MHB Zero Integral Implies Zero Function

[Julio1]

If $h: \Omega\to \mathbb{R}$ is continuous and $\displaystyle \int\limits_{B(x_0,r)}h(x)\, dx=0$, for all $B(x_0,r)\subseteq \Omega$, then $h(x)=0, \, \forall x\in \Omega.$Hello MHB, is an problem that I don't solve :(. Help me please or any hints :). Thanks.

 

[Ackbach]

This looks to me a lot like the Fundamental Theorem of the Calculus (FTC). You're given information about the integral of a continuous function, and asked to prove something about the function. Could you try to get the "if" part of the FTC satisfied, maybe?

 

[Euge]

Hi Julio,

Prove the contrapositive, i.e., suppose $h$ is not identically zero, and prove that there is an $x_0\in \Omega$ and $r > 0$ such that $\int_{B(x_0,r)} h(x)\, dx$ is nonzero.

If $h$ is not identically zero, there is an $x_0$ such that $h(x_0) \neq 0$. Thus $|h(x_0)|/2 > 0$; by continuity of $h$ at $x_0$, there exists $r > 0$ such that $|h(x) - h(x_0)| < |h(x_0)|/2$ for all $x \in B(x_0,r)$. Thus

$$h(x_0) - \frac{|h(x_0)|}{2} < h(x) < h(x_0) + \frac{|h(x_0)|}{2}$$

for all $x\in B(x_0,r)$. If $h(x_0) > 0$, then the latter inequalities give $h(x_0)/2 < h(x) < 3h(x_0)/2$ for $x\in B(x_0,r)$. Hence

$$0 < \int_{B(x_0,r)} \frac{h(x_0)}{2}\, dx < \int_{B(x_0,r)} h(x)\, dx.$$

If $h(x_0) < 0$, you can show

$$0 > \int_{B(x_0,r)} \frac{h(x_0)}{2}\, dx > \int_{B(x_0,r)} h(x)\, dx.$$