TITLE: Show polynomial is a Lipschitz function
QUESTION [8 upvotes]: If $A\subseteq \mathbb{R}$ is a bounded set and $p$ is a polynomial, then show that $p:A\to \mathbb{R}$ is a Lipschitz function.

REPLY [10 votes]: Let $f: [a,b] \rightarrow \mathbb{R}$ be a $C^1$-function, i.e., $f'$ exists at every point and is continuous.   Since $f': [a,b] \rightarrow \mathbb{R}$ is continuous, it is bounded: there is a constant $M$ such that $|f'(x)| \leq M$ for all $x \in [a,b]$.  By the Mean Value Theorem, for any $a \leq x \leq y \leq b$, 
$|f(y)-f(x)| = |f'(c)||y-x| \leq M|y-x|$,
so $M$ is a Lipschitz constant for $f$ on $[a,b]$.
Finally, note that all polynomials are $C^1$ (indeed infinitely differentiable) on any interval and that any bounded set is contained in a closed, bounded interval.