2
$\begingroup$

This point has been giving me a lot of trouble. I am skipping around in learning functional analysis, and I've directly gone to the study of bounded operators without studying topology and basic Banach Space theory, in hopes to pick it up along the way. I'm doing okay with this....

Anyways, I am trying to show that given two Banach Spaces $X$ and $Y$, the operator $T$ in $L(X,Y)$ and its adjoint $T^{\ast}$ in $L(Y^{\ast},X^{\ast})$, then the map $T \mapsto T^\ast$ is an isometric isomorphism of $L(X,Y)$ into $L(Y^\ast,X^\ast)$. The proof is very straightforward, with one important detail, which is

$\Vert Tx\Vert_Y = \sup_{\Vert l \Vert \leq 1} |l(Tx)| \qquad \text{for } l \in Y^{\ast}$

the authors justify this as follows: "this equality uses a corollary of the Hahn-Banach Theorem".

I just don't see this! How does this follow?

1 Answers 1

3

One important consequence of the Hahn-Banach theorem is the following:

If $x \in X$ is a non-zero vector then there exists a functional $\phi:X \to \mathbb{R}$ of norm $1$ such that $\phi(x) = \|x\|$. (This is of course trivial for $x = 0$)

To see this, take the linear subspace $U$ generated by $x$, define $\varphi: U \to \mathbb{R}$ by $\varphi(x) = \|x\|$, $\varphi(\lambda x) = \lambda\|x\|$. Note that $\varphi:U \to \mathbb{R}$ is a functional satisfying $\|\varphi\| = 1$ and extend it linearly and norm-preservingly to a functional $\phi: X \to \mathbb{R}$.

One application of this observation is that the canonical inclusion $X \to X^{\ast\ast}$ is an isometry.

I interpret your formula as $\|Tx\|_Y = \sup_{\|\ell\| \leq 1} |\ell(Tx)|$ which is then an immediate consequence of the highlighted comment above.

  • 0
    @Jonas: Thanks! I knew there was a certain threshold to bumping, but I wasn't sure anymore (I thought it was two votes on the answers in total). Moreover, it was the third question he didn't vote on nor accept, so I thought I might point out the site's workings.2011-04-22