Contraction operator.

Question

Let $\psi : \mathbb{R} \rightarrow \mathbb{R}$ a function derivable so that, for all $t \in \mathbb{R}$, $\vert {\psi'(t)} \vert \leq \alpha < 1$.

How do I prove that $\psi$ is a contraction?

How can I generalize for functions of $\mathbb{R}^n$ on $\mathbb{R}^m$?.

By definition we have; Let $(X,d)$ and $(Y,D)$ metric spaces. A function $A:X \rightarrow Y$ is a contraction if there is a constant $0 \leq \alpha < 1$ such that, for all $\xi, \eta \in X$, $D(A(\xi), A(\eta)) \leq \alpha d(\xi, \eta)$.

Answer 1

Well, suppose for the moment that $s \le t$; then since $\psi (t) \in C^1(\Bbb R)$,

$\psi(t) - \psi(s) = \displaystyle \int_s^t \psi'(u) \; du, \tag 1$

whence

$\vert \psi(t) - \psi(s) \vert = \left \vert \displaystyle \int_s^t \psi'(u) \; du \right \vert \le \displaystyle \int_s^t \vert \psi'(u) \vert \; du \le \displaystyle \int_s^t \alpha \; du = \alpha(t - s) = \alpha \vert t - s \vert, \tag 2$

since both $\alpha$ and $t - s$ are non-negative; if $t \le s$, we still have

$\vert \psi(t) - \psi(s) \vert = \vert \psi(s) - \psi(t) \vert = \left \vert \displaystyle \int_t^s \psi'(u) \; du \right \vert$ $\le \displaystyle \int_t^s \vert \psi'(u) \vert \; du \le \displaystyle \int_t^s \alpha \; du = \alpha(s - t) = \alpha \vert t - s \vert, \tag 3$

so thus for all $s, t \in \Bbb R$ we have

$\vert \psi(t) - \psi(s) \vert \le \alpha \vert t - s \vert; \tag 4$

since $\alpha < 1$, (4) shows $\psi(t)$ is a contraction.

Now let

$\Psi(x) \in C^1(\Bbb R^m, \Bbb R^n) \tag 5$

such that

$\Vert D\Psi(x) \Vert \le \alpha < 1, \; \forall x \in \Bbb R^n; \tag 6$

then if

$\gamma:[0, 1] \to \Bbb R^m \tag 7$

is the line segment defined by

$\gamma(t) = (1 - t)x + ty = x + t(y - x), \tag 8$

we have

$\Psi(y) - \Psi(x) = \Psi(\gamma(1)) - \Psi(\gamma(0)) = \displaystyle \int_0^1 \dfrac{d}{dt}\Psi(\gamma(t)) \; dt$ $=\displaystyle \int_0^1 D\Psi(\gamma(t)) \gamma'(t) \; dt = \int_0^1 D\Psi(\gamma(t)) (y - x) \; dt; \tag 9$

therefore,

$\Vert \Psi(y) - \Psi(x) \Vert = \left \Vert \displaystyle \int_0^1 D\Psi(\gamma(t)) (y - x) \; dt \right \Vert \le \displaystyle \int_0^1 \Vert D\Psi(\gamma(t)) (y - x) \Vert \; dt;$ $\le \displaystyle \int_0^1 \Vert D\Psi(\gamma(t)) \Vert \Vert (y - x) \Vert \; dt \le \displaystyle \int_0^1 \alpha \Vert (y - x) \Vert \; dt = \alpha \Vert y - x \Vert; \tag{10}$

(10) shows $\Psi(t)$ is a contraction mapping by virtue of the fact that $\alpha < 1$, thus generalizing (4) to functions $\Psi(x) \in C^1(\Bbb R^n, \Bbb R^m)$.

Answer 2

Let $\psi:\mathbb R^m \to \mathbb R^n$ be continuously differentiable.

Then the result is still true if you require that the (operator) norm of $\psi'$ has norm strictly less than one: assume $\|\psi'(x)\|\le \alpha<1$ for all $x$. Let $x,y$ be given. Then $$ \|f(x)-f(y)\| = \sup_{t\in[0,1]}\|f'(x+t(y-x)\| \cdot \|x-y\|\le\alpha \|x-y\|, $$ where I used the mean-value theorem for vector-valued functions.