# What does it mean for the total derivative to be bounded?

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?

• 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. – yoshi Jun 7 at 2:09
• I've edited a typo, if that helps. – gtoques Jun 7 at 2:11

## 1 Answer

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.