Suppose a bounded sequence $(x_n)$ converges to $x$ in the Cesaro sense (i.e., $\frac{1}{n}(x_1 + x_2 + \dots + x_n)\rightarrow x$) in a separable Hilbert space $H$. How to prove that some subsequence $(x_{n_k})$ converges weakly to $x$?
[Math] Cesaro convergence implies weak convergence of a subsequence
fa.functional-analysishilbert-spaces
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No. In $\ell_r \oplus_r \ell_r$ let $y_n = (z_n, x_n)$ where $x_n$ are disjoint, $\|x_n\|_r=4^n$, $\|x_n\|_p \to 0$, $\|z_n\|_r = 1$, and $\|z_n-y\|_r \to 0$ for some non zero $y$.
Since my comments above didn't evince any response, I have decided to elaborate them to an answer, especially as it is part of a longer and interesting story which goes back to the distinguished mathematicians Arens and Eells in (1955) (Pac. J. Math, vol. 78). They showed that a metric space can be embedded into a Banach space in such a way that a suitable universal property is satisfied (the special case of the unit interval is not particularly relevant here---the natural setting is that of a general metric space). In the language of category theory (which they did not use), this means that you can reflect a suitable category of metric spaces into one of Banach spaces. In order to simplify the notation, I shall assume that the metric space has a base point and its radius is at most $1$---then every Lipschitz function which vanishes at the base point is automatically bounded. It turns out that this space is precisely that predual of the space of Lipschitz functions in question which I mentioned in my comment (more or less a direct consequence of the universal property). Then the remarks in my comments apply. The sequence in your question will converge weak $\ast$ but not weakly, in general.
Your question of when it will converge weakly is a bit vague for me to answer but the following considerations suggest that it won't be easy to give a simple necessary and sufficient condition. One can embed your space in a natural way into the bounded, continuous functions on the square of the metric space minus the diagonal set. So your question can be reinterpreted as one concerning weak convergence in a $C^b(S)$. But the latter is the same as $C(\beta S)$ (the Stone-Čech compactification) and so this convergence is equivalent to uniform boundedness (no problem here) and pointwise convergence on $\beta S$. Interpreting this in your original setting might be a problem.
Best Answer
If we take $x_n = (-1)^n x$ then $x_n$ converges to $0$ in Cesaro sence. But no subsequence of $x_n$ converges weakly to $0$. $x_n$ is also a bounded sequence. Hence your statements seems wrong.