Consider a Riemannian manifold $\mathcal{M}$ modeled on a (possibly infinitely dimensional) Hilbert space. Suppose that $\{p\_i\} \subset \mathcal{M}$ and $p\_i \to p$ .


We say that a sequence of covectors $v^\*\_i \in T^\*\_{p\_i}\mathcal{M}$ is weakly converges to $v \in T\_p^\*\mathcal{M} $ 
if for every smooth function $f$ on $\mathcal{M}$ we have 
$$\langle v^\*\_i, df(p\_i) \rangle \to \langle v^\*, df(p) \rangle. $$


**Question:** Is it true that if $\{v\_i^\*\}$ is bounded then it has a weakly convergent subsequence?



Weak convergence and bounded sequence on a Riemannian manifold
Sorry, I did not read it carefully enough. Here is the revised version.


The answer is yes, if it is a strong Riemannian manifold modelled on Hilbert space (strong: generated the topology on each tangent space).


Namely, since $p\_i \to p$ we may assume without loss that all $p\_i$ are in one chart, so we are in a Hilbert space $H$, and the Riemannian metric is bounded by a contant times the standard one on an open subset containing all $p\_i$.


By assumption, the $v\_i^\*$ are bounded, thus contained in a closed ball which is compact in the weak topology, thus there is a weakly convergent subsequence. 
Note that for each $(p,\alpha)\in T^\*\_p H$ there exists a smooth functions $f$ such that $df(p) = \alpha$. So we have the the weak topology to deal with.


