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Does there exist a function that is differentiable everywhere with everywhere discontinuous partial derivatives?

I just had this doubt, talking about first order partials.

YoTengoUnLCD
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  • One could try to generalize one dimensional argument to say that it is impossible. – A.Γ. Aug 23 '15 at 00:01
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    Loosely speaking, yes. One can construct a function or a series of functions that converges to something that experiences a structure similar to Brownian motion. It's been awhile since I've worked on something like this and when I did, I had a 10 page proof. Look into Brownian motion. Hopefully that will guide you in the right direction. Hope I helped :) –  Aug 23 '15 at 02:09

2 Answers2

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It is not possible for such function to exist.
The differentiability of $f$ implies $f$ is continuous , therefore $\frac{\partial{f}}{\partial{x}}$ and $\frac{\partial{f}}{\partial{y}}$ are both of Baire class one, which means they have a dense set of points of continuity .

By the way, I am also curious about if there exists a function with both partial derivative exist everywhere but are continuous nowhere. Since such function is only of Baire class one and its partial derivatives are of Baire class two (not necessarily of Baire class one), it seems to have a little possibility.

Antimonius
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Yes it does. Take $$ f\left( {x,y} \right) = \left\{ \begin{array}{l} x^2 \sin \left( {\frac{1}{x}} \right) + y^2 \sin \left( {\frac{1}{y}} \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,xy \ne 0 \\ x^2 \sin \left( {\frac{1}{x}} \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \ne 0,y = 0 \\ y^2 \sin \left( {\frac{1}{y}} \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0,y \ne 0 \\ 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0,y = 0 \\ \end{array} \right.. $$ Then, $f$ is differentiable everywere however its partial derivatives are discontinuous $$f_x= \left\{ \begin{array}{l} 2x\sin \left( {\frac{1}{x}} \right) - \cos \left( {\frac{1}{x}} \right),\,\,\,\,\,\,\,\,x \ne 0 \\ 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0 \\ \end{array} \right. $$ and $$f_y=\left\{ \begin{array}{l} 2y\sin \left( {\frac{1}{y}} \right) - \cos \left( {\frac{1}{y}} \right),\,\,\,\,\,\,\,\,y \ne 0 \\ 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y = 0 \\ \end{array} \right.$$ which are discontinuous at the origin $(0,0)$. This example given in the book of B. Gelbaum and J. Olmsted, "Counterexample in Analysis", 3rd ed., page 119.

Mohammad W. Alomari
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    I asked for "everywhere discontinuous" partials, both of those seem to be continuous everywhere but at $(0,0)$... – YoTengoUnLCD Aug 22 '15 at 23:32
  • Easily, you can take a sequence of points $a_0,a_1,...,a_n \in \mathbb{R}$ and define $f_{n}(x,y)=f(x-a_n,y-a_n)$. After that define $F(x,y)=\sum{f_{n}(x,y)}$. – Mohammad W. Alomari Aug 22 '15 at 23:36
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    Your comment doesn't seem to resolve the problem. That would make it discontinuous at finitely many points. At best you could do countably many this way. The OP wants discontinuity at every single point. – Matt Samuel Aug 22 '15 at 23:43
  • I didn't say finite sequence. Since every real number can be a member of a closed bounded interval (then compact) there is a sequence of points $a_n$ converges to that real number. And this is true for every real number. – Mohammad W. Alomari Aug 22 '15 at 23:47
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    How exactly is that relevant? – Matt Samuel Aug 22 '15 at 23:48