Lipschitz continuity of $x^n (1-x^n)^m$ on $[0,1]$ for integers $n, m \geq 3$

Question

For integers $n, m \geq 3$, consider the function $f: [0.75,1] \to [0,1]$ defined as \begin{equation} f(x) = f_{n,m}(x) := x^n (1 - x^n)^m. \end{equation} The first-order derivative of $f$ can be written as \begin{equation} L(x) = L_{n,m}(x) := n x^{n-1} (1 - x^n)^{m-1} (1 - (m+1)x^n). \end{equation} For $1 \geq x_2 > x_1 \geq 0.75$, if we define \begin{equation} L = L_{n,m} := \sup_{x \in [0.75,1]} |L(x)|, \end{equation} then \begin{equation} |f(x_2) - f(x_1)| \leq L_{n,m} |x_2 - x_1|. \end{equation} The question is: How to estimate $L_{n,m}$ and compare it with $1/m$, when $n$ and $m$ are large; or equivalently, does there a constant $c > 0$ such that $L_{n,m}$ can be further upper bounded by $c/m$, i.e., \begin{equation} L_{n,m} \leq \frac{c}{m}. \end{equation} In particular, I want to consider the case where $n = n(N) = \log_2(N)$ and let $m = m(N)$ ranges from $3$ to $N$, where $N \in \mathbb{N}$. This question has the following observation: For any $m$ and any fixed $x \in [0.75,1]$, it holds that \begin{equation} \lim_{N \to \infty} L_{n(N), m(N)}(x) = 0. \end{equation} E.g., when $N = 2^{21}$, $m = m(N) = \sqrt{N}$ and $x=0.9$, the term $n x^{n-1}$ appearing in $L_{n,m}(x)$ is $2.5531$, the term $(1 - x^n)^{m-1}$ is roughly $10^{-73}$ and $1 - (m+1)x^n$ is $-157.565$. In total $L_{21, \sqrt{2^{21}}}(0.9) \approx 10^{-71}$, which is considerably smaller than $1 / m = 1 / \sqrt{2^{21}} \approx 7 \times 10^{-4}$. Anyone has an idea? Thanks very much.

Answer 1

Note that $|L(x)|$ is a continuous function on the closed interval $[\frac{3}{4},1]$ and also $\sup_{x \in [\frac{3}{4},1]}|L(x)|>0$ thus exists $y \in (\frac{3}{4},1)$ such that $$\sup_{x \in [\frac{3}{4},1]}|L(x)|=|L(y)|$$

Let $m \in \Bbb{N}$. We can find $n \in \Bbb{N}$ large enough such that $1-(m+1)y^n<1$ and $ny^{n-1}<\frac{1}{m}$

Thus you have the conclusion for $C=1$