Let $E$ denote a real Hilbert space and suppose $G \subset E$ is a nonempty closed convex set. I know that in this case, there is a unique nearest point in $G$ to each $x \in E$. Call this point $P_G(x)$.
I am trying to prove the following proposition:
Let $\bar x \in E \setminus G$, and consider the hyperplane $H = \{x \in E : \langle x - P_G(\bar x), \bar x - P_G(\bar x)\rangle \leq 0\}$.Then $G \subset H$ but $\bar x \not \in H$.
It is clear that $\bar x \not \in H$, for else $\bar x = P_G(\bar x)$ and then $\bar x \in G$.
To show $G \subset H$, I came up with the following proof that works in $\mathbf{R}^d$, but I am not sure how to justify (e.g. the differentiation) in the setting of a Hilbert space.
Suppose that $x' \in G$ but $x' \not \in H$; hence $\langle x' - P_G(\bar x), \bar x - P_G(\bar x)\rangle > 0$. Consider the point $$ x_t := P_G(\bar x) + t\big(x' - P_G(\bar x)\big), \quad t \in \mathbf{R} $$ And also consider $$ f(t) = \|x_t - \bar x\|^2 - \|P_G(\bar x) - \bar x\|^2 $$ Note that $$ \dot{f}(t) = 2\Big\langle P_G(\bar x) - \bar x + t\big(x' - P_G(\bar x)\big), x' - P_G(\bar x)\Big\rangle $$ In particular $\dot{f}(0) < 0$ by hypothesis. Therefore for some $0 < \tau < 1$, we see that $f(\tau) < f(0) = 0$. This implies that $$ \|x_\tau - \bar x\| < \|P_G(\bar x) - \bar x\|, $$ but on the other hand $x_\tau \in G$, which is a contradiction.
Let $p=P_G(\bar{x})$.
Since $\bar{x} \notin G$ you know that$ \|\bar{x}-p\| >0$. Now note that $\langle \bar{x} - p, \bar x - p \rangle = \|\bar{x}-p\|^2 >0$ and so $\bar{x} \notin H$.
For any $x \in G$ we have $\|x-\bar{x}\|^2 = \|x-p+p-\bar{x}\|^2 \ge \|p-\bar{x}\|^2$, and so$\|x-p\|^2 + 2 \langle x-p,p-\bar{x} \rangle \ge 0$ for all $x \in G$.
Replacing $x$ by $tx+(1-t)p$ with $t \in [0,1]$, we get $t^2 \|x-p\|^2 + 2 t\langle x-p,p-\bar{x} \rangle \ge 0$ for all $x \in G$ and $t \in [0,1]$. Hence $\langle x-p,p-\bar{x} \rangle \ge 0$ for all $x \in G$.
In particular, if $x \in G$ then $x \in H$.