$L^1(μ)$ is finite dimensional if it is reflexive

Question

If $(X,\Omega,\mu)$ is a $\sigma -$ finite measure space, show that if $L^1(X,\Omega,\mu)$ is reflexive then it is finite dimensional.

My attempt: I want to show there is a copy of $\ell^1$ in $L^1(X,\Omega,\mu)$. For this suppose $L^1(X,\Omega,\mu)$ is infinite dimensional. there is a sequence $\{x_n\}$ of disjoint points of $X$. For every n, Put $\chi_n:=\chi(x_n)$ (characteristic function ). define $\phi:\ell^1\to L^1(X,\Omega,\mu)$ such that $\phi(\{a_n\})=\Sigma a_n \chi_n$. But I can not show that $\phi$ is an isometry.

I think I did not in a correct way. Please help me. Thanks in advance.

Answer 1

Call $A\in\Omega$ an atom if $\mu(A)>0$, but if $B\subset A$ is measurable, then $\mu(B)=0$ or $=\mu(A)$. Since $\mu$ is $\sigma$-finite, we can cover all atoms by at most countably many atoms $A_n$. Let $Y=X\setminus \bigcup A_n$. Then $\mu(Y)=0$ if $L^1(X,\mu)$ is a dual space because the unit ball of $L^1(Y,\mu)$ has no extreme points: if $\|f\|=1$, we can split $\{x\in Y: f(x)\not=0\}$ into two disjoint sets of positive measure (since we're away from the atoms) and write $f$ as a corresponding convex combination, which would contradict Krein-Milman + Banach-Alaoglu (we're applying KM to the relative topology induced by the weak $*$ topology on the subspace $L^1(Y)$).

However, if $\mu(Y)=0$, then $L^1(X)\cong \ell^1$ via the isomorphism $f\mapsto\int_{A_n}f\, d\mu$.

Answer 2

This is less smart solution than Christian's.

For the begining, every measure space $(\Omega,\mu)$ have continuous $(\Omega_c,\mu_c)$ and atomic part $(\Omega_a,\mu_a)$. Note, for $\sigma$-finite spaces, all atoms are of finite measure. See this question.

Case 1. If atomic part contains infinite amount of atoms we may choose a countable collection of of disjoint atoms $(A_n)_{n\in\mathbb{N}}$. The desired embedding then would be $$ i:\ell_1\to L_1(\Omega,\mu):(a_n)_{n\in\mathbb{N}}\mapsto \left(\omega\mapsto\sum_{n\in\mathbb{N}}a_n\mu(A_n)^{-1}\chi_{A_n}(\omega)\right) $$ Therefore $L_1(\Omega,\mu)$ can't be reflexive as it contains a copy of $\ell_1$ which is not reflexive.

Case 2. If $(\Omega_a,\mu_a)$ consist of finite amount of atoms and continuous part is non empty, then choose any set $A\subset\Omega_a$ of positive finite measure. By Sierpinski theorem, we can divide $A$ onto two parts of equal measure. Each of these aprts we can divide onto two parts of equal measure and etc. In the end we get a sequence of disjoint sets of positive measure. The desired embedding is $$ i:\ell_1\to L_1(\Omega,\mu):(a_n)_{n\in\mathbb{N}}\mapsto \left(\omega\mapsto\sum_{n\in\mathbb{N}}a_n\mu(A_n)^{-1}\chi_{A_n}(\omega)\right) $$ Therefore $L_1(\Omega,\mu)$ can't be reflexive as it contains a copy of $\ell_1$ which is not reflexive.

Case 3. If continuous part of measure space is empty, and atomic part consist of finite amount of disjoint atoms $(A_n)_{n\in\mathbb{N}_m}$, then the $L_1(\Omega,\mu)=\operatorname{span}\{\chi_{A_n}:n\in\mathbb{N}_m\}$. So $\dim (L_1(\Omega,\mu))=m$, and $L_1(\Omega,\mu)$ is finite dimensional.

Answer 3

Here is another proof that does not require any knowledge about atoms or the Banach-Alaoglu or Krein-Milman theorems.

Lemma. Let $(X, \Omega , \mu )$ be a measure space such that $L^{1}(X, \mu )$ is infinite-dimensional. Then there is some $E\in \Omega$ such that $\mu (E) > 0$ and $L^{1}(X\setminus E, \mu )$ is infinite-dimensional.

