# Homework Help: Does A Finite Integral Over The Plane Imply A Function Is Bounded?

1. Jul 3, 2012

### Daron

Suppose I have a C function, which I wish to prove attains its maximum/minimum. First I must prove that the function is bounded at all. If I determine R, the region (of the plane in this case) where the function is strictly positive, and integrate over R to find a finite answer, can I say the function is bounded from above over R?

I am sure this is true for R having finite measure, but am not sure exactly what the condition is for an infinite region, which leads to an improper Riemann integral.

Once I am sure the function is bounded, I can just take some closed set about the sup and min, and use continuity to prove the function attains the extrema.

The function in question is (x2 - y2)e(x2 + y2) but that is probably not important.

2. Jul 3, 2012

### algebrat

This reminds of a cool trick, not sure if it applies here without thinking about it more, but here goes. Give some notation to the sets where f is betwen 1/n and 1/(n+1). Then bound the integral below by the sum of them. Hmm, or take balls of radius n. Something like that, but I think the idea is you want to make the argument countable in a sense, so that you have a better control on it, like a sequence, or sequences easily become series don't they?

Or let's see, if the function were not bounded... or rather, it has too converge to zero right? So for some radius out, it must stay below dome value, so it is definitely bounded outside of some ball. Then it is bounded inside the ball by extreme value theorem. Oh wait, this is analysis, so we don't assume continuity. So yeah, I think you need some sort of... Wait, you said the function was continuous, so yeah, there you go.

3. Jul 3, 2012

### LCKurtz

I don't think so. In $\mathcal R^1$ think about a non-negative function f(x) where the area under the graph on [0,1] is $\frac 1 2$, on [1,2] is $\frac 1 4$ etc. On each interval make the function skinnier and taller to get the area and smooth it out. The improper Riemann integral will exist and total area will be 1 and the function is unbounded above. A similar idea should work for a function of 2 variables.

4. Jul 3, 2012

### algebrat

I believe $C^\infty$ implies continuity. A good analogue to your example is 1/√x inside the unit disk, glued to 1/x2 outside the unit disk. It's integral is a finite value, but it is not continuous at the origin, thus this type of counterexample is excluded. You can find an outline of the intuition behind the truth of statement buried in my ramble above.

5. Jul 3, 2012

### LCKurtz

My example is not discontinuous and can easily be $C^\infty$ (that's what I meant by smooth it out). Rotate it about the y axis and you can make a 2-D example. Am I missing something?

6. Jul 3, 2012

### algebrat

Dang. I guess we need bounded slope or Lipschitz or something. I guess the result from series does not carry over. You could have got me to bet a dollar.

Last edited: Jul 3, 2012
7. Jul 3, 2012

### LCKurtz

Here's another way to look at it. Consider ripples in a pond. The height of the ripples are all zero on circles of integer radius. The first ripple for $r \in (0,1)$ has volume 1/2. The second for $r\in (1,2)$ has volume 1/4 etc. The ripples are $C^\infty$ smooth and get skinnier and taller as they get farther from the origin. Same idea.

8. Jul 4, 2012

### Daron

I'm not sure exactly what you mean here by "smooth it out". But all of the functions of this type that I've seen go to infinity at the origin, which is prohibited by this one being C.

This function is bounded from above by 1/2, unless you are proposing to make it 1 on
[1/4, 1/2]; 2 on [1/8. 1/4] etc. But then it is similarly undefined at the origin. It is also not continuous. But we can just find a decreasing function joining all the corner points, so that's probably not a problem.

Is this differentiable on the unit sphere? Sorry, to clarify C means the function is continuous, and infinitely differentiable with continuous derivatives, which can be all zero after some point.

I have found a solution to this problem, by the way. First we switch to polar coordinates:

(r2Cosθ2 - r2Sin2)er2.

Then we observe that the function is positive from θ = -∏/4 to ∏/4, negative over the next quarter plane, positive over the next, and positive over the next. Furthermore it is antisymmetric over the lines x = ±y. So we just have to integrate over one wedge.

When we integrate (r2Cosθ2 - r2Sin2)er2 first from -∏/4 to ∏/4, the trigonometric terms drop out, leaving us with 4r2er2 to be integrated. But since this function is Schwartz, we know the integral is finite.

9. Jul 4, 2012

### LCKurtz

You could do it with tall skinny triangles. By "smooth it out" I mean do the same thing with infinitely smooth functions.
My functions are all 0 at the origin. They get taller and skinnier for large values [n,n+1].

10. Jul 6, 2012

### Ray Vickson

That function is *very important*: it has no max or min in the *open* set $R=\{(x,y) \in \mathbb{R}^2\,: f(x,y) > 0 \}.$ For $f(x,y) = (x^2-y^2)\exp(x^2+y^2),$ we have $R = \{|x| > |y|\},$ which is a union of two open cones with vertices at (0,0). The line y=0 lies in R. For y = 0 the function is $x^2 \exp(x^2),$ which has no maximum in $\mathbb{R}.$ For any (x,y) in R we may write $y = kx,$ with $0 < k < 1,$ giving $f = (1-k^2) x^2 \exp((1+k^2)x^2).$ This is minimized at x = 0 for any $|k| < 1,$ but that value is not in the set R. (However, if we make R closed, by asking for f ≥ 0 on R, then the minimum would be attained.)

RGV