Check proof of contraction mapping

Question

let When X=[-1,1] endowed with standard distance. I want to show that if $T: X \rightarrow X$, defined by $T(x) = sin(x)$ is a contraction mapping?

Here's what I've done:

Suppose $T(x)$ is a contraction mapping, then $|T(x) - T(y) \leq c|x-y|$ and so $|(T(y) - T(x))/(y-x)| \leq c$ where$ c \in [0,1) $and all $x,y \in [-1,1]$. Since T is differentiable, by mean value theorem, we can have $\lim_{y\rightarrow x}|(T(y)-T(x))/(y-x)| \leq c$ and so$ |T(x)| \leq c <1$ for all $x \in [-1,1] $

Then I take$ T'(0) = cos(0) =1$. So I got contradiction and it's not contraction mapping.

Is this correct?

Answer 1

Note that $\lim_{x \to 0} {f(x)-f(0) \over x-0} = \cos 0 = 1$, hence it is not a contraction map.

In particular, for any $K \in (0,1)$ there is some $x$ such that $|f(x)-f(0 ) |> K|x-0|$.

Addendum: If $f = \cos$ then $f:[-1,1] \to [-1,1]$ and with $K=\max_{|x| \le 1} |\cos' x| = \sin 1 <1$, we see that $f$ is a contraction map.

Answer 2

In mathematics, a contraction mapping, or contraction or contractor, on a metric space $$ (M,d)$$ is a function $$f : M\to M$$ with the property that there is some nonnegative real number$$ {\displaystyle 0\leq k<1} $$such that for all $x$ and $y$ in $M$

$${\displaystyle d(f(x),f(y))\leq k\,d(x,y).}$$

The function $$ T(x) = sin(x)$$ defined on $[-1,1]$ does not satisfy the contraction mapping condition.

Note that the ratio $$ |f(x)-f(y)|/|x-y|=\cos (t)$$ does not have a uniform bound $k<1$ on the interval $[-1,1]$.