I have been trying to solve the followig problem:
Problem: Show that $f(x)=x^{n}$ is Lipschitz if $f: X \rightarrow \mathbb{R}$ where X is a bounded set.
What have i tried so far?
Well, since X is bounded, there exists a constant $M > 0$ such that $|\mathrm{x}| < M$, $\forall x \in X$. With that in mind, i tried writing:
$$ |f(x)-f(y)|=\left|x^{n}-y^{n}\right|= $$ $$\left|(x-y)\left(x^{n-1}+x^{n-2} y+\cdots+x y^{n-2}+y^{n-1}\right)\right| \leq $$
$$ \leq |x-y| (\left|x^{n-1}\right|+\left|x^{n-2} \cdot y\right|+\ldots+\left|x y^{n-2}\right|+\left|y^{n-1}\right|) $$
Now, since terms such as $|x^{n-1}|$ satisfy $|x|^{n - 1} \leq M^{n-1}$ i should be able to bound the power terms and obtain an inequality of the form: $$|f(x) - f(y)| \leq c|x-y|$$
Is the ideia correct? Can someone help me write the answer more precisely?
Thanks in advance, Lucas