# Find function with inequality for derivative

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

$$L$$ can be arbitrarily large, for any given $$C>0$$. Indeed, take any real $$c:=C>0$$ and $$L>0$$. Let $$$$g(x):=\frac{1+e^{c L}-e^{c (L-x)}}{c};$$$$ everywhere here, $$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.