# Proof involving integrals equal to 0.

• Feb 23rd 2010, 08:17 PM
amm345
Proof involving integrals equal to 0.
Let
f be a bounded integrable non-negative function continous on [a, b].
Prove:
integral from a to b f(x) dx = 0 implies f(x) = 0 for every x

I believe I understand why it is true, I'm just not sure how to write the formal proof.

• Feb 23rd 2010, 08:30 PM
Drexel28
Quote:

Originally Posted by amm345
Let
f be a bounded integrable non-negative function continous on [a, b].
Prove:
integral from a to b f(x) dx = 0 implies f(x) = 0 for every x

I believe I understand why it is true, I'm just not sure how to write the formal proof.

What have you tried? Here is a weird approach.

Suppose that $f\ne 0$ then there exists some $x\in[a,b]$ such $f(x)\in(0,\infty)$ and since this set is open there exists some open ball $B_{\varepsilon}(f(x))\subseteq(0,\infty)$ and by continuity there exists some $B_{\delta}(x)$ such that $f\left(B_{\delta}(x)\right)\subseteq B_{\varepsilon}(f(x))\subseteq(0,\infty)$. In other words, there exists some open ball around the point where $f(x)\ne0$ which is entirely positive. Take the concentric closed ball $B_{\frac{\delta}{2}}$. This is a closed and bounded interval and thus, by the Heine-Borel theorem, compact. Thus, since $f:B_{\frac{\delta}{2}}[x]\mapsto\mathbb{R}$ is continuous we know that $f$ assumes a minimum on $B_{\frac{\delta}{2}}[x]$. In particular, $\inf_{x\in B_{\frac{\delta}{2}}[x]}f(x)=\varepsilon>0$. Thus, $\int_a^b\text{ }f=\int_a^{x-\frac{\delta}{2}}\text{ }f+\int_{x-\frac{\delta}{2}}^{x+\frac{\delta}{2}}\text{ }f+\int_{x+\frac{\delta}{2}}^b \text{ }f\geqslant 0+\frac{\varepsilon \delta}{2}+0>0$. This of course is a contradiction.
• Feb 23rd 2010, 08:33 PM
amm345
That was sort of my original approach, but when I checked with my instructer he said we could not use that theorem and I can't think of anything else.
• Feb 23rd 2010, 08:39 PM
Drexel28
Quote:

Originally Posted by amm345
That was sort of my original approach, but when I checked with my instructer he said we could not use that theorem and I can't think of anything else.

Which theorem is that?
• Feb 23rd 2010, 08:40 PM
amm345
Oh sorry, the Heine-Borel theorem.
• Feb 23rd 2010, 08:42 PM
Drexel28
Quote:

Originally Posted by amm345
Oh sorry, the Heine-Borel theorem.

Are you serious? What's the point of having tools if you can't use them! What if I proved it first? (Wink)
• Feb 23rd 2010, 08:44 PM
amm345
I have no idea.. I couldn't think of anything else it is kinda ridiculous!
• Feb 23rd 2010, 08:48 PM
Drexel28
Quote:

Originally Posted by amm345
I have no idea.. I couldn't think of anything else it is kinda ridiculous!

Ok. I can think of one or two other ways....what are you currently doing in class?
• Feb 23rd 2010, 08:57 PM
amm345
We started integration last week and we've mostly just covered the riemann integral, basic principles, the fundamental theorem and continuity.
• Feb 23rd 2010, 09:03 PM
Drexel28
Quote:

Originally Posted by amm345
We started integration last week and we've mostly just covered the riemann integral, basic principles, the fundamental theorem and continuity.

Hmm...Ok. I see an angle. Make a modified argument of mine and use the FTC.
• Feb 24th 2010, 12:48 AM
Moo
Hello,

Why can't we just say that (without HB theorem) :
if there exists $m\in[a,b]$ such that $f(m)>0$, then by definition of the continuity, $\forall \epsilon>0, \exists \delta>0,\forall x\in[a,b], (|x-m|<\delta)\Rightarrow (|f(x)-f(m)|<\epsilon)$

In particular, if we let $\epsilon=\frac{f(m)}{2}$, $|f(x)-f(m)|<\epsilon \Leftrightarrow -\tfrac{f(m)}{2}

So there exists $\delta>0$ such that $\forall x\in]m-\delta,m+\delta[$, $f(x)>\tfrac{f(m)}{2}>0$

Hence $\int_a^b f(x) ~dx=\int_a^{m-\delta} \underbrace{f(x)}_{\geqslant 0} ~dx+\int_{m+\delta}^b \underbrace{f(x)}_{\geqslant 0} ~dx+\int_{m-\delta}^{m+\delta} \underbrace{f(x)}_{> \frac{f(m)}{2}} ~dx$

So $\int_a^b f(x) ~dx>0+0+2\delta \cdot\tfrac{f(m)}{2}=\delta f(m)>0$

So m can't exist, and hence $f(x)=0 ~,~ \forall x\in[a,b]$