Let $M$ be a closed subspace of a Hilbert space $H$. Take $x_0 \in H\setminus M$ and let $d = \text{dist}(x_0,M)$. Show without using Hahn-Banach that there exists a $y \in H$ such that $\langle x_0,y\rangle = 1 $, $\langle x,y\rangle = 0$ for all $x \in M$, and $\|y\|= 1/d$.
I am sure that this is an easy exercise, but I cannot quite complete it. My initial idea is to take $y = x_0 – z$ where $z$ is the orthogonal projection of $x_0$ onto $M$. Indeed, this $y$ would satisfy $\langle x,y\rangle = 0$ for all $x \in M$, and we have $$\langle x_0,y\rangle = \langle y,y\rangle = \|y\|^2,$$ after normalizing $y$ we would get $\langle x_0,y\rangle =1$. But in the act of normalizing $y$ we could not have $\|y\| = 1/d$ … Is there another approach to this?
Best Answer
Your approach is perfectly valid:
Let $\Pi_M$ be the orthogonal projection from $H$ to $M^\top$($M$'s orthogonal complement) that vanishes $M$, the distance $d=\text{dist}(x_0,M)$ can be expressed as $||\Pi_M(x_0)||$.
From the $\langle x,y \rangle = 0,\forall x\in M$, one knows $y$ would be in $ M^\top$. Let $$y=\frac{\Pi_M(x_0)}{d^2}$$ Instantly follows from that, $||y||=d^{-2}||\Pi_M(x_0)||=d^{-1}$
Using $\Pi_M \Pi_M=\Pi_M$ and $\Pi_M{}^\dagger=\Pi_M$(properties of orthogonal projections) we can verify $$\langle x_0,y \rangle=\langle x_0,\Pi_M(y) \rangle=\langle \Pi_M{}^\dagger(x_0),y \rangle=\langle \Pi_M(x_0),d^{-2}\Pi_M(x_0) \rangle=d^{-2}||\Pi_M(x_0)||^2=1$$ thus being the required y.