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I am trying to show the following:

Denote $L^2=L^2(\mathbb{R}^d)$ and $H^1=H^1(\mathbb{R}^d)=\{f\in{\cal S}';\sqrt{1-\Delta}f\in L^2\}$. Let $\{f_n\}_{n\in\mathbb{N}}\subset H^1$ and let $f\in H^1$. If $$f_n\rightharpoonup f\ {\rm in}\ H^1\ {\rm as}\ n\to\infty,$$ then $$f_n\to f\ {\rm in}\ L^2\ {\rm as}\ n\to\infty.$$

Here is my attempt. Since $L^2\subset H^{-1}$, I can get $f_n\rightharpoonup f$ in $L^2$ as $n\to\infty$. Then, by the uniqueness of the limit, if I can see that $\{f_n\}$ is Cauchy in $L^2$, then I can conclude the proof. But I could not show that $\|f_n- f_m\|_{L^{2}}\to0$ as $n,m\to\infty$.

I appreciate any advice.

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  • $\begingroup$ ??? You say you're trying to show $f_n\to f$ in $L^2$. Then you say you can get $f_n\to f$ in $L^2$. $\endgroup$ – David C. Ullrich Oct 24 '17 at 20:53
  • $\begingroup$ Anyway: Plancherel. $\endgroup$ – David C. Ullrich Oct 24 '17 at 20:54
  • $\begingroup$ Yes, I want to show $f_n \to f$ in $L^2$. Thank you for your comment. $\endgroup$ – Tom TJ Oct 24 '17 at 20:56
  • $\begingroup$ It would be good if you tried to explain the confusion. You say you can get exactly what you say you're trying to prove. So what's the problem? $\endgroup$ – David C. Ullrich Oct 24 '17 at 20:58
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    $\begingroup$ You could have just said it means weak convergence. I've never seen that notation. Whatever. It's very easy to show that $||f||_{L^2}\le||f||_{H^1}$ using the Fourier transform. $\endgroup$ – David C. Ullrich Oct 24 '17 at 21:36
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$\newcommand\tow\rightharpoonup$ (In case I'm not the only one who was not familiar with the notation: We're using "$\tow$" to denote weak convergence.)

The suggested implication is false. It's true on the torus $\mathbb T^d$. We give the counterexample and the proof in one dimension; the higher-dimensional case is the same except for notation.

Say $f_0\in H^1(\mathbb R)$, $f_0\ne0$. Define $$f_n(t)=f_0(t-n).$$ The sequence $(f_n)$ is certainly not (norm) convergent in $L^2$, but it's clear that $f_n\tow0$ in $H^1$.

Or if it's not clear, note that $H^1$ is a Hilbert space with inner product $$<f,g>=\int f\overline g+\int f'\overline{g'}$$(where $f,g,f',g'\in L^2$ and the derivatives are weak derivatives in any of various senses.)

Suppose now that $f_n\tow0$ in $H^1(\mathbb T)$. Uniform Boundedness implies that $||f_n||_{H^1}$ is bounded; in particular $$\sum_k k^2|\hat f_n(k)|^2\le c$$for all $n$. Since $\lim_n\hat f_n(k)=0$ for all $k$ this makes it clear that $||f_n||_{L^2}\to0$.

In more detail: Let $\epsilon>0$. Choose $A$ so $c/A^2<\epsilon/2$. Then $$\sum_{|k|\ge A}|\hat f_n(k)|^2<\epsilon/2$$for every $n$. But $\lim_n\hat f_n(k)=0$ for every $k$, so there exists $N$ such that $$\sum_{|k|<A}|\hat f_n(k)|^2<\epsilon/2\quad(n>N).$$

Oops Just realized that $\lim_n\hat f_n(k)=0$ is not quite as obvious as I thought, because the inner product is not just $\int f\overline g$. But it's ok: Let $e_k(t)=e^{ikt}$. Then $$<f_n,e_k>=\hat f_n(k)-ik\widehat{f_n'}(k)=(1+k^2)\hat f_n(k),$$so $\lim_n<f_n,e_k>=0$ implies $\lim_n\hat f_n(k)=0$.

Conjecture The result is true for $H^1(\Omega)$, if $\Omega$ is a bounded open set in $\mathbb R^d$.

This can't be quite as simple as the argument above, since we don't have those Fourier coefficients to play with. But:

Hand-Waving If not there exists a sequence with $f_n\tow0$ in $H^1( \Omega)$ but $||f_n||_{L^2}\ge1$. This is impossible. Roughly, if $f_n\tow0$ then either $|f_n|^2$ must have some mass wandering off to the boundary or $f_n$ must have a lot of oscillation; in either case $||f_n'||_{L^2}$ must be unbounded, contradicting $f_n\tow0$.

I wouldn't be surprised if that could be made into a proof. I've got class soon...

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