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If $u_n=1_{[n,n+1]}$, then for any real $t$ $$\|u_n\|^2_{L^2_t}=g(t)e^{tn},$$ where $g(t):=\frac{e^t-1}t$ if $t\ne0$ and $g(0):=1$. So, $\|u_n\|_{L^2_t}/\|u_n\|_{L^2_s}\to\infty$ as $n\to\infty$ if $s …

$\newcommand{\R}{\mathbb R}$Yes, this is true for any Lipschitz compactly supported function $f$. Indeed, we have $f(x)=0$ for some real $R>0$ and all $x\in B_R^c$, where $B_R^c:=\R^n\setminus B_R$ an …

Another way to do this (again assuming that $\Omega$ is open and connected): $$\nabla\frac{w_1}{w_2}=\frac{w_1\nabla w_2-w_2\nabla w_1}{w_2^2} =\frac{w_1}{w_2}\Big(\frac{\nabla w_2}{w_2}-\frac{\nabla …

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 …

We have $$d(t):=\int dx\Big(\frac{f(x+t)-f(x)}{t}-f'(x)\Big) =\int dx\int_0^1 ds\,(f'(x+st)-f'(x)),$$ whence $$|d(t)|\le\int_0^1 ds\, J_t(s),$$ where $$J_t(s):=\int dx\,|f'(x+st)-f'(x)|\underset{t\to0 …

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 …

$\newcommand\de\delta$The answer is no. E.g., let $X$ be the linear span of the set $\{\de,\de',\de'',\dots\}$, where $\de$ is the Dirac delta distribution supported on $\{0\}$. Then $X$ satisfies you …

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 …

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\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, …

$\newcommand\la\lambda\newcommand\al\alpha$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 …

$\newcommand{\R}{\mathbb R}\newcommand{\ep}{\varepsilon}$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 …

$\newcommand{\al}{\alpha}$In leo monsaingeon's answer, for the the value $J(\al)$ of the infimum it was shown that \begin{equation*} J(\al)\le9 \end{equation*} and conjectured that \begin{equation …

$\newcommand{\R}{\mathbb R}$Let $F:=W^{1,\infty}(\R^n)$, with $\|f\|_{1,\infty}:=\|f\|_\infty+L(f)$ for $f\in F$, where $L(f)$ is the Lipschitz constant of $f$. Take any $f\in F$ with $q:=\|f\|_{1,\in …

$\newcommand{\na}{\nabla}\newcommand{\R}{\mathbb R}$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 …