1. Let $(X,\mu,M)$ be a measure space with $X=[0,1]$, $\mu$=counting measure and $M$ is the $\sigma$-algebra contains all the subset $E$ of $[0,1]$ such that either $E$ or $E^c$(complement) is countable. Since $X$ is not $\sigma$-finite in $\mu$, $(L^1)^* \neq L^{\infty}$. The proof is as follow:

Let $g(x)=x$ which is not measurable. Let $\Lambda f=\int fg d\mu$ for all $f\in L^1$, the integration here makes sense since we can prove $fg$ is measurable. We check that $\|\Lambda \|\leq1$ hence $\Lambda $ is a linear bounded functional. Suppose there is $h \in L^{\infty}$ such that $\Lambda f=\int fh d\mu=\int fg d\mu$, then we can choose some special $f$ to obtain $h=g$ a.e.. Thus we have a contradiction here since $h$ is measurable but $g$ is not.

2 Let $(X,\mu,M)$ be a measure space with $X=[0,1]$, $\mu$=Lebesgue measure. It is well known that $(L^{\infty})^* \neq L^{1}$. The proof is as follow:

Let $C(X)$ be the space containing all continuous functions in $X$, which is proper closed subspace of $L^{\infty}$. Thus, by Hahn-Banach theorem, there is $\Lambda \in (L^{\infty})^*$ such that $\Lambda =0$ on $C(X)$ and $\Lambda \neq 0$ in $L^{\infty}$. Suppose there is $g\in L^1$ such that $\Lambda f= \int fgd\mu $. But $\Lambda f= \int fgd\mu=0 $ for all $f \in C(X)$, we can prove that $g=0$ which implies $\Lambda = 0$ contradicts to $\Lambda \neq 0$ in $L^{\infty}$.

In both of the proof, it seems that they assume that the space $X$ is the dual of $Y$ if and only if they have a Riesz representation. But I am not sure if it is true. The above proof shows that the map $g \mapsto \Lambda$ by $\Phi(g)=\Lambda_g(f)=\int fg d\mu$ is not a isometry (not onto). But it does not mean that there exist no other isometry between the space.

  • $\begingroup$ You are right. The proofs do not show that $(L^{1})^{*}$ is not isometrically isomorphic to $L^{\infty}$ $\endgroup$ – Kavi Rama Murthy Feb 14 '18 at 6:33
  • $\begingroup$ You mean that my proof is correct but does not show the required result? I think the above is the required answer since i have seen it many many times. $\endgroup$ – mnmn1993 Feb 14 '18 at 7:35
  • $\begingroup$ The statement "$(L^1)^\ast = L^\infty$" (if given without further explanation) is always taken as meaning that the map $\Phi$ that you write at the end of your post is an isometric isomorphism. In this sense, your proof proves the claim. $\endgroup$ – PhoemueX Feb 14 '18 at 7:45
  • $\begingroup$ You mean that the map in my proof is the only possible isometry between the space? $\endgroup$ – mnmn1993 Feb 14 '18 at 9:45
  • $\begingroup$ I do not know if an isometry exist or not. It is even possible that the answer is not known. $\endgroup$ – Kavi Rama Murthy Feb 14 '18 at 10:10

The point of the dual is to use its elements (you don't get anything useful from saying that the dual of $\mathbb C^4$ is isomorphic to $M_2 (\mathbb C) $, but you do from saying that it is isomorphic to $\mathbb C^4$ with the duality given by the inner product).

In the case of $L^1$ and $L^\infty $, "integration against" gives you natural isometric embeddings $$L^\infty\subset (L^1)^*,\ \ \ \ \ L^1\subset (L^\infty)^*. $$The examples you quoted show that in general both embeddings fail to be onto.


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