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The Wiener algebra $\mathcal W$ is defined as $\text{Fourier}(L^1(\mathbb R))$, i.e. the image by the Fourier transform of $L^1(\mathbb R)$. Riemann-Lebesgue's lemma ensures that $$ \mathcal W\subset C^0_{(0)}(\mathbb R)=\{\phi\text{ continuous on }\mathbb R, \lim_{\vert \xi\vert\rightarrow+\infty}\phi(\xi)=0\} . $$

  1. I believe that the injection $\mathcal W\subset C^0_{(0)}(\mathbb R)$ is not onto. Is it due to Hardy? Gaier? Both at different times?

  2. Is there an "explicit" function $\phi\in C^0_{(0)}(\mathbb R)$ whose inverse Fourier transform (say in the distribution sense) does not belong to $L^1(\mathbb R)$?

  3. Is there a functional analytic reason for why the Banach spaces $L^1(\mathbb R)$ and $C^0_{(0)}(\mathbb R)$ cannot be isomorphic?

Ricardo Andrade
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Bazin
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2 Answers2

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For (2), the answers to Does there exist a continuous function of compact support with Fourier transform outside L^1? may be helpful.

For (3), the answer is yes, although one always needs to rely on some theoretical background. My preferred argument is to note that every bounded linear map $C_0({\bf R}) \to L^1(X,\mu)$ ($X$ and $\mu$ arbitrary) is weakly compact, hence if the two spaces were isomorphic then the identity map on $C_0({\bf R})$ would be weakly compact, hence $C_0({\bf R})$ would be a reflexive Banach space, which it isn't.

I don't know to what extent these remarks answer (1) -- are you looking for an actual reference as to who showed the FT is not surjective?

Community
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Yemon Choi
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    As for (3), $C^0_{(0)}$ contains a copy of $c_0$ (take the linear span of a sequence disjointly supported functions), whereas for $L_1(\mu)$ this is clearly impossible for many reasons (e.g. because $L_1(\mu)$ is weakly sequentially complete). – Tomasz Kania Apr 21 '15 at 17:29
  • Thanks for your answer. Question (1) was only intended to clarify the priorities in a reference list. – Bazin Apr 22 '15 at 11:34
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I want to add that all functions in Wiener algebra is uniformly continuous. So This gives examples to (2).

In addition, https://mathoverflow.net/a/203514/514950 gives a link to a function which is uniformly continuous but not in $\mathcal W$.

FLiite_coro
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    Regarding your first line: every function in $C^0_{(0)}({\mathbb R})$ is uniformly continuous, so your observation doesn't seem to address point (2). Regarding your second line: you seem to just be repeating what I wrote 8 years ago... – Yemon Choi Oct 12 '23 at 21:23