0
$\begingroup$

Here's what I'm doing:

Prove that if $f: [a,b] \to [0,+\infty)$ is a continuous nonnegative function with $\int_{a}^{b} f(x) \ dx = 0$, then $f(x) = 0$ for all $x \in [a,b]$.


Proof Attempt:

Suppose that $f(x) \neq 0$ for some $x \in [a,b]$. Let $c$ be one of those points in $[a,b]$ where $f(c) \neq 0$. Let $\epsilon > 0$ be given. Since $f$ is Riemann Integrable, there exists a $\delta > 0$ such that for all partitions $P$ with partition norm $|P| < \delta$ and all associated evaluation sets $T$:

$$|R(f,P,T)-0| = |R(f,P,T)| < \epsilon$$

where $R(f,P,T)$ is shorthand for the Riemann Sum taken over some partition and evaluation set. Now, we know that $R(f,P,T) \geq 0$ and, in fact, if $c \in T$, then $R(f,P,T) > 0$.

So, then, pick $T$ above so that $c \in T$ and let $\epsilon = R(f,P,T)$. Then, we get:

$$R(f,P,T) < R(f,P,T)$$

which is impossible. Hence, $f(x) = 0$ for all $x \in [a,b]$. $\Box$

Does the proof above work? If it doesn't, then why? How can I fix it?

$\endgroup$
1
$\begingroup$

An usual and simple way would be to say, like you did, if there exists $c\in [a,b]$ such that $f(c)>0$ than by continuity, given $\epsilon =f(c)/2$, there exists $\delta >0$ for which for every $x\in (c-\delta,c+\delta)$, $f(x)\ge \frac{f(x)}{2}$. So we have this: $$\int_a^bf(x)dx \ge \int_{c-\delta}^{c+\delta}f(x)dx\ge \int_{c-\delta}^{c+\delta}\frac{f(c)}{2}dx=\frac{f(c)}{2}2\delta>0\mbox,$$ which goes against our hipoyhesys

$\endgroup$
3
  • $\begingroup$ Ah so my argument wasn't correct? Or was it just too complicated? $\endgroup$
    – Abhi
    Nov 28 '20 at 12:08
  • $\begingroup$ For what i followed it made sense to me, but you made it way to complicated for me $\endgroup$
    – Measure me
    Nov 28 '20 at 12:09
  • $\begingroup$ Ah okay I see. Your proof sounds a lot more clever than mine, thank you very much! $\endgroup$
    – Abhi
    Nov 28 '20 at 12:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.