# Finding a real-valued function not differentiable in any $B_{r}(\mathbf{x})$

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.

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, $$|f'_y(x,y)-f'_y(a,b)|\le M|x-a|+M|y-b|.\quad\Box$$