# example of nonexpansive mappings

As we know :

### $T:X\longrightarrow X$ is a nonexpansive mapping iff $\|Tx-Ty\|\leq\|x-y\|,$ $\forall x,y\in X$

So my question is I want some nonexpansive mappings, I know $\sin(x)$ , $\cos(x)$ and if I'm not wrong $\frac{1}{x}$ for $x\geq1$

Thanks.

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Any differentiable function on $\mathbb R$ whose derivative is bounded by $1$ is an example. – Srivatsan Sep 22 '11 at 13:58
Thanks, but this is obvius from definition,only thing that I want is a nice and beautiful examples.. – Vahid Sep 22 '11 at 14:05
@Vahid..I don't understand your objection: you want something nice... ok let's thik about something differentiable at least... but then we fall in what Srivatsan said.. I can propose you arctan(x) as a nice example.. but how do you prove that this function works?? Uh it is differentiable and $\arctan'(x)=\frac{1}{1+x^2}\leq 1$... :) – uforoboa Sep 22 '11 at 14:09
Uh ok... what about $x\mapsto |x|?$ This is not differentiable... but it seems to me as obvious as any 1-Lipschitz Function... – uforoboa Sep 22 '11 at 14:11

I guess that $X$ is a normed space. Among nonexpansive mappings there are just nonexpansive (as you defined) and contractions: $$\|Tx-Ty\|\leq\alpha\|x-y\|$$ for some $\alpha<1$. The last case usually is much more desired since it allows to use e.g. the Banach Fixpoint (or Contraction Mapping) Theorem.

The idea how to imagine such function is the following: you define $$\alpha = \sup\limits_{x,y\,\,\in X}\frac{\|Tx-Ty\|}{\|x-y\|}$$ which is called the Lipschitz Constant and then if $\alpha<1$ you have the contraction mapping, if $\alpha = 1$ then the mapping is nonexpansive.

E.g. if $X = \mathbb R$ and $T$ (i.e. the function $T:\mathbb R\to \mathbb R$) is differentiable then $\alpha = \sup\limits_{x\in \mathbb R}|T\,\,'(x)|$ which explaines an example by @Srivatsan.

Other nice and beautiful example from probability. For a Markov process $\xi$ the shift operator $$\mathcal Pf(y) = \mathsf E[f(\xi_1)|\xi_0 = y]$$ is widely used and exactly non-expansive (usually it's not a contractive operator).

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Well.. If $X$ is any normed space, the function $$x\mapsto \|x\|$$ works fine, since by the triangular inequality

$$\|\|x\|-\|y\|\|\leq\|x-y\|.$$ Note that this example provides a function not everywhere differentiable on $X$ which is not expansive.

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Generalizing user15453's example you could take $x\mapsto \phi(\|x\|)$ for any nonexpansive mapping from $\mathbb{R}\to\mathbb{R}$.

I don't know for what purpose you would like to know more about nonexpansive mappings but if it is for convergent algorithms based on fixed point iterations, it is good to keep in mind that rotations and reflections are also nonexpansive operators. These are very easy examples that show that the simple fixed point iteration $x\mapsto Rx$ does not have to be converging. However, relaxing this to $x\mapsto \lambda x + (1-\lambda) Rx$ converges to a fixed point.

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One important class of examples is the collection of bounded linear operators $T : X \to X$ (where $X$ is a normed linear space) whose operator norm is at most $1$. (Proving this claim is less than a challenge because it amounts to just expanding the definitions of the terms "linear operator", "bounded" and "operator norm".)

A subclass of examples is the set of $n \times n$ unitary matrices with complex entries, seen as a linear map from $\mathbb C^n$ to itself. Note that these matrices are isometric (i.e., they preserve distances: $\| Tx- Ty \| = \| x-y \|$), and hence they are clearly non-expansive.

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@Vahid A $n \times n$ matrix $A$ can be seen as an element in $\mathbb C^{n^2}$ space. But it can be seen as a linear function $T : \mathbb C^{n} \to \mathbb C^n$ given by $T(x)=Ax$. – Srivatsan Sep 22 '11 at 14:55
you mean $\mathbb{C}^{n^2}$ but I didn't undrestand why it is isometirc what is your map ? – Vahid Sep 22 '11 at 14:56