Let $H$ be a Hilbert space(or a reflexive Banach space) and $(x_n)$ a sequence in $H$. Is the following proposition true?

If every subsequence of $(x_n)$ has a subsequence converging weakly to $x$ then $x_n$ converges weakly to $x$.

I think this is true for bounded sequences since bounded sets are weakly sequencially compact. But I couldn't prove it.


This is in general true for any sequence in a topological space.

Let $X$ be a topological space, $\left\{x_n\right\}\subset X$ a sequence, and $x\in X$. Then the following are equivalent:

  1. $x_n$ converges to $x$;

  2. any subsequence of $\left\{x_n\right\}$ admits a subsequence which converges to $x$.

Proof: $(1)\Rightarrow (2)$ is obvious. Assume $(2)$ and suppose by contradiction that $x_n\not \to x$. Then there is a neighbourhood $\mathcal{U}$ of $x$ and a subsequence $\left\{x_{n_k}\right\}\subset \left\{x_n\right\}$ such that $\left\{x_{n_k}\right\}$ stays away from $\mathcal{U}$. But by assumption $\left\{x_{n_k}\right\}$ should have a subsequence converging to $x$, which is impossible.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.