Lipschitz Continuous Polynomials Let f : R → R be a polynomial of degree d ≥ 2. Show that f is not Lipschitz continuous. Where exactly should I begin with this problem?
 A: If it were Lipchitz continuous, it would have a bounded derivative, as it is differentiable. 

THM Suppose $f:A\to\Bbb R$ is differentiable on $A$. Then $f$ is Lipschitz continuous $\iff$ $f'$ is bounded on $A$. 

P Suppose $f'$ is bounded. Let $M=\sup \{f'(x):x\in A \}$. Then, by the mean value theorem, $$|f(x)-f(y)|\leq M|x-y|$$ for $x,y\in A$. Now assume $f'$ is unbounded on $A$. Then for any $K>0$ we can pick $\xi\in A$ such that $|f'(\xi)|>K$.  Let $\{\xi_n\}$ be a sequence of points in $A$ converging to $\xi$, but different from it. Since $$\left|\frac{f(\xi)-f(\xi_n)}{\xi-\xi_n}\right|\to f'(\xi)$$
there must exist $N>0$ such that $$\left|\frac{f(\xi)-f(\xi_N)}{\xi-\xi_N}\right|>\frac K2$$
and $f$ is not Lipschitz continuous.
A: Let $P(x)=a_n x^n+\cdots+a_0$ a polynomial with $n\geq 2$. You should show that there's no $k$ such that
$$|P(x)-P(y)|\leq k|x-y|\quad\forall x,y\in\mathbb R.$$
A: Intuition first: For large values of $x\in \mathbb R$, a polynomial of degree $n$ behaves much like $f(x)=x^n$ does. The latter, $n\ge 2$, is not Lipschitz. 
To turn this intuition into a proof, you might first want to convince yourself of the validity of the two claims, and how that will prove what you are looking for. 
So:
1) Campare $|P(x)-P(y)|$ to $|x^d-y^d|$ in a sensible way. 
2) Prove that $f(x)=x^d$, when  $d\ge 2$, is not Lipschitz. 
3) Use a suitable inequality you found in step 1 to conclude $P(x)$ is also not Lipschitz. 
