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Let $X \subseteq \mathbb{R}^n$ open, $f : X \to \mathbb{R}$ and $\mathbf{x} \in X$.

If the second derivatives mixed $f_{x_i x_j}(\mathbf{x})$, $f_{x_j x_i}(\mathbf{x})$ exist and are equal, then they are continuous in $\mathbf{x}$.

This proposition, in my opinion, is false. But I can not find a counterexample.

Do you have any idea?

Thank you!

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    My idea would be to start from the example instead of Peano (https://en.wikipedia.org/wiki/Symmetry_of_second_derivatives#Requirement_of_continuity) and making sure that the second derivatives coincide although not continuous. – 2017-01-12

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I suspect you want the derivatives to agree everywhere on $X$. I was confused because I reserve the notation $f(x)$ to refer to a specific value.

In any case consider $f:\mathbb{R}^{2}\to\mathbb{R}$ defined by $f(x,y)=x^{2}y^{2}\sin\left(\frac{1}{x}\right)\sin\left(\frac{1}{y}\right)$ and extended continuously to include the $x$ and $y$-axis. Then $f_{xy}=f_{yx}$ agree as functions everywhere. However, these functions are not continuous at $(0,0)$.