2 cleaning up the mathematical markup

It's not germane to your question, but I can't resist pointing out that it is very hard to exhibit any continuous linear bijection from $L^1$(sensible measure space) onto $C_0$(sensible topological space), and in fact if either space is infinite then I suspect this is never possible, just for reasons of Banach space geometry. Thus, although it doesn't help with what you want to look at, I thought it might be worth mentioning that one can know the answer to "is the FT onto?" must be "no", before looking for an example or using properties of the Fourier transform.

(My caveats are because I don't want to categorically state it can't be done, but in all cases I can think of no such bijection will exist. However, both my general measure theory and my general topology are not what they should be, so I can't remember how to do things precisely in the most general settings.)

Anyway. I claim that there is no continuous linear bijection between $L^1(R)$ L^1({\mathbb R}^d)$and$C_0(X)$, where$X$is locally compact Hausdorff (e.g. a metric space). The reason is that we have big powerful results telling us that (i) every bounded linear operator from$C_0(X)$to$L^1(R)$L^1({\mathbb R}^d)$ is weakly compact;

(ii) if the identity map on a Banach space $E$ is weakly compact, then $E$ is reflexive;

(iii) $L^1(R)$ L^1({\mathbb R}^d)$is not reflexive (ibid). Unfortunately I can't locate a self-contained proof of the key fact (i). (It can be deduced as a corollary of a rather powerful, fundamental and beautiful result - due to some promising former student of Dieudonné and Schwartz, not sure if he ever went on to do anything important...) 1 It's not germane to your question, but I can't resist pointing out that it is very hard to exhibit any continuous linear bijection from$L^1$(sensible measure space) onto$C_0$(sensible topological space), and in fact if either space is infinite then I suspect this is never possible, just for reasons of Banach space geometry. Thus, although it doesn't help with what you want to look at, I thought it might be worth mentioning that one can know the answer to "is the FT onto?" must be "no", before looking for an example or using properties of the Fourier transform. (My caveats are because I don't want to categorically state it can't be done, but in all cases I can think of no such bijection will exist. However, both my general measure theory and my general topology are not what they should be, so I can't remember how to do things precisely in the most general settings.) Anyway. I claim that there is no continuous linear bijection between$L^1(R)$and$C_0(X)$, where$X$is locally compact Hausdorff (e.g. a metric space). The reason is that we have big powerful results telling us that (i) every bounded linear operator from$C_0(X)$to$L^1(R)$is weakly compact; (ii) if the identity map on a Banach space$E$is weakly compact, then$E$is reflexive; (iii)$L^1(R)\$ is not reflexive (ibid).

Unfortunately I can't locate a self-contained proof of the key fact (i). (It can be deduced as a corollary of a rather powerful, fundamental and beautiful result - due to some promising former student of Dieudonné and Schwartz, not sure if he ever went on to do anything important...)