Let f be a differentiable function from R to R such that |f(x) -...

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}$<br /><br />$⇒ \left|\dfrac{f\left(y + h\right) - f ( y )}{h}\right| \le 2{\left|h\right|}^{1 / 2}$<br /><br />$⇒ \lim_{h \to 0} \left|\dfrac{f\left(y + h\right) - f\left(y\right)}{h}\right| \le \lim_{h \to 0} 2{\left|h\right|}^{1 / 2}$<br /><br />$⇒ \left|{f}^{′}\left(y\right)\right| \le 0$<br /><br />$⇒ \left|{f}^{′}\left(y\right)\right| = 0$<br /><br />$⇒ f′\left(y\right) = 0$<br /><br />$⇒ f\left(y\right) = c$<br /><br />$⇒ f\left(y\right) = 1$(since$f\left(0\right) = 1$)<br /><br />Now${∫}_{0}^{1}{f}^{2}\left(x\right)dx = {∫}_{0}^{1}1dx = 1$.