Prove that if $f:U \rightarrow \mathbb{R}$ is differentiable on $U$ then it is Lipschitz on $U$.

Let $U \subset \mathbb{R}$ be a closed and bounded set. Prove that if $f:U \rightarrow \mathbb{R}$ is continuously differentiable on $U$ then it is Lipschitz on $U$.

Could anyone show how to prove this (I am given a hint to use Mean Value Thm and Extreme value theorem)?

• $f$ needs to be continuously differentiable, otherwise this is false. – J.R. Feb 12 '14 at 19:19
• thanks for pointing it out...I just changed it.3 – afsdf dfsaf Feb 12 '14 at 19:26

Let us assume that $f$ is not only differentiable, but continously differentiable. Otherwise the conclusion does not necessarily hold. Consider the example $f(x)=x^{3/2}\sin(1/x)$ and $f(0)=0$.

Since $U$ is compact and $f^\prime$ is continuous, there exists $C>0$ such that

$$|f^\prime(x)|\le C\tag{1}$$

for all $x\in U$ (I suppose this is what is in your source called the extreme value theorem?).

Now let $x\not=y\in U$. By the mean value theorem, there exists $\xi\in U$ (actually in the interval spanned by $x$ and $y$ but we don't need that detail) such that

$$f(x)-f(y)=f^\prime(\xi)(x-y)$$

Taking absolute values and applying $(1)$ we obtain

$$|f(x)-f(y)|=|f^\prime(\xi)| |x-y|\le C |x-y|$$

That is, $f$ is Lipschitz on $U$.

• Do we need Extreme value theorem? – afsdf dfsaf Feb 12 '14 at 19:28
• @afsdfdfsaf: Being Lipschitz on $U$ means having a bounded derivative on $U$. How we obtain the bound on the derivative doesn't really matter. It may for example be given a priori or we exploit continuity and compactness as in this case (i.e. using the extrem value theorem). – J.R. Feb 12 '14 at 19:29
• so if the question is stated as "Let $U \subset \mathbb{R}$ be a closed and bounded set. Prove that if $f:U \rightarrow \mathbb{R}$ is differentiable on $U$ then it is Lipschitz on $U$." Can we use extreme value thm to solve it? – afsdf dfsaf Feb 12 '14 at 19:51
• @afsdfdfsaf No. See the counterexample above. The derivative need not be bounded, if $f$ is not continuosly differentiable. – J.R. Feb 12 '14 at 20:48
• The continuous function I am talking about is the derivative. That is why we need the derivative to be continuous. – J.R. Feb 12 '14 at 22:22