The problem statement is the same as this post: $L^{\infty *}$ is not isomorphic to $L^1$ .
Let $L^\infty = L^\infty(m)$, where $m$ is Lebesgue measure on $I=[0,1]$ . Show that there is a bounded linear functional $G \neq 0$ on $L^\infty$ that is $0$ on $C(I)$ , and that therefore there is no $g∈L^1(m)$that satisfies $G(f) = \int_I fg$ for every $f \in L^\infty$. Thus $(L^\infty)^{*} \neq L^1$.
I have no problem in constructing such $\lambda$ and then deduce that it cannot have an integral representation by an integrable function $g$. My question is that I don't think this conclusion is strong enough to conclude that the two spaces are not isomorphic as it was answered in that post.
The accepted answer there stated that $L^1$ is a $\textit{subspace}$ of $(L^\infty)^{*}$ be considering the embedding map $\lambda :L^1 \to (L^\infty)^{*} :h \mapsto \lambda_h$, where $$ \lambda_h(f) : = \int hf \enspace ,\forall f \in L^\infty. $$ The functional $G$ constructed above is an example such that $ G\in (L^\infty)^{*} \backslash\lambda(L^1)$, but this doesn't imply the non-existence of an isomorphsim $\phi$ between $(L^\infty)^{*} $ and $L^1$. It just simply says that such isomorphism, if exists at all, cannot be $\lambda$. The problem with the answer is that it really let $L^1$ be a subspace of $(L^\infty)^{*}$, instead of just an embedding.
Do I get anything wrong? If this example is really sufficient to prove that the two spaces are not isomorphic, please provide a proof.
I'll give a proof here, which uses Hahn-Banach theorem and fills out the holes in the answer of the post mentioned above.
We consider the collection $\{H_x\}_{x\in I}$ of functionals on $L^\infty$, where each $H_x$ is the Hahn-Banach extension of the functional $G_x: C(I) \to \mathbb{R}$ defined by $$ G_x(f) = f(x) \enspace \forall x \in I, f\in C(I). $$ Let $ 1 >\delta >0$ be given. For each $x\neq y$, we can always find an $f_{xy} \in C(I)$ such that $||f_{xy}||_\infty = 1$ and $|H_x(f) - H_y(f)| = |f(x) - f(y) | \geq \delta >0$. Thus, $||H_x - H_y|| \geq \delta$ for all $ x\neq y$. Then $ \{H_x\}_x \subseteq (L^\infty)^{*}$, hence $(L^\infty)^{*}$ cannot be separable, which in turn implies that $(L^\infty)^{*}$ cannot be isometrically isomorphic to $L^1$, since the later is separable.