# Thread: [SOLVED] Lipschitz

1. ## [SOLVED] Lipschitz

Problem:
Suppose that $\displaystyle p:R \rightarrow R$ is a polynomial. Show that $\displaystyle p:R \rightarrow R$is Lipschitz if and only if the degree of the polynomial is less than 2.

Attempt:
By definition of Lipschitz, then we need to prove that the function p is Lipschitz for degree less than 2 or equivalently $\displaystyle \leq 1$ then
$\displaystyle | p(x_{1}) - p(x_{2}) | \leq K |x_{1} - x{2}|$ for $\displaystyle K \geq 0$ and for all $\displaystyle x, x_{1} \in R$

Here I'm not sure how to approach this. Since p is a polynomial, then it must be continuous. I was thinking of showing that if p is a polynomial of less than 2, then its derivative is a constant. Thus, its derivative is bounded and therefore a Lipschitz function.

Thank you for your time.

2. Originally Posted by Paperwings
Problem:
Suppose that $\displaystyle p:R \rightarrow R$ is a polynomial. Show that $\displaystyle p:R \rightarrow R$is Lipschitz if and only if the degree of the polynomial is less than 2.
obviously every polynomial of degree at most 1 is Lipschitz. so we only need to show that a polynomial of degree

at least 2 is not Lipschitz. so suppose p(x) is a polynomial of degree at least 2. see that $\displaystyle \lim_{x\to\infty}p'(x) = \pm \infty. \ \ \ (1)$

now suppose p(x) is Lipschitz. then for some constant $\displaystyle K \geq 0: \ \left|\frac{p(t)-p(x)}{t-x}\right| \leq K,$ for all $\displaystyle t \neq x.$ taking limit of

both sides as $\displaystyle t \rightarrow x$ gives us $\displaystyle |p'(x)| \leq K,$ which will contradict (1) if we let $\displaystyle x \rightarrow \infty. \ \ \ \square$

3. Thank you NonCommAlg. I was planning to prove by contradiction, but didn't know how to show it mathematically. By dividing both sides by |t-x| and taking the limit, we get the definition of a derivative.