1
$\begingroup$

$H$ Hilberspace, $K_1$, $K_2$ convex, closed subset of H. $K_1\subset K_2$

I would like to show that for all $x\in H$: $ \|P_{K_1}(x)-P_{K_2}(x)\|^2 \leq 2(d(x,K_1)^2-d(x,K_2)^2) $

My first idea was to add and substract x on the left side $ \|P_{K_1}(x)-P_{K_2}(x)-x+x \|^2 \leq \| P_{K_1}(x)-x\|^2+\|x-P_{K_2}(x)\|^2 $

Can I do that? How is it possible to continue?

1 Answers 1

1

You don't even need the factor $2$. In general, if $\operatorname{Re} \langle v, w \rangle \leq 0$ then $||v||^2 \leq ||v - w||^2 - ||w||^2$. Now $K_1 \subset K_2$ and so $P_{K_1}(x) \in K_2$. In particular $\operatorname{Re} \langle P_{K_1}(x) - P_{K_2}(x), x - P_{K_2}(x) \rangle \leq 0$ since $K_2$ is convex. Take $v = P_{K_1}(x) - P_{K_2}(x)$ and $w = x - P_{K_2}(x)$.

  • 0
    Thank you very much for your answer.Maybe you can take a look at http://math.stackexchange.com/questions/227673/orthogonal-projection-p-k8x2012-11-05