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Let $H$ be a real Hilbert space and $\bar{B_r} = \{x\in H :\Vert x\Vert \leq r\}$ with $r>0$. Define $f:H \rightarrow \bar {B_r}$ by

$ f(x)=\left\{ \begin{array} xx,& \Vert x \Vert \leq r\\ rx/\Vert x \Vert,& \Vert x \Vert >r \end{array}\right. $

I want to prove that f is a nonexpansive map(i.e. $\Vert f(x)-f(y) \Vert \leq \Vert x-y \Vert$).

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  • $\begingroup$ Your thoughts on the problem? $\endgroup$ – tattwamasi amrutam Jul 17 '17 at 13:10
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First we need the following Hilbert space inequality:

Lemma 1: For every nonzero $x, y\in H$, $ \left\| \frac{x}{\|x\|}-\frac{y}{\|y\|} \right\| \leq \frac{\|x-y\|}{\min\{ \|x\|, \|y\|\}}$.

Proof: Without loss of generality we may assume that this minimum is attained at $\|x\|$, so we are trying to show that $\left\|x-\frac{\|x\|}{\|y\|}y\right\| \leq \|x-y\|$. Taking squares in both sides and using the inner product properties we get that this inequality is equivalent to $$2\langle x, y\rangle \left(1-\tfrac{\|x\|}{\|y\|}\right) \leq \|y\|^2-\|x\|^2, $$ which is true (just apply the Cauchy–Schwarz inequality on the left hand side of it).

To return to your question, there are three cases we need to examine, regarding where $x,y$ lie: If $x, y\in B(0,r)$, then the result is obvious. If $\|x\|, \|y\|>r$, then it follows from the previous lemma. If $\|x\|\leq r$ and $\|y\|>r$, then we need to show that $$\left\|\|x\|-\tfrac{ry}{\|y\|}\right\| \leq \|x-y\|.$$ I will leave the details of this case to you, just follow the steps of the proof of Lemma 1.

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  • $\begingroup$ how do you prove that $2\langle x, y\rangle \left(1-\tfrac{\|x\|}{\|y\|}\right) \leq \|y\|^2-\|x\|^2, $ because after applying cauchy schwarz inequality on left side I get $2\langle x, y\rangle \left(1-\tfrac{\|x\|}{\|y\|}\right) \leq 2\left(\|y\|^2-\|x\|^2\right) $ @tree detective $\endgroup$ – ankit kumar Jul 20 '17 at 9:28
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    $\begingroup$ @ankitkumar If $\langle x, y\rangle < 0$, then the inequality is clearly true. If $\langle x, y\rangle \geq 0$, then after applying $\langle x, y\rangle\leq \|x\|\|y\|$, the inequality becomes $2\|x\|\|y\| -2\|x\|^2\leq \|y\|^2-\|x\|^2$, which is equivalent to $\|y\|^2 +\|x\|^2-2\|x\|\|y\|\geq 0$. $\endgroup$ – tree detective Jul 20 '17 at 11:25
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$f$ is in general not nonexpansive: Let $r=3$, fix $x_0 \in H$ with $||x_0||=3$ and take $x=\frac{1}{2}x_0$ and $y=2x_0$.

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