Let $C$ be a closed, convex set in a Hilbert space $H$. $f(x)=\inf_{y \in C} \frac{1}{2}||x-y||^2$
WTS:
$$\lim_{\alpha \to 0^{+}} \frac{||x+\alpha y-P_c(x+\alpha y)||^2-||x-P_c(x)||^2}{2\alpha}= \langle y | \textbf{I}(x)-P_c(x) \rangle $$
Thus $\nabla f=\textbf{I}-P_c$
We have this characterization of the projection operator,
$\forall x \in H$, we have $P_c(x)=p$ and $p \in C$. and $\forall y \in C$, we have that
$$\langle y-P_c(x)|x-P_c(x) \rangle \leq 0$$
My attempt:
From this expression, I get that $$\langle P_c(x+\alpha y)-P_c(x)|x-P_c(x) \rangle \leq 0$$ $$\langle P_c(x)-P_c(x+\alpha y)|x+\alpha y -P_c(x+\alpha y) \rangle \leq 0$$ Since $P_c(x+\alpha y) \in C$ and $P_c(x) \in C$.
$$=\frac{||x+\alpha y||^2 + ||P_c(x+\alpha y)||^2- 2\langle x+\alpha y | P_c(x+\alpha y) \rangle -||x||^2-||P_c(x)||^2+2\langle x | P_c(x) \rangle }{2\alpha} $$
$$=\frac{||x||^2+\alpha^2 ||y||^2 +2\alpha\langle x|y \rangle+ ||P_c(x+\alpha y)||^2- 2\langle x+\alpha y | P_c(x+\alpha y) \rangle -||x||^2-||P_c(x)||^2+2\langle x | P_c(x) \rangle }{2\alpha} $$
$$=\frac{\alpha^2 ||y||^2 +2\alpha\langle x|y \rangle+ ||P_c(x+\alpha y)||^2- 2\langle x+\alpha y | P_c(x+\alpha y) \rangle -||P_c(x)||^2+2\langle x | P_c(x) \rangle }{2\alpha} $$
$$ =\frac{\alpha ||y||^2}{2}+\langle x | y \rangle+\frac{||P_c(x+\alpha y)||^2- 2\langle x+\alpha y | P_c(x+\alpha y) \rangle -||P_c(x)||^2+2\langle x | P_c(x) \rangle }{2\alpha} $$
$$ =\langle x | y \rangle+\frac{||P_c(x+\alpha y)||^2- 2\langle x+\alpha y | P_c(x+\alpha y) \rangle -||P_c(x)||^2+2\langle x | P_c(x) \rangle }{2\alpha} $$
$$ =\langle x | y\rangle + \frac{\langle P_c(x+\alpha y) | P_c(x+\alpha y)\rangle - 2\langle x+\alpha y | P_c(x+\alpha y) \rangle -\langle P_c(x) | P_c(x) \rangle + 2\langle x | P_c(x) \rangle}{2\alpha} $$
$$ =\langle x | y\rangle + \frac{\langle P_c(x+\alpha y) | P_c(x+\alpha y)\rangle - 2\langle x| P_c(x+\alpha y) \rangle-2 \langle \alpha y | P_c(x+\alpha y)\rangle -\langle P_c(x) | P_c(x) \rangle + 2\langle x | P_c(x) \rangle}{2\alpha} $$
$$ =\langle x | y\rangle - \langle y | P_c(x+\alpha y) \rangle +\frac{\langle P_c(x+\alpha y) | P_c(x+\alpha y)\rangle- 2\langle x| P_c(x+\alpha y)\rangle-\langle P_c(x) | P_c(x) \rangle + 2\langle x | P_c(x) \rangle}{2\alpha} $$
I cannot figure out a way to proceed. Can someone help?