The Mean Value Theorem for Multivariable Maps
Point of Post: In this post we state and prove the multidimensional analogue of the mean value theorem.
Anyone who has taken a basic analysis course knows that the mean value theorem (MVT) is a very important, and widely used tool. In fact, we’ve had several occasions to use it in our study of multidimensional analysis. So, the obvious question is “is there a multidimensional analogue?” Well, in the strictest sense, there isn’t. By this I mean if one literally transposes the usual MVT to higher dimensions one arrives at something like “Let be everywhere differentiable on and let . Then, there exists some (where is the line segment connecting and ) such that ” Unfortunately, this is wildly false as the map clearly shows. So, what then is the correct formulation? We play our old trick of thinking of a map as secretly by first mapping for some vectors and then evaluating our map there (more explicitly something of the form ). From there if we could find some way of going back into we’d have a map which, being an honest to god real valued real variable map, can have the one dimensional case of the MVT applied to it. So, the question remains as to what kind of maps we want to consider? Well, we know from the above that we’re going to have to apply the chain rule to find the derivative of the map and so we don’t want to pick something so crazy that we are left knowing nothing new. No, we’ll restrict our maps into to be the simplest (in terms of derivatives), namely we’ll consider and so our map really looks like (where .
Mean Value Theorem
We begin with the theorem:
Theorem: Let , with open, be differentiable and . Suppose further that the line for some . Then, there exists some such that .
Proof: As indicated in the motivation consider given by . Note then that since and is open we actually have that for some . So, consider then the function . Now, since (recalling that constant maps and multillinear maps are differentiable) each of these functions are differential be know from the chain rule that is differentiable on . Thus, by the one-dimenionsal mean value theorem there exists such that
That said, we first note that . Next, by the chain rule we have that
Note though that a quick check shows that (this is clear since is, up to an irrelevant constant, just the linear map and thus its derivative is itself) and so . Next, we note that since is linear we know that . Thus, finally letting we may conclude that
Thus, putting this all together
lastly noting that finishes the argument.
Remark: I believe it is important to point out precisely why we conclude from the one-dimensional mean value theorem that . This is because we (thinking of as a real valued function from analysis, i.e. that the derivative is a number) know from the mean value theorem that that said, we have from first principles that from where that mysterious inclusion should now seem clear.
Where this most often comes up (this is actually taken as the usual multidimensional mean value theorem, but if we recall the canonical isomorphism between an inner product space and its dual we realize that these two theorems are actually equivalent):
Corollary: Let , with open , and . Then, if for some then there exists such that .
We now prove a theorem which, in all calculus textbooks, immediately follows the multivariable mean value theorem. Namely, it is the generalization that the only functions with zero derivatives on open connected subsets of the real numbers are the constants. Namely:
Theorem: Let , where is open and connected, is differentiable everywhere with for all . Then, is constant on .
Proof: Let be arbitrary. It is a common fact that every open connected subspace of Euclidean space is polygonally line connected, so let be points for and . We evidently have then by the mean value theorem that there exists some such that but by assumption this right hand side is zero. Thus,
but this sum telescopes to and so we see that and so . Since were arbitrary the conclusion follows.
From this we get the obvious corollary:
Corollary: Let , with open, be such that for every . Then, there exists such that for every .
Proof: We merely note that is zero everywhere on and so by the previous theorem for some constant . The conclusion follows.
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.