# Good example of a non-continuous function all of whose partial derivatives exist

What's a good example to illustrate the fact that a function all of whose partial derivatives exist may not be continuous?

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I removed the tag "math-education" (and replaced it by "examples"). Remember: this whole site is math-education in the sense that people are asking math questions and hoping to learn from the answers. The tag should be saved for questions with an explicit pedagogical component. –  Pete L. Clark Mar 25 '10 at 14:43

The standard example I have seen is: $f(x,y)=\frac{2xy}{x^2+y^2}$.

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But surely that's continuous on its domain of definition. –  Dyke Acland Mar 25 '10 at 14:23
Define it to be 0 at (0,0) and it's discontinuous there, although the partial derivatives exist. –  Mark Meckes Mar 25 '10 at 14:25
However you define f at the origin, it will be discontinuous since its limit along the coordinate axes is zero, whereas it is 1 along the diagonal of $\mathbb R^2$. You get wilder example by starting with any antisymmetric function on the unit circle (say non-measurable) and extending it linearly on all (vector)lines of the plane i.e. defining $f(ra)=rf(a)$ for $r \in \mathbb R$ and $a$ on the circle. –  Georges Elencwajg Mar 25 '10 at 15:58
In George's "wilder example", further conditions on f will be needed to make the partial derivatives exist everywhere (is that what the proposer wanted?) –  Bjorn Poonen Mar 25 '10 at 16:36
Bjorn is right, of course: the example was only meant to show that a function can be quite pathological and yet have directional derivatives in all directions at the origin. The construction is the source of a few amusing exercises: e.g. the extended function is continuous at the origin iff the function on the circle is bounded. On the other hand even if you start with a $C^{\infty}$ function on the circle (seen as submanifold of $\mathbb R^2$) the extended function on the plane will NOT be differentiable at the origin in general ( think bump function on the circle). –  Georges Elencwajg Mar 25 '10 at 18:13
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I was travelling when this came up.

With the same results as Georges Elencwajg's comment and Marco's recent answer, take $$f(x,y) = \frac{2 x^2 y}{x^4 + y^2}$$ and set to $0$ at the origin $(0,0).$ Along any line through the origin $x = a t, \; y = b t$ the limit is 0, as $$| f(a t, b t) | \leq a^2 | t / b | .$$ However, along the parabola $y = x^2$ the value is 1, and along the parabola $y = - x^2$ the value is $-1.$

To get "directional derivative" 0 in every direction through the origin switch to $$g(x,y) = \frac{2 x^3 y}{x^6 + y^2}$$ as $$| g(a t, b t) | \; \leq \; t^2 \; | a^3 / b |$$ when $b \neq 0,$ but then if $y = \pm x^3$...the directional derivative generalizes to the Gateaux derivative in other settings.

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Consider $f(x,y)=1$ on the set $\lbrace{(x,y):y=x^2}\rbrace\backslash\lbrace(0,0)\rbrace$ and $f(x,y)=0$ on any other point.

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Unfortunately, that doesn't work: we also need the function to tend to zero for fixed y as x tends to zero. But at least this has given us the idea that we would like to write f as a quotient. What properties would we need of g and h if we tried f(x,y)=g(x,y)/h(x,y)? We would want g(x,0)=g(0,y)=0, g(x,x)=h(x,x), and h(x,y) is never 0 (except that we don't mind what happens at (0,0). We also want g and h to be nice so that the partial derivatives will obviously exist. The simplest function that vanishes only if (x,y)=(0,0) is $x^2+y^2$. The simplest function that vanishes when x=0 or y=0 but not when $x=y\ne 0$ is $xy$. Multiply that by 2 to get 1 down the line x=y and there we are.
Now suppose we had wanted the value to be, say, $e^{1/x}$ at (x,x), so that the function is wildly unbounded near (0,0). Then we could just multiply the previous function by $\exp((2/(x^2+y^2))^{1/2})$.
The main point I want to make is that we could just as easily have chosen many other functions. For example, $\sin(x)\sin(y)/(x^4+y^4)$ vanishes on the axes and clearly does not tend to zero down the line x=y (in fact it tends to infinity).