About Lipschitz condition on polynomial mapping from $\mathbb R$ to $\mathbb R$

Question

Let $p$ be a polynomial. Show that $p\colon\Bbb R \to \Bbb R$ is Lipschitz iff the degree of $p$ is less than $2$.

Answer 1

We can use the following facts:

• A $C^1$ function on $\Bbb R$ is Lipschitz continuous if and only if its derivative is bounded.
• A polynomial is bounded if and only if it's a constant one (hence of degree $0$ or $-\infty$).
• The derivative of a polynomial of degree $d\geq 1$ is of degree $d-1$.

Answer 2

One direction: let $p$ have degree less than $2$. Then $p(x) = ax + b$ for some $a,b \in \mathbb R$. Then $\dot{p}(x) = a$ that is, the first derivative is bounded. Hence $p$ is Lipschitz continuous.

Here we have used the following theorem:

---

A continuous and everywhere differentiable function $f: \mathbb R \to \mathbb R$ is Lipschitz continuous if and only if its first derivative is bounded, that is, there is $K \in \mathbb R$ such that $|f'(x)| \leq K$ for all $x$ in $\mathbb R$.

Proof:

$\Longrightarrow$: Let $f$ be Lipschitz, that is, there is $K \in \mathbb R$ such that $\frac{|f(x) - f(y)|}{|x-y|} \leq K$ for all $x$ in $\mathbb R$. Then in particular, $\lim_{\delta \to 0} \frac{|f(x) - f(x + \delta)|}{|\delta|} = |f'(x)| \leq K$ for all $x$.

$\Longleftarrow$: Let $|f'(x)| \leq K$ for all $x$. By the mean value theorem we know that for any $x,y \in \mathbb R$ we have that there exists $x_0$ such that $ \frac{f(y) - f(x)}{y - x} = f'(x_0)$. Hence we have that for all $x,y \in \mathbb R$: $$ \frac{|f(y) - f(x)|}{|y - x|} = |f'(x_0)| \leq K$$ that is, $f$ is Lipschitz. $\Box$

---

Now try the other direction.