Showing $ \left\lVert P_C(x) - x \right\rVert_2^2 \le \left\lVert x - c \right\rVert_2^2 - \left\lVert P_C(x) - c \right\rVert_2^2 $

Question

$C$ is a convex set and $ x \in R^d. P_c(x) $ is the orthogonal project of $x$ onto $C$ and $c \in C$. I want to show that

$ \left\lVert P_C(x) - x \right\rVert_2^2 \le \left\lVert x - c \right\rVert_2^2 - \left\lVert P_C(x) - c \right\rVert_2^2 $

My approach is using triangular inequality for $ \left\lVert P_C(x) - x \right\rVert_2^2$

$ \left\lVert P_C(x) - x \right\rVert_2^2 \ge \left\lVert P_C(x) \right\rVert_2^2 - \left\lVert x \right\rVert_2^2$.

We can multiply this by minus as all terms are $ \ge0 $. So

$ \left\lVert P_C(x) - x \right\rVert_2^2 \le \left\lVert x \right\rVert_2^2- \left\lVert P_C(x) \right\rVert_2^2$.

Adding $\left\lVert c \right\rVert_2^2$ and substracting it.

$ \left\lVert P_C(x) - x \right\rVert_2^2 \le \left\lVert x \right\rVert_2^2- \left\lVert c \right\rVert_2^2 - ( \left\lVert P_C(x) \right\rVert_2^2 - \left\lVert c \right\rVert_2^2)$.

But I am cannot progress after this step.

Is there any other way to show this inequality using properties of the orthogonal projection on to convex sets.

Answer 1

Hint: Use

$$ \|x-c\|_2^2 = \|P_C(x)-c\|_2^2+ \|P_C(x)-x\|_2^2 -2(P_C(x)-x,P_C(x)-c). $$