0
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

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

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.