Clairaut's theorem and smooth functions

1. Sep 6, 2007

Simfish

Hello
So, my advanced calculus book (Folland) has this theorem...
If f is of class $$C^k$$ on an open set S, then...

$$\partial i_1 \partial i_2 ... \partial i_k f = \partial j_1 \partial j_2 \partial j_k f$$ on an open set S whenever the sequence $${j_1 ,..., j_k}$$ is a reordering of the sequence $${i_1 ,..., i_k}$$, which defines a smooth function when $$k = \infty$$

So my question is, is Clairaut's Theorem a special case of this? (when k = 2?). Also, is the existence of derivatives in the nxn Hessian matrix logically equivalent to the conclusion of Clairaut's Theorem? Does this theorem even have a name? (I can't find it on Wikipedia or Mathworld anywhere).

2. Sep 6, 2007

quasar987

You can see Clairaut as a special case but if you look at the proof of the theorem you stated, I suspect that it begins by proving Clairaut and then shows how your thm is a direct generalization (corollary) of it.

For your second question, I'm not sure I understand what you're asking, but if the Hessian matrix is well defined, it only means that the second partial derivatives all exist. It does not say anything about their being equal when they are locally continuous.

3. Sep 7, 2007

matt grime

generalization and corollary aren't the same thing, quasar.