13
$\begingroup$

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

$\endgroup$
2
  • 12
    $\begingroup$ $T^{\ast\ast} = T$ $\endgroup$
    – t.b.
    Oct 18, 2011 at 15:40
  • $\begingroup$ Ah, of course, of course. Thanks a lot! $\endgroup$
    – caligula
    Oct 18, 2011 at 16:04

1 Answer 1

18
$\begingroup$

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

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .