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Let $f \in L^{p}(\mathbb{R}^{n})$ with $p > \max\{n/2,1\}$. Suppose $u \in L^{2}(\mathbb{R}^{n})$ and let $\{h_{n}\}_{n\in \mathbb{N}}$ be a bounded family of $H^{1}(\mathbb{R}^{n})$ functions converging weakly to $h_{0}$ in $L^{2}(\mathbb{R}^{n})$. This means that, for every $g \in L^{2}(\mathbb{R}^{n})$ one has $\langle h_{n},g\rangle_{L^{2}} \to \langle h_{0},g\rangle_{L^{2}}$ as $n \to \infty$.

My question is: under these hypothesis, does the following condition $$\int |(h_{n}-h_{0})(x)|^{2}(|u|^{2}*f)(x)dx \to 0 $$ holds true as $n \to \infty$?

I am trying to prove this but I did not get anywhere. I think I should use Hölder or maybe Young's inequality but I am having trouble applying it.

Idontgetit
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1 Answers1

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Break the integral into the integral over a ball $B_R$ and its complement. Let’s call the first integral $A$ and the second $B$. By Young’s inequality for convolutions you have that $g=u^2*f$ belongs to $L^p$, and so by Holder’s inequality $$A\le \|( h_n-h_0)^2\|_{L^q(B_R)}\|g\|_{L^p(B_R)}$$ Now you have to divide the proof in three cases.

  • If $2=n$, then you know that $H^1$ is embedded in $L^r$ for every $2\le r<\infty$ and so by the Rellich-Kondrachov theorem you have strong convergence in $L^{2q}(B_R)$.
  • If $n<2$, by Ascoli Arzela you have uniform convergence and so strong convergence in $L^{2q}(B_R)$.
  • If $n>2$, then $2^*=2n/(n-2)>q=p/(p-1)$ exactly for $p>n/2$ and so again by the Rellich-Kondrachov theorem you have strong convergence in $L^{2q}(B_R)$.

This proves that $A$ goes to zero.

Now for $B$ you still apply Young and Holder except that now you integrate over the complement of the ball. Now you have to use the same three embedding theorems to prove that $\|( h_n-h_0)^2\|_{L^q(\mathbb{R^n})}\le M$ for every $n$ for some constant $M$.

By the Lebesgue dominated convergence theorem you have that $\|g\|_{L^p(B_R^c)}\to 0$ as $R\to \infty$ and so given $\epsilon>0$ you can find $R$ so large that $\|g\|_{L^p(B_R^c)}<\epsilon/M$. Hence $B<\epsilon$.

Daniele Tampieri
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Gio67
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  • Thanks!! That is a great answer. Only thing I still could not work out the details isthe Arzela Ascoli argument. I am assuming you take your compact space $X=L^{2}(B_{R})$ and a sequence like $T_{n}(g) = h_{n}-g$ defined for $g\in X$. This implies uniform convergence of a subsequente of $T_{n}$. But (a) why this implies convergence to zero and (b) why is convergence along subsequences enough? – Idontgetit Jul 26 '24 at 10:44
  • Morrey’s embedding theorem tells you that the sequence $h_n-h_0$ is equibounded and equicontinuous. Take a subsequence. By Ascoli Arzela, a further subsequence converges uniformly (and so weakly in $L^2(B_R)$) to some function $w$, but since the entire sequence converges weakly in $L^2$ in the ball to zero, by the uniqueness of the weak limit $w=0$. This implies that the entire sequence converges uniformly to zero in the ball. – Gio67 Jul 27 '24 at 03:27
  • We use this fact https://math.stackexchange.com/questions/397978/every-subsequence-of-x-n-has-a-further-subsequence-which-converges-to-x-the – Gio67 Jul 27 '24 at 03:29
  • By Morrey's embedding Theorem you mean the inclusion of Sobolev spaces in the space of Hölder continuous functions $H^{1}(\mathbb{R}) \hookrightarrow C^{0,1/2}(\mathbb{R})$? I don't see how this theorem can be used because $h_{0}$ is not necessarily in the Sobolev space $H^{1}(\mathbb{R})$ since I am assuming $h_{n}$ converges weakly in $L^{2}(\mathbb{R})$. Only the elements $h_{n} \in H^{1}(\mathbb{R})$, right? – Idontgetit Jul 29 '24 at 08:43
  • you said that the family is bounded in $H^1$, and since this space is reflexive, $h_0$ is also in $H^1$ – Gio67 Jul 30 '24 at 09:23