Let $c>0$ and let $Y$ be the space of all distributions of compact support in $(-1,1)$ with singular support at $\{0\}$. Let $X$ be subspace of $Y$ such that for any $\phi \in X$ there holds: $$ \int_{t^2>r^2} \phi'(t)^2\,dt \leq c\,\int_{t^2>r^2-r^4} \phi(t)^2\,dt \quad \forall\, r\in (0,1).$$ Is it true that $X$ must be finite dimensional?
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$\begingroup$ If $\phi$ is a distribution, how are the displayed integrals defined? Also, what do you mean by "of compact support in $(-1,1)$ with singular support at $\{0\}$", and what role can the "singular support at $\{0\}$" possibly play in the displayed inequality, involving only values of $\phi(t)$ for $t$ away from $0$? $\endgroup$– Iosif PinelisCommented May 10, 2023 at 3:19
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$\begingroup$ Also, by "Let $X$ be subspace of $Y$ such that for any $\phi \in X$ there holds", do you mean "Let $X$ be the subspace of $Y$ consisting of all $\phi \in Y$ such that the following holds"? $\endgroup$– Iosif PinelisCommented May 10, 2023 at 3:26
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2$\begingroup$ @IosifPinelis 1) The integrals are taken over the sets of regular points, where $\varphi$ is an ordinary function (perhaps, even $C^\infty$). 2) No, the meaning is as written: we take some linear subspace of $Y$ with the described property and ask if it can be infinite-dimensional. 3) There should be a straightforward counterexample, but it is too late here for me to check the details. $\endgroup$– fedjaCommented May 10, 2023 at 3:34
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1$\begingroup$ @fedja : Is 1) your guess, or how did you figure that out? And, if that is so, why mention distributions at all? $\endgroup$– Iosif PinelisCommented May 10, 2023 at 3:52
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2$\begingroup$ @IosifPinelis Singular point of a distribution is usually defined as the point in no neighborhood of which $\varphi$ is a (nice) function (how nice may depend of the context). Not every function with a singularity at $0$ can be extended to a distribution. $\endgroup$– fedjaCommented May 10, 2023 at 3:55
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$\newcommand\de\delta$The answer is no. E.g., let $X$ be the linear span of the set $\{\de,\de',\de'',\dots\}$, where $\de$ is the Dirac delta distribution supported on $\{0\}$. Then $X$ satisfies your condition but is infinite dimensional.
answered May 10, 2023 at 4:38
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$\begingroup$ The one comment about this example is that the restrictions of X away from the origin is finite dimensional. Maybe that should have been my question so as to make this kind of counter example not possible. $\endgroup$– AliCommented May 10, 2023 at 12:14