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On page 210 of Analysis by by Elliott H. Lieb, Michael Loss (GSM 14), there is a theorem stating that if $f_j \to f$ weakly in $H^1(\mathbb R^n)$ then $\mathbf 1_A f_j \to \mathbf 1_A f$ strongly in $L^p(\mathbb R^n),$ where $A$ is any set with finite measure. Here $p<2^* := \frac{2n}{n-2}.$ So the injection $H^1 \to L^p_{\text{loc}}$ is continuous even if $H^1$ is equipped with weak topology.

In the proof below, it is shown by convolution mollification that $\|\mathbf 1_A (f_j-f)\|_{L^2} \to 0.$ But I cannot find a proof in the book that $\|\mathbf 1_A (f_j-f)\|_{L^p} \to 0.$

Here is my reasoning to fill in this gap: since $f_j$ converges weakly in $H^1,$ it is uniformly bounded in norm in $H^1.$ Therefore, by Sobolev embedding, $f_j$ is uniformly bounded in norm in $L^q,$ for all $2\leq q \leq 2^*.$ Write $\|f_j\|_{L^q} \leq M_q.$ We have, by interpolation, $$ \|\mathbf 1_A (f_j-f)\|_{L^p} \lesssim \|\mathbf 1_A (f_j-f)\|_{L^2}^{\theta} \|\mathbf 1_A (f_j-f)\|_{L^q} ^{1-\theta} \leq 2M_q^{1-\theta}\|\mathbf 1_A (f_j-f)\|_{L^2}^{\theta}. $$ Hence $\mathbf 1_A f_j\to \mathbf 1_A f$ in $L^p$ iff $\mathbf 1_A f_j \to \mathbf 1_A f$ in $L^2.$

Is this the originally intended way to complete this proof?

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    $\begingroup$ Yes (to your final question). $\endgroup$
    – Willie Wong
    Commented Apr 28, 2021 at 20:55
  • $\begingroup$ You can also do it from the fact that in a space of finite measure, a family of functions uniformly bounded in $L^{1+\epsilon}$ is uniformly integrable. This tells us that $1_A |f_j|^p$ are uniformly integrable, and it follows that they converge in $L^p$. $\endgroup$
    – Nate Eldredge
    Commented Apr 28, 2021 at 21:18

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