Element of minimal distance to a convex, closed cone: orthogonality?

Question

I'm trying to prove the following theorem:

Let $H$ be a Hilbert space and $C\subseteq H$ a convex, norm-closed cone. Let $\xi \in H$. There is a unique $\eta \in C$ such that $\|\xi-\eta\| = d(\xi, C)$ (= the minimal distance of $\xi$ to $C$). Moreover, $\xi-\eta \perp \eta$.

I can show the first part of the theorem, namely that $\eta \in C$ exists with $\|\xi-\eta\| = d(\xi, C)$. I am not able to show that $\xi-\eta \perp \eta$ though, i.e. that $$0 = \langle \xi-\eta, \eta\rangle \iff \langle \xi,\eta\rangle =\|\eta\|^2.$$ Any hints or answers for this will be highly appreciated!

Answer 1

If you consider cones inside complex Hilbert spaces (which is often not done because of pathologies like this), then the result is not necessarily true. For example, if $H=\mathbb C$ (as complex Hilbert space with inner product $\langle z,w\rangle=\overline z w$), $C=i\mathbb R_+$ and $\xi=1+i$, then $\eta=i$ and $\langle \xi-\eta,\eta\rangle=i\neq 0$.

Answer 2

It is straightforward to show that the nearest point $\eta \in C$ to $\xi$, where $C$ is a closed convex set, satisfies $\langle \xi-\eta, c-\eta \rangle \le 0$ for all $c \in C$.

If $C$ is also a cone, then $c=\lambda \eta \in C$ for all $\lambda \ge 0$ and so $(\lambda -1) \langle \xi-\eta, \eta \rangle \le 0$ for all $\lambda \ge 0$. Hence $\langle \xi-\eta, \eta \rangle = 0$.