Higher Order Partial Derivatives and the Equality of Mixed Partials (Pt. II)
Point of Post: This is a continuation of this post.
Remark: For notational convenience we assume that . It’s clear the only place this is implied is the writing of the tuples and thus is inconsequential.
Proof: We begin by showing that it suffices to prove this for mappings . We do this in two steps, first showing that our theorem follows if it’s proven for mappings and then show that this is implied by the result for mappings . To prove this first claim we suppose that the result is true for all mappings with and let satisfy the hypotheses. We then note if then from a previous result we have that and , we have by assumption that is continuous on that are continuous on for each . Similarly, since exists on and is continuous at we have that exists on and is continuous at for each . Thus, by assumption that the theorem holds for we have that exists for and moreover and thus by first principles exists and .
So, now let’s show that the result being true for , where is open, which satisfies the same hypotheses implies the result for . Indeed, since is open there exists some open ball and there of course sits some symmetric -cube with center . We then define given by
and to be . Fix then some . We note then if we define where is some interval for which and define similarly we have by previous remark that exists if only if is differentiable at zero. But, noting that
we have by assumption that exists at that is differentiable at zero and moreover then we see that . Thus, since were arbitrary we have that exists everywhere on and Moreover, it’s clear that is continuous since and are continuous (the latter because it’s an isometry). Using the same tricks one shows that is defined everywhere on , is continuous, and . Moreover, using the same idea, defining we see the same method applies and shows that exists on and moreover that which in particular implies that is continuous at . Thus, if we assume the theorem is true for mappings with we see that exists and . But, this clearly implies that exists and
Thus, we now assume without loss of generality that and let satisfy the hypotheses of the theorem with . Since the following argument (as you shall see) is independent of whether or we assume the latter.So, what we want to show is that
exists and equals . Now, define for where is chosen to be sufficiently large so that for all (this can be done for sufficiently small ). We note then by our favorite observation that is differentiable on and . We note then that may be rewritten as
where we’ve used the mean value theorem on (where we know for sufficiently small ). Using a similar approach we see that may be rewritten as
It’s not quite as easy as it looks “hey! We know that and and so the result follows from continuity”, the problem with this is that actually depends on as well–so we don’t get this totally for free (although the idea is right). Well, from continuity at we know that for any there exists some such that implies . So, choosing enables us to conclude that . In particular, if we choose and let tend to zero we get that
Thus, since was arbitrary we have the desired result.
1. Spivak, Michael. Calculus on Manifolds; a Modern Approach to Classical Theorems of Advanced Calculus. New York: W.A. Benjamin, 1965. Print.
2. Apostol, Tom M. Mathematical Analysis. Reading, MA: Addison-Wesley Pub., 1974. Print.