Let f be any function defined on R satisfying |f(x)-f(y)|(x-y)^2 x,yR.

Question

Let $f$ be any function defined on $\mathbb{R}$ satisfying $|f(x)-f(y)|\le(x-y)^2\ \forall x,y\in\mathbb{R}$. If $f(0)=1$, then:

Accepted Answer

$$\left|\frac{f(x)-f(y)}{x-y}\right|\le|x-y|$$
Taking $x\to y$, $|f'(y)|\le0\Rightarrow f'(y)=0$ for all $y$. So $f$ is constant: $f(y)=f(0)=1$ for all $y$. Since $f(x)=1>0$ everywhere, the answer is D.

Solutions · 1