Proof. Since $L^{1}(X, \mu )$ is infinite-dimensional and the subspace $\mathcal{S}(X, \mu )$ of $L^{1}(X, \mu )$ consisting of simple functions each of which vanish outside a set of finite measure is dense in $L^{1}(X, \mu )$, we have that $\mathcal{S}(X, \mu )$ is infinite-dimensional. Furthermore, as the linear span (taken in $L^{1}(X, \mu )$) of all characteristic functions of members of $\Omega$ with finite measure is equal to $\mathcal{S}(X, \mu )$, there exist $E_{1}, E_{2}\in \Omega$ such that $1_{E_{1}}$ and $1_{E_{2}}$ are linearly independent in $L^{1}(X, \mu )$. The linear independence implies that $\mu (E_{1}) > 0$ and $\mu (E_{2}) > 0$ as well as that $\mu (X\setminus E_{j}) > 0$ for some $j\in \{1, 2\}$ which we now fix. The map $\phi \colon L^{1}(X, \mu ) \to L^{1}(E_{j}, \mu ) \times L^{1}(X\setminus E_{j}, \mu )$ defined by $\phi (f) := (f\vert_{E_{j}}, f\vert_{X\setminus E_{j}})$ is an isometric isomorphism when $L^{1}(E_{j}, \mu ) \times L^{1}(X\setminus E_{j}, \mu )$ is given the norm $\|(f_{1}, f_{2})\| = \int_{E_{j}} |f_{1}| \, d\mu + \int_{X\setminus E_{j}} |f_{2}| \, d\mu$. Hence as $L^{1}(X, \mu )$ is infinite-dimensional, we must have that one of $L^{1}(E_{j}, \mu )$ or $L^{1}(X\setminus E_{j}, \mu )$ is infinite-dimensional. Taking $E$ as one of $E_{j}$ or $X\setminus E_{j}$ so that $L^{1}(X\setminus E, \mu )$ is infinite-dimensional, we see that $E$ has the desired properties.

Proposition. Let $(X, \Omega , \mu )$ be a measure space such that $L^{1}(X, \mu )$ is infinite-dimensional. Then there is a sequence $(E_{n})_{n\in\mathbb{N}}$ of pairwise disjoint members of $\Omega$ with $0 < \mu (E_{n}) < \infty$ for all $n\in\mathbb{N}$.

Proof. As $L^{1}(X, \mu )$ is infinite-dimensional, there exists a sequence $(f_{n})_{n\in\mathbb{N}}$ of linearly independent members of $L^{1}(X, \mu )$. Then the set $X_{0} := \bigcup_{n\in\mathbb{N}} \{x\in X : f_{n}(x) \neq 0\}$ is $\sigma$-finite and $(f_{n}\vert_{X_{0}})_{n\in\mathbb{N}}$ is a linearly independent sequence in $L^{1}(X_{0}, \mu )$. Hence we may assume without loss of generality that $(X, \Omega , \mu )$ is $\sigma$-finite.

We first construct a sequence $(F_{n})_{n\in\mathbb{N}}$ consisting of pairwise disjoint members of $\Omega$ with $\mu (F_{n}) > 0$ for all $n\in\mathbb{N}$. By the Lemma there is some $F_{1} \in \Omega$ such that $\mu (F_{1}) > 0$ and $L^{1}(X\setminus F_{1}, \mu )$ is infinite-dimensional. Suppose $n\in\mathbb{N}$ and $F_{1}, \ldots , F_{n} \in \Omega$ have been chosen to be pairwise disjoint such that $\mu (F_{j}) > 0$ for each $j\in \{1, \ldots , n\}$ and that $L^{1}(X\setminus \bigcup_{j=1}^{n} F_{j}, \mu )$ is infinite-dimensional. Then by the Lemma there is some $F_{n+1}\in \Omega$ such that $\mu (F_{n+1}) > 0$ and that $L^{1}(X\setminus \bigcup_{j=1}^{n+1} F_{j}, \mu )$ is infinite-dimensional. By recursion we obtain a sequence $(F_{n})_{n\in\mathbb{N}}$ of pairwise disjoint members of $\Omega$. As $(X, \Omega , \mu )$ is $\sigma$-finite, for each $n\in\mathbb{N}$ we can find some $E_{n}\in \Omega$ such that $E_{n} \subseteq F_{n}$ and $0 < \mu (E_{n}) < \infty$. The sequence $(E_{n})_{n\in\mathbb{N}}$ has the desired properties.

Theorem. Let $(X, \Omega , \mu )$ be a measure space such that $L^{1}(X, \mu )$ is infinite-dimensional. Then $L^{1}(X, \mu )$ is not reflexive.

Proof. By the Proposition there is a sequence $(E_{n})_{n\in\mathbb{N}}$ of pairwise disjoint members of $\Omega$ with $0 < \mu (E_{n}) < \infty$ for all $n\in\mathbb{N}$. If $(x_{n})_{n\in\mathbb{N}} \in \ell^{1}$, then it is straightforward to check that the series $(\sum_{k=1}^{n}x_{k} \mu (E_{k})^{-1} 1_{E_{k}})_{n\in\mathbb{N}}$ converges in $L^{1}(X, \mu )$ and that the norm of its limit in $L^{1}(X, \mu )$ is $\|(x_{n})_{n\in\mathbb{N}}\|_{\ell^{1}}$. With this in mind, we define $\Phi \colon \ell^{1} \to L^{1}(X, \mu )$ by \begin{equation} \Phi ((x_{n})_{n\in\mathbb{N}}) := \sum_{n\in\mathbb{N}} x_{n} \mu (E_{n})^{-1} 1_{E_{n}} . \end{equation} Then $\Phi$ is an isometry. As $\ell^{1}$ is a non-reflexive Banach space, this implies that $L^{1}(X, \mu )$ has a closed subspace that is not reflexive. Therefore $L^{1}(X, \mu )$ is not reflexive.