I was reading Proof of Proposition 3.5.c of Haim Brezis I have done proof without using Uniform Boundedness Principal But Author mentioned that.SO I was thinking I had done some Wrong .Please Help me to understand this proof enter image description here

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$x_n$ converges weekly to $x$ this implies $\forall f\in E^*$ $f(x_n)\to f(x)$

$|f(x_n)|\leq \|f\|\|x_n\|$

As $n\to \infty $ $|f(x)|\leq \|f\|$liminf $\|x_n\|$

THis lead to $\|x\|\leq $ liminf $\|x_n\|$

Where is I am wrong Please Help me

Any Help will be apprecited


Up to $|f(x)| \leq \|f\|\lim \inf \|x_n\|$ your argument is fine. But how did you conclude that $\|x\| \leq \lim \inf \|x_n\|$?. Though this is true it requires something more than basic properties of norms. If you know Banach Alaoglu Theorem this would follow. (You can also prove it using Hahn-Banach Theorem). Perhaps the book hasn't gone that far yet, so it is using UBP.

  • $\begingroup$ Brezis establishes that $\| x \|=\sup_{\|f\| \le 1} |f(x)|$ using a corollary of Hahn-Banach. This is Corollary 1.4. So I don't think this explains where the UBP is used. $\endgroup$ – Theoretical Economist Sep 9 '19 at 6:44
  • $\begingroup$ Dear Sir |f(x)|/||f||<lim inf ||x_n|| Form this we conclude that $||x||\leq $limin f||x_n|| There is no use of UBP here I think $\endgroup$ – MathLover Sep 9 '19 at 6:50
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    $\begingroup$ @SRJ Note that you need Hahn-Banach (or, perhaps, UBP) to conclude that $\sup_{f}|f(x)|/\|f\| = \|x\|$. Otherwise, all you have is that $\sup_{f}|f(x)|/\|f\| \le \|x\|$. See the proof of Corollary 1.4. $\endgroup$ – Theoretical Economist Sep 9 '19 at 6:58

You need the uniform boundedness principle to conclude that the collection $\{\|x_n\|\}$ is bounded.

Corollary 2.4 reads:

Let $G$ be a Banach space and let $B$ be a subset of $G$. Assume that for every $f\in G^*$ the set $f(B) = \{ \langle f,x\rangle:x\in B\}$ is bounded. Then, $B$ is bounded.

This corollary is a consequence of the uniform boundedness principle. You can now use the fact that $B=\{x_n\}$ converges weakly along with corollary above to conclude that $\{\|x_n\|\}$ is bounded.


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