Prove that the function is not non-expansive.
T is non-expansive if
$$ \forall x,y \hspace{3mm} ||Tx-Ty|| \leq ||x-y|| $$
$T(x)= \frac{x sin(x)}{2} \hspace{2mm} 0\leq x \leq \pi $ and $0$ elsewhere. I want to prove this without taking derivatives. My attempt: \begin{align} |\frac{x sin(x)}{2}-\frac{y sin(y)}{2}|&=\frac{1}{2}|xsinx-xsiny+xsiny-ysiny| \\ &\leq\frac{1}{2}(|xsinx-xsiny|+|xsiny-ysiny|) \\ &= \frac{1}{2}|x||sinx-siny|+|siny|(|x-y|)\\ &\leq \frac{1}{2}|x||x-y|+\frac{1}{2}|x-y| \end{align}
I am unable to proceed further, Can someone please help?
As mentioned by Mengchun Zhang, this is not true. In fact, we have $T'(x) < -1$ for $x \in [3,\pi]$ and, therefore, $T$ is not non-expansive.