Let f: R rightarrow R be a differentiable fun | vedclass

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(D) Given that $|f(x) - f(y)| \leq 2|x - y|^{\frac{3}{2}}$.Dividing both sides by $|x - y|$ (where $x 
eq y$),we get $\left|\frac{f(x) - f(y)}{x - y}

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$1$

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(D) Given that $|f(x) - f(y)| \leq 2|x - y|^{\frac{3}{2}}$.Dividing both sides by $|x - y|$ (where $x 
eq y$),we get $\left|\frac{f(x) - f(y)}{x - y}ight| \leq 2|x - y|^{\frac{1}{2}}$.Taking the limit as $x ightarrow y$,the left side becomes the definition of the derivative $|f'(y)|$.Thus,$|f'(y)| \leq 2 \lim_{x ightarrow y} |x - y|^{\frac{1}{2}} = 0$.Since the absolute value cannot be negative,we must have $|f'(y)| = 0$,which implies $f'(y) = 0$ for all $y \in R$.This means $f(x)$ is a constant function,$f(x) = c$.Given $f(0) = 1$,we have $c = 1$,so $f(x) = 1$.Therefore,$\int_0^1 f^2(x) dx = \int_0^1 (1)^2 dx = \int_0^1 1 dx = [x]_0^1 = 1 - 0 = 1$.

Let $f: R \rightarrow R$ be a differentiable function such that $|f(x) - f(y)| \leq 2|x - y|^{\frac{3}{2}}$ for all $x, y \in R$. If $f(0) = 1$,then $\int_0^1 f^2(x) dx = $

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