2

In Rudin's book <> Page 66, it says "If $X$ is a infinite dimensional topology vector space, then $X$ under the weak topology is not locally bounded" . Hence I think the topology of any (infinite)Banach space $X$ is different from the weak topology, for $X$ is locally bounded. Since the topology determines the convergence, can I get that weak convergence is equivalent to original convergence in any (infinite)Banach spaces? Thank you.

Siminore
  • 35,136
Shijie Dong
  • 71
  • 2
    For nets, in infinite-dimensional normed spaces, weak convergence is never equivalent to norm-convergence. Norm-convergence always implies weak convergence, and there are some spaces ($\ell^1(\mathbb{N})$ for example) in which every weakly convergent sequence is also norm-convergent, but that is a rare exception. – Daniel Fischer Oct 31 '14 at 12:13
  • Thank you. I'll appreciate it very much if you could add more details. – Shijie Dong Nov 01 '14 at 04:49

1 Answers1

1

Here is a good counter-example to keep in mind, only using basic Hilbert analysis.

Let $E$ be an infinite-dimensional Hilbert space, with $(e_{n})_{n \in \mathbb{N}}$ a Hilbert basis. Consider $x \in E$ : then, $$||x||^{2} = \sum_{n \in \mathbb{N}} |\langle x, e_{n} \rangle |^{2}$$ and therefore, the real sequence $(\langle x, e_{n} \rangle )$ converges to $0$ (keep in mind that $E' = E$ since $E$ is a Hilbert space), which proves the fact that the sequence $(e_{n})$ is weakly convergent to $0$. However, since $||e_{n}|| = 1$ for every $n$, this sequence does not (strongly) converge to $0$.

Tlön Uqbar Orbis Tertius
  • 3,194
  • I'm sorry, but I want to ask "weak convergence is never equivalent to original convergence in any (infinite)Banach spaces?". My mistake. – Shijie Dong Nov 01 '14 at 04:45