Isometry in Riesz's representation theorem

668 Views Asked by At

Riesz's representation theorem states that if $H$ is a Hilbert space and $H^*$ its dual space, then the map $\Phi$ which maps $x\in H$ to $x^*\in H^*, x^*y:=\langle x, y\rangle \,\forall\,y\in H$ is an antilinear isometric isomorphism. The fact that it is isometric is supposed to be trivial, and is presented as "manifestly true" in all proofs I could find, but the only thing I can tell is that by the Cauchy Schwartz inequality

$$||\Phi(x)||=\sup\left\{|\langle x, y\rangle|, ||y||\leq 1\right\}\leq\sup\left\{||x||\,||y||, ||y||\leq 1\right\}=||x|| $$

But why do we have equality?

2

There are 2 best solutions below

0
On BEST ANSWER

We have $\|\Phi(x)\|\geq\left|\left<x,\dfrac{x}{\|x\|}\right>\right|=\dfrac{\|x\|^{2}}{\|x\|}=\|x\|$ for nonzero $x$.

0
On

Note that $$\|\Phi(x)\| \ge \frac{|\Phi(x)(x)|}{\|x\|} = \frac{\langle x, x \rangle}{\|x\|} = \frac{\|x\|^2}{\|x\|} = \|x\| $$