I am currently trying to understand the following proposition:
Let $\mu$ be a Radon measure on $\mathbb{R}^n$. Then:
(1) $L^{\infty}(\mu)$ is neither reflexive nor separable
(2) $L^{1}(\mu)$ is not reflexive
I have thought about a counterexample and came up with one, but I have the feeling it is incorrect. I would appreciate it if someone could tell me if this counterexample makes any sense. I have seen different posts about this question but wanted to find a solution that is based on the topics we have discussed in my class.
Let $X=L^{\infty}([0,1],\mathcal{B}([0,1]),\lambda_{|[0,1]})$. I was able to construct a bounded sequence of operators $(T_k)_k$ in $X'$, which has no weakly* convergent subsequence.
Since every sequence in the dual Space of a separable Banach space has a weakly* convergent subsequence it follows that $X$ is not separable.
Now suppose that X is reflexive then it follows that $X'$ is reflexive, so every bounded sequence in $X'$ has a weakly convergent subsequence. Since weak convergence in $X'$ implies weak* convergence in $X'$, it follows that every bounded sequence in $X'$ has a weakly* convergent subsequence. This is a contradiction to the above. So $X$ and $X'$ can't be reflexive.
Since the Lebesgue-measure on $[0,1]$ is $\sigma$-finite the Riesz-Represantation-Theorem shows that there is an isomorphism $I_{1}: L^1(\lambda_{|[0,1]})\to X'=L^\infty(\lambda_{|[0,1]})'$. Since $X'=L^\infty(\lambda_{|[0,1]})'$ is non reflexive it follows that $L^1(\lambda_{|[0,1]})$ is non reflexive.
Best Answer
An additional assumption about the support of the measure $\mu$ is needed for the statement made by the OP to hold. If $\operatorname{supp}(\mu)$ is finite, then the $L_p$ spaces are finite dimensional and equivalent and so, density and reflexivity trivially hold.
For the rest of this posting, we assume that $\operatorname{supp}(\mu)$ is an infinite set.
To address the separability in question (1) we may proceed as follows:
There is a countable sequence of $\mu$-measurable sets $A_n\subset\mathbb{R}^n$ such that $0<\mu(A_m)<\infty$ and $A_k\cap A_m=\emptyset$ if $k\neq m$, and $\mu(\mathbb{R}^n\setminus\bigcup_mA_m)=0$. This can be obtained for example, by splitting $\mathbb{R}^n$ in unit boxes $(m_1,m_1+1]\times\ldots\times(m_n,m_n+1]$ with integer vertices and then bisecting each side of the boxes if necessary. For each $B\subset\mathbb{N}$, define $f_B=\mathbb{1}_{\bigcup_{m\in B}A_m}$. For any $B,B'\subset \mathbb{N}$ with $B\neq B'$, $\|\mathbb{1}_B-\mathbb{1}_{B'}\|_\infty=1$. This precludes the existence of a countable dense set in $L_\infty(\mu)$ since $\{f_B:B\subset\mathbb{N}\}$ is uncountable.
As for the reflexivity (or lack thereof) of the spaces $L_\infty(\mu)$ and $L_1(\mu)$, one requires some heavier machinery.
Suppose $(\mathbb{R}^n,\mathcal{M},\mu)$ is a Radon measure ($\mathscr{B}(\mathbb{R}^d)\subset\mathcal{M})$ and that $\operatorname{sump}(\mu)$ is an infinite set.
First notice that $\mu$ is $\sigma$-finite and so $(L_1(\mu))^*=L_\infty(\mu)$.
It is a well known fact that $(L_\infty(\mu))^*$ is the space of all finitely additive charges $m$ on $\mathscr{B}(\mathbb{R}^n)$ of finite variation such that $m(A)=0$ whenever $\mu(A)=0$. (A charge satisfies the properties of a a complex measure (or real finite), except that countable additivity is substituted by finite additivity).
Observe that for any $f\in L_1(\mu)$, $m_f:=f\,d\mu$ defines a countable additive complex (or real finite) measure.
Though no one has given an explicit example of a charge, their existence can be obtained through the Hanh-Banach extension theorem. To construct a (finite) charge the is absolutely continuous with respect to $\mu$, suppose there is a cluster point $x_0$ of $S_\mu$ (i.e. $B(x_0;r)\cap S_\mu\setminus\{x_0\}\neq\emptyset$ for all $r>0$), and define $$Y=\{\tilde{f}\in L_\infty(\mu): \exists f\in \tilde{f},\,\text{such that}\, \lim_{x\rightarrow x_0}f(x)\quad\text{exists}\}$$ It is easy to see that if $f',f\in\tilde{f}$ and $\lim_{x\rightarrow x_0}f'(x)=\alpha'$ and $\lim_{x\rightarrow x_0}f(x)=\alpha$, then $\alpha=\alpha'$. Indeed, given $\varepsilon>0$, there is an open neighborhood $U$ of $x_0$ such that \begin{align} |f'(x)-\alpha'|&<\varepsilon/2\\ |f(x)-\alpha|&<\varepsilon/2 \end{align} for all $x\in U$. As $\mu(\{f\neq f'\})=0$ and $x_0$ is a cluster point of $S_\mu$, there is $x\in U\setminus\{x_0\}$ such that $f(x)=f'(x)$; hence $|\alpha-\alpha|\leq |\alpha-f(x)|+|f'(x)-\alpha'|<\varepsilon$. Letting $\varepsilon\searrow0$ yields $\alpha=\alpha'$. Define $$\Lambda \tilde{f}=\lim_{x\rightarrow x_0}f(x)$$ for $\tilde{f}\in Y$. This is a well define linear functional on $Y$ and $|\Lambda\tilde{f}|\leq\|\tilde{f}\|_{L_\infty}$ for all $\tilde{f}\in Y$. The Hanh-Banach theorem provides an extension of $\Lambda$ to a bounded functional on $L_\infty(\mu)$ such that $$\|\Lambda\|_Y=\sup_{f\in Y:\|f\|_\infty=1}|\Lambda f|=\sup_{f\in L_\infty(\mu):\|f\|_\infty=1}|\Lambda f|=\|\Lambda\|$$ It is easy to check that $\Lambda$ is additive but not countable additive, for $\Lambda \mathbb{1}_{B(x_0;1/n)}=1$ but $\Lambda\mathbb{1}_{\{x_0\}}=0$.
The case when $S_\mu$ is discrete is left to the OP.
The point of all this is that $L_1(\mu)\subsetneq (L_\infty(\mu))^*$. The lack of reflexivity of $L_1(\mu)$ follows.
From the lack of reflexivity of $L_1(\mu)$ the lack of reflexivity of $L_\infty(\mu)$ follows since for ant Banach space $X$, $X$ is reflexive iff $X^*$ is reflexive.
Comments:
An interesting discussion of this construction of $\Lambda$ made above is in the posting by Yousuf Soliman here which mentions that if $\Lambda$ is restricted to $\mathcal{C}_0(\mathbb{R}^d)$, then extension obtained from the Hahn-Banach theorem coincides with $\delta_{x_0}$ measure on $(\mathbb{R}^d,\mathscr{B}(\mathbb{R}^d))$ by virtue of the Riesz-represention theorem. This extension is not useful for our purposes. In other words, the choice of the domain $Y$ for $\Lambda$ is important.
The idea of splitting the space $\mathbb{R}^n$ as a countable union of pairwise disjoint sets of finite and positive measure to construct an uncountable family $\mathcal{F}$ of functions in $L_\infty$ with $\|f-f'\|_\infty=1$ for all $f,f'\in \mathcal{F}$ follows an idea of Andreas Blass here