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Let $\Omega\subset \mathbb R^d$ be a smooth, bounded domain. Let $(e_n)_{n\geq 0}\subset L^2(\Omega)$ be the Hilbert basis generated by the Dirichlet-Laplacian eigenfunctions, i-e $-\Delta e_n=\lambda_n e_n$ with zero boundary conditions. We know that $\lambda_n\to+\infty$.

For $s\geq 0$ let me denote the "spectral" $H^s$ norm $$ \|f\|^2_{H^s}=\sum\limits_{n\geq 0} \lambda_n^{s}|f_n|^2 \hspace{1cm}\mbox{for }f=\sum\limits_{n\geq 0} f_ne_n, $$ and finally for $N\geq 1$ let me denote the projector onto $E_N=span(e_0,\dots,e_{N-1})$ as $$ P_N f=\sum\limits_{n\leq N-1} f_n e_n. $$


Fact 1: for any fixed $f\in H^s$ there holds $(1-P_N)f\to 0$ in $H^s$, which can be stated as "$1-P_N\to 0$ pointwise on $H^s$". This is easy to check, since $$ \|(1-P_N)f\|_{H^s}^2=\sum\limits_{n\geq N}|\lambda_n|^{s}|f_n|^2\to 0 $$ as the remainder of a convergent series.


Fact 2: for any $r<s$ there holds $$ \|(1-P_N)f\|_{H^r}\leq \lambda_{N}^{(r-s)/2}\|f\|_{H^s}, $$ which can be stated as "$(1-P_N)\to 0$ in the $H^r$ norm uniformly on any $H^s$ ball." Indeed, one can write immediately $$ \|(1-P_N)f\|_{H^r}^2 =\sum\limits_{n\geq N}|\lambda_n|^{r-s}|\lambda_n|^s|f_n|^2 \leq |\lambda_N|^{r-s}\sum\limits_{n\geq N}|\lambda_n|^s|f_n|^2 \leq |\lambda_N|^{r-s} \|f\|_{H^s}^2 $$


Question: can we extend the uniform convergence on arbitrary $H^r$-compact sets? More explicitly, is it true that $$ \sup\limits_{f\in K}\|(1-P_N)f\|_{H^r}\to 0 \qquad\mbox{as }N\to\infty $$ for any $H^r$-compact set $K$? Fact 2 ecactly guarantees that this holds at least if $K$ is an $H^s$ ball, which is indeed $H^r$-compact classical Sobolev embedding since $r<s$. What about more generic compact sets?

I suspect this is well-known but for some reason I could not find anything on the subject (me not being a specialis in spectral analysis certainly does not help).

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  • $\begingroup$ The Banach-Steinhaus theorem seems promising. Have you tried this? $\endgroup$
    – Leo Moos
    Commented Jul 20, 2021 at 13:21
  • $\begingroup$ If $(T_n)$ is a sequence of uniformly bounded, linear operators in a Banach space $X$ and $T_n x \to 0$ for every $x \in X$, then the convergence is uniform on a compact set $K$. Just fix $\epsilon>0$ and cover $K$ with a finite number of balls $B(x_i, \epsilon)$ and $\|T_n x\| \leq \|T_n(x-x_i)\|+\|T_n x_i\|$. $\endgroup$ Commented Jul 20, 2021 at 13:31
  • $\begingroup$ @LeoMoos: yes, of course this is the first thing that came to mind, but I couldn't make it work. Giorgio Metafune's comment seems to do the job, though. $\endgroup$ Commented Jul 20, 2021 at 13:37
  • $\begingroup$ @GiorgioMetafune : that works, I can't believe how simple the answer was... Please make this a proper answer so I can accept it? $\endgroup$ Commented Jul 20, 2021 at 13:37
  • $\begingroup$ It is fine that you get an answer to the question. No problem if it stays in the comments. $\endgroup$ Commented Jul 20, 2021 at 13:45

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If $(T_n)$ is a sequence of uniformly bounded, linear operators in a Banach space $X$ nd $T_nx→0$ for every $x∈X$, then the convergence is uniform on a compact set $K$. Just fix $ϵ>0$ and cover $K$ with a finite number of balls $B(x_i,ϵ)$ and use $∥T_nx∥≤∥T_n(x−x_i)∥+∥T_n x_i∥$.

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  • $\begingroup$ Just for the record: I realized this morning that this is not limited to linear equibounded operators: the proof works exactly the same for an equi-continuous family of possibly nonlinear functions. And I also realized that this is actually a variant of the Ascoli-Arzela theorem, where we already assume pointwise convergence (instead of pointwise relative compactness) and want to upgrade to uniform convergence. $\endgroup$ Commented Jul 21, 2021 at 13:05

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