Let $T(x)=\frac{x}{\sqrt{1+x^2}}$ be a function from $R$ to $R$. Prove that the function is firmly non-expansive.
Definition:
$$||Tx-Ty||^2 + ||(Id-T)x-(Id-T)y||^2 \leq ||x-y||^2$$ If the space is Hilbert, this simplifies to,
$$||Tx-Ty||^2 \leq \hspace{3mm} <Tx-Ty, x-y> $$
My attempt:
Since $R$ is a Hilbert space, I used the second inequality to simplify the expression into solving for the non-negativity of the following expression.
Prove that the following expression is non-negative:
$$ (\frac{x}{\sqrt{1+x^2}}-\frac{y}{\sqrt{1+y^2}})(x-y) - (\frac{x}{\sqrt{1+x^2}}-\frac{y}{\sqrt{1+y^2}})^2 \geq 0 $$
$$ \frac{x^2\sqrt{1+y^2}-xy\sqrt{1+x^2}-xy\sqrt{1+y^2}+y^2\sqrt{1+x^2}}{\sqrt{1+x^2}\sqrt{1+y^2}} - \frac{x^2+y^2-2xy\sqrt{1+x^2}\sqrt{1+y^2}}{(1+x^2)(1+y^2)} \geq 0$$
$$ \frac{x^2(1+y^2)\sqrt{1+x^2}-xy(1+x^2)\sqrt{1+y^2}-xy\sqrt{1+x^2}(1+y^2)+y^2\sqrt{1+y^2}(1+x^2)}{(1+x^2)(1+y^2)} - \frac{x^2+y^2-2xy\sqrt{1+x^2}\sqrt{1+y^2}}{(1+x^2)(1+y^2)} \geq 0$$
I tried to prove using algebra without using any inequalities, and I got this. I am unable to proceed further.Can someone help?
Since
$$T'(x) = \frac{\sqrt{1+x^2} - x\times \frac{x}{\sqrt{1+x^2}}}{1+x^2} = \frac{1}{\left(1 + x^2\right)^{\frac32}}.$$ This proves that $0 \le T'(x) \le 1$
If $y < x$ taking the integral from $y$ to $x$,
$$0 \le T(x) - T(y) \le x - y$$
multiplying by $T(x) - T(y)$ and you will the inequality you are looking for.