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

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Aicharya · Accepted Answer

$ \begin{aligned} & f^{\prime}(x)=\lim _{h ightarrow 0} \frac{f(x+h)-f(x)}{h} \ & \left|f^{\prime}(x)ight|=\lim _{h ightarrow 0}\left|\frac{f(x+h)-f(x)}{h}ight| \leq \lim _{h ightarrow 0}\left|\frac{(h)^2}{h}ight| \ & \Rightarrow\left|f^{\prime}(x)ight| \leq 0 \Rightarrow f^{\prime}(x)=0 \quad \Rightarrow f(x)=	ext { constant } \ & 	ext { As } f(0)=0 \Rightarrow f(1)=0 . \end{aligned} $

If $f$ is a real-valued differentiable function satisfying $|f(x)-f(y)| \leq(x-y)^2, x, y \in R$ and…

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