# On subspaces of Hilbert spaces

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.

• Some things that could be useful here: Cauchy-Schwarz inequality and the orthogonal decomposition $x = x_S + x_{S^\perp}$.
– 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.

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

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