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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 continous2011-10-07

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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.

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    @Kofi: You accidentally created a duplicate account; with fewer than 50 reputation points, [an account can only comment on its own questions and answers](http://meta.stackexchange.com/questions/19756/how-do-comments-work/19757#19757). I have now merged your duplicate account into your current account (the one used to ask the question), so you will now be able to comment here (and anywhere, once you have >50 points). If you have trouble logging in, or if you accidentally create duplicate accounts, simply flag one of your own questions for moderator attention, and we will help out.2011-10-11