5
$\begingroup$

Is the following statement $$\frac{\partial^2 f}{\partial x \, \partial y}=\frac{\partial^2 f}{\partial y \, \partial x}$$ always true? If not what are the conditions for this to be true?

$\endgroup$
4
  • $\begingroup$ Very nice result here: math.stackexchange.com/a/47885/27978 $\endgroup$
    – copper.hat
    Dec 20, 2014 at 17:15
  • 2
    $\begingroup$ Just a remark: Note that in a broader sense, if you replace the derivatives with covariant derivatives on a Riemannian manifold (concider that everything is happening on a sphere, for example), the difference between $f_{xy}$ and $f_{yx}$ is the curvature (at least when differentiating vector fields). So, you can interpret the above identity as saying that $\mathbb{R}^n$ is "flat". $\endgroup$ Dec 20, 2014 at 17:21
  • $\begingroup$ Here, you can find the more generalized version of the above theorem. math.stackexchange.com/questions/3439436, mathoverflow.net/questions/346161 $\endgroup$
    – Kumar
    Sep 3, 2020 at 4:49
  • 1
    $\begingroup$ And here's the standard counterexample with $f_{xy}(0,0) \neq f_{yx}(0,0)$: math.stackexchange.com/questions/219759/… $\endgroup$ Oct 21, 2021 at 7:20

2 Answers 2

13
$\begingroup$

Second order partial derivatives commute if $f$ is $\mathcal{C}^2$ (i.e. all the second partial derivatives exist and are continuous). This is sometimes called Schwarz's Theorem or Clairaut's Theorem; see here.

$\endgroup$
0
5
$\begingroup$

This is true in general if $ f \in \mathcal{C}^2 $. This has a name: symmetry. More formally, it is known as Clariut's Theorem or Schwarz's theorem.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .