# Characterizations of Wiener algebra

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?

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.
• 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). – Tomek Kania Apr 21 '15 at 17:29