Good evening, I'm trying to solve the below problem from an old exam.
Let $H$ be a $\mathbb R$-Hilbert space and $v \in H$. We denote by $\mathbb {\overline B}(0,1)$ the closed unit ball in $H$ and $K = \{x \in \mathbb {\overline B}(0,1) \mid \langle x,v\rangle = 0\}$. For each $x \in H$, prove that
There is a unique $T_x \in K$ such that $$\forall y\in K: \|x - T_x\| \le \|x-y\|$$
$T_x = -T_{-x}$
Could you please verify whether my attempt is fine or contains logical gaps/errors? Any suggestion is greatly appreciated!
My attempt:
1.
It is not hard to verify that $K$ is a nonempty closed convex subset of $H$. Then the orthogonal projection of $x$ on $K$ is our required $T_x$.
2.
If $y \in K$ then $y \in \mathbb {\overline B}(0,1)$ and $\langle y,v\rangle = 0$. This is equivalent to $\|y\| \le 1$ and $\langle y,v\rangle = 0$, which is in turn equivalent to $\|-y\| \le 1$ and $\langle -y,v\rangle = 0$. Hence $y \in K \iff -y \in K$.
We have $\forall y\in K: \|x - T_x\| \le \|x-y\|$ and thus $\forall y\in K: \|(-x) - (-T_x)\| \le \|(-x)+y\|$. It follows from $y \in K \iff -y \in K$ that $\forall y\in K: \|(-x) - (-T_x)\| \le \|(-x) - y\|$. As such, $T_{-x} = -T_x$.