Let $f:\mathbb R^d \to \mathbb R^m$ be a map of class $C^1$. That is, $f$ is continuous and its derivative exists and is also continuous. Why is $f$ locally Lipschitz?


Such $f$ will not be globally Lipschitz in general, as the one-dimensional example $f(x)=x^2$ shows: for this example, $|f(x+1)-f(x)| = |2x+1|$ is unbounded.

  • 4
    $\begingroup$ Well, $x \mapsto x^2$ is not particularly Lipschitz, is it? You seem to miss either a locally or that $df$ must be bounded. $\endgroup$
    – t.b.
    Oct 17 '11 at 15:37
  • $\begingroup$ @t.b. In some contexts $C^1$ implies that $\sup |f| + \sup |df| < \infty$. (At least, when I write $C^1(M,\mathbb{R}^k)$ with $M$ being non-compact, that's what I would mean.) Of course, this is not what the OP wrote in the parenthetical. $\endgroup$ Oct 17 '11 at 15:41
  • $\begingroup$ yes you're right. I meant locally Lipschitz. $\endgroup$
    – bass
    Oct 17 '11 at 15:41
  • $\begingroup$ @bass: use that continuous functions are locally bounded. $\endgroup$ Oct 17 '11 at 15:43
  • $\begingroup$ First reduce to the scalar-valued case (this is easy). Then note that $\left|f\left(x\right)-f\left(y\right)\right|=\left|\int_{0}^{1}\frac{d}{dt}f\left(tx+\left(1-t\right)y\right)dt\right|$. Then use the chain rule and... $\endgroup$
    – Mark
    Oct 17 '11 at 17:20

If $f:\Omega\to{\mathbb R}^m$ is continuously differentiable on the open set $\Omega\subset{\mathbb R}^d$, then for each point $p\in\Omega$ there is a convex neighborhood $U$ of $p$ such that all partial derivatives $f_{i.k}:={\partial f_i\over \partial x_k}$ are bounded by some constant $M>0$ in $U$. Using Schwarz' inequality one then easily proves that $$\|df(x)\|\ \leq\sqrt{dm}\>M=:L$$ for all $x\in U$. Now let $a$, $b$ be two arbitrary points in $U$ and consider the auxiliary function $$\phi(t):=f\bigl(a+t(b-a)\bigr)\qquad(0\leq t\leq1)$$ which computes the values of $f$ along the segment connecting $a$ and $b$. By means of the chain rule we obtain $$f(b)-f(a)=\phi(1)-\phi(0)=\int_0^1\phi'(t)\>dt=\int_0^1df\bigl(a+t(b-a)\bigr).(b-a)\>dt\ .$$ Since all points $a+t(b-a)$ lie in $U$ one has $$\bigl|df\bigl(a+t(b-a)\bigr).(b-a)\bigr|\leq L\>|b-a|\qquad(0\leq t\leq1)\>;$$ therefore we get $$|f(b)-f(a)|\leq L\>|b-a|\ .$$ This proves that $f$ is Lipschitz-continuous in $U$ with Lipschitz constant $L$.

  • $\begingroup$ I have a little doubt, isn't $df$ with respect to $t$ , then how can we justify the inequality ? $\endgroup$
    – Theorem
    Nov 18 '13 at 10:05
  • $\begingroup$ @Theorem: $df\bigl(a+t(b-a)\bigr)$ is the derivative ("Jacobian") of $f$, evaluated at the point $a+t(b-a)\in U$. $\endgroup$ Nov 18 '13 at 10:14
  • $\begingroup$ Does this hold for just differentiable functions f, not continuously differentiable. $\endgroup$
    – kam
    Mar 8 '20 at 0:44
  • 1
    $\begingroup$ @kam:Consider the function $f(x):=x^2\sin(1/x^3)$. With $f(0):=0$ it is differentiable on all of ${\mathbb R}$, but it is not Lipschitz continuous near $x=0$. $\endgroup$ Mar 8 '20 at 9:55

Maybe this can help. The Lipschitz condition comes many times from the Mean Value Theorem. Search the link for the multivariable case. The fact that $f$ is $C^1$ helps you to see that when restricted to a compact set the differential is bounded. That's why you only have local Lipschitz condition.


A function is called locally Lipschitz continuous if for every x in X there exists a neighborhood U of x such that f restricted to U is Lipschitz continuous. Equivalently, if X is a locally compact metric space, then f is locally Lipschitz if and only if it is Lipschitz continuous on every compact subset of X. In spaces that are not locally compact, this is a necessary but not a sufficient condition.The function $f(x) = x^2$ with domain all real numbers is not Lipschitz continuous. This function becomes arbitrarily steep as x approaches infinity. It is however locally Lipschitz continuous.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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