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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?

Remark

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.

Rodrigo de Azevedo
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bass
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    Well, $x \mapsto x^2$ is not particularly Lipschitz, is it? You seem to miss either a locally or that $df$ must be bounded. – t.b. Oct 17 '11 at 15:37
  • @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. – Willie Wong Oct 17 '11 at 15:41
  • yes you're right. I meant locally Lipschitz. – bass Oct 17 '11 at 15:41
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    @bass: use that continuous functions are locally bounded. – Willie Wong Oct 17 '11 at 15:43
  • 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... – Mark Oct 17 '11 at 17:20

3 Answers3

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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 $L$.

Christian Blatter
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  • I have a little doubt, isn't $df$ with respect to $t$ , then how can we justify the inequality ? – Theorem Nov 18 '13 at 10:05
  • @Theorem: $df\bigl(a+t(b-a)\bigr)$ is the derivative ("Jacobian") of $f$, evaluated at the point $a+t(b-a)\in U$. – Christian Blatter Nov 18 '13 at 10:14
  • Does this hold for just differentiable functions f, not continuously differentiable. – kam Mar 08 '20 at 00:44
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    @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 $x=0$. – Christian Blatter Mar 08 '20 at 09:55
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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.

Beni Bogosel
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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.

Fatemeh Shiravand
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