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$f(x,y)$ is real-valued function on $R^2$

$f$ is of class $C^1$ and $\dfrac{\partial ^2 f}{\partial x \,\partial y}$ exists and is continous.

how to show $\displaystyle{\partial ^2 f \over \partial x \, \partial y}$ exists and same as $\displaystyle{\partial ^2 f \over \partial y \, \partial x}$ ??

ashku
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    The standard approach uses Green's theorem, but I'm not sure if your hypotheses are strong enough to go there directly. In other words, nothing you've written gives me a guarantee that $\frac{\partial^2 f}{\partial y \partial x}$ exists, unless I'm forgetting a theorem. – Emily May 06 '14 at 21:20
  • i editted the question again sorry – ashku May 06 '14 at 21:23
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    @Arkamis The standard approach actually just uses the mean value theorem twice. – Ryan Reich May 06 '14 at 21:27
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    Uh oh, now we need to convene a standards committee. – Emily May 06 '14 at 21:29
  • Ashku, I see little reason to think this works, but I'm not sure a counterexample is so easy either. I suggest you get http://store.doverpublications.com/0486428753.html Counterexamples in Analysis, by Gelbaum and Olmsted – Will Jagy May 06 '14 at 21:40
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    Related – Giuseppe Negro May 06 '14 at 21:53
  • @Min There are two considerations here: first, that the desired quantity $\frac{\partial^2 f}{\partial y \partial x}$ exists (and is continuous), and second, that if it exists, it is equal to $\frac{\partial^2 f}{\partial x \partial y}$. The second part is easy to show if the first part is satisfied. Satisfying the first part is the problem (existence + continuity), and there is nothing that is written that suggests it is the case. – Emily May 06 '14 at 23:08

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