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Let $\Omega$ be an open bounded subset of $\mathbb{R}^n$ and let $ p\geq 1$. Let $(u_n)_n$ be a bounded sequence of $W_0^{1, p}(\Omega)$ and let $\Omega_n\subset\Omega$ be a subset of $\Omega$ which depends only 0n $n$ and such that $$meas(\Omega_{n})\longrightarrow 0 \quad \mbox{ as } n\to +\infty.$$ Could I conclude that $$\int_{\Omega_{n}} \vert\nabla u_n\vert^{p} dx\longrightarrow 0 \quad \mbox{ as } n\to +\infty?$$ If not, what additional assumptions I need?

Could anyone please help? Thank you in advance!

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Consider $n=1$, $\Omega=(0,1)$ and a sequence $u_{n}$ of piecewise linear bumps of height $1$ on $(0,\frac{1}{n})$: $$ u_{n}(x) = \begin{cases} 2nx, & \text{for } 0<x\le \frac{1}{2n}\\ 1-2n\cdot(x-\frac{1}{2n}), & \text{for } \frac{1}{2n}\leq x\leq \frac{1}{n}\\ 0, & \text{for } \frac{1}{n} \le x < 1 \end{cases} $$ The sequence is bounded in $W^{1,1}(\Omega)$, as $||u_{n}||_{L^{1}} \le \frac{1}{n}$ and $||u_{n}'||_{L^{1}} = 2$ for all $n$.

(Note that $|u_{n}'(x)|=2n$ on $(0,\frac{1}{n})$ and $0$ everywhere else).

Now choose $\Omega_{n}=(0,\frac{1}{n})$.

Then $vol(\Omega_{n}) \rightarrow 0$, but $||u_{n}'||_{L^{1}(\Omega_{n})} = 2$ for all $n \in \mathbb{N}$.

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  • $\begingroup$ What does a linear bump look like. Can you give an explicit example? $\endgroup$
    – Medo
    Commented Jul 2, 2020 at 12:17
  • $\begingroup$ How is it obvious that $\|u_n\|=1/n$ ? if $u_{n}$ has height 1 and supported on $]0,1/n[$ then $\|u_n\|\leq 1/n$. If by a bump you mean a hut-shaped function $f$ with $f(0)=f(1/n)=0$, then $\|u^{\prime}_n\|\neq 1/2$...so please describe your "linear" bump $\endgroup$
    – Medo
    Commented Jul 2, 2020 at 12:28
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    $\begingroup$ @Medo The function is now written down explicitely above. Thanks for pointing out the calculation error... it should be $||u_{n}'||_{L^{1}} = 2$ of course. But that doesn´t change the argument. $\endgroup$ Commented Jul 2, 2020 at 12:39
  • $\begingroup$ Thanks. Good answer! $\endgroup$
    – Medo
    Commented Jul 2, 2020 at 15:10
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    $\begingroup$ @C.Bishop It´s all about which conditions you are willing to relax. One possibility: If you change to $W^{1,\infty}(\Omega)$, it works because every $\nabla{u_{n}}$ is bounded and hence the integral vanishes for $vol(\Omega_{n}) \rightarrow 0$ $\endgroup$ Commented Jul 3, 2020 at 8:29

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