Problem 5

Exercise 7.5 (Weak convergence is unique in a reflexive space)  

1. If $X$ is reflexive, show that a weakly converging sequence converges to a point.
2. Show that the conclusion need not be true if $X$ is not reflexive.

Proof. For part (a), suppose $x\in X$ satisfies the property that $\lim \phi(x_n)$ exists for each $\phi \in X^*$. Define a linear functional on the dual space

$$x^* : X^* \to \mathbb{R}$$ (7.19)

$$\phi \mapsto \lim_{n\to\infty} x_n^*(\phi)$$ (7.20)

Then $x=J^{-1}(x^*)$ is the unique candidate limit because $X$ is reflexive, completing the proof.

For part (b), consider the space $B=C([0,1])$ and the functions $f_n(x)=x^n$ with norm $\Vert f_n\Vert=1$. The dual space is given by the measures on $[0,1]$, so that any linear functional equals

$$\phi(f) = \int f d\nu$$ (7.21)

for some measure $\nu$. Then $\phi(f_n)\to \phi(0)$. But this limit is not unique because $\phi(0)=\phi(1_E)$ for any set $E$ of measure zero, say $E=\mathbb{Q}\cap[0,1]$. $\qedsymbol$