A continuously differentiable map is locally Lipschitz 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.
 A: 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: 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 $L$.
A: 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.
