This question is from Haim Brezis' functional analysis exercise 3.13, but generalizes it a little bit. Two related questions are posted here: closed convex hull of weak convergent sequence and here: Closed convex hull of a weakly convergent sequence has empty interior in an infinite-dimensional Banach space. The first one is only with Hilbert space, and the second one is unsolved.
The question states as follows:
Let $(V,\|\ \cdot\|)$ be a complex normed space and let $\{x_{k}\}_{k=1}^{\infty}\subset V$ that converges weakly to $x$. For each $n$, we define $$K_{n}:=\overline{conv\cup_{k\geq n}x_{k}}$$ be the norm closure of the convex hull of the tail sequence $\{x_{k}\}_{k=n}^{\infty}$. Define $K:=\cap_{n}K_{n}$. Show that $\{x\}=K$.
I have proved that $x\in K$ so that $\{x\}\subset K$. Recall that if $C$ is a convex subset of a normed vector space $E$, then it is weakly closed if and only if it is strongly closed.
It is clear that $K_{n}$ is strongly closed and convex, so it is weakly closed. Fixing $n$, by construction $x_{n+p}\in K_{n}$ for all $p\in\mathbb{N}$ and thus taking $p\rightarrow\infty$, we see that $x_{n+p}$ weakly converges to $x$, but $K_{n}$ is weakly closed, so $x\in K_{n}$. As this holds for arbitrary $n$, we know that $x\in K_{n}$ for all $n$ and thus belong to their intersection.
However, I got stuck in $\{x\}\supset K$. I have a hint, suggesting me show that given a point $y\neq x$ and a linear functional such that $\Re f(x)<\Re f(y)$, show that there exists $N$ large enough such that $\Re f(z)<\Re f(y)$ for all $z\in K_{N}$.
I don't quite understand how to prove it and why this can lead to the desired conclusion.
Any idea?? Thanks in advance for any help!
Best Answer
If $f$ is a continuous linear functional then $f(x_n) \to f(x)$. Let $0<\epsilon <\Re f(y)-\Re f(x)$. There exists $N$ such that $|f(x_k)-f(x)| <\epsilon$ for all $ k \geq n$. This implies that $|f(z)-f(x)| \leq \epsilon$ for all $z \in K_N$. Hence, $\Re f(z)<\epsilon +\Re f(x) <\Re f(y)$ if $z \in K_N$.
We get the desired contradcition by taking $z=y$.