Let $H$ be Hilbert space and $u_1,u_2,...u_n \in H$ (vectors dont have to be orthogonal)
$V=span\{u_1,u_2,...u_n\}\subset H$
and $S$ is unit sphere in $V$. $P_V$ is orthogonal projection on V.

Now lets take some $h\in H$ $$ \sup_{v\in S}|\langle h,v \rangle| = \|P_Vh \| $$

How can I show this property? I know that $\|P_Vh-h \|= \inf_{v\in V}\|v-h\|$ for $V$ closed and convex , $(P_Vh-h) \perp V $ , uniqueness , $P^2=P$ ... but supremum confused me. Is there some short elegant way to show this?

P.S. give hint if its easy and short


Hint: From $(P_Vh-h) \perp V$ you know $\langle h,v \rangle=\langle P_Vh,v \rangle$. Which $v\in S$ maximizes this expression?

  • $\begingroup$ now cauchy-schwartz $|\langle P_V h,v \rangle| \le \|P_V h\| \| v \|$ equality of $v=\alpha P_V h $ , $\|v\|=1$ so $\alpha = 1/ \| P_V h \| $. correct? $\endgroup$ – jack Feb 21 '16 at 23:21
  • 1
    $\begingroup$ @jack: Yes, except that's $\alpha^{-1}$ (and you need to treat $h=0$ separately). $\endgroup$ – joriki Feb 21 '16 at 23:24

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.