# for every $f\in H^*$ there is a unique $y\in H$ such that $f(x)=\langle x│y\rangle$ , moreover $\|y\|=\|f\|$

I'm trying to prove a theorem on folland real analysis book

let $H$ be a Hilbert space

for every $f\in H^*$ there is a unique $y\in H$ such that $f(x)=\langle x│y\rangle$ , moreover $\|y\|=\|f\|$

I can prove the exsistance of $y$ that is $y=\frac{\overline {f(z)}z}{\|z\|}$ such that $f(x)=\langle x│y\rangle$

here $z$ non zero element of orthogonal complement of $\ker f$.

how can I prove $\|y\|=\|f\|$ hint is given that use cauchy schwarz inequality

$f( \frac y {||y||}) =\frac {||y||^{2}} y=||y||$ so $||f||\geq ||y||$. Cauchy Schwartz inequality says $|<x,y>| \leq ||x|| ||y||$ and this gives $||f|| \leq ||y||$.

• does the second inequity holds because in operator norm ||x||=1 ? – Bad English Dec 27 '17 at 7:52
• Yes, I am just using the definition of operator norm. – Kavi Rama Murthy Dec 28 '17 at 12:34

To prove uniqueness, assume there exist two vectors $y, z \in H$ such that $$f(x) = \langle x, y \rangle = \langle x, z \rangle, \text{ for all } x \in H$$

Rearranging gives:

$$0 = \langle x, y \rangle - \langle x, z \rangle = \langle x, y - z\rangle , \text{ for all } x \in H$$

Hence $y - z \perp H$, which implies $y - z = 0$ so $y = z$.

Notice that this makes sense because $\dim \,(\ker f)^\perp = 1$, so there is only one vector $v \in (\ker f)^\perp$ with $\|v\| = 1$. You have shown that $f(x) = \left\langle x , \overline{f(v)}v\right\rangle$ for all $x \in H$..

Regarding the norm, we have:

$$\|f\| = \sup_{\|x\|= 1} \left|f(x)\right| = \sup_{\|x\|= 1} \left|\langle x, y\rangle\right| = \|y\|$$

because

$$\|y\| = \left|\left\langle \frac{y}{\|y\|}, y\right\rangle\right| \le \sup_{\|x\|= 1} \left|\langle x, y\rangle\right| \stackrel{CSB}{\le}\underbrace{\|x\|}_{=1}\|y\| = \|y\|$$