Monsieur Ejemplé

Give an example of a function f:R^2 --> R where both partial derivatives df/dx and df/dy exist non-trivially, but neither are continuous at (0,0).

f(x,y) = sqrt(x^2 + y^2).

It's a cone with the vertex at (0,0).

>>2

Won't that give a sort of double-cone, along which gradients are constant all the way, even through (0,0)?

>>3
Yes.

Also, the function itself must be continuous, though this is obvious from the question, since you're using derivatives.

I'd assume some function of the form

f(x,y) = xsin1/x + ysin1/y

should suffice.

I'm extrapolating from my knowledge of functions of a single parameter, but xsin1/x and ysin1/y are both continous functions, so I'd assume the addition would be continous.

However df/dx|y isn't continous and neither is df/dy|x.

Am I right?

>>5

Not quite. You've got a basic idea that will work with some refinement, though. Remember (I should have made this clearer) the function is continuous at (0,0) and both partial derivatives exist at (0,0).

>>6

Are you sure I've got the right idea?

The function needs to have a real value at 0, but not be continous, that means functions of the form sin(1/x) won't work, because whilst they're not continous at (0,0) they don't have a defined value.

Is using the dirac delta function cheating? :p
Is that even continous, I doubt it.

It'd be x^2sin(1/x) + y^2sin(1/y) in my last example anyway, thinking about it. Not that it matters :p

>>7

Oh but it does matter. Keep thinking, your thoughts are just slightly muddy at the moment.

And no physics bollocks like the delta function. Why does Firefox spellcheck highlight "bollocks"!?

>>3, 4
What do you mean by double cone?  I phail to see a 2nd cone.

>>6
By that requirement, >>2 would also fail due to the 0/0 singularity in the partial derivatives at (0,0).

>>7
The delta function is definitely not continuous.  The limit of the singularity is 0, but the actual "value" of the singularity is "infinite".

>>3

No; the sqrt() function is single-valued.

You are thinking of something like z^2 = x^2 + y^2.

>>8


Looked at my old notes from analysis last year.

Apparently I have proven that if f(x)  = x + 2x^2sin(1/x) if x!=0 and f(0)=0

then f'(0) = 1, but there is no interval around 0 where f is increasing, which means it can't be continous.

So I define my function as

f(x,y) = x + y + 2x^2sin(1/x) + 2y^2sin(1/y)
f(0,y) = y + 2y^2sin(1/y)
f(x,0) = x + 2x^2sing(1/x)
f(0,0) = 0

And if my result is true then that should satisfy your conditions :p

I know it sounds a bit weak, but I could prove that first result if I could be arsed, I hate typing maths into textboxes like thes :p

>>9
Sorry. You're right. As you said, the cone still isn't a valid solution though.

>>12
I think I'll give you that... yeah.

The example I was thinking of was
f(x,y) = x^(3/2)sin(1/x) + y^(3/2)sin(1/y)
f(0,y) = y^(3/2)sin(1/y)
f(x,0) = x^(3/2)sin(1/x)
f(0,0) = 0

If you take f(x) = x^(3/2)sin(1/x) then, by taking the required limit, you find the derivative at 0 to be 0. However, the derivative is 3/2x^(1/2)sin(1/x) - x^(-1/2)cos(1/x) which gives an unbounded value as you approach zero.

>>13

Is the power of 3/2 really needed?
This makes it not continous at 0 by not having a bounded value, but with a power of 2 you get the same result, but instead of it being unbounded it just does not approach a limit.

Also I can get rid of the x and y terms in my answer, then the derivative at (0,0) would be 0, but the function would oscillate wildy between 1 and -1 as it approached 0.

Meh, I just think x^2sin(1/x) + y^2sin(1/y) is somehow a bit more....aesthetically pleasing than x^(3/2)sin(1/x) + y^(3/2)sin(1/y). No idea why.

I think the boundedness pleases me :p

>>14

Goddamn, next time I'm REALLY going to think about the solutions before posing the question.

Any power in (1,2] will do the trick, amirite?

>>15

Almost certainly right.

pooooooooooooooooooooooooooo