TITLE: Example that in a normed space, weak convergence does not implies strong convergence.
QUESTION [15 upvotes]: The book "Introductory Functional Analysis with Applications" (Kreyszig) presents the following definitions.

A sequence $(x_n)$ in a normed space $X$ is said to be strongly convergent if there is an $x\in X$ such that $\lim\|x_n-x\|=0$. (page 256)
A sequence $(x_n)$ in a normed space $X$ is said to be weakly convergent if there is an $x\in X$ such that $\lim f(x_n)=f(x)$ for every $f\in X'$. (page 257)

Then is proved that strong convergence implies weak convergence but the converse is not generally true (unless that $X$ is finite-dimensional). To prove it, he gives an example in a Hilbert space and uses the Riesz Representation Theorem and the Bessel inequality (page 259).
I would like an exemple that weak convergence does not implies strong convergence in a normed space $X$ that is not a Hilbert space. Is there one?
Thanks.

REPLY [4 votes]: Another easy non-Hilbert example: Take $X = C_0(\mathbb{R})$ with the uniform norm. $X^*$ is the Radon measures on $\mathbb{R}$. The sequence $\{ \chi_{[n, n + 1]} \}_n$ goes to $0$ weakly but is not Cauchy in norm.