If $f$ is differentiable on $[a,b]$, then it is also Lipschitz on it. He guys, I am trying to show that a differentiable function defined on a closed interval is also Lipschitz on it. I managed to weave the below proof, but I have a feeling that it may be just a tad too general for this purpose:
Theorem. If $f$ is differentiable on $[a,b]$, then it is also Lipschitz on it.
Recall that $f:A\to\mathbb{R}$ is Lipschitz on $A$ if there exists an $M>0$ such that$$\left|\frac{f(x)-f(y)}{x-y}\right|\leq M$$for all $x,y\in A$.
Proof. Let $f$ be differentiable on $[a,b]$. Because $f$ is continuous and $[a,b]$ is compact, by the Extreme Value Theorem, it follows that $f$ attains a maximum value $M$. Moreover, since $f$ is differentiable on $[a,b]$,$$f'(y)=\lim_{x\to y}\left|\frac{f(x)-f(y)}{x-y}\right|\leq M,$$for all $x,y\in[a,b]$, as required. $\square$
What do you guys think?
Edit: What if we were to add that $f'$ is also continuous on $[a,b]$?
 A: It is not true in general. Consider function
$$
f(x)=
\begin{cases}
x^2 \sin\frac{1}{x^2}\qquad x\neq 0\\
0\qquad\quad\qquad x=0
\end{cases}
$$
It is differentiable on $[0,1]$ and
$$
f'(x)=
\begin{cases}
2 x \sin\frac{1}{x^2}-\frac{2}{x} \cos\frac{1}{x^2},  &x\neq 0\\
0,  &x=0
\end{cases}
$$ 
But this derivative is unbounded, since
$$
\lim\limits_{n\to\infty} f'\left(\frac{1}{\sqrt{ {\pi} +2\pi n}}\right)=+\infty
$$
On the other hand if require $f'\in C([0,1])$ then by Weierstrass theorem there exist 
$$
M=\sup\limits_{t\in [0,1]}|f'(t)|<+\infty.
$$
This $M$ is a Lipschitz constant you are looking for.
A: You need to make the slightly stronger assumption that $f$ is $C^1$ or continuously differentiable on $[a,b]$, in order to avoid counterexamples like Norbert's. In this case, we have that $f'(x)$ attains a maximum $M$ on $[a,b]$, and if $$\left|\frac{f(x)-f(y)}{x-y}\right|>M$$ for some $x,y\in [a,b]$ then by the mean value theorem we would have some $z\in (a,b)$ such that $f'(z)>M$, a contradiction, hence $f$ is Lipshitz.
A: You would need to let $f\in C^1$ (i.e., its derivative is continuous). Simply use the mean value theorem.
A: Actually, if f is differentiable in $[a,b]$, then it is also continuous there. I think this disqualifies Norbert's example, which is not differentiable at $0$
Since f is continuous on $[a,b]$ , we have that there are reals m,m with:
$m\leq f(x) \leq M$ , then you have that :
$\frac{|f(x)-f(y)}{x-y}\leq\frac{M-m}{x-y}$, and I think you can show this value is
finite by the mean value theorem, i.e., there is a $ c$ in $[x,y]$ with $f'(c)=(M-m)/(x-y)$;
since f is differentiable, the ratio is finite.
EDIT: My answer is incorrect. We need to have f' bounded, which is not guaranteed by the conditions of the problem. Also, as pointed out by Alex B, Norbert's function is differentiable at x=0.
A: your original proof would be sufficient if you add the condition that f' is bounded on the interval I. 
