Proof involving weak and weak-* convergence

Question

"Prove that if $(f_n)_{n=1}^\infty\subset X'$ converges strongly in $X'$ and a sequence, $(x_n)_{n=1}^\infty\subset X$ converges weakly in $X$, then, $f_n(x_n)\to f(x),\,n\to\infty$."

My attempt:

Consider $$0\le|f_n(x_n)-f(x)|$$

$$=|f_n(x_n)-f(x)+f(x_n)-f(x_n)|$$

$$\le|f_n(x_n)-f(x_n)|+|f(x_n)-f(x)|$$

Taking the first expression on the last line, since $(f_n)_{n=1}^\infty\subset X'$ converges strongly in $X'$ to say $f\in X'$ for all $x\in X$ this tends to zero.

For the second expression on the last line, since we know that $(x_n)_{n=1}^\infty\subset X$ converges weakly in $X$, to say $x\in X$, it follows that from the definition of weak convergence that, for all $f\in X'$, $f(x_n)\to f(x),\,n\to\infty$.

We than have that,

$$0\le|f_n(x_n)-f(x)|\le0$$

And using the Sandwich Theorem, we can deduce that $|f_n(x_n)-f(x)|\to0,\,n\to\infty$

Is this correct? I am unsure because in another proof on weak/weak-* convergence I had to use the fact that every weak/weak-* convergent sequence is bounded.

Answer 1

This is correct, but you have to use boundedness of $(x_n)$ to prove $|f_n(x_n) - f(x_n)| \to 0$ in the following way:

$$ |f_n(x_n) - f(x_n)| = |(f_n-f)(x_n)| \leq \|f_n-f\|_{X'}\|x_n\|_X \to 0 $$

because $f_n \to f$ strongly and $(x_n)$ is bounded.

Answer 2

I take it that $X$ is a Banach space and $X{^\prime }$ its dual. The first expression in the last line can be estimated according to \begin{equation*} |<x_{n},f_{n}-f>|\leqslant \parallel x_{n}\parallel \parallel f_{n}-f\parallel \end{equation*} It seems that here the boundedness of $\{\parallel x_{n}\parallel \}$ comes in.