Let f be a differentiable function from R to R such that |f(x)-f(y)|≤ 2|x-y|^3/2, for all x,y∈ R . If f(0)=1 then ∫ _0^1 f^2(x)dx is equal to

Given functional equality is

$$|f(x)-f(y)|\leq 2 {|x-y|}^{3/2}$$∣f(x)−f(y)∣≤2∣x−y∣3/2

Put $x=y+h,$x=y+h, we get

$$|f(y+h)|-f(y){|\leq 2. |h|}^{3/2}$$∣f(y+h)∣−f(y)∣≤2.∣h∣3/2

$$\Rightarrow |\frac{f(y+h)-f(y)}{h}|\leq 2 {|h|}^{1/2}$$⇒∣hf(y+h)−f(y)​∣≤2∣h∣1/2

$$\Rightarrow \underset{h\rightarrow 0}{\mathrm{lim}}⁡|\frac{f(y+h)-f(y)}{h}|\leq \underset{h\rightarrow 0}{\mathrm{lim}}⁡2 {|h|}^{1/2}$$⇒h→0lim​⁡∣hf(y+h)−f(y)​∣≤h→0lim​⁡2∣h∣1/2

$$\Rightarrow |{f}^{'}(y)|\leq 0$$⇒∣f′(y)∣≤0

$$\Rightarrow |{f}^{'}(y)|=0$$⇒∣f′(y)∣=0

$$\Rightarrow f'(y)=0$$⇒f′(y)=0

$$\Rightarrow f(y)=c$$⇒f(y)=c

$\Rightarrow f(y)=1$⇒f(y)=1 (since $f(0)=1$f(0)=1 )

Now $\displaystyle {\int }_{0}^{1}{f}^{2}(x)dx= {\int }_{0}^{1}1 dx=1$∫01​f2(x)dx=∫01​1dx=1.