Question

Let $f$ be differentiable function from R to R such that $|f\left(x\right)-f\left(y\right)|\le 2|x-y{|}^{\frac{3}{2}}$ , for all $x,y$ $\epsilon$ R.

If $f\left(0\right)=1$ then ${\int }_0^1{f}^2\left(x\right)dx$ is equal to :

• A. $0$
• B. $\frac{1}{2}$
• C. $2$
• D. $1$

Answer 1

Answer: (D)

$1$

$|f\left(x\right)-f\left(y\right)|\le 2|x-y{|}^{\frac{3}{2}}$ divide both side by $|x-y|$
$\left|\frac{f\left(x\right)-f\left(y\right)}{x-y}\right|\le 2|x-y{|}^{\frac{1}{2}}$
Apply limit $x\to y$

$|f^{\prime }\left(y\right)|\le 0\Rightarrow f^{\prime }\left(y\right)=0\Rightarrow f\left(y\right)=c\Rightarrow f\left(x\right)=1$

${\int }_0^{1}1.dx=1$