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Is there a function $f:O\to \mathbb{R}$, $O$ is an open subset in $\mathbb{R}^2$, which satisfies both $(1)$ and $(2)$ ?

$(1)$ All of its second partial derivatives are defined on $O$ and continuous at $(x_0,y_0)\in O$ ;

$(2)$ It is not differentiable in any neighborhood of $(x_0,y_0).$


Obviously, $(1)$ involves the differentiability of $f$ at $(x_0,y_0)$.

The original post come from here. In order to express my purpose clearly, I simplify $n$-variables to two-variables.

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1 Answer 1

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The answer is no. Indeed, if your condition (1) holds, then all of the second-order partial derivatives of $f$ are bounded on an open neighborhood $U$ of $(x_0,y_0)$. So, by Lemma 1 below, both of the first-order partial derivatives of $f$ are Lipschitz and hence continuous on $U$, and therefore your condition (2) cannot hold.

Lemma 1. If $|g''_{xx}|+|g''_{yy}|+|g''_{xy}|+|g''_{yx}|\le M$ on $U$, then $|f'_x(x,y)-f'_x(a,b)|\le M|x-a|+M|y-b|$ and $|f'_y(x,y)-f'_y(a,b)|\le M|x-a|+M|y-b|$ for all $(x,y)$ and $(a,b)$ in $U$.

Proof. For any $(x,y)$ and $(a,b)$ in $U$, $$|f'_x(x,y)-f'_x(a,b)|\le|f'_x(x,y)-f'_x(a,y)|+|f'_x(a,y)-f'_x(a,b)| \\ \le M|x-a|+M|y-b|.$$ Similarly, $$|f'_y(x,y)-f'_y(a,b)|\le M|x-a|+M|y-b|.\quad\Box$$

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