Problem 5

Exercise 3.5 (Closed unit ball in the weak topology)   Let $X$ be a Banach space and $B=\{x\in X \;\vert\; \Vert x\Vert \leq 1\}$. Show that $B$ is closed in the weak topology. Is the unit sphere closed in the weak topology?

Proof. Let $x\in X$ be a limit point of $B$. We may assume $x\neq 0$, so that there exists $\phi : X \to \mathbb{R}$ such that $\phi(x)=\Vert x\Vert$ and $\Vert\phi\Vert=1$. Then for each $\epsilon>0$, select $x_\epsilon$ in the neighborhood

$$\{y\in X \;\vert\; \vert\phi(y)-\phi(x)\vert<\epsilon\}\cap B$$ (3.38)

Then we have $\phi(x) < \phi(x_\epsilon) + \epsilon$. Then

$$\Vert x\Vert < \phi(x_\epsilon)+\epsilon \leq \Vert\phi\Vert\vert x_\epsilon\vert + \epsilon \leq 1 + \epsilon$$ (3.39)

Since $\epsilon>0$ is arbitrary, this means $\Vert x\Vert\leq 1$.

To see that the sphere is not necessarily closed in the weak topology, consider the Banach space $B=C([0,1])$ and the sequence of functions $f_n(x)=x^n$. The linear functional

$$\int : C([0,1]) \to \mathbb{R}$$ (3.40)

is bounded, but

$$\int f_n = \left.\frac{1}{n+1}x^{n+1}\right\vert _0^1 = \frac{1}{n+1} \to 0$$ (3.41)

and $\Vert\Vert\neq 1$. $\qedsymbol$