I encountered the following example, which is supposed to show, that "weak convergence" and "convergence with respect to the norm" do not have to be the same.
Consider the normed space $(C[0,1],||\cdot||_2)$. Let $f_n = \sin(nt)$, then $$||f_n||_2^2 = \int_0^1 |sin(nt)|^2 = 1/2.$$ We see, that $f_n$ goes strongly to $1/2$. It then claims without demonstrating, that $f_n$ goes weakly to zero.
I know that for a sequence $(x_n)_{n\in \mathbb{N}} \subset E$ in the normed space $E$, weak convergence to $x\in E$ holds, if $$ f(x_n)\rightarrow f(x) \quad \forall f \in E^* $$ I would like to check on that. But how do I find $E^*$?
Note that $[0,1]$ is compact and hence, $C[0,1]$ is a dense subspace of $L^2[0,1]$. Hence, $L^2[0,1] = (L^2[0,1])^* \subset (C[0,1],||.||_2)^*$ .
But density of $C[0,1]$ implies that any element of $(C[0,1],||.||_2)^*$ extends to an element of $(L^2[0,1])^*$ uniquely! So $(C[0,1],||.||_2)^*=L^2[0,1]$
Note that this argument follows in the more general setup: If $(N,||.||)$ is a normed linear space with $(\bar N^{||.||},||.||)$ as its norm closure then $N^*= (\bar N^{||.||})^*$