Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $H$ is a Hilbert space and let $T \in B(H,H)$ where in our notation $B(H,H)$ denotes the set of all linear continuous operators $H \rightarrow H$.

We defined the adjoint of $T$ as the unique $T^* \in B(H,H)$ such that $\langle Tx,y \rangle = \langle x, T^*y\rangle$ for all $x,y$ in $H$. I proved its existence as follows:

Fix $y \in H$. Put $\Phi_y: H \rightarrow \mathbb{C}, x \mapsto \langle Tx,y \rangle$. This functional is continuous since $|\langle Tx, y\rangle | \leq ||Tx||\; ||y|| \leq ||T||\; ||x||\; ||y||$. Therefore we can apply the Riesz-Fréchet theorem which gives us the existence of a vector $T^*y \in H$ such that for all $x \in H$ we have $\langle Tx, y\rangle = \langle x, T^* y\rangle$.

I now have to prove that $||T^*|| = ||T||$. I can show $||T^*|| \leq ||T||$:

Since the Riesz theorem gives us an isometry we have $||T^*y|| = ||\Phi_y||_{H*} = \sup_{||x||\leq 1} |\langle Tx, y\rangle| \leq ||T||\;||y||$ and thus $||T^*|| \leq ||T||$.

However, I do not see how to prove the other inequality without using consequences of Hahn-Banach or similar results. It seems like I am missing some quite simple point. Any help is very much appreciated!

Regards, Carlo

share|improve this question
2  
$T^{\ast\ast} = T$ –  t.b. Oct 18 '11 at 15:40
    
Ah, of course, of course. Thanks a lot! –  caligula Oct 18 '11 at 16:04
add comment

1 Answer

up vote 3 down vote accepted

Why don't you look at what is $T^{**}$ ...

share|improve this answer
add comment

Your Answer

 
discard

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.