# Partial differentiability and Continuity

If a function has, say, partial derivatives up to order n, can you conclude continuity of some or all derivatives of lower order?

Especially, if a function has partial derivatives of any order is it automatically smooth?

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If a function has (locally) bounded partial derivatives then it's continous – Jose27 Oct 7 '11 at 18:32

Consider the function $f:\mathbb{R}^2\to \mathbb{R}$ defined by $$f(x,y):=\frac{xy}{x^2+y^2}$$ for $(x,y)\ne (0,0)$ and $f(0,0):=0$.
This function has partial derivatives of any order anywhere in $\mathbb{R}^2$ (note that on the $x$- and $y$- axes it is just 0), but it is not even continuous.
Edit: I see now that this is not a counterexample because $\frac{\partial f}{\partial x}$ is no longer zero on the $y$-axis, so $\frac{\partial^2 f}{\partial x\partial y}$ does not exist. So here is a corrected counterexample: $$f(x,y):=e^{-\frac{(x^2+y^2)^2}{x^2 y^2}}$$ for $x\ne 0, y\ne 0$ and $f(x,y)=0$ on the coordinate axes. This function is not continuous at 0 (consider the restriction to the line $x=y$), but it is smooth outside of 0, and all derivatives still have the property that they are 0 on the coordinate axes. Hence all partial derivatives of any order also exist in 0.
This function is no counterexample, since $\partial^2/\partial x \partial x$ of the function doesn't exist. The question is: If any reapeated partial derivatives (in any order) exist, is the function smooth? I am looking for a counterexample. Once again: If the function is on $U \subset\mathbb{R}^2$, then it has to be differentiable w.r.t. to $x$, then to $y$, then again to $x$ etc., in any order. The question is, if a function fulfills this, is it smooth? – Kofi Oct 11 '11 at 11:33