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Consider a function $f:\mathbb{R}^2 \to \mathbb{R}$. If $f$ is twice derivable we can consider $\frac{\partial^2 f}{\partial x \partial y}$ and $\frac{\partial^2 f}{\partial y \partial x}$. If $\frac{\partial^2 f}{\partial x \partial y}$ is continuous, then $$\frac{\partial^2 f}{\partial x \partial y} =\frac{\partial^2 f}{\partial y \partial x}$$

My question is: can we find some intuition behind this? In a more general context ($f:\mathbb{R}^n\to\mathbb{R}^m$), does the continuity of $\frac{\partial^n f}{\partial x_{i_1}\dots \partial x_{i_n}}$ ensure that the order of derivation does not matter?

ted
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  • The Schwarz lemma usually refers to a result in complex analysis https://en.wikipedia.org/wiki/Schwarz_lemma. – copper.hat Nov 28 '21 at 20:12
  • @copper.hat I thought adding '(partial derivatives)' would avoid confusion. Should I delete the question so I can change the title to 'Symmetry of partial derivatives' or something similar? – ted Nov 28 '21 at 20:23
  • @ted: Just [edit] your question. – Prasun Biswas Nov 28 '21 at 20:26
  • I think a change of title might be better, I never realised that it is known as Schwarz's theorem https://en.wikipedia.org/wiki/Symmetry_of_second_derivatives. – copper.hat Nov 28 '21 at 20:27
  • Similar: https://math.stackexchange.com/questions/868199/clairauts-equality-of-mixed-partial-derivatives-theorem-interpretation, https://math.stackexchange.com/questions/942538/geometric-interpretation-of-mixed-partial-derivatives – Hans Lundmark Nov 28 '21 at 20:28
  • I changed lemma -> theorem in the title. Hope that is not disturbing anyone. – copper.hat Nov 28 '21 at 20:29
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    Usually the notation for second derivative has a $\partial^2 f$ instead of $\partial f^2$ – podiki Nov 28 '21 at 20:35
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    @copper.hat thanks for the help and the links – ted Nov 28 '21 at 21:33
  • @podiki true, I will edit it now – ted Nov 28 '21 at 21:33

1 Answers1

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The intuition is in the proof (page 49 here https://mtaylor.web.unc.edu/wp-content/uploads/sites/16915/2018/04/analmv.pdf), which proceeds by approximating $\partial_k$ with the finite difference quotient operator $\Delta_{k, h}$ defined by $\Delta_{k, h}f(x) = \frac{f(x + he_k) - f(x)}{h}$. I will write the proof here. We have \begin{align} \Delta_{k,h}\Delta_{j, h}f(x) &= \partial_k(\Delta_{j, h}f)(x + \theta_khe_k) \\ &= \Delta_{j, h}(\partial_k f)(x + \theta_khe_k) \\ &= \partial_j\partial_k f(x + \theta_khe_k + \theta_jhe_j), \end{align} where $\theta_k, \theta_j \in (0, 1)$. Thus if $\partial_j\partial_kf$ is continuous at $x$, then $$\lim_{h \to 0}\Delta_{k,h}\Delta_{j, h}f(x) = \partial_j\partial_kf(x).$$ Then note that $\Delta_{k, h}\Delta_{j, h} = \Delta_{j, h}\Delta_{k, h}$, so by swapping the roles of $j$ and $k$, it follows that if $\partial_k\partial_jf$ is also continuous at $x$, then $\partial_k\partial_jf(x) = \partial_j\partial_kf(x)$. You can extend the result to higher indices by induction, or by repeating the proof.

Mason
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