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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 $[a,b]$?

wjmolina
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    If you require the derivative $f'$ to be continuous, then it's true. Probably this is what you are looking for. – Stephen Feb 29 '12 at 00:43

5 Answers5

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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.

David Mitra
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Norbert
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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.

Alex Becker
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  • Just to complement what you say in your first sentence: if I am not mistaken, it is possible for a function to be both Lipschitz and differentiable on $[a,b]$ without being $C^1$ on $[a,b]$. –  Feb 29 '12 at 08:17
  • @YemonChoi It is, but this is the natural condition on the derivative IMHO and the one I usually see cited/invoked. – Alex Becker Feb 29 '12 at 08:38
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You would need to let $f\in C^1$ (i.e., its derivative is continuous). Simply use the mean value theorem.

Stuck_pls_help
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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.

AQP
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  • Any explanation for the downvote, please? – AQP Feb 29 '12 at 02:41
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    I downvoted this answer because it is wrong. Norbert's example is differentiable at $0$, but the derivative is not continuous at $0$. In order to apply MVT you need an upper bound for the derivative, which is only guaranteed if the derivative is continuous. – Alex Becker Feb 29 '12 at 02:42
  • I agree my answer is incorrect, but I'm not sure that a continuous derivative is necessary, tho clearly it is sufficient. – AQP Feb 29 '12 at 03:06
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    Continuous derivative is not necessary, merely a bounded one, which is implied by continuous since $[a,b]$ is compact. But the natural condition and the one usually given on the derivative is that it be continuous. Regardless, the fact still stands that your answer is incorrect and you should edit it to correct it. – Alex Becker Feb 29 '12 at 03:09
  • O.K, I will change it by tomorrow, see if I can make more general the necessary conditions. I will add a comment to that effect. I'm just not clear that Norbert's function is differentiable at 0. – AQP Feb 29 '12 at 03:25
  • Take the limit of $x^2\sin(1/x^2)/x=x\sin(1/x^2)$ as $x\to 0$. This is $0$ be squeeze theorem. Therefore the derivative at $0$ is $0$. – Alex Becker Feb 29 '12 at 03:29
  • Just to explain myself better: 1/x^2 is not itself differentiable at 0, so we cannot just conclude [f(g(x))]'=f'(g(x)g'(x) , since g'(x) is not defined at x=0; I'm trying to do a different analysis. – AQP Feb 29 '12 at 03:32
  • Yes, but that's not the definition of derivative, and only necessarily works if the derivative is continuous. That's why I applied the definition of derivative. – Alex Becker Feb 29 '12 at 03:34
  • Yes; I assumed Norbert was using this, which necessitates that g be differentiable in the range of f(x). But , yes, under the def. of derivative, it is differentiable. I edited to reflect my error. Maybe others will learn a possible pitfall when seeing my post, and the edit will make sure they know it is wrong. – AQP Feb 29 '12 at 03:38
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your original proof would be sufficient if you add the condition that f' is bounded on the interval I.

user92536
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