If f is a real valued differentiable function satisfying |f(x)-f(y)|(x-y)^2, x,yR, and f(0…

Question

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

Accepted Answer

$$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} {{f\left( {x + h} \right) - f\left( x \right)} \over h}$$

$$\,\,\,\,\,\,\,\,\,\left| {f'\left( x \right)} \right| = \mathop {\lim }\limits_{h \to 0} \left| {{{f\left( {x + h} \right) - f\left( x \right)} \over h}} \right| \le \mathop {\lim }\limits_{h \to 0} \left| {{{{{\left( h \right)}^2}} \over h}} \right|$$

$$ \Rightarrow \left| {f'\left( x \right)} \right| \le 0 \Rightarrow f'\left( x \right) = 0$$

$$ \Rightarrow f\left( x \right) = $$ constant

As $$f\left( 0 \right) = 0 \Rightarrow f\left( 1 \right) = 0$$

Solutions · 1