
> 
> Let $C\geq 2$ and $L>0$. Does there exist $g \in C^1([0,L])$ such that
>  \begin{equation\*}
> g(x)>0, \qquad g'(x)>0, \qquad g'(x) > (g(L) - g(x))C
> \end{equation\*}
>  holds for any $x \in [0,L]$? How large can the length $L$ of the interval be chosen?
> 
> 
> 


**First example:** if 
\begin{equation\*}
g(x) = ax + b
\end{equation\*}
with $a, b>0$, then $$g'(x) > (g(L) - g(x)) C$$ is equivalent to $L < \frac{1}{C}$.


**Second example:** let $f \colon [0,L] \rightarrow [\varepsilon, \frac\pi2]$ be defined by 
\begin{equation\*}
f(x) = \frac{\pi x}{2L} + \varepsilon \Big(1 - \frac{x}{L}\Big). 
\end{equation\*}
Then, if we choose 
\begin{equation\*}
g(x) = -\cot(f(x)) + \cot(\varepsilon) + \delta
\end{equation\*}
with $\delta>0$ and $\varepsilon \in (0, \frac\pi2)$, we have $$g'(x) = \Big(\frac{\pi}{2} - \varepsilon\Big) \frac{1}{L \sin^2(f(x))},$$ while $$g(L) - g(x) = \cot(f(x)).$$ Hence $g'(x) > (g(L) - g(x)) C$ is equivalent to $L < \frac{\pi - 2\varepsilon}{C}$.


But is there a function $g$ allowing $L$ to be larger?


This question is also posted on Math Stack Exchange: <https://math.stackexchange.com/questions/3280984/find-function-with-inequality-for-derivative> 



Find function with inequality for derivative
$L$ can be arbitrarily large, for any given $C>0$. Indeed, take any real $c:=C>0$ and $L>0$. Let 
\begin{equation}
 g(x):=\frac{1+e^{c L}-e^{c (L-x)}}{c}; 
\end{equation}
everywhere here, $x\in[0,L]$. 
Then $g(x)\ge1/c>0$, $g'(x)=e^{c (L-x)}>0$, and $g'(x)=1+(g(L)-g(x))c>(g(L)-g(x))c$, as desired.


