It is well known that the fourier transform map $\mathcal F:L^1(\mathbb R^n)\to C_0(\Bbb R^n)$ is an injective continuous map (actually, it's a morphism between $(L^1,*)$ and $(C_0,\cdot)$) which is not surjective (if it were, the inverse would be continuous and this is easily seen not to be the case). However, this does not settle the question of wether the two Banach spaces (respectively, Banach algebras) can be isomorphic (take isomorphic here to mean either isometrically isomorphic or simply linearly isomorphic; an answer to either of those questions would be interesting).
My first idea on how to solve this would be to try and find a Banach space property which is invariant under isomorphism that $L^1$ has and $C_0$ does not. Unfortunately, both spaces are separable, their unit balls have no extremal point, both have the DPP (Dunford-Pettis property) and non-separable dual, so I'm stuck.
My question hence is: Are $L^1$ and $C_0$ isomorphic as Banach spaces? If so, are they as Banach algebras and if not, is $L^1$ isomorphic to some $C(K)$ or $C_0(X)$ (with $K$ compact and $X$ locally compact respectively)?
Notice that in the negative case, the second part of the question becomes basically a sort of Gelfand duality result, with the important difference from the standard result that we require bijectivity (which is not guaranteed by Gelfand for generic Banach algebras. In particular, in the case of $L^1$ the standard Gelfand transform is not surjective).
This result feels to me like the kind of result that has been thoroughly investigated and it's probably a corollary of some well-known result that such a morphism does not exist. However, my literature research didn't produce any interesting result; if somebody knows of a book where I can find similar results, please share.
Best Answer
It is known that $L_1(\mu)$ with $\mu$ a probability measure (or even just $\sigma$-finite, I believe) is weakly sequentially complete. A proof of this can be found in Section 5.2 of Kalton and Albiac's Topics in Banach Space Theory (see also the comments above). As an infinite dimensional $C(K)$ or $C_0(X)$, as above, contain $c_0$, they are not weakly sequentially complete and thus not isomorphic to a $L_1$ space (a norm-norm linear isomorphism is a weak-weak isomorphism).
Aside: $L_1$ being weakly sequentially complete implies it does not contain a copy of $c_0$ (which also implies the spaces in question are not isomorphic). Kalton and Albiac show this using $L_1$'s weak sequential completeness and results concerning weakly unconditionaly convergent series. I wonder if this, or the OP's question, can be shown with simpler machinery.