A sequence that converges weakly but not in the Cesàro sense

Solution 1:

I am currently not sure about your sequence, but I will edit the answer if I see something.

Meanwhile, the following example works: Define $$ x_n := e_k \text{ for } 2^k \leq n < 2^{k+1}, $$ where $e_k$ is the $k$th member of the standard basis of $\ell^2$.

Then we have $x_n \rightharpoonup 0$, but $$ \bigg(\frac{1}{2^{k+1} - 1} \sum_{\ell = 1}^{2^{k+1} - 1} x_\ell \bigg)_{k} = \frac{(2^{k+1} - 1) - 2^k + 1}{2^{k+1}-1} = \frac{2^{k}}{2^{k+1} - 1} \rightarrow \frac{1}{2}. $$ In particular, this shows $$ \limsup_k \bigg \Vert \frac{1}{2^{k+1} - 1} \sum_{\ell = 1}^{2^{k+1} - 1} x_\ell \bigg\Vert \geq \limsup_k \bigg| \bigg(\frac{1}{2^{k+1} - 1} \sum_{\ell = 1}^{2^{k+1} - 1} x_\ell \bigg)_{k} \bigg| \geq 1/2 $$ and thus $(x_n)_n$ does not converge Cesaro to $0$.

The intuition here is that if we are at $2^{k+1}$, then still about half of all the sequence members we will have seen are equal to $e_k$.

Note though, that even though Cesaro convergence does not hold in general, Mazur's lemma (https://en.wikipedia.org/wiki/Mazur%27s_lemma) shows that we can always find some convex combination which converges strongly.