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.

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- Feb 23rd 2010, 07:17 PMamm345Proof 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, 07:30 PMDrexel28
What have you tried? Here is a weird approach.

Suppose that $\displaystyle f\ne 0$ then there exists some $\displaystyle x\in[a,b]$ such $\displaystyle f(x)\in(0,\infty)$ and since this set is open there exists some open ball $\displaystyle B_{\varepsilon}(f(x))\subseteq(0,\infty)$ and by continuity there exists some $\displaystyle B_{\delta}(x)$ such that $\displaystyle f\left(B_{\delta}(x)\right)\subseteq B_{\varepsilon}(f(x))\subseteq(0,\infty)$. In other words, there exists some open ball around the point where $\displaystyle f(x)\ne0$ which is entirely positive. Take the concentric closed ball $\displaystyle B_{\frac{\delta}{2}}$. This is a closed and bounded interval and thus, by the Heine-Borel theorem, compact. Thus, since $\displaystyle f:B_{\frac{\delta}{2}}[x]\mapsto\mathbb{R}$ is continuous we know that $\displaystyle f$ assumes a minimum on $\displaystyle B_{\frac{\delta}{2}}[x]$. In particular, $\displaystyle \inf_{x\in B_{\frac{\delta}{2}}[x]}f(x)=\varepsilon>0$. Thus, $\displaystyle \int_a^b\text{ }f=\int_a^{x-\frac{\delta}{2}}\text{ }f+\int_{x-\frac{\delta}{2}}^{x+\frac{\delta}{2}}\text{ }f+\int_{x+\frac{\delta}{2}}^b \text{ }f\geqslant 0+\frac{\varepsilon \delta}{2}+0>0$. This of course is a contradiction. - Feb 23rd 2010, 07:33 PMamm345
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, 07:39 PMDrexel28
- Feb 23rd 2010, 07:40 PMamm345
Oh sorry, the Heine-Borel theorem.

- Feb 23rd 2010, 07:42 PMDrexel28
- Feb 23rd 2010, 07:44 PMamm345
I have no idea.. I couldn't think of anything else it is kinda ridiculous!

- Feb 23rd 2010, 07:48 PMDrexel28
- Feb 23rd 2010, 07:57 PMamm345
We started integration last week and we've mostly just covered the riemann integral, basic principles, the fundamental theorem and continuity.

- Feb 23rd 2010, 08:03 PMDrexel28
- Feb 23rd 2010, 11:48 PMMoo
Hello,

Why can't we just say that (without HB theorem) :

if there exists $\displaystyle m\in[a,b]$ such that $\displaystyle f(m)>0$, then by definition of the continuity, $\displaystyle \forall \epsilon>0, \exists \delta>0,\forall x\in[a,b], (|x-m|<\delta)\Rightarrow (|f(x)-f(m)|<\epsilon)$

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

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

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

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

which is a contradiction.

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