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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

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1 Answer 1

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$L$ 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.

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