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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.

Pelota
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    $L_1$ does not contain a subspace isomorphic to $c_0$ (or maybe $L_1$ being Weakly sequentially complete suffices; I can't recall if $C(K)$ is wsc or not). – David Mitra Apr 16 '23 at 13:20
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    $C(K)$ is not weakly sequentially complete. – David Mitra Apr 16 '23 at 13:37
  • See here for the weak sequential completeness of $L^1$: https://math.stackexchange.com/questions/40920/weak-limit-of-an-l1-sequence see here for the claim regarding $C(K)$: https://math.stackexchange.com/questions/866650/c0-1-is-not-weakly-sequentially-complete – PhoemueX Apr 16 '23 at 13:41
  • @DavidMitra Thank you for the answer. Ineed, I should have thought about the embeddability of $c_0$ in $L_1$ and that settles everything (of course, the sequential completion works as well). If you want to write it down as an answer, I'll be happy to accept it. – Pelota Apr 16 '23 at 17:38
  • Another isomorphic property that $C(K)$ has but $L^1$ does not is Pelczynski's property (V). – Onur Oktay Apr 19 '23 at 21:06

2 Answers2

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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.

David Mitra
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  • Concerning the hypothesis on the measure, I think one can discard them altogether: clearly, a sequence of elements in $L_1$ has $\sigma-$finite support, hence the $\sigma-$finite result implies the general one. On whether there's a simpler proof, I guess it depends on whether you consider the concept of cotype to be simpler (I don't). – Pelota Apr 16 '23 at 19:18
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A different proof of the result (inspired by the answer of David Mitra) is as follows: notice that $L_1$ has cotype $2$, while $C_0$ does not have nontrivial cotype (since it contains $c_0$), hence the two cannot be isomorphic.

Pelota
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