that $\|f\|_{H^1}=a+b$; see e.g. Wikipedia for the definition of $H^k$. Without loss of generality, $x=0$. For all $y\in[0,1]$, we have $$|f(y)-f(0)| …

$\newcommand{\na}{\nabla}\newcommand{\R}{\mathbb R}$Without loss of generality, $C=1$, so that \begin{equation} f(x)\equiv\exp\Big\{-\frac{\pi |x|^2}{2a^2}\Big\}, \end{equation} where it is assum …

By the Euler–Maclaurin formula (with $p=4$, $m=0$, and $g(x):=\frac1n\,f^2(\frac xn)$ in place of $f(x)$ there in the formula), $$d_n(f):=\int_0^1 f^2(y) \, dy - \frac{1}{n}\sum_{i=1}^n f^2(i/n) \\ =- …

$\newcommand\la\lambda\newcommand\al\alpha$If $C$ is allowed to depend on $\lambda$, just take $C=2\lambda^p$. If $C$ is not allowed to depend on $\lambda$, take any nonzero $f\in H_{0}^{1}(0,1)$ and …