Find The Supremum

Clearly $f(x) = 0$ gives us $c = 0$ . Since we want to find the maximum value of $c$ , $\text{WLOG}$ $c> 0$ .

For any solution $f$ , let $\delta= \inf\left\{ \frac{f(x+1)}{f(x)} : x > 0 \right\}$

Note $f(x)$ is strictly increasing, $\frac{ f(x+1) } { f(x) } > 1$ . Thus, $\delta$ exists and it is greater than or equal to $1$ .

Now note $f'(x) = cf(x+1) \ge c\delta f(x) \Rightarrow (\log f(x))' \ge c\delta \Rightarrow \log f(x+1) - \log f(x) \ge \int_x^{x+1} c\delta dx = c\delta$

And so $\delta= \inf\left\{ \frac{f(x+1)}{f(x)} : x > 0 \right\} \ge e^{c\delta} \rightarrow c \le ye^{-y} \text{ where } y=c\delta>0$

From this, $c \le \sup\big\{ y e^{-y} : y > 0 \big\} = \frac{1}{e}$

The answer is $e^{-1}=0.36787944...$ , with the equality being true when $f(x)=e^x$ .