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Is it true that the space of hyperfunctions on $\mathbb{R}^n$ supported at 0 coincides with the space of Schwartz distributions supported at 0?

More explicitly, is it true that any hyperfunction supported at 0 is a finite linear combination of various partial derivatives of the delta-function?

If the answer is yes, a reference would be helpful.

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    $\begingroup$ See Schlichtkrull's book on page 6, where he gives the example $e^{1/x}$ as a hyper function (clearly supported at zero) which is not a distribution. $\endgroup$
    – user1688
    Commented Aug 24, 2015 at 9:11
  • $\begingroup$ I don't understand why $e^{1/x}$ is supported at zero. It seems to me that its support is equal to $\mathbb{R}$. $\endgroup$
    – asv
    Commented Aug 24, 2015 at 9:22
  • $\begingroup$ Because it is holomorphic outside zero. See the first pages of Schlichtkrull's book on how hyper functions are described as boundary values of holomorphic functions. $\endgroup$
    – user1688
    Commented Aug 24, 2015 at 9:32
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    $\begingroup$ Look at the last bullet under Examples. $\endgroup$
    – Deane Yang
    Commented Aug 24, 2015 at 18:51
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    $\begingroup$ @DeaneYang: Right! I missed it. $\endgroup$
    – asv
    Commented Aug 25, 2015 at 6:14

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No, this is not true. Schwartz distributions with compact support inovlve only finitely many differentiations (every such distribiution is a derivative of some order of a usual, integrable function). Hyperfunctions may involve infinitely many differentiations. For example, in dimension $1$ you can take any entire function $f(x)=\sum a_nx^n$, and then $F=\sum a_n\delta^{(n)}$ is a hyperfunction with support at $0$.

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  • $\begingroup$ Thank you! Is there an explicit description of hyperfunctions supported at zero or it is too large? $\endgroup$
    – asv
    Commented Aug 25, 2015 at 6:16
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    $\begingroup$ Distributions were invented by Laurent SCHWARTZ (not Schwarz). $\endgroup$
    – Jochen Wengenroth
    Commented Aug 25, 2015 at 7:47
  • $\begingroup$ @Jochen Wengenroth: Thanks for spotting this misprint. I corrected. $\endgroup$
    – Alexandre Eremenko
    Commented Aug 25, 2015 at 12:08
  • $\begingroup$ Sorry for commenting on a really old answer, but just to avoid misinforming whoever stumbles upon it - the entire function $f$ has to be of strictly slower-than-exponential growth. Equivalently, $\forall \varepsilon>0: |a_n| \ll \varepsilon^n / n!$. When $f$ is exactly an exponential, you get a delta at a different point. $\endgroup$
    – Alexander Shamov
    Commented Apr 30 at 6:58

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