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The following is a repeat from the following question on MathStack Exchange. I am just hoping to have more success here.

This is a follow-up from that question. The question is this: I want to determine a subspace (as large as possible) of $H^1(\mathbb{R})$ or $L^2(\mathbb{R})$ on which there exists a constant $C$ such that $$ \|f'\|_2 \leq C \|f\|_2,\text{ for any } f \text{ on that subspace}. \tag{1}\label{1} $$ I have learned from a comment from @whpowell96 to my question that I should be looking at the Poincaré inequality, which led me to the reverse Poincaré inequality. I looked and found out that the solutions of certain 2-D (i.e. $\mathbb{R}^2$) PDEs satisfy that inequality. However, I am looking for a subspace in one dimension.

Looking at the norm of the Fourier transform of the derivative $$ \|f'\|_2 = \||\xi| \hat{f}(\xi)\|_2, $$ then one can just look at the set of functions whose Fourier transform has a compact support, such as $[-a,a] \subset \mathbb{R}$, and the $C$ in (1) can be chosen as $C = a^2$. Then I found this question concerning what it means to have a compactly supported Fourier transform. However, I do not quite understand what the condition given means.

Is there a more general way to define a subspace satisfying \eqref{1}? If not, is there a better one than the one here to explain what it means to look at functions with compact Fourier transform?

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  • $\begingroup$ Morally, compact support of the Fourier transform is the right sort of condition to look for. Note that (1) fails if $f$ is permitted to become highly oscillatory. Compact support of the Fourier transform is a standard (but not exhaustive) way to enforce limited oscillation. $\endgroup$ Commented Aug 13 at 4:52
  • $\begingroup$ @BenJohnsrude I really appreciate your comment! I am particularly grateful for the reassuring words "morally" and "standard"! Thanks. If you happen to know another way to limit oscillations so that (1) is satisfied, please let me know. Also, if you know some works where limiting oscillations is done that way, I would appreciate it. I would dig from there just to have some justification for what I am doing. $\endgroup$ Commented Aug 13 at 10:29
  • $\begingroup$ I think the theorems of type Paley-Wiener are a good start. In the book by Yosida Functional analysis there is a section on those theorems (also for distributions with compact support). EDIT sorry if that doesnt help, since they already talk in your linked question about Paley Wiener. $\endgroup$ Commented Aug 13 at 11:36
  • $\begingroup$ @Jens Fischer Thank you, I overlooked the comment about that theorem, and now I looked at it, and it seems very helpful. I have that book in my office. So your comment was very useful! Thanks. $\endgroup$ Commented Aug 13 at 21:33
  • $\begingroup$ Another possibility is the image of the spectral projection $E_{\lambda}$ of a Hamilton operator of the form $p^2 + V(x)$ with $V \geq 0$ for every positive $\lambda$. $\endgroup$ Commented Aug 14 at 18:34

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What you seem to be asking is a "converse Bernstein-type inequality":

Theorem. Let $S$ be a subspace of $L^2(\mathbf{R})$ on which the differentiation is defined as a bounded operator $S\to S$: $$\| f'\|\leq a\| f\|,\quad f\in S.\quad\quad\quad (1)$$ Then the supports of Fourier transforms of all functions $f\in S$ are contained in $[-a,a]$.

Proof. Let $f\in S$, and suppose that $\mathrm{essupp}\hat{f}$ contains some points outside $[-b,b]$ where $b>a$. Then $$2\pi\| f^{(n)}\|^2=\| (f^{(n)})\hat{}\|^2=\int s^{2n}|\hat{f}(s)|^2ds$$ $$\geq\int_{|s|\geq{b}}s^{2n}|\hat{f}(s)|ds\geq b^{2n}c,$$ where $$c=\int_{|s|\geq b}|\hat{f}|^2ds>0.$$ This is a contradiction since (1) implies $\| f^{(n)}\|^2\leq a^{2n}\| f\|^2$ for all functions in $S$.

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    $\begingroup$ This argument presumes that S is closed under differentiation. But there are spaces (e.g. finite dimensional ones) on which differentiation is bounded as an operator in $L^2$ but does not map back to S. $\endgroup$ Commented Aug 14 at 14:51
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    $\begingroup$ @Michael Renardy: I don't think that the question makes sense for such subspaces. On a finite dimensional space, every linear transformation is bounded, So you can construct many infinite dimensional spaces on which derivation is bounded. $\endgroup$ Commented Aug 14 at 15:29
  • $\begingroup$ @AlexandreEremenko Thank you! I really would like that result of yours to be true but I do not see your contradiction. Since c<\| f \|^2, I don't see how your inequality is in contradiction with \|f'\|^2\leq a^2\|f\|^2. $\endgroup$ Commented Aug 14 at 19:15
  • $\begingroup$ @AlexandreEremenko To my understanding, I would start your proof like that. I would take the same $f$ as you but then I would define $\tilde{f}$ as the difference between $f$ and the function whose Fourier transform is the restriction of $\hat{f}$ to $[-a,a]$. Then the FT of $\tilde{f}$ has support in $\{s>|b|\}$ and thus, for $\tilde{f}$, that $c$ in the end is $\|\tilde{f}\|^2$, and that gives the contradiction. $\endgroup$ Commented Aug 14 at 23:28
  • $\begingroup$ @Gateau au fromage: I also corrected few misprints, hope it is more readable now. $\endgroup$ Commented Aug 15 at 13:41

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