Isn't this just the definition of weakly convergence?

Question

Hi I come across this question when reviewing my notes.

But isn't this just the definition of weak convergence? Am I failing to understand the problem? The definition given on my notes is

A more interesting question is this one: Where the $x^*$ is in a dense subset of the unit ball in $X^*$

Notes by Scott Morrison.

Thanks in advance.

Answer 1

You are right, what you are asked to show in the first exercise is just the definition of weak convergence.

The second exercise is indeed more interesting. What you are asked to show in the second exercise is wrong. There is an assumption of boundedness missing, a bounded sequence $(x_n)$ in $X$ converges weakly to $x$ if and only if $x^\ast(x_n) \to x^\ast(x)$ for all $x^\ast$ in a dense subset of the unit ball of $X^\ast$.

If you don't find a counterexample without the boundedness assumption yourself:

Let $X = c_0 \subset \ell^\infty$, the space of sequences converging to $0$. Then $X^\ast$ is isometrically isomorphic to $\ell^1$ under the pairing $\langle c, x\rangle = \sum_{n=0}^\infty c_n\cdot x_n$, and for the dense subset, we take the subset $D = \left\{x\in \ell^1 : \lVert x\rVert < 1,\, \{n : x_n \neq 0\} \text{ is finite}\right\}$ of the unit ball of $\ell^1$. If $e_n$ denotes the $n$-th standard unit vector in $c_0$, then $x_n = n\cdot e_n$ has the property that $x^\ast(x_n) \to 0$ for all $x\ast \in D$, but $(x_n)$ is not bounded, hence does not converge weakly.