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.
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$\begingroup$ Do you know the Hahn-Banach Theorem? The least you can do is say that $\sup_{T\in B'}\frac{|T(x)|}{||T||}\leq ||x||$ $\endgroup$– Mr.Gandalf SauronCommented May 4, 2023 at 14:13
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$\begingroup$ Yes I know H-B Theorem $\endgroup$– Elise9Commented May 4, 2023 at 14:16
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2 Answers
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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).
answered May 4, 2023 at 14:18
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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 $\sup ||Tx|| / || T ||$.
That two implies equality.
