Weak Convergence Lemma - Is Banach needed?

Question

Lemma. Let $X$ be a normed space.

• If $x_n \rightharpoonup x$ in $X$ and $x_n^* \to x^*$ in $X^*$, then $\lim_{n \to \infty} x_n^*(x_n) = x^*(x)$.

• If $X$ is even Banach, then $x_n \to x$ in $X$ and $x_n^* \overset{*}\rightharpoonup x^*$ in $X^*$ implies $\lim_{n \to \infty} x_n^*(x_n) = x^*(x)$.

The lecture notes I work with, had this wrong at first - it didn't include Banach for the second statement. (The proof uses boundedness of $(x_n^*)$ which relies on Banach-Steinhaus!)

Is Banach really needed, though? Does anyone have a counterexample?

Answer 1

Here is a counter-example for the second statement without $X$ being complete. The first statement is valid for non-complete $X$ (by means of embedding into the Banach space $X^{**}$).

Take $X=c_{00}$ provided with the $l^2$-norm. Define $x_n = n^{-1}e_n$ and $$ x_n^*(y):= n y_n. $$ Then $x_n^*\rightharpoonup^*0$, $x_n\to0$, but $x_n^*(x_n)=1 $ for all $n$.