Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $(X,\|\cdot\|)$ be a infinite dimensional normed vector space, and Suppose that the weak topology in $X$ is metrizable by a metric $d$. How the opens of $\tau_d $ should be the same as the weak topology, we have that for every $n$ the ball $B^d(0,\frac{1}{n})$ mus be have a non trivial subspace we have the following construction:

Choose in each $B^d(0,\frac{1}{n})$ a $x_n$ such that $\|x_n\|=n$, so we have $x_n\rightharpoonup x$ but $\|x_n\|\to \infty$. Wich is an absurd, because we know that the sequence $(\|x_n\|)$ is bounded.

My question lies in the fact that in the Brezis book:

exercice 3.8 there is a proof script, which uses up to Baire's theorem! I can not conceive why to use such a complicated demonstration when there is so much simpler proof, if my proof is correct

share|cite|improve this question
There's a big hammer used in your proof also. Namely the Uniform Boundedness Principle, which is used to show weakly convergent sequences are norm bounded. – David Mitra May 29 '14 at 21:15
Incidentally, most proofs of the Uniform Boundedness principle use Baire; though, the use of Baire can be avoided, see this paper. – David Mitra May 29 '14 at 21:24
I had not looked through this point of view, well observed! Thanks for the reference. – O Empalador de Cabras May 29 '14 at 21:30

The proof seems correct. The assumption that $X$ is infinite dimensional is used to show that $B^d(0,1/n)$ contains a non-trivial subspace and the argument should be detailed a little more because it is not straightforward. The set $B^d(0,1/n)$ contains an open subset of the form $O:=\bigcap_{j=1}^N\{x,|f_j(x)|<\delta\}$ for some integer $N$, $f_j\in X^*$ and $\delta >0$. There is $y\neq 0$ such that $f_j(y)= 0$ for each $j\in \{1,\dots,N\}$ because $X$ has a dimension $\geqslant N+1$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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