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$f\geq 0$ and $\int_a^b f=0$ implies $f=0$ everywhere on $[a,b]$
Is this a correct solution?
Thanks for your help
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Miguel Mora Luna
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3It is correct, and the usual way to show this problem. – Calvin Lin Jan 09 '13 at 03:40
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4Looks correct and pretty nice. – DonAntonio Jan 09 '13 at 03:41
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1Was going to ask for the source, but then I realized this was from baby Rudin. – Gyu Eun Lee Jan 09 '13 at 03:45
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1A nitpick, $\delta > 0$ and $\delta < \min(x_0-a, b-x_0)$ to be completely sure. – Jan 09 '13 at 03:46
1 Answers
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Here is another approach (I prefer 'constructive' proofs, albeit they tend to be longer, as is the case here):
Since $\int f = 0$, and $f$ is continuous, we have $\int f = \sup_{\pi \in {\cal P}} L(f,\pi) = 0$, where ${\cal P}$ is the set of partitions of $[a,b]$, and $L(f,\pi)$ is the Riemann sum. Since $[a,b]$ is compact, $f$ is uniformly continuous.
Let $\epsilon>0$. Choose $\delta>0$ so that if $|x-y| < \