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In the proof of a compactness theorem involving fractional derivatives in Temam's Navier-Stokes Equations, an argument as the following is made.

Suppose $X_0,X,X_1$ are Hilbert spaces such that $ X_0\overset{\textrm{cpt}}{\hookrightarrow} X\hookrightarrow X_1. $ Suppose $\gamma>0$. Suppose $v_m$ is a sequence all supported in a bounded subset $K$ of $\mathbb{R}$ such that

  • $v_m\to 0$ in $L^2(\mathbb{R};X_0)$ weakly; (2.29)
  • $|\xi|^\gamma\hat{v_m}\to 0$ in $L^2(\mathbb{R};X_1)$ weakly. (2.30)

Then for (fixed large enough) $M>0$, $$ \lim_{m\to\infty}\int_{|\xi|\leq M}\|\hat{v_m}(\xi)\|_{X_1}^2\ d\xi\to 0.\tag{1} $$

Here is my question: Would anybody elaborate why this is true?


[Added:] For $u:\mathbb{R}\to X_1$, $\hat{u}$ denotes the Fourier transform (when exists) $$ \hat{u}(\xi)=\int_\mathbb{R}e^{-2\pi i \xi x}u(x)\ dx. $$


[Added later 11/26/16] In order to show (1) by the Lebesgue dominated convergence theorem, one essentially needs the following:

  • Show that for each $\xi$ with $|\xi|\leq M$, $\|\hat{v_m}(\xi)\|_{X_1}^2\to 0$ as $m\to\infty$. This is the same as saying that $\hat{v_m}(\xi)\to 0$ in $X_1$ strongly and it is done by showing that $\hat{v_m}(\xi)\to 0$ in $X_0$ weakly thanks to the elaboration in Fabian Wirth's answer.
  • Show that the map $\xi\mapsto 1_{|\xi|\leq M}\|\hat{v_m}(\xi)\|^2_{X_1}$ is dominated by some integrable function. (This step has not been elaborated yet.) The argument for this given by the proof is as follows:

    The assumption regarding the supports of $\{v_m\}$, we have $v_m=1_Kv_m$ and thus $$ \hat{v_m}(\xi)=\int_{\mathbb{R}}e^{-2\pi it\xi}1_K(t)v_m(t)\ dt, $$ which implies by Cauchy-Schwartz that for each $\xi$, $$ \|\hat{v_m}(\xi)\|_{X_1}\leq\|v_m\|_{L^2(\mathbb{R};X_1)}\|e^{-2\pi it\xi}1_K\|_{L^2(\mathbb{R};X_1)}. $$ The proof in the book immediately concludes that $$ \|\hat{v_m}(\xi)\|_{X_1}\leq C $$ for some constant $C$. Here is my follow-up question: How does one know that $C$ is independent of $m$?


[Added:] Here is the author's original argument:

enter image description here

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  • $\begingroup$ Is the hat a Fourier transform, or something else? $\endgroup$ – Nate Eldredge Oct 11 '16 at 4:35
  • $\begingroup$ @NateEldredge: Yes, it is a Fourier transform. $\endgroup$ – Jack Oct 11 '16 at 12:05
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The underlined sentence is a general statement about your triple $ X_0\overset{\textrm{cpt}}{\hookrightarrow} X\hookrightarrow X_1 $. If I understand the question correctly, you are satisfied that you have weak convergence of the sequence $\hat{\nu}_\mu$. The rest plays on general properties.

You have a weakly convergent sequence $\{\nu_\mu\}$ in $X_0$. By the principle of uniform boundedness this sequence is bounded in $X_0$. As the embedding into $X$ is compact, the embedded sequence has subsequences which are Cauchy in the norm of $X$. With this is mind it is not hard to show that $\{\nu_\mu\}$ cannot have a subsequence which does not converge to $0$ in the norm of $X$. (Take a subsequence which does not, this has a subsequence converging in $X$-norm to some $x\neq 0$. This gives a contradiction to weak convergence to $0$ in $X_0$.) In other words, the sequence converges to $0$ in the norm of $X$.

Finally, as the emebdding from $X$ to $X_1$ is continuous the sequence will also converge strongly to $0$ in $X_1$ because it already does in $X$.

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  • $\begingroup$ Thank you for your answer. Now I see that (2.30) is not needed at all in this argument. $\endgroup$ – Jack Oct 16 '16 at 14:24
  • $\begingroup$ I am very embarrassed that I just found later your helpful answer explains only one part of the argument in order to get (1). By any chance, would you bother to take a look at the other part of it: why is $\{\|v_\mu\|_{X_1}\}$ bounded? $\endgroup$ – Jack Nov 26 '16 at 16:04
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    $\begingroup$ Hi Jack. Not sure what the question is here: ${\nu_\mu}$ is already bounded in $X_0$. As the embeddings are continuous, bounded sequences are mapped to bounded sequences. Does this solve your problem? $\endgroup$ – Fabian Wirth Nov 28 '16 at 10:16
  • $\begingroup$ Thank you very much for your comment! I think you meant "$v_{\mu}$ is already bounded in $L^2(\mathbb{R};X_0)$", which is implied by the assumption of weak convergence. (And I should have asked instead in the previous comment "why is $\{\|v_\mu\|_{L^2(\mathbb{R};X_1)}\}$ bounded") $\endgroup$ – Jack Nov 28 '16 at 13:25

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