Proof that $(L^1)\neq(L^\infty)^\ast$

Question

I have seen a "proof" that $L^1\neq(L^\infty)^\ast$ which goes as follows: show that there is an element of $(L^\infty)^\ast$ which is not in the image of the canonical map $L^1\rightarrow(L^\infty)^\ast$. From this they conclude that $(L^1)\neq(L^\infty)^\ast$, how does this follow? I mean simply because the canonical map is not an isomorphism it does not follow that they are not isomorphic (isometrically), isn't it? We must somehow distinguish them by properties (reflexive, separable etc) isn't it?

Answer 1

We know that, $L_\infty$ is not separable, so neither does its dual $L_\infty^*$. It is remains to recall that $L_1$ is separable.