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I have been asked to show that if (where $B_r(a)=B_r=\{(x,y)|x^2+y^2<r^2\}$ - ie open ball at a) to prove (that if f is continuous):

$$\iint_{B_r}f(x,y)dxdy=0$$ for all $r$

that $f(x,y)=0$ always.

Okay if $f(x,y)=0$ then the result follows trivially, so the iff thing doesn't matter.

But I don't agree with the next part.

$\iint_{B_{r_1}-B_{r_2}}f(x,y)dxdy=\iint_{B_{r_1}}f(x,y)dxdy-\iint_{B_{r_2}}f(x,y)dxdy$ where $0<r_1<r_2$

By the continuity of $f$ the integrals are continuous (bounded interval, result is bounded and cont)

Now think about it, all we know is that over a disk shape that the integral is 0, now we know that over a washer shape it is 0, all well and good.

Now imagine a function that goes up and down radially, like a cupcake wrapper sort of shape, crinckled. If there's the same number of "ups" as "downs" the integral around will be zero, but the cup cake wrapper is not flat!

So have I missed something or is the question wrong?

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    $\begingroup$ isn't it for all $a$ also ? Because if $a=0$ is fixed, I think you're right $\endgroup$ – yago Apr 25 '14 at 9:47
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    $\begingroup$ Are you sure the balls to be considered are all centered at $0$? $\endgroup$ – Henning Makholm Apr 25 '14 at 9:48
  • $\begingroup$ @YannHamdaoui that's a really good point, if a isn't fixed how would the proof go? (I stumbled on this contradiction trying to find a way) $\endgroup$ – Alec Teal Apr 25 '14 at 9:49
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    $\begingroup$ If the balls are all $0$-centered, then a simple counterexample would be $f(x,y)=x$. $\endgroup$ – Henning Makholm Apr 25 '14 at 9:49
  • $\begingroup$ The balls are not all centred, not sure what to do now. $\endgroup$ – Alec Teal Apr 25 '14 at 9:50
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If the balls are not centered (see comments), then :

if $f$ is not null, there exists $a$ such that $f(a) \neq 0$. Without lost of generality, we will assume $f(a) > 0$. By continuity, you can find $\epsilon > 0$ such that $\forall x \in B_a(\epsilon), f(x) > 0$. Then $\int_{B_a{\epsilon}} f > 0$. So if the integral is null for all ($a,\epsilon$) then $f$ must be null.

If the ball must be centered on a fixed $a$, then you're right, the statement is false.

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  • $\begingroup$ Of course! The non vanishing lemma! I even used balls to prove that (they just were not very round :P) thanks. $\endgroup$ – Alec Teal Apr 25 '14 at 9:55
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I think there must be a lost assumption that your integral is to be $0$ for any ball centered anywhere, not just for ones centered at $0$.

With the assumption you have, you can at least prove that $f(0)=0$. Namely, if $f(0)=c>0$, then because $f$ is continuous there's a ball $B_r(0)$ inside which $f$ is always at least $c/2$. And then the integral over that ball is clearly $\ge\frac{cr^2\pi}{2}$.

If all balls are considered, this argument can be repeated for other centers.

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