Projection onto closed convex set Show that the function defined by $f(t)=|P_{D}(x+td)-x|$ is nondecreasing, where $D$ is closed convex, $x\in D$, $t\geq 0$, $d\in \mathbb{R}^{n}$ and $P_{D}$ is projection onto D.
I tried to solve this question in a lot of ways, for example, if we use the polarization identity, we get $$\frac{1}{2}(|P_{D}(x+td)-x|^{2}-|P_{D}(x+sd)-x|^{2})= \\ \langle P_{D}(x+td)-x+P_{D}(x+sd)-x,P_{D}(x+td)-x-(P_{D}(x+sd)-x)\rangle$$ where $\langle,\rangle$ stands for the usual scalar product on $\mathbb R^{n}$. I tried to explore this equality, but i got nothing.
Here is some inequalities that maybe can help: if D is closed and convex then: $$<x-P_{D}(x),v-P_{D}(x)>\ \leq \ 0 \ \forall v\in D$$ $$<x-v,v-P_{D}(x)>\ \leq \ 0 \ \forall v\in D$$
I appreciate some help, this is a problem from my homework. Thanks
 A: Suppose $v, w \in \mathbb{R}^n$ and let $z = P_D(w) - P_D(v)$.   Then (using your first inequality) we have
$$
\langle v, z \rangle \leq \langle P_D(v), z \rangle \leq \langle P_D(w), z \rangle \leq \langle w, z \rangle.
$$
Now take $v = x + t_1d$ and $w = x + t_2d$ for some $t_2 > t_1 > 0$.  Then these inequalities can be rewritten as
$$
t_1 \langle d, z \rangle \leq \langle P_D(v) - x, z \rangle \leq \langle P_D(w) - x, z \rangle \leq t_2 \langle d, z \rangle.
$$
This implies that $\langle d, z \rangle \geq 0$ and in particular
$$
\langle P_D(v) - x, z \rangle \geq 0.
$$
Then
$$
||P_D(w) - x||^2 = ||P_D(v) - x + z||^2 = ||P_D(v)-x||^2 + 2 \langle P_D(v) - x, z\rangle + ||z||^2 \geq ||P_D(v)-x||^2.
$$
A: Simplifications: we can assume that $x=0$ (since we could shift the whole picture by $-x$), so $0\in D$.
Denote $P_t:=P_D(td)$ and $P_s:=P_D(sd)$, $\ 0<t<s$. We want to prove that $|P_s|^2 > |P_t|^2$, i.e. $$\langle P_s - P_t, P_s+P_t\rangle \ge 0$$
That's where I could get, also $t=1$ can be a simplification, but doesn't matter much. Now we can use that $0$, $P_s$, $P_t$ and $\displaystyle\frac{P_s+P_t}2$ are all in $D$.
I think we are very near...
