Limit Continuity And Differentiability Ques 52
Let $R$ be the set of real numbers and $f: R \rightarrow R$ be such that for all $x$ and $y$ in $R, |f(x)-f(y)|^2 \leq(x-y)^3$. Prove that $f(x)$ is a constant.
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Solution: Because,
$\Rightarrow \quad \frac{|f(y)-f(x)|^2}{(y-x)^2} \leq(x-y)^2$
$\Rightarrow \quad\left|\frac{f(y)-f(x)}{y-x}\right|^2 \leq |x-y|$
$\Rightarrow \quad \lim _{y \rightarrow x}\left|\frac{f(y)-f(x)}{y-x}\right|^2 \leq \lim _{y \rightarrow x}|y-x|$
$\Rightarrow \quad\left|f^{\prime}(x)\right|^2 \geq 0$
which is only possible, if $\left|f^{\prime}(x)\right|=0$
$\therefore \quad f^{\prime}(x)=0$
or $f^{\prime}(x)=$ Constant
BETA
