So here is my problem,
I want to show the following,
Let $X$ be a normed $\mathbb K$-Vectorspace.
And let $(x_n)_{n\in\mathbb N}\subset X$ be such that it converges weakly to some $x\in X$.
Then,
$$\frac{1}{n}\sum_{k=1}^{\infty}x_k\rightharpoonup x\;,n\rightarrow\infty$$ i.e the sum also converges weakly to $x$.
I think by linearity it is sufficent to show that if $(x_n)_{n\in\mathbb N}$ converges weakly to $0$ then the some does.
After a couple attempts trying to bound $\sum_{k=1}^{\infty}f(x_k)$ for some arbitrary $f\in X^*$ I gave up.
So I wanted to ask if someone wants to give me hint how to proceed? Thank you!
If $(a_n)_{n\geqslant 1}$ is a sequence of complex numbers converges to some $a$, then so does the sequence $(n^{-1}\sum_{j=1}^na_j)_{n\geqslant 1}$.
For a fixed linear continuous functional, use this result with $a_k:=\langle f,x_k\rangle_{X',X}$.
Not completely related, but if $X$ was a Hilbert space, then we could extract a subsequence $(x_{n_k})_k$ of $(x_k)_k$ such that $(N^{-1}\sum_{j=1}^Nx_{n_j})_N$ converges strongly to $x$.