# Decomposition of distributions

Can we write every (tempered) distribution $\psi$, say on $\mathbb{R}$, as the sum of two distributions

$\psi = \psi_1 + \psi_2$

such that $\psi_1$ and the Fourier transform of $\psi_2$ are actually measurable functions of moderate growth. If so, under which additional conditions are the choices $\psi_1$ and $\psi_2$ unique?

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## 2 Answers

The Dirac Comb, an infinite sum of delta functions, is an example of a tempered distribution that cannot be thusly decomposed (its Fourier transform is another Dirac Comb).

[Added:] There is a positive result in this direction that I (among others) only partly-remembered: Any distribution can be written as a locally finite sum of derivatives of continuous functions. If the distribution has finite order, then the sum is finite. See Rudin's Functional Analysis, Theorem 6.28.

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Oh, that fact follows from Poisson summation. Thank you! – Marc Palm Nov 2 '11 at 16:03
You are welcome! It would be interesting to know how common this situation is, i.e. are "most" tempered distributions either functions or the Fourier transform of functions, or is the situation like that of continuous functions, where most of our examples are a.e. differentiable even though the vast majority are nowhere differentiable. – B R Nov 2 '11 at 16:16
I meant to add: But I don't think there is necessarily a sensible answer here, as the Baire Category theorem does not apply. – B R Nov 2 '11 at 16:51

And any tempered distribution can be written as a finite sum of derivatives of continuous moderate growth functions.

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