If $f$  is a real-valued differentiable function satisfying $\left|f\left(x\right)-f\left(y\right)\right|\le {\left(x-y\right)}^2,x,y\in R$  and $f\left(0\right)=0$ , then $f\left(1\right)$  equals

$\left|f\left(x\right)-f\left(y\right)\right|\le {\left|x-y\right|}^2$ $\Rightarrow \left|\frac{f\left(x\right)-f\left(y\right)}{x-y}\right|\le \left|x-y\right|$

$\underset{x\to y}{lt}\left|\frac{f\left(x\right)-f\left(y\right)}{x-y}\right|\le \underset{x\to y}{lt}\left|x-y\right|$ $\Rightarrow \left|f\text{'}\left(x\right)\right|\le 0\Rightarrow f\text{'}\left(x\right)=0$

$\Rightarrow f\left(x\right)$  is constant function $f\left(0\right)=0\Rightarrow f\left(1\right)=0$ .