3
$\begingroup$

Let $H$ be a Hilbert space and $S$ be a subspace of $H.$ Let $x \in H$ and $\left \|x \right \| = 1.$ Prove that $$\inf\limits_{z \in S^{\perp}} \left \|x - z \right \| = \sup \left \{\left \lvert \left \langle x , y \right \rangle \right \rvert\ \big |\ y \in S, \left \|y \right \| \leq 1 \right \}.$$

My attempt $:$ Let $L = \inf\limits_{z \in S^{\perp}} \left \|x - z \right \|$ and $M = \sup \left \{\left \lvert \left \langle x , y \right \rangle \right \rvert\ \big |\ y \in S, \left \|y \right \| \leq 1 \right \}.$ If $x \in S^{\perp}$ then clearly $L = 0$ and $M = 0$ (because if $x \in S^{\perp}$ then for any $y \in S$ we have $\left \langle x,y \right \rangle = 0$). Also if $x \in S$ then we have \begin{align*} L & = \inf\limits_{z \in S^{\perp}} \sqrt {\|x\|^2 + \|z\|^2} \\ & = \inf\limits_{z \in S^{\perp}} \sqrt {1 + \|z\|^2} \\ & = \sqrt {1 + \inf\limits_{z \in S^{\perp}} \|z\|^2} \\ & = 1 \end{align*} and for all $y \in S$ with $\|y\| \leq 1$ we have by Cauchy Schwarz's inequality $$\left \lvert \langle x,y \rangle \right \rvert \leq \|x\| \|y\| \leq 1.$$ This shows that $M \leq 1.$ Also since $x \in S$ with $\|x\| = 1$ we have by taking $y = x$ $$\langle x,x \rangle = \|x\|^2 = 1.$$ So $M = 1.$ Therefore $L = M$ holds if $x \in S \cup S^{\perp}.$ Now $H = S \oplus S^{\perp}.$ So every element of $H$ can be written as $x = u + v,$ where $u \in S$ and $v \in S^{\perp}.$ For this case \begin{align*} \|(u+v) - z \|^2 & = \|u+v\|^2 + \|z\|^2 - \langle v , z \rangle - \langle z , v \rangle \\ & = \|u+v\|^2 + \|z\|^2 - 2 \mathfrak {R} \left ( \langle v,z \rangle \right ) \\ & \geq \|u+v\|^2 + \|z\|^2 - 2 \left \lvert \langle v , z \rangle \right \rvert \\ & \geq \|u+v\|^2 + \|z\|^2 - 2\|v\| \|z\| \\ & = \left (\|u+v\|^2 - \|v\|^2 \right ) + \left (\|z\| - \|v\| \right )^2 \\ & \geq \|u+v\|^2 - \|v\|^2 \end{align*} So by taking $z = v$ we have $$L = \sqrt {\|u+v\|^2 - \|v\|^2} = \sqrt {\|u\|^2 + 2 \mathfrak {R} \langle u,v \rangle} = \|u\|\ \ (\text {since}\ u \perp v).$$ Now for any $y \in S$ with $\|y\| \leq 1$ we have \begin{align*} \left \lvert \langle u + v , y \rangle \right \rvert & = \left \lvert \langle u , y \rangle + \langle v , y \rangle \right \rvert \\ & = \left \lvert \langle u,y \rangle \right \rvert\ \ \ \ \ \ \ \ (\text {Since}\ v \perp y ) \\ & \leq \|u\| \|y\| \\ & \leq \|u\| \end{align*} Now if $u = 0$ then $x = v \in S^{\perp}$ in which case we have already proved that $L = M.$ So WLOG we may assume that $u \neq 0.$ Then by taking $y = \dfrac {u} {\|u\|}$ we have $M = \|u\|.$ So in this case also we have $L = M,$ as required.

QED

Does my proof hold good? Please check it.

Thanks in advance.

EDIT $:$ I don't think that what I did is correct. Because Hilbert space can't have such decomposition unless $S$ was given to be closed.

