Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is $f:\mathbb{R}^n\rightarrow\mathbb{R}^n$ globally Lipschitz continuous, i.e. there exists an $L>0$ such that

$\frac{|f(x)-f(y)|}{|x-y|}\leq L$ for all $x,y\in\mathbb{R}^n$,

and $\mathcal{C}^1$ if and only if $f$ is $\mathcal{C}^1$ and its total derivative is bounded?

Based on intuition alone, I'm strongly inclined to believe that the answer is yes. However I'm having trouble coming up with a proof (probably because my grasp of multivariable calculus is far from great). Could someone give one if the statement is true, or provide a counter example if it is false?


share|cite|improve this question
up vote 6 down vote accepted

If $f$ is $\mathscr{C}^1$, then $f(x) - f(y) = \int_0^1 Df(y + t(x-y)).(x-y) dt$, by the fundamental theorem of calculus.

Hence, $$\begin{aligned} \| f(x) - f(y) \| &\le& &\int_0^1 \|Df(y+t(x-y)).(x-y) \| dt& \\ &\le& &\left( \int_0^1 \| Df( y + t(x-y) )\| dt \right) \| x-y \|& \le \sup_{z \in \mathbb{R}^n} \, \|Df(z) \| \; \| x-y \| \end{aligned}$$

If $\sup_{z \in \mathbb{R}^n} \, \| Df(z) \| = C$ is finite, we get $\| f(x) - f(y) \| \le C \|x - y \|$ for all $x,y$.

Reciprocally, suppose that your function is $\mathscr{C}^1$ and that it's globally Lipschitz, with constant $C$.

Then, for all $x \in \mathbb{R}^n$, and all $h \in \mathbb{R}^n$, we know that $$Df(x).h = \lim_{t \to 0} \frac{f(x+th) - f(x)}{t}$$

But, by assumption, $\| f(x +th) - f(x) \| \le C \|th \| = C |t| \|h\|$, and we finally get $\|Df(x).h \| \le C \|h \|$ for all $h$, which by definition implies $\| Df(x) \| \le C$. Hence the total derivative is bounded all over $\mathbb{R}^n$.

All of this works also on an open set of $\mathbb{R}^n$, instead of the whole space.

Remark also that you don't need to assume $f$ to be $\mathscr{C}^1$, but only differentiable. The second part of my proof works as well, and for the first part, instead of applying fundamental theorem of calculus, you can use the mean value theorem.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.