Let $ f $ be a differentiable real-valued function defined on $ [ 0 , 1 ] $, stisfying the following conditions: $$ f ( 1 ) = e f ( 0 ) $$ $$ \int _ 0 ^ 1 \left( \frac { f ' ( x ) } { f ( x ) } \right) ^ 2 \mathrm d x \le 1 $$ Prove that there is a constant real number $ c $ such that $ f ( x ) = c e ^ x $.
I think we first need to prove $ f ' ( x ) = f ( x ) $.
On
Hint: By Jensen's Inequality, $$ \begin{align} \log\!\left(\frac{f(1)}{f(0)}\right)^2 &=\left(\int_0^1\frac{f'(x)}{f(x)}\,\mathrm{d}x\right)^2\\ &\le\int_0^1\left(\frac{f'(x)}{f(x)}\right)^2\,\mathrm{d}x\\[9pt] &\le1 \end{align} $$
Since $f(1)=e\,f(0)$, we also have $\log\!\left(\frac{f(1)}{f(0)}\right)^2=1$. Since $x\mapsto x^2$ is strictly convex, equality in Jensen's Inequality means that $\frac{f'(x)}{f(x)}$ must be constant.
use that $$\ln(f(x))'=\frac{1}{f(x)}f'(x)$$