Show polynomial is a Lipschitz function

Question

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.

Answer 1

We have $$x^k-y^k=(x-y)\sum_{i=0}^{k-1}x^iy^{k-1-i}$$ and $A$ is bounded then there's $M>0$ such that $$\forall x\in A\quad |x|\leq M$$ hence $$|x^k-y^k|\leq|x-y|k M^{k-1}$$ then if $$P(x)=\sum_{k=0}^n a_k x^k$$ we have $$|P(x)-P(y)|\leq \sum_{k=1}^n|a_k||x^k-y^k|\leq|x-y|\sum_{k=1}^nk|a_k| M^{k-1}=C|x-y|$$

Answer 2

• It's enough to show that $x\mapsto x^p$ is Lipschitz for any integer $p\geqslant 1$.

• Use the identity $$x^p-y^p=(x-y)\sum_{j=0}^{p-1}x^jy^{p-1-j},$$ and show that $\left|\sum_{j=0}^{p-1}x^jy^{p-1-j}\right|$ can be bounded independently of $x$ and $y$.

Answer 3

Let $P(x)=a_0x^n+a_1x^{n-1}+\cdots+a_n$. Calculate $P'(x)$. Using the Triangle Inequality, we find that $$|P'(x)|\le n|a_0||x|^{n-1}+(n-1)|a_1||x|^{n-2} +\cdots +|a_n|.$$

Now use the fact that there is a $B$ such that $|x|\le B$.

Answer 4

We can write $P(x)-P(y)=(x-y)Q(x,y)$ for some bivariate polynomial $Q$ (because we can do so at least when $P$ is a monomial). As continuous function, $|Q(x,y)|$ is bounded on the compact set $A\times A$, by $L$ say. Then $L$ is a Lipschitz constant for $P$.

Answer 5

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.