# It is true that this integral converges to $0$?

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?

Consider $$n=1$$, $$\Omega=(0,1)$$ and a sequence $$u_{n}$$ of piecewise linear bumps of height $$1$$ on $$(0,\frac{1}{n})$$: $$u_{n}(x) = \begin{cases} 2nx, & \text{for } 0 The sequence is bounded in $$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 $$n \in \mathbb{N}$$.

• What does a linear bump look like. Can you give an explicit example?
– Medo
Jul 2, 2020 at 12:17
• 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
– Medo
Jul 2, 2020 at 12:28
• @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. Jul 2, 2020 at 12:39
• @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 $vol(\Omega_{n}) \rightarrow 0$ Jul 3, 2020 at 8:29