Let $H$ be a Hilbert space and $S$ be a subspace of $H$. Let $x\in H$ and $\lvert\lvert x\rvert\rvert=1$. Prove that $$\inf_{z\in S^\perp}\lvert\lvert x-z\rvert\rvert=\sup\{\lvert\langle x,y\rangle\rvert :y\in S, \lvert\lvert y\rvert\rvert\leq1\}.$$

I am not a functional analysis person but I need to solve a few problems for some tests. Any hints would be helpful. If someone can point me the underlying philosophy and mechanics of such a problem, it shall be great. As this problem stands now, I do not have a feel for what is actually going on here.

Thanks in advance!

  • $\begingroup$ Some things that could be useful here: Cauchy-Schwarz inequality and the orthogonal decomposition $x = x_S + x_{S^\perp}$. $\endgroup$
    – TSF
    Apr 12 '19 at 12:18

Let's start with the left hand side but lets replace the norm with the norm squared, since if we minimize the norm squared we also minimize the norm. We can decompose $x$ on the two subspaces $S$ and $S^\perp$ in the following way,

$$x = x_S + x_{S^\perp},\quad x_S\in S,\quad x_{S^\perp}\in S^\perp$$

Substituting that into the l.h.s gives,

$$\inf\limits_{z\in S^\perp}\Vert x_S+x_{S^\perp} - z\Vert^2$$

Now, there is a theorem (in fact, it is the Pythagorean theorem for Hilbert spaces) which says $\Vert x+y\Vert^2 = \Vert x\Vert^2 + \Vert y\Vert^2$ if $x$ and $y$ are orthogonal. We know that $x_{S^\perp}-z$ is an element of $S^\perp$ because it is a subspace and $z\in S^\perp$. We also know, since $x_S\in S$, that $x_S$ is orthogonal to $x_{S^\perp}-z$. Thus we can write the following,

$$\inf\limits_{z\in S^\perp}\Vert x_S+x_{S^\perp} - z\Vert^2 = \inf\limits_{z\in S^\perp}\Big(\Vert x_{S^\perp} - z\Vert^2 + \Vert x_S\Vert^2\Big) = \Vert x_S\Vert^2 + \inf\limits_{z\in S^\perp}\Vert x_{S^\perp} - z\Vert^2$$

Since $x_{S^\perp}\in S^\perp$ we can take $z=x_{S^\perp}$ to get,

$$\inf\limits_{z\in S^\perp}\Vert x-z\Vert^2 = \Vert x_S\Vert^2\implies \inf\limits_{z\in S^\perp}\Vert x-z\Vert =\Vert x_S\Vert.$$

Now let's focus on the right hand side. We will use the same decomposition for $x$ as above but now we will use orthogonality and then the Cauchy Schwarz inequality.

$$\sup\limits_{y\in S, \vert y\vert\leq 1}\vert \langle y,x_S+x_{S^\perp}\rangle\vert = \sup\limits_{y\in S, \vert y\vert\leq 1}\vert \langle y,x_S\rangle\vert$$

The term we are taking the supremum of has an upper bound given by the CS inequality,

$$\vert \langle y, x_S\rangle \vert\leq \Vert y\Vert \Vert x_S\Vert\leq\Vert x_S\Vert$$

where the last inequality is due to our assumption that $\Vert y \Vert \leq 1$. Equality for the CS inequality is attained if the two vectors are linearly dependent, i.e. $y=\alpha x_S$ for some scalar $\alpha$. Since we are choosing $y$ to be in the unit ball, take $y=\frac{x_S}{\Vert x_S\Vert}$ which gives,

$$\sup\limits_{y\in S, \vert y\vert\leq 1}\vert \langle y, x_S\rangle\vert = \Vert\frac{x_S}{\Vert x_S\Vert}\Vert\Vert x_S\Vert = \Vert x_S\Vert$$

Thus the two sides are equal.

  • $\begingroup$ Wow! This was very neat. Thank you very much! $\endgroup$
    – WhySee
    Apr 12 '19 at 17:39
  • $\begingroup$ You don't get the decomposition when $S$ is not closed. $\endgroup$ Apr 13 '19 at 5:27
  • $\begingroup$ You're right, it holds in finite dimension at least and in arbitrary dimension if the subspace is closed. $\endgroup$
    – TSF
    Apr 13 '19 at 10:56

By continuity of the inner product, we may assume that $S$ is closed.

Write $x=Px+Qx$ where $P$ and $Q$ are the canonical projections on $S$ and $S^{\perp}$ respectively.

Then, $\inf_z\|x-z\|=\|Px\|.$

Now, the Hahn Banach theorem (applied to $S$) shows that

$\|Px\|=\sup\{|L(Px)|:L\in \mathscr L(S,\mathbb C);\ \|L\|=1\}.$

And, by the Riesz theorem, there is a $w\in S$ such that $Ls=\langle s,w\rangle$, and $\|w\|=1$ so

$\|Px\|=\sup\{|\langle Px,w\rangle :w\in S;\ \|w\|=1\}=$

$\sup\{|\langle Px+Qx,w\rangle :w\in S;\ \|w\|=1\}=$

$\sup\{|\langle x,w\rangle :w\in S;\ \|w\|=1\},$

as desired.

  • $\begingroup$ Thank you for your answer. The only thing is that you have used more technical results which I aren't extremely comfortable with. $\endgroup$
    – WhySee
    Apr 12 '19 at 20:22

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.