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Lemma. Let $\mu$ be a measure in $\mathcal{M}(\Omega)$ and let $(v_{n})$ be a sequence of functions in $W^{1,p}_{0}(\Omega)\cap L^{\infty}(\Omega)$ converging to a function $v$ in the weak topology of $W^{1,p}_{0}(\Omega)$ and almost everywhere in $\Omega$. If there exists $c>0$ such that $||v_{n}||_{L^{\infty}(\Omega)}\leq c$ for every $n\in\mathbb{N}$, then $(v_{n})$ converges to $v$ in the strong topology of $L^{2}(\Omega,\mu)$.

I don't know how the last statement of this Lemma can be true in the unbounded case?

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    $\begingroup$ What is denoted by $\Omega$ and $\mathcal M(\Omega)$? What is $p$? How can convergence in $L^2(\Omega)$ make sense if $\mu$ is mutually singular with respect to Lebesgue measure? Why the statement should hold in the bounded case (and what do you mean by bounded case)? $\endgroup$
    – Skeeve
    Commented Dec 10, 2019 at 14:30
  • $\begingroup$ $\Omega$ is a bounded open subset of $\mathbb{R}^{N}$, $\mathcal{M}(\Omega)$ is the space space of absolutely continuous Radon measures and $1<p\leq N$. The bounded case concerns the case where $\Omega$ have finite measure ($\mu(\Omega)<\infty$ where $\mu\in\mathcal{M}(\Omega)$). $\endgroup$
    – M.A
    Commented Dec 10, 2019 at 15:53
  • $\begingroup$ $\mu$ is not singular with respect to Lebesgue measure. $\endgroup$
    – M.A
    Commented Dec 10, 2019 at 16:06
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    $\begingroup$ but what if for some smooth compactly supported nonzero $\phi\colon \mathbb R^N \to \mathbb R$ we consider a sequence of the form $\phi(x-n)$, $x\in \mathbb R^N$? It should converge weakly to 0 (at least for $p=2$), but not strongly. $\endgroup$
    – Skeeve
    Commented Dec 10, 2019 at 19:57
  • $\begingroup$ How can I verify that $\phi:x\mapsto \phi(x-n)$ converges weakly to $0$ but not strongly (excuse but I'm not familiar with functional analysis properties)? $\endgroup$
    – M.A
    Commented Jan 17, 2020 at 14:55

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