Question

Let $f$ be a continuous function defined on $[0,1]$ such that $\underset{0}{\overset{1}{\int }}{f}^2\left(x\right)dx={\left(\underset{0}{\overset{1}{\int }}f\right(x\left)dx\right)}^2$. Then the range of f

• A. has exactly two points
• B. has more than two points
• C. is the interval $[0,1]$
• D. is a singleton

Answer 1

The correct option is D is a singleton

By Cauchy Schwarz inequality,
${\left[\underset{a}{\overset{b}{\int }}f\right(x\left)g\right(x\left)dx\right]}^2\le \underset{a}{\overset{b}{\int }}{f}^2\left(x\right)dx\underset{a}{\overset{b}{\int }}{g}^2\left(x\right)dx$
$g\left(x\right)=1$ and $a=0$, $b=1$

$\Rightarrow {\left[\underset{0}{\overset{1}{\int }}f\right(x\left)dx\right]}^2\le \underset{a}{\overset{b}{\int }}{f}^2\left(x\right)dx$
Equality hold when $\frac{f\left(x\right)}{g\left(x\right)}=\lambda$
So, $f\left(x\right)$ is constant.