# In Proposition 3.5.c of Haim Brezis Where did Author Use Uniform Boundedness theorem

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

$$x_n$$ 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 $$\|x_n\|$$

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.

• 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. – Theoretical Economist Sep 9 '19 at 6:44
• 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 – MathLover Sep 9 '19 at 6:50
• @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. – Theoretical Economist Sep 9 '19 at 6:58

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

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.