I'm having trouble finding the lipschitzian constant from the exercise below for $n$ even or odd, the constant will depend on $n$ and also on the limitation of $f$. Can someone help me please ?
Let $f : I \rightarrow \mathbb{R}$ defined by $f(x) = x^n$, where $n$ is a number natural fixed. Show that if $I$ is a bounded interval then f is lipschitzian.
We have the well known formula
$$x^{n+1}-y^{n+1}=(x-y)(x^n+x^{n-1}y+x^{n-2}y^2+\cdots+xy^{n-1}+y^n)$$
Denote the expression on the right by $p(n,x,y)$. Note that for a fixed $n$ our $p(n,x,y)$ is a polynomial in two variables. Then we have
$$|x^{n+1}-y^{n+1}|\leq L\cdot |x-y|$$
where $L=\sup\big(\big\{|p(n,x,y)|\ \big|\ x,y\in I\big\}\big)$. Note that if $I$ is bounded then $L$ is finite, because then $\overline{I\times I}$ is compact, and moreover $|p(n,x,y)|$ achieves $L$ on $\overline{I\times I}$. And of course this is the best (smallest) Lipschitz constant for $x\mapsto x^{n+1}$.
I'm not sure if we can say more about $L$ in such general setup. You've mentioned $n$ being even or odd, but I don't see how this helps us here unfortunately.
