Say we have a function $f: \mathbb{R}^n \to \mathbb{R}^m$ which has differentiable, and its total derivative at some point $x \in \mathbb{R}^n$ is given by $Df(x)$. What does it mean for $|Df|(x)$ to be bounded. Obviously, in one dimension we can view the derivative as a scalar valued function, so we have an idea of what boundedness is. How do we understand boundedness in higher dimensions?

My point being: how can we understand (by looking at the total derivative) when a multivariable function is Lipschitz continuous?

  • $\begingroup$ One guess that came to my mind is the boundedness of each entry of $Df(x)$ - without context of how this was used, it might be tough. I've never encountered this convention. $\endgroup$ – yoshi Jun 7 '19 at 2:09
  • $\begingroup$ I've edited a typo, if that helps. $\endgroup$ – gtoques Jun 7 '19 at 2:11

For convenience, denote $V = \mathbb{R}^n$, and $W = \mathbb{R}^m$, and let $\mathcal{L}(V,W)$ be the vector space of linear transformations from $V$ into $W$. For each $x \in V$, $Df(x)$ is a linear transformation from $V$ into $W$; or said differently, $Df(x) \in \mathcal{L}(V,W)$. On $\mathcal{L}(V,W)$, we can define a norm via the rule \begin{equation} \lVert T \rVert = \sup \{ \lVert T(\xi) \rVert: \lVert \xi\rVert \leq 1 \}. \end{equation} (There are 3 different norms in the equation above, all of which I've denoted by $\lVert \rVert$, so hopefully you know which norm is acting on which space).

So, one possible notion of boundedness of the derivative is the following: there is an open ball $B \subset V$, and a number $M > 0$, such that for all $x \in B$, we have $\lVert Df(x) \rVert \leq M$. Under this assumption, you can prove via the mean-value theorem$^1$ that $f$ is Lipschitz continuous on $B$.

One special case is the following: suppose all the partial derivatives $\partial_jf_i$ are continuous. Then, $Df: V \to \mathcal{L}(V,W)$ is continuous (see the book linked below for a proof of this). Then, given a compact convex set $K \subset V$ (such as a closed ball or a closed rectangle), since $\lVert Df(\cdot) \rVert$ is continuous, by the extreme value theorem, this function attains a maximum value $M$ on $K$. Then, you can apply the MVT as in the previous paragraph to conclude $f$ is Lipschitz on $K$.

Of course, you can generalize these to situations with weaker hypotheses etc, but this relatively simple formulation (or a minor variant of it) has served me well for many applications

[1.] For a statement and proof of the MVT, see Loomis and Sternberg's Advanced Calculus, Chapter 3, Theorem 7.4.

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