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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?

user2850514
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  • Very nice result here: http://math.stackexchange.com/a/47885/27978 – copper.hat Dec 20 '14 at 17:15
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    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". – Peter Franek Dec 20 '14 at 17:21
  • Here, you can find the more generalized version of the above theorem. https://math.stackexchange.com/questions/3439436, https://mathoverflow.net/questions/346161 – Kumar Sep 03 '20 at 04:49
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    And here's the standard counterexample with $f_{xy}(0,0) \neq f_{yx}(0,0)$: https://math.stackexchange.com/questions/219759/show-that-both-mixed-partial-derivatives-exist-at-the-origin-but-are-not-equal – Hans Lundmark Oct 21 '21 at 07:20

2 Answers2

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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.

Michael Albanese
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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.

Ivo Terek
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Ahaan S. Rungta
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