# $C([0,1])$ is not weakly sequentially complete.

I'm studying Functional Analysis by myself. For a counterexample of every Banach space is not weakly sequentially complete, I was suggested to check $C([0,1])$ is not weakly sequentially complete. For this, let $f_n(t)=1-nt$ if $0\leq t\leq \frac{1}{n}$ and $f_n(t)=0$ if $\frac{1}{n}\leq t \leq 1$. But I do not know how to use monotone convergence theorem to show $\{f_n\}$ is weakly Cauchy and also why it is not weakly sequentially complete. Please help me. Thanks in advance.

Let $x^*\in C([0,1])^*$ be any continuous functional. Then there is a finite measure $\mu$ on $[0,1]$, such that $$\int_{[0,1]} f\, d\mu = x^*(f), \qquad f \in C[0,1]$$ Now, by dominated convergence, we have $$\int_{[0,1]} (f_n - f_m) \, d\mu \to 0 ,\qquad n,m \to \infty$$ Edit: By looking at $\mu = \delta_x$, $x \in [0,1]$, show that $(f_n)$ cannot converge weakly to a continuous function.

• I thought $f_n \rightharpoonup 0$ too for a moment, but consider $\delta_0 \in C([0,1])^\ast$. – Daniel Fischer Jul 14 '14 at 9:42
• @DanielFischer Thanks. – martini Jul 14 '14 at 10:00

Fact 1. Note that $c_0$ is not weakly sequentially complete. Just consider sequence $\left(\sum_{k=1}^n e_k\right)_{n\in\mathbb{N}}$.

Fact 3. The property of being weakly sequentially complete is preserved by subspaces. Indeed, let $Y$ be a closed subspace of weakly sequentially complete space $X$. Consider weakly Cauchy sequence $(y_n)_{n\in\mathbb{N}}$ in $Y$, then obviously it is weakly Cauchy in $X$. Since $X$ is weakly sequentially complete then $(y_n)_{n\in\mathbb{N}}$ weakly converges to some $x\in X$. Thus $x$ lies in the weak closure of $Y$. Since $Y$ is closed and convex, then by Mazur's theorem weak closure coincides with $Y$. Thus weak limit of $(y_n)_{n\in\mathbb{N}}$ $\in Y$, and $Y$ is weakly sequentially complete.

Conclusion. Any $C(K)$ is not weakly sequentially complete.

It is standard trick to show that a Banach space contains a copy of $c_0$ and then conclude that this space is not weakly sequentially complete.

• I think C(K) is a subspace of $L^\infty(K)$ and $L^\infty(K)$ is weakly sequentially complete while C(K) is not. it's a counter example for fact3. isn't it? – niki Jul 14 '14 at 11:18
• Why is $L_\infty(K)$ is weakly sequentially complete? I think it is not. – Norbert Jul 14 '14 at 11:22
• Oh, In the old version of Conway's Functional Analysis it's an exercise. but checking in new version, I found $L^\infty$ is weak* sequentially complete. – niki Jul 14 '14 at 11:40

The monotone convergence theorem also works for decreasing sequences $\{f_n\}$ provided $f_1\in L^1$ (cf. Rudin's Real and Complex Analysis, exercises to chap.1), which is the case in your problem. For any regular Borel measure $\mu$ on $[0,1]$, $\int_0^1 f_n d\mu \to \mu(\{0\})$, so the sequence is weakly convergent and hence weakly Cauchy. Here we are using the basic property of finite measures that $\lim_{n\to\infty}\mu([0,1/n])= \mu(\cap_{n=1}^\infty[0,1/n])$. Check that $f_n$ converges weakly to the (discontinuous) characteristic function $\chi_{\{0\}}$.