Let $f$ be a function $f:[0,\pi]\to\mathbb{R}$ such that: $$f(x)=\left\{ \begin{array}{ll} \frac{\sin(x)}{x} & \mbox{if } x \neq 0 \\ 1 & \mbox{if } x = 0 \end{array} \right.$$ I want to prove that $f$ is a contraction, i.e., $\exists\alpha\in(0,1)\forall x,y\in[0,\pi]:\lvert f(x)-f(y)\rvert\leq\alpha\lvert x-y\rvert$
I wanted to prove that $f'(x)=\frac{x\cos(x)-\sin(x)}{x^2}$ attains a minimum in $(0,\pi)$ which is greater than $-1$, hence $\lvert f'\rvert$ has a maximum $\alpha=\lvert f'(m)\rvert\in(0,1)$. Finally by the M.V.T for all $x,y\in[0,\pi]$ there'd be a $c\in(x,y)$ such that $$\lvert f(x)-f(y)\rvert=\lvert f'(c)\rvert\lvert x-y\rvert\leq\lvert f'(m)\rvert\lvert x-y\rvert=\alpha\lvert x-y\rvert$$ However when studying the second derivative of $f$, I got to a point where $m$ must satisfy the equation $(2-m^2)\sin(m)=2m\cos(m)$ and from here I have not figured out a way to get what I want.
So can you give me a hint on how to proceed? or if there's an easier way to prove that $f$ is a contraction, can you give an outline of the proof? (Thanks in advance).
$$x\cos x - \sin x = \int_0^x (\cos x -\cos t)\, dt = \int_0^x \int_t^x (-\sin s)\,ds\, dt.$$
Crash through with absolute values to see the above, in absolute value, is less than
$$\int_0^x \int_t^x 1\,ds\, dt = \int_0^x (x-t)\,dt = x^2/2.$$
Divide by $x^2$ and we get $|f'(x)| \le 1/2$ everywhere on $\mathbb {R}.$ That leads to the desired solution.