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Problem. Let $X$ be a Hilbert space and $\emptyset \neq K \subseteq X$ be closed and convex. Then, $$ \|P_Kx - P_Ky \| \leq \|x-y \|$$ for all $x,y \in X$. Here, $P_K$ is the projection from $X$ onto $K$; that is the unique nearest element in $K$.

It doesn't look too difficult but I can't really see how to prove it. Geometrically (in low-dimensional cases at least) it makes sense but I'm not sure how to connect these objects. How should I approach this?

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  • $\begingroup$ An answer was posted already. Unfortunately, the author deleted it, as (I assume that was the reason) it was slightly flawed. I'll keep the user anonymous but will sketch the idea. If $K$ was even a subspace of $X$, then we can apply the projection theorem. In this case, the inequality makes great geometric sense and we can discover Pythagoras' theorem in it and finish off. The flaw was that the projection theorem, of course, does not work for arbitrary convex closed sets. $\endgroup$ – Qi Zhu Feb 6 '19 at 11:49
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Hint: The projection is characterized by the variational inequality (VI) $$ ( P_K(x) - x, z - P_K(x) ) \ge 0 \qquad\forall z \in K.$$ (If you do not have seen this formula, you should try to prove it.) Now, you have to use this VI for $P_K(x)$ and for $P_K(y)$ and you have to use clever choices for $z$.

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  • $\begingroup$ Ahh, thank you! Indeed, it was actually a Corollary (left as an exercise) of VI in our lecture notes. So by putting $z = P_Ky$ and $z= P_Ky$ each, adding those inequalities and applying Cauchy-Schwarz gives the result for me. (And of course, it works the same for complex numbers - where the variational inequality becomes $\operatorname{Re}(\langle P_K(x)-x, z-P_K(x) \rangle) \geq 0$.) I'll write an additional answer for other people who might be interested. $\endgroup$ – Qi Zhu Feb 6 '19 at 13:00
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Let the ground field be $\mathbb{R}$. By the variational inequality $$ \langle P_K(x)-x, P_K(y)-P_K(x) \rangle \geq 0 \quad \text{and} \quad \langle P_K(y) - y, P_K(x) - P_K(y) \rangle \geq 0. $$ Summing them up gives $$ \langle x-y, P_K(x)-P_K(y) \rangle \geq \|P_K(x)-P_K(y) \|^2 $$ Cauchy-Schwarz finishes. $$ \|x-y \| \cdot \|P_K(x)-P_K(y) \| \geq \langle x-y, P_K(x)-P_K(y) \rangle \geq \|P_K(x)-P_K(y) \|^2. $$

Remark. The proof is identical on $\mathbb{C}$. Only the variational inequality is slightly different.

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  • $\begingroup$ What is the complex version of the variational inequality? $\endgroup$ – JustDroppedIn Feb 6 '19 at 15:14
  • $\begingroup$ It's almost the same: $\operatorname{Re}(\langle P_K(x)-x, z-P_K(x) \rangle \geq 0$. The motivation is that angles can be measured in this way in $\mathbb{C}$ by the law of cosines. $\endgroup$ – Qi Zhu Feb 6 '19 at 16:37

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