$\endgroup$
6
  • 1
    $\begingroup$ Note that $\sup \left \{\left \lvert \left \langle x , y \right \rangle \right \rvert\ \big |\ y \in S, \left \|y \right \| \leq 1 \right \}=\sup \left \{\left \lvert \left \langle x , y \right \rangle \right \rvert\ \big |\ y \in \overline S, \left \|y \right \| \leq 1 \right \}$, where $\overline S$ denotes the closure of $S$ in $H$. $\endgroup$ – Mathlover Sep 10 '20 at 18:29
  • 1
    $\begingroup$ To prove this note that $\sup \left \{\left \lvert \left \langle x , y \right \rangle \right \rvert\ \big |\ y \in S, \left \|y \right \| \leq 1 \right \}\leq \sup \left \{\left \lvert \left \langle x , y \right \rangle \right \rvert\ \big |\ y \in \overline S, \left \|y \right \| \leq 1 \right \}$ as supremum of larger set is larger. $\endgroup$ – Mathlover Sep 10 '20 at 18:29
  • 1
    $\begingroup$ For other direction, choose $\varepsilon>0$, and then find $z\in\overline S$ with $\sup \left \{\left \lvert \left \langle x , y \right \rangle \right \rvert\ \big |\ y \in \overline S, \left \|y \right \| \leq 1 \right \}-\varepsilon<\left \lvert \left \langle x , z \right \rangle \right \rvert=\lim\left \lvert \left \langle x , z_n \right \rangle \right \rvert\leq \sup \left \{\left \lvert \left \langle x , y \right \rangle \right \rvert\ \big |\ y \in S, \left \|y \right \| \leq 1 \right \}$, where $S\ni z_n\to z\in\overline S$. Since, $\varepsilon>0$ is arbitrary, we are done. $\endgroup$ – Mathlover Sep 10 '20 at 18:29
  • 1
    $\begingroup$ So, without loss of generality, you may assume $S$ to be closed as $\big(S^\perp\big)^\perp=\overline S$ for a subspace $S$. math.stackexchange.com/questions/1761685/… $\endgroup$ – Mathlover Sep 10 '20 at 18:33
  • $\begingroup$ While writing down a proof for this in the entrance do I have to write down the proof of $\left (S^{\perp} \right )^{\perp} = \overline {S}$ or it is fine to just mention the result and proceed? I'm asking this question because of time bounds @Sumanta. $\endgroup$ – Anacardium Sep 10 '20 at 19:36
1
$\begingroup$

It's not clear why you first take the case $x\in S$, as it is fairly particular.

When $x\in S^\perp$, one gets directly that $L=M=0$. So we may assume $x\not\in S^\perp$. Also, neither $L$ nor $M$ change if we replace $S$ with its closure, so we may assume that $S$ is closed.

What you have is, since $H=S\oplus S^\perp$, that $x=x_S+x_{S^\perp}$. As $S^\perp$ is a subspace, for $z\in S^\perp$ we have $x-z=x_S-(z-x_{S^\perp})$. Then $$ L=\inf\{\|x_s-z\|:\ z\in S^\perp\}=\|x_S\|, $$ since $\|x_s-z\|^2=\|x_s\|^2+\|z\|$ for any $z\in S^\perp$. Now, for any $y\in S$ with $\|y\|=1$, we have $$ |\langle x,y\rangle|=|\langle x_S,y\rangle|\leq\|x_S\|\,\|y\|=L, $$ so $M\leq L$. And $$ M\geq\Bigg|\bigg\langle x,\frac{x_S}{|x_S\|}\bigg\rangle\Bigg|=\|x_S\|=L. $$

$\endgroup$
4
  • $\begingroup$ I have started with the the easier subcases just to warm up. No other intention was there. BTW $H$ can't have a decomposition as $S \oplus S^{\perp}$ unless $S$ was a closed subspace of $H$ which I clearly mentioned in my edit section. One needs to show that there will be no harm in considering $S$ to be closed subspace of $H$ as Sumanta has rightly pointed out in his comment above. $\endgroup$ – Anacardium Sep 11 '20 at 4:46
  • $\begingroup$ Another remark I want to make which is $:$ You can't divide any vector by $\|x_S\|$ to make it a unit vector unless you know well in advance that $\|x_S\| \neq 0.$ For $x_S = 0$ you need to go back to one of the fairly particular cases I have considered namely the case when $x \in S^{\perp}.$ (See the last paragraph of my proof) Please feel free to correct me if I'm wrong. $\endgroup$ – Anacardium Sep 11 '20 at 4:58
  • $\begingroup$ You are right, but all one needs is to mention, as you did, that $L=M=0$ in that case. As for $S$ closed or not I missed that, but neither $L$ nor $M$ change when you replace $S$ with its closure. $\endgroup$ – Martin Argerami Sep 11 '20 at 5:46
  • $\begingroup$ Yeah one needs to show that WLOG we may assume that $S$ is closed. Thank you so much for your prompt response. Accepted your answer. $\endgroup$ – Anacardium Sep 11 '20 at 6:04

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.