Mathematics - Continuity - Differentiability Question with Solution | TestHub

Given functional equality is

$\left|f\left(x\right)-f\left(y\right)\right|\le 2 {\left|x-y\right|}^{3/2}$

Put$x=y+h,$we get

$\left|f\left(y+h\right)\right|-f\left(y\right){\left|\le 2. |h\right|}^{3/2}$

$\Rightarrow \left|\frac{f\left(y+h\right)-f\left(y\right)}{h}\right|\le 2 {\left|h\right|}^{1/2}$

$\Rightarrow \underset{h\to 0}{\lim }\left|\frac{f\left(y+h\right)-f\left(y\right)}{h}\right|\le \underset{h\to 0}{\lim }2 {\left|h\right|}^{1/2}$

$\Rightarrow \left|f^{\prime }\left(y\right)\right|\le 0$

$\Rightarrow \left|f^{\prime }\left(y\right)\right|=0$

$\Rightarrow f\prime \left(y\right)=0$

$\Rightarrow f\left(y\right)=c$

$\Rightarrow f\left(y\right)=1$(since$f\left(0\right)=1$)

Now${\int }_0^1{f}^2\left(x\right)dx= {\int }_0^{1}1 dx=1$.