Proof: integral of continuous function is 0 so function is 0

1. Feb 10, 2016

cmkluza

I've just encountered this somewhere and I need some sort of formal proof for why a continuous function $f(x)$ can equal zero because its integral is zero. Are there any out there? I've seen similar forum posts on places like Stack Exchange and one here, but I can't exactly follow the logic they always use in their proof. Can anyone point me to a full/formal proof and/or explain how it is that others arrive at their proof?

Thanks for any help!

2. Feb 10, 2016

Staff: Mentor

What does the fundamental theorem of calculus tells you?

3. Feb 10, 2016

cmkluza

Fundamental theorem of calculus is $$\int^B_A f(x) dx = F(B) - F(A)$$ right? So, $\int^B_A f(x) dx = 0 = F(B) - F(A)$? This suggests that F(B) = F(A), but doesn't tell me anything about $f(x) = 0$, right? What am I missing?

4. Feb 10, 2016

micromass

Staff Emeritus
Well, it's not true. We have
$$\int_0^{2\pi} \sin(x) = 0$$
but the sine is not the zero function.

5. Feb 10, 2016

Staff: Mentor

Since you haven't made any assumptions on the domain or initial values I assumed it would be valid to all of them.

6. Feb 10, 2016

cmkluza

In that case, could you help me explain this in a specific situation? I'm looking at this resource for my research, specifically the end of page 6 and start of page 7. It shows equation 11, which I've been able to follow up to, and then it has a paragraph explaining the logic it uses to produce equation 13, which is just equation 11 without the integral. Do you have any idea about why this is possible? If this isn't correct in general, why would it be correct in this situation?

Thanks!

7. Feb 10, 2016

micromass

Staff Emeritus
OK, so the statement is: Let $f$ be continuous on $[a,b]$ and such that for all $A,B\in [a,b]$ holds that $\int_A^B f(x)dx=0$, then $f=0$. This is true indeed. And this seems to be how @fresh_42 interpreted it. His proof using the fundamental theorem should work.

8. Feb 10, 2016

cmkluza

I'm sorry, I don't think I'm understanding this at a fundamental level yet. Could you tell me why that statement is true on a more conceptual level? That might allow me to understand the proof using the fundamental theorem.

9. Feb 10, 2016

pasmith

Let $f$ be continuous on $[a,b]$ such that $\int_A^B f(x)\,dx = 0$ for every $a \leq A < B \leq b$.

We show, by contradiction, that under these conditions it cannot be the case that $f(c) \neq 0$ for any $c \in (a,b)$:

Suppose $c \in (a,b)$ is such that $f(c) > 0$. Then, by continuity of $f$ at $c$, there exists a $\delta > 0$ such that if $A=\max\{a, c - \delta\} < x < \min\{b, c + \delta\}=B$ then $f(x) > 0$. But then by a basic property of integrals we have
$$\int_A^B f(x)\,dx > \int_A^B 0\,dx = 0,$$ which is a contradiction. Thus such a $c$ cannot exist. A similar argument shows that there cannot exist any $c \in (a,b)$ such that $f(c) < 0$.

Hence $f$ is constantly zero on $(a,b)$ and by continuity $f(a) = f(b) = 0$ also.

This is a more "conceptual" proof than one using the fundamental theorem, and one can rephrase it thus:

"$\int_a^b \frac{\partial \rho}{\partial t} + \frac{\partial q}{\partial x}\,dx = 0$ says only that, on average, cars are neither created nor destroyed on the stretch of highway between $a$ and $b$. But suppose there exists a $c \in [a,b]$ where $\frac{\partial \rho}{\partial t} + \frac{\partial q}{\partial x} \neq 0$. Then, if $\frac{\partial \rho}{\partial t} + \frac{\partial q}{\partial x}$ is continuous, there is around $c$ a stretch of highway where cars are being spontaneously created if $\frac{\partial \rho}{\partial t} + \frac{\partial q}{\partial x} > 0$ or destroyed if $\frac{\partial \rho}{\partial t} + \frac{\partial q}{\partial x} < 0$. Both of these we reject as unphysical."

Last edited: Feb 10, 2016
10. Feb 10, 2016

PeroK

Conceptually, if f is not zero at a point and you integrate near that point the integral cannot be zero. That's the basic idea that needs to be formalised.

11. Feb 10, 2016

Svein

There is a theorem that states: if $\int_{L}\left\lvert f \right\rvert=0$ then $f=0$ a. e. on L.

Explanation: a.e. (almost everywhere) means "except on a subset of measure zero".

12. Feb 11, 2016

Edgardo

Here is a proof mentioned on math.stackexchange using the "sign-preserving property" of continuous functions:
http://math.stackexchange.com/a/543800

And two more links for the proof of the sign-preserving property:
https://math.la.asu.edu/~dajones/class/371/ch4.pdf
http://zimmer.csufresno.edu/~mnogin/math171spring05/test2-sol.pdf

Intuitively it says the following. Suppose you wanted to construct a continuous function f that is positive at some point a, i.e. f(a) > 0. So you draw a dot somewhere above the x-axis at x=a for its graph. Now, if you draw the graph a little to the left of that dot and to the right of it, then that graph segment lies above the x-axis.