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The space $H^{-s}$ consists of all $f$ such that $$\|f\|_{H^{-s}} = \int (1+|k|^2)^{-s/2}\hat{f}(k) dk < \infty.$$ What about this definition implies that $f$ may not be a function but a distribution? How can I see that this must be the dual space to $H^s$?

Mathematics
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    Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Another User Dec 31 '24 at 23:36
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    No. The definition of $H^s(\Bbb{R}^n)$ is the set of all tempered distributions $f$ on $\Bbb{R}^n$ such that $\widehat{f}$ (as a tempered distribution) is in $L^1_{\text{loc}}(\Bbb{R}^n)$ (or rather has a representative) and a certain integral is finite. So, of course it is conceivable that $H^s$ contains distributions for suitable values of $s$. – peek-a-boo Jan 01 '25 at 00:45
  • @peek-a-boo So distributions are present by definition and not as a consequence of the norm? Maybe I am wrong but I thought the $H^{-s}$ norm allows for distributions while the $H^s$ norm does not. If its a matter of definition why do we not allow distributions in the definition of $H^s$ for positive $s$? – Mathematics Jan 01 '25 at 01:22
  • In my comment I made no assumption about the sign of $s$. The definition is right as I wrote it. But now clearly if $s\geq 0$ it only contains functions (since the bound on the norm implies $\widehat{f}\in L^2$, so $f\in L^2$, hence a function). So you should just take another look at the definitions, I think that will clear it up. – peek-a-boo Jan 01 '25 at 02:28
  • @peek-a-boo Thanks. Does $H^{-s}$ being the dual space to $H^s$ follow from their norms? – Mathematics Jan 01 '25 at 05:12

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The point is that $H^s$ for negative $s$ contains distributions that are not “functions”, by which I mean a distribution that has a representative in $L^1_{loc}$. For example, consider the Dirac delta, which lies in $H^s$ if and only if $s<-1/2$ and is not the action of integration against a locally integrable function. This is not the case for. $s\ge 0$ since then $H^s$ embeds in $L^2$ which embeds in $L^1_{loc}$

Diffusion
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  • Thanks. Can the fact that $H^{-s}$ is the dual space to $H^s$ be seen from the norm? – Mathematics Jan 01 '25 at 05:10
  • As shown here for dimension $n=1$ (https://math.stackexchange.com/questions/4033589/sobolev-space-with-negative-index/4046938#4046938), whenever s is negative, the space $H^s$ contains elements that are "irregular" distributions, i.e., "true" distributions that are not given by locally integrable functions. – PhoemueX Jan 02 '25 at 21:35