# Proving exp function is bounded and not extended continuously.

1. Oct 20, 2009

1. The problem statement, all variables and given/known data

The function

exp[ (x2 + y2 - xy)/(x2 + y2) ] = f(x,y)

is continuous on the open first quadrant.

Prove it is bounded there.

Prove f cannot be extended continuously to the closed first quadrant.

3. The attempt at a solution

Since f is a real-valued function defined and continuous on a compact set (open first quadrant...I think...), then f is bounded..? So I have to show that the open first quadrant is a compact set maybe?

I think I have to show that f is not uniformly continuous on the closed first quadrant. So, there exists an epsilon > 0 where a number d >0 cannot be found such that |f(p) - f(q)| < epsilon whenever p and q are in the closed first quadrant and |p-q|<d.

I have so much trouble starting my proofs properly. I don't know how to set them up at the beginning, even when I know the rough outline.

2. Oct 20, 2009

### aPhilosopher

You need to check your definition of compact.

can you bound the rational function on the open first quadrant?

3. Oct 21, 2009

### Staff: Mentor

Try working with this expression:
$$\frac{x^2 + y^2 - xy}{x^2 + y^2}$$
I can't give you much more of a hint than that without essentially doing the problem.

4. Oct 22, 2009

|(x2 + y2 - xy)/(x2 + y2)| = |(r2 - rcosArsinA)/r2)| = |1-cosAsinA| which is bounded by 0 and 1

So f(x,y) is bounded by 1 and e.

Is it enough to show that? Or do I need to use definition of limits...

Also, for the second part, it cannot be extended because x=0=y is a singularity. Does it suffice just to say that or does that not count as a full proof?

5. Oct 22, 2009

### Staff: Mentor

You don't need to convert to polar form. How did you conclude that f(x, y) is bounded by 1 and e? Are you saying that all values of f(x, y) on the open first quadrant are between 1 and e? That's not true.

Think division.

6. Oct 22, 2009

Ok let me try this again...

|(x2 + y2 - xy)/(x2 + y2)|<or= 1 because you are subtracting xy from the top so something smaller divided by something larger is a fraction less than one.

Then, since the exponent is bounded by m=1, can you say f(x,y) is bounded by 0 and e, i.e. when you plug in 1 in the exponent you get e^1 = e?

For the second part, I THOUGHT I could use the discontinuity x=y=0 but since S is the open first quadrant, can I do that?

7. Oct 22, 2009

### HallsofIvy

Staff Emeritus
I would be inclined to change to polar coordinates.