# Is there a discontinuous function on the plane having partial derivatives of all orders?

If one requires simply the existence of partial derivatives of first order rather than all orders, then a standard example is the function

$$f(x,y) = \left\{\begin{array}{l l} \frac{2xy}{x^2+y^2} & \quad \text{if (x,y)\neq(0,0),}\\ 0 & \quad \text{if (x,y)=(0,0).} \end{array} \right.$$

However, this does not constitute an answer to my question since the partial derivative of $\frac{\partial f}{\partial x}$ with respect to $y$ does not exist at the origin.

PS: This question rose out of my wonder as to whether, in the definition of a smooth function, continuity of partials is an essential requirement or not.

• If all partial derivatives are continuous in a neighborhood of $x_0$, then the function is differentiable at $x_0$. – dls Jul 31 '13 at 17:32
• See this sci.math post for a discontinuous function whose partial derivatives of all orders exist. – Dave L. Renfro Jul 31 '13 at 18:15

For ease of reading, I TeXified a part of sci.math post by Dave L. Renfro (this post is a CW).

What follows is from The American Mathematical Monthly 67 #8 (October 1960), 813-814.

Discontinuous Function with Partial Derivatives Everywhere 4876 [1959, 921]. Proposed by Naoki Kimura, University of Washington

If a real valued function $f(x,y)$ of two real variables possesses all of its partial derivatives $$\frac{\partial^{m+n} f(x,y)}{\partial x^m \partial y^n}$$ at every point, is it necessarily continuous?

Solution by John Burr, University of New England, Australia.

The following example shows that the function need not be continuous. The function $$f(x,y) = \begin{cases} \exp\left(-\frac{x^2}{y^2} - \frac{y^2}{x^2}\right) \quad & xy \ne 0 \\ 0 & \text{otherwise}\end{cases}$$ is discontinuous at $(0,0)$, since $\lim_{t\to 0}f(t,t) = e^{-2}\ne f(0,0)$. Suppose that it has been proved that a particular partial derivative $\phi(x,y)$ has the properties

1. $\phi(0,y) = \phi(x,0) = 0$
2. if $xy \ne 0$ then $\phi(x,y) = R(x,y) f(x,y)$ where $R(x,y)$ is a rational function with denominator of the form $x^p y^q$.

Then, by 1, $\phi_y (0,y) = \phi_x (x,0) = 0$; by 2, $\phi_x(0,y) = 0$ when $y \ne 0$, since when $x\to 0$, $f(x,y) \to 0$ more rapidly than any power of $x$. Similarly $\phi_y (x,0) = 0$. Hence $\phi_x (x,y)$ and $\phi_y (x,y)$ both have the property 1, and it is clear from 2 that they also have the property 2. Since $f(x,y)$ has these properties, it follows by induction that all the partial derivatives have them, and that these derivatives exist at every point.

It may be noted that this function satisfies conditions more stringent than those prescribed in the problem; namely the additional conditions that the mixed partial derivatives all exist, and are independent of the order in which the differentiations are performed.

Now, to the question

whether, in the definition of a smooth function, continuity of partials is an essential requirement or not.

I'll say that defining smoothness via partials is a wrong approach to begin with. The definition should be in terms of the total derivative. Then one can inquire into its relation with partial derivatives and obtain results such as: existence and continuity of partials implies the existence of the total derivative of that order.

@DavidL.Renfro's comment's link gives a very nice example... But/and there may be some reason to give a complementary sort of answer. Namely, if we look at (let's say tempered for simplicity, to avoid worrying about smooth truncations and such) distributions, first (as warm-up to the point I want to make) Clairault's theorem about interchangeability of second mixed partials becomes always-true, because Fourier transform (of tempered distributions) converts $\partial/\partial x$ to mere multiplication by $ix$, and similarly for $y$, and these multiplication operators certainly commute! In particular, any failure in counter-examples has to be in a form that integration against test functions cannot detect.

Similarly, with the question at hand, at a just-slightly more sophisticated level, a Sobolev imbedding theorem says that if $f\in L^2(\mathbb R^2)$ and $\Delta f\in L^2(\mathbb R^2)$, then $f$ is continuous. Here $\Delta$ is the sum of the second pure partials, as usual. (Further, $f$ can be smoothly truncated so that any obstacle to square-integrability is not at infinity, but just local.)

Thus, once again, for $f$ locally square integrable (!) compactly supported and $\Delta f$ locally square-integrable, $f$ is continuous.

Thus, counter-examples must have other pretty pathological features, too.