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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) \\ =- …

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 …

$\newcommand{\R}{\mathbb R}$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 …

$\newcommand{\R}{\mathbb R}$ 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 …

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

$\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 …

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 …

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)| …

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* …

Your conjecture is true. Indeed, let $g_n:=\sqrt{f_n}$ and $g:=\sqrt{f}$. Let $$v:=\|g\|, $$ where $\|h\|:=\|\,h|_J\,\|_{L^2(J)}$ for $h\in L^2(\mathbb R^2)$, $J:=I^2$, $I:=[-u,u]$, and $u\in(0,1/20 …

$\newcommand{\al}{\alpha}$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 …

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 …