# Prove that weak convergence does not necessarily imply strong convergence without counterexample.

Here is the set of original problems.

Let $\{x_n\}$ be a sequence in a normed linear space $X$. Prove that:

1. Strong convergence implies weak convergence with the same limit.
2. The converse of 1. is not generally true.
3. If $\mathrm{dim}(X) < \infty$, then weak convergence implies strong convergence.

As the professor suggests, use the basis, go to Banach space and then use dual basis.

I am done with the first part. Now, I need to work on the second and third parts. Here is my start:

Assume the weak convergence implies strong convergence. Let $f\in X^*$. By definition of weak convergence, if $x_n$ converges weakly to $x$, then $f(x_n) \rightarrow f(x)$ aka $\|f(x_n) - f(x) \| \rightarrow 0$.

I consider the dimension of $X$ - either $\dim(X)$ is infinite or $\dim(X) < \infty$ [which is for the third part of the problem]. For infinite dimension, set $X:\{e_1, e_2, \dots\}$. Otherwise, for finite dimension, set $X:\{e_1, \dots, e_n\}$, where $n$ is finite. For this part, it seems that I hit the roadblock or have no way to approach this proof.

Let $X=\ell^2(\mathbb N)$, and let $\{e_n\}$ be the canonical basis. Then $e_n\to0$ weakly, while $\|e_n\|=1$ for all $n$.
There are several ways to do 3. One would be to note that $X^*$ is finite-dimensional, and use a basis of $X^*$ to define a norm that determines the weak topology.