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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: $\ \leq \ 0 \ \forall v\in D$ $\ \leq \ 0 \ \forall v\in D$

I appreciate some help, this is a problem from my homework. Thanks

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    @DonAntonio I guess from the title that $P_D$ is the orthogonal projection onto $D$.2012-09-23

2 Answers 2

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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. $

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    Nice answer, thanks WimC.2012-10-01
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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. 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...

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    The first property is an immediate consequence of the definition of projection. You can find it, for example, in the book of Brezis Functional Analysis. The other one is an exercise (consequence of the first one).In fact they are equivalent.2012-09-24