weak convergence and weak * convergence criterium

Question

I have to solve the following problem

Let $X$ be a Banach space. Prove that $x_n\rightharpoonup x$ in $X$ if and only if

1. $\sup||x_n||<+\infty$
2. there exists a dense subset $E'$ of $X'$ (the dual of $X$) such that $f(x_n)\rightarrow f(x)$ for all $f\in E'$.

Moreover determine what are the necessary changes to state a similar result concerning weak * convergence and prove it.

For the first part, the forward direction is easy: the first point is a consequence of Banach-Steinhaus Theorem (if $x_n$ weakly converges to $x$, then $\{||x_n||\}$ is bounded) and the second one follows by definition of weak convergence. My problem is to show the converse (must the subset $E'$ be total in $X'$?). Give me some helps!! Thanks

Answer 1

Let $M = \sup \|x_n\| + \|x\|$.

Select any $f \in X'$ and let $g \in E'$. Then \begin{align*}|f(x_n) - f(x)| & \le |f(x_n) - g(x_n)| + |g(x_n) - g(x)| + |g(x) - f(x)|\\ &\le 2M \|f-g\|_{X'} + |g(x_n) - g(x)|.\end{align*} Let $n \to \infty$ to conclude $$\limsup_{n \to \infty} |f(x_n) - f(x)| \le 2M \|f-g\|_{X'}.$$ Since $E'$ is dense in $X'$ the quantity on the right can be made arbitrarily small. Thus $$\limsup_{n \to \infty} |f(x_n) - f(x)| \le 0$$ which implies $f(x_n) \to f(x)$.