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I am self studying PDEs from Evans' "Partial Differential Equations" textbook. Currently, I am going through Theorem 1 from Section 5.7 (Rellich-Kondrachov compactness theorem) and am having difficulty obtaining an inequality from the third step.

The author uses an $L^p$-interpolation identity (which I still follow) to obtain $$ \left\Vert u_m^\epsilon - u_m \right\Vert_{L^q(V)} \leq \left\Vert u_m^\epsilon - u_m \right\Vert_{L^1(V)}^\theta \left\Vert u_m^\epsilon - u_m \right\Vert_{L^{p^\ast}(V)}^{1-\theta} $$ where $p^{\ast}$ is the Sobolev-conjugate to $p$.

He then invokes the Gagliardo-Nirenberg-Sobolev inequality to yield $$ \left\Vert u_m^\epsilon - u_m \right\Vert_{L^q(V)} \leq C \left\Vert u_m^\epsilon - u_m \right\Vert_{L^1(V)}^\theta $$ for all $m$ and all $\epsilon$. This last step is where I get stuck.

The constant $C > 0$ seems to come from the fact that the sequence $(u_m)_m$ is uniformly bounded in $W^{1,p}(V)$. However, for the inequality to hold, wouldn't the collection $\left( u_m^\epsilon \right)$ also need to be uniformly bounded? If so, I have no idea how to go about proving this.

ASSUMPTIONS:

  1. We are working in $\mathbb{R}^n$ and it is assumed that $1\leq p < n$.
  2. $V$ is a pre-compact subset of $\mathbb{R}^n$, the functions $u_m$ and $u_m^\epsilon$ all have compact support in $V$.
  3. $u_m^\epsilon = \eta_\epsilon \ast u_m$ where $\eta_\epsilon$ are molifiers. I have already shown that $u_m^\epsilon \to u_m$ in $L^1(V)$ uniformly in $m$.
  4. It holds that $\sup_m \left\Vert{u_m}\right\Vert_{W^{1,p}(V)} < \infty$.
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    $\begingroup$ $\|u_m^\epsilon\|_{W^{1,p}(V)}\le C\|u_m\|_{W^{1,p}(V)}$. $\endgroup$
    – Fan Zheng
    Jul 17, 2018 at 23:30
  • $\begingroup$ @FanZheng How did you obtain this inequality? $\endgroup$
    – Quoka
    Jul 17, 2018 at 23:33
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    $\begingroup$ $\|u_m^\epsilon\|_{L^p}\le \|\eta_\epsilon\|_{L^1}\|u_m\|_{L^p}$ and $\|\nabla u_m^\epsilon\|_{L^p}\le \|\eta_\epsilon\|_{L^1}\|\nabla u_m\|_{L^p}$, because $\nabla u_m^\epsilon=\eta_\epsilon * \nabla u_m$. These are quite standard yoga with mollifiers. $\endgroup$
    – Fan Zheng
    Jul 17, 2018 at 23:45
  • $\begingroup$ Didn't think of that. Thank you :-) $\endgroup$
    – Quoka
    Jul 17, 2018 at 23:47

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