Question

$f:R\to R$ be such that ${\left|\begin{array}{c}f\left(x\right)-f\left(y\right)\end{array}\right|}^2\le {\left|\begin{array}{c}x-y\end{array}\right|}^3$ for all $x,y\in R$ then the value of $f^{\prime }\left(x\right)$ is

• A. $f\left(x\right)$
• B. constant possibly different from zero
• C. ${\left(f\left(x\right)\right)}^2$
• D. $0$

Answer 1

The correct option is D $0$

${|f\left(y\right)-f\left(x\right)|}^2\le {(x-y)}^3\Rightarrow \frac{{|f\left(y\right)-f\left(x\right)|}^2}{{(y-x)}^2}\le (x-y)$
$\Rightarrow {\left|\frac{f\left(y\right)-f\left(x\right)}{y-x}\right|}^2\le x-y\Rightarrow \underset{y\to x}{lim}{\left|\frac{f\left(y\right)-f\left(x\right)}{y-x}\right|}^2\le \underset{y\to x}{lim}(x-y)\Rightarrow {\left|f^{\prime }\left(x\right)\right|}^2\le 0$
which is only possible if $\left|f^{\prime }\left(x\right)\right|=0$
$\therefore \left|f^{\prime }\left(x\right)\right|=0$ and $f\left(x\right)=$ constant