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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 $gāˆˆL^1(m)$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.

Nazono Sumiko
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    Since $L^\infty$ is not separable, its dual can't be separable. See e.g. this. But $L^1$ is separable. ā€“ Robert Israel Mar 24 '24 at 04:39
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    This may be a problem of title phrasing. It is not clear if the OP in the post you link is quoting Rudin (to whom they attribute the statement), but "$(L^{\infty})^* \neq L^1$" makes sense if by "$L^1$" one means the "copy" of $L^1$ that lives inside $(L^{\infty})^$ via the canonical embedding of a Banach space in its double dual. In that context, the statement summarized by the inequality is that that* embedding is not an isomorphism, not (as the author of the linked post suggests in that post's title) that the spaces are not abstractly isomorphic (also true, but unrelated to the exercise). ā€“ leslie townes Mar 24 '24 at 07:22
  • @leslietownes I understand what you're saying. It seems like that the answer depends on the way of interpreting the statement of problem(even though I personally think the way the how the problem formulate is quite confusing). Thanks! ā€“ Nazono Sumiko Mar 25 '24 at 07:21

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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.

Nazono Sumiko
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