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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

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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?

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  • $\begingroup$ 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
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    $\begingroup$ @jack: Yes, except that's $\alpha^{-1}$ (and you need to treat $h=0$ separately). $\endgroup$ – joriki Feb 21 '16 at 23:24

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