# About Norm of an element $x$ and Norm of an element in its Dual

We have Banach Space $$B$$ and choose $$x\in B$$ then $$||x|| _ B = {\sup}_{T \in B'} \frac {|T(x)|}{||T|| _ {B'}}$$

I know I should write something about question but I couldn't move. Suggestion will be appreciated.

• Do you know the Hahn-Banach Theorem? The least you can do is say that $\sup_{T\in B'}\frac{|T(x)|}{||T||}\leq ||x||$ Commented May 4, 2023 at 14:13
• Yes I know H-B Theorem Commented May 4, 2023 at 14:16

Hint:-Define $$f:\Bbb{K}x\to\Bbb{K}$$ by $$f(\lambda x)=\lambda||x||$$ . Then $$||f||=1$$. Now what would the Hahn-Banach Extension of $$f$$ do?
Also notice that $$||\frac{T}{||T||}||=1$$ (the norms are operator norms).
We always have $$|| Tx || \le ||T|| ||x||$$ in other words this is what stated in comment.
For the other part, just use the fact that can be obtained from Hahn-Banach Theorem. More detaily, look at hint in answer and that $$f$$ whose norm is $$1$$ is certainly smaller than $$\sup ||Tx|| / || T ||$$.