Why do partial derivatives commute

But first partials being differentiable functions at a,b does say a lot about the 2nd partials. Existence of partial derivatives of a function says nothing about continuity of the function since partial derivatives only involve function values on horizontal and vertical lines.

It implies existence of all the second partials at a,b but says nothing about the existence of any 2nd partials away from a,b. The 2 theorems are different,neither is stronger. Another reference for all this perhaps better is Apostol ,Tom -Mathematical Analysis 2nd edition.

This is an excellant reference for alot of things. In the 's the Russian mathematician Georgii Pavlovich Tolstov [Tolstoff] published several papers on this topic. One of his examples was a function f x,y with continuous first order partial derivatives everywhere and such that both mixed second order partial derivatives exist everywhere, but the two mixed second order partial derivatives differ on a set of positive measure.

Another example was a function g x,y with continuous first order partial derivatives such that the mixed second order partial derivatives exist and differ almost everywhere i. The difference between the first and second examples is that in his second example, at least one of the mixed partial derivatives fails to exist at some of the points that belong to the exceptional set of measure zero.

In the positive direction, Tolstov proved that if all the partial derivatives up to and including order n exist throughout an open connected set, then: a each mixed partial derivative of order less than n is independent of the order of differentiation at each point in the domain; b each mixed partial derivative of order n is independent of the order of differentiation at almost every point in the domain; c each partial derivative of order up to and including n is a Baire one function.

These results hold for functions having an arbitrary finite number of variables. I'm sure these results must have been improved in several different directions by now pun intended , but I don't know any of the literature on this topic. The results I gave for Tolstov are given on p. Kolmogorov, V. As with most "proof-by-picture" demonstrations, the point of this is not rigor, but rather to show intuitively what's going on. We've ignored a number of details here, such as the difference between the second derivatives at the rear corner and the front corner.

To produce a rigorous proof of this requires explicitly evaluating the limits used to define the derivatives to be sure that nothing we ignored made a difference, and the algebra gets a bit messy. Nor does it mean that the geometric structure is not co-ordinate-free. The study of Riemannian geometry does not depend on co-ordinates.

Of course, even the study of submanifolds in Euclidean space does not, too. Peter Michor Peter Michor 24k 1 1 gold badge 56 56 silver badges bronze badges. It is possible to explain this in simpler terms "soldering form" seems to be advanced from the basic place in diff. Or do you think a true understanding of the torsion tensor requires more advanced concepts?

ClassicalPhysicist ClassicalPhysicist 6 6 bronze badges. First , you fix a metric. Then you fix a connection compatible with the metric. If you are interested only in geodesic then, as E. Cartan observed, there are several connections compatible with the metric that give the same geodesics. If you are interested in more than geodesics, then curvature matters, and the curvature does depend on the choice of connection.

Thanks for clarifying. Sign up or log in Sign up using Google. Sign up using Facebook. Connect and share knowledge within a single location that is structured and easy to search.

This is sometimes called Schwarz's Theorem or Clairaut's Theorem. This theorem is in my textbooks, yet I cannot seem to find the proof in them.

I have tried proving the theorem yet I have gotten stuck. So, what is the proof of Clairaut's Theorem, or why do partial derivatives commute? If the proof is too long, a link to the proof with an intuitive explanation will be sufficient. There's this little proof, which uses differentiation under the integral sign.

Another one "TVM" stands for "mean value theorem" :. Note that this second proof also shows that we only need that both partial derivatives exist, with only one of them being continuous. It is a stronger version. Here's a nice proof online,but I'm puzzled by the fact you can't find a full proof. Most good multivariable calculus books I've seen, like Bandaxall and Liebeck, have detailed proofs.