I'm doing Ex 3.22 in Brezis's book of Functional Analysis.
Let $E$ be an infinite-dimensional Banach space satisfying one of the following assumptions:
- (a) $E'$ is separable
- (b) $E$ is reflexive.
Prove that there exists a sequence $\left(x_{n}\right)$ in $E$ such that
$$ \left\|x_{n}\right\|=1 \quad \forall n \quad \text { and } \quad x_{n} \rightarrow 0 \text { weakly } \sigma\left(E, E^{\star}\right)
$$
I'm able to solve with (a). For (b), I can only show there is a net satisfying the condition. Could you elaborate on I go on to show there is such a sequence for (b).
a. The dual space $E'$ is separable, so the subspace topology $\tau$ that the weak topology $\sigma(E, E')$ induces on the closed unit ball $B_E$ is metrizable. Let $d$ be a metric on $B_E$ that induces $\tau$. Let $U_n := \{x\in B_E \mid d(x,0) < 1/n\} \in \tau$. Because $\dim X \ge \aleph_0$, there exists $0 \neq x_n \in U_n$ such that $\mathbb Rx_n \subseteq U_n$. Let $y_n := x_n / |x_n|$. Then $y_n \in U_n$ and $|y_n|=1$. Clearly, $y_n \rightharpoonup 0$ in $\tau$.
b. The space $E$ is reflexive, so $B_E$ is compact in $\sigma(E, E')$, i.e., $\tau$ is a compact topological space. For each neighborhood $U$ of $0$ in $\tau$, there is $0 \neq x_U \in U$ such that $\mathbb Rx_U \subseteq U$. Let $y_U := x_U/|x_U|$. Then $y_U \in U$ and $|y_U| = 1$. We order the set $\mathcal U$ of neighborhoods of $0$ in $\tau$ by reverse inclusion. Then $(y_U)_{U \in \mathcal U}$ is a net in $B_E$. Then there is a subnet $(x_{\varphi(d)})_{d\in D}$ such that $x_{\varphi(d)} \rightharpoonup 0$.
Best Answer
Since $E'$ is infinite dimensional there is some sequence of unit norm vectors that does not admit a norm convergent sub-sequence. By Banach-Alaoglu there is a weak* limit point of this sequence, but $E$ is reflexive so the weak* limit point is actually a weak limit point. Finally by Eberlein Smulian you can find a subsequence converging weakly to this limit point, denote this sub-sequence by $x_n$ and the limit point by $x$.
Remember that $x_n$ does not admit a norm convergent subsequence, so there is an $\epsilon$ with $\|x_n-x\|>\epsilon$, it follows that $$\frac{x_n-x}{\|x_n-x\|}$$ converges weakly to $0$, but this is a sequence of norm one vectors.