Lipschitz condition and continuity I was wondering, if a function of one real variable is bounded on a compact interval and is $C^2$, is it necessarily true that the function is Lipschitz on that interval?
Thanks
 A: As the function is $C^2$, its derivative is bounded on the compact:
$|f'(.)|\le M$.
Now use the mean value theorem:
$$
|f(x)-f(y)| = |f'(c)||x-y| \le M|x-y|
$$
hence $f$ is $M$-lipschitz.

Remark:
 $C^1$ is enough.
A: Solution:
Let $f: [a,b] \to \mathbb{R}$. Say $f$ is differentiable, and suppose $f'$ is continuous. This we know since we are given that $f$ is $C^2$. Since $[a,b]$ is compact, then $f'$ is bounded in $[a.b]$. In particular, we can find $K > 0 $ such that $| f'(x) | \leq K $ for all $x \in [a,b] $. Second, Suppose $x,y \in [a,b]$ are arbitrary. Apply the Mean value theorem to obtain $\xi $ between $a$ and $b$ such that:
$$ \frac{ |f(x) - f(y) | }{|x- y| } = |f'( \xi) | \leq K $$
$$ \therefore |f(x) - f(y) | \leq K | x - y| $$
Since $x,y$ were arbitrarily chosen, we see that $f$ is in fact Lipschitz.
A: If only $f$ is bounded on that compact interval and $f^{\prime}$ is not bounded, then it is not necessarily true. Here is an example, $f(x)=\sqrt[3]{x}$ on $[-1,1]$ and its derivative $f^{\prime}(x)=\frac{1}{3\sqrt[3]{x^2}}$. Although $[-1,1]$ is compact, $f^{\prime}(x)$ is unbounded on $[-1,1]$ as $\lim_{x\rightarrow0}f^{\prime}(x)=+\infty$.
