Question

Let $R$ be the set of all real numbers. If, $f:R\to R$ be a function such that $|f\left(x\right)-f\left(y\right){|}^2\le |x-y{|}^3,\forall x,y\epsilon R$, then $f^{\prime }\left(x\right)$ is equal to

• A. $f\left(x\right)$
• B. $1$
• C. $0$
• D. $x^2$
• E. $x$

Answer 1

The correct option is C $0$

Given, $f:R\to R$ and $|f\left(x\right)-f\left(y\right){|}^2\le |x-y{|}^3$
$\Rightarrow \frac{|f\left(x\right)-f\left(y\right){|}^2}{|x-y{|}^2}\le |x-y|$
Let, $y=x+h,h\to 0$
$\Rightarrow \underset{h\to 0}{lim}\frac{|f\left(x\right)-f(x+h{)}^2|}{|x-(x+h){|}^2}\le \underset{h\to 0}{lim}|x-(x+h)|$
$\Rightarrow \underset{h\to 0}{lim}\frac{|f(x+h)-f\left(x\right){|}^2}{|h{|}^2}\le \underset{h\to 0}{lim}|h|$
$\Rightarrow |f^{\prime }\left(x\right){|}^2\le 0\Rightarrow |f^{\prime }\left(x\right){|}^2=0$
Therefore, $f^{\prime }\left(x\right)=0$