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Let $H$ be a Hilbert space, $e\in H$ with $\|e\|=1$. Let $L=\{x\in H\mid \langle x,e\rangle=0\}$. For $L$ we consider the sublinear function $d_L\colon H\to \mathbb{R}$ $$d_L(h):=\inf\{\|h-y\| : y\in L\}.$$ Claim: $d_L(h)=|\langle h,e\rangle|$ for all $h\in H$.

I first thought about applying Riesz representation theorem for Hilbert spaces but here we have sublinear functions instead of linear functionals (i.e. instead of elements in the dual $H'$) and I'm not sure how to define a suitable functional to apply this theorem.

An attempt: Is is $min\{\|x-y\|:y\in L\}=max\{|\langle x,y\rangle|:y\in L^\perp , \|y\|=1\}$ (see for example Prove that $min\{\|x-y\|:y\in M\}=max\{|\langle x,y\rangle|:y\in M^\perp , \|y\|=1\}$ ) We have that $e\in L^\perp$. If it is possible to show that $max\{|\langle x,y\rangle|:y\in L^\perp , \|y\|=1\}=|\langle x,e\rangle|$, we are done. But I don't know how to prove the last equlity.

How to prove $$max\{|\langle x,y\rangle|:y\in L^\perp , \|y\|=1\}=|\langle x,e\rangle|$$ or how else to prove the claim? I appreciate any help and hints.

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  • $\begingroup$ Prove that you can decompose every $h$ in an orthogonal part and a normal part w.r.t. the plane $L$. Then apply Pythagoras' theorem $\endgroup$ – Del Jan 4 '18 at 11:25
  • $\begingroup$ Thank you. I already know that I can decompose $h$ like this. Pythagoras: $\|h\|^2=\|h_1\|^2+\|h_2\|^2$ for $h=h_1+h_2$ with $h_1\in L$ and $h_2\in L^\perp$. But I don't know how to use ist. I got an idea in the meantime and edited my question. $\endgroup$ – user471056 Jan 4 '18 at 13:41
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Notice that $L = (\operatorname{span}\{e\})^\perp$ so $L^\perp = (\operatorname{span}\{e\})^{\perp\perp} = \operatorname{span}\{e\}$.

Therefore we have:

\begin{align} \max\Big\{\left|\langle x, y\rangle\right| : y \in L^\perp, \|y\| = 1\Big\} &= \max\Big\{\left|\langle x, y\rangle\right| : y \in \operatorname{span}\{e\}, \|y\| = 1\Big\} \\ &= \max\Big\{\left|\langle x, y\rangle\right| : y = \lambda e \text{ with } |\lambda| = 1\Big\} = \left|\langle x, e\rangle\right| \end{align}

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Decompose $x \in H$ into the orthogonal sum $$ x = \langle x,e\rangle e+(x-\langle x,e\rangle e). $$ For any $l\in L=\{ x\in H : \langle x,e\rangle = 0 \}$, the following is an orthogonal decomposition: $$ x-l = \langle x,e\rangle e+\{(x-\langle x,e\rangle e)-l \}. $$ Therefore, $$ \|x-l\|^2 = |\langle x,e\rangle|^2+\|(x-\langle x,e\rangle e)-l\|^2, $$ which has a unique minimum where $l=x-\langle x,e\rangle e$. The minimum distance is $|\langle x,e\rangle|$.

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  • $\begingroup$ very nice solution, thank you! $\endgroup$ – user471056 Jan 7 '18 at 17:59

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