I'm studying functional analysis, and is trying to understand the underlying consistency of the definition of weak (or weak $\star$) convergence for sequence of operators v.s. viewing them as a sequence in their suitable topo. space. I came up with a crucial problem that I need to confirm.
To be clear, let $X$ be a topological space (or locally convex space), $x_n,~a\in X$.
- Does $\forall f\in X^\star,~\lim_{n\to\infty}f(x_n)=f(a)$ imply $\lim_{n\to\infty}x_n=a$?
- Let $T_n,~T\in X^\star$. If $\forall f\in~X^{\star\star},~\lim_{n\to\infty}|f(T_n)-f(T)|=0$, then $T_n$ is said to be weakly convergent to $T$. How does this terminology relate to weak topology or weak star topology?
Answer for (1):
The answer is no. Weak convergence does not imply convergence in $X$. Here is a canonical counterexample,
Take $e_j \in l^2$ where $e_j$ is a the canonical basis vector of $0$s except for $1$ in the j-th position. Then it is clear that $e_j \to 0$ weakly but $e_j\to 1$ strongly.
Strong convergence is easy: $||e_j-0||_{l^2}=||e_j||_{l^2}=(\sum_j |e_j|^2)^{1/2}=1$.
For weak convergence notice that the dual space of $l^2$ is $l^2$ since $1/2 + 1/2 = 1$. Then notice that,
$$|\langle f,e_j\rangle|^2=\sum_j|f_je_j|^2\leq ||f||^2\to 0$$
Because, if $f\in l^2$ then $\sum_i |f_i|^p<\infty$ which implies that $|f|^2$ must eventually go to zero in order for the sum to be finite.
Answer for (2):
Recall that convergence in the weak topology of $X$ implies that for $\{x_n\}\subset X$ converging to $x\in X$ we have,
$$\forall f \in X^* \lim_{n\to\infty} f(x_n)=f(x)$$
So if instead, we have $x_n,x\in X^*$ then we would say that $x_n$ converges in the weak topology of $X^*$ if,
$$\forall f \in X^{**} \lim_{n\to\infty} f(x_n)=f(x)$$
Hence that should answer your question about (2).