Schur’s property in $L^1( \mathbb X )$ where $\Bbb X$ is a mearsure space

Question

We know $\ell^1( \mathbb N )$ enjoys the Schur property, I want to know if $L^1( \mathbb X )$ also enjoys the Schur property.

A Banach space is said to have Schur’s property if weak convergence implies strong convergence

Any help would be greatly appreciated.

Answer 1

Well, of course not in this generality.

Example. Consider the space $L^1(0, 2\pi)$ and the sequence $$f_n(x):=\cos(nx).$$ This sequence is such that $f_n\rightharpoonup 0$, and $\|f_n\|_{L^1(0, 2\pi)}=\int_0^{2\pi}|\cos(x)|\, dx\ne 0$, so $f_n$ does not converge strongly.

To see that $f_n$ converges weakly to $0$, it suffices to observe that

1. $\|f_n\|_{L^1(0, 2\pi)}$ is uniformly bounded (actually, $\|f_n\|_{L^1(0, 2\pi)}=4$ for all $n\in\mathbb N$);
2. for all interval $(a, b)$, $$ \int_a^b f_n(x)\, dx \to 0.$$

These properties are easy to prove by direct computation.

The proof is completed by a standard density argument, which proves that, for all $g\in L^\infty(0, 2\pi)$, $$\tag{1} \int_0^{2\pi} f_n(x)g(x)\, dx \to 0.$$ Sketch of the density argument: for every $\epsilon>0$ there exists a step function $\sum_{j=1}^m c_j \mathbf 1_{(a_j, b_j)}$ such that $$\left\lVert g-\sum_{j=1}^m c_j \mathbf 1_{(a_j, b_j)}\right\rVert_{L^{\infty}(0, 2\pi)}\!\!\!\le \epsilon.$$ Adding and subtracting in (1), we see that$$ \left\lvert\int_0^{2\pi} f_n(x)g(x)\, dx\right\rvert\le \sum_{j=1}^m\left\lvert c_j\int_{a_j}^{b_j} f_n(x)\, dx\right\rvert + \epsilon\|f_n\|_{L^1},$$ and the second summand on the right hand side is uniformly bounded because of 1, while the first tends to 0 because of 2. The proof is complete.

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In a previous version of this post I considered another example which turned out to be wrong. It is interesting.

(WRONG). For example, $L^1(\mathbb R)$ does not satisfy the Schur property; if $f\in L^1(\mathbb R)$ is not null, then the sequence $f_n(x):=f(x-n)$ converges weakly to the null function, but it does not converge strongly, because $\|f_n\|_{L^1}=\|f\|_{L^1}\ne 0$.

This is not true, because $L^1(\mathbb R)^\star =L^\infty(\mathbb R)$; now, if $g\in L^\infty(\mathbb R)$, then $$\int_{\mathbb R} f(x-n)g(x)\, dx \text{ needs not tend to }0;$$ take for example $g=1$. Thus, it is not true that $f_n\rightharpoonup 0$.

Answer 2

Another classic from Probability.

Certainly $\ell_2$ (or any infintie-dimensional reflexive space) lacks the Schur property. Let us take a sequence of independent identically distributed Bernoulli (with $p=1/2$) random variables, say $(\xi_k)_{k=1}^\infty$. Then by Kchinchine's inequality there exist constants $A,B>0$ such that for any scalars $x_1, \ldots, x_N$ and any $N>0$ we have $$A\left( \sum_{n=1}^N |x_n|^2 \right)^{1/2} \leqslant \operatorname{E}\, \left|\sum_{n=1}^N \xi_n x_n\right| \leqslant B \left(\sum_{n=1}^N |x_n|^2\right)^{1/2}.$$ In functional-analytic terms, this means that the closed linear span of $(\xi_n)_{n=1}^\infty$ in $L_1$ is isomorphic to $\ell_2$. Consequently, $L_1$ fails the Schur property as so does $\ell_2$.

Note that any measure space $(\Omega, \mu)$, where $\mu$ is not purely atomic, admits a sequence of random variables like above, so you may prove the following.

Proposition. Let $(\Omega, \mu)$ be a measure space. Then $L_1(\mu)$ has the Schur property if and only if $\mu$ is purely atomic.