I would like to ask a question which is related to Find function with inequality for derivative
Let $C\geq 2$ and $L>0$ be fixed. Let $n \in \{1, 2, \ldots\}$. Does there exist a polynomial $g$ of degree $n$ such that \begin{equation*} 0 < g(0) < g(L), \qquad g'(x) > (g(L) - g(x))C, \ \ \text{for all }x \in [0,L] \end{equation*} holds at least for $n$ big enough?
In Find function with inequality for derivative, a function $g$ fulfilling such conditions was found, but its expression involved the exponential function. Here, I would like to know if this problem can also be solved with polynomials.
You can see already that for $n = 1$, one has $g(x) = ax + b$, $g'(x) = a$, and consequently the conditions become \begin{equation*} 0 < b < aL+ b, \qquad (L-x)C < 1, \ \ \text{for all }x \in [0,L], \end{equation*} which adds a constrain on $C$ and $L$.
For any fixed $L,C>0$, pick $n$ sufficient large so that $L<\frac{n}{C}$. Now define the polynomial, $$p(x):=-\left(-\frac{C}{2n}x+\frac{1}{2}\right)^{2n}+1$$ Note that $p(x)$ satisfies the points $(-n/C,0)$, $(0,1-(1/2)^{2n})$, $(n/C,1)$, $(2n/C,1-(1/2)^{2n})$, which implies that $p(x)$ has a vertex at $n/C$ (the absolute maximum of $p$), and that $p(x)$ is increasing from $(-\infty,n/C]$ (moreover, $p(x)$ has positive derivative and is in fact concave down on that same region); additionally we see that $0<p(x)< 1$ for $x\in[0,L]$. Thus $0 < -\frac{C}{2n}x+\frac{1}{2} < 1$ for $x\in[0,L]$. Additionally, we clearly see that $0<p(0)<p(L)<1$. So for any $x\in[0,L]$ we have $$\begin{array}\ p'(x) &=& -2n\left(-\frac{C}{2n}x+\frac{1}{2}\right)^{2n-1}\cdot\frac{-C}{2n} \\ &=& C\left(-\frac{C}{2n}x+\frac{1}{2}\right)^{2n-1} \\ &>& C\left(-\frac{C}{2n}x+\frac{1}{2}\right)^{2n} \\ &>& C\left(\left(-\frac{C}{2n}x+\frac{1}{2}\right)^{2n}-\left(-\frac{C}{2n}L+\frac{1}{2}\right)^{2n}\right) \\ &=& C\left(p(L)-p(x)\right) \\ \end{array}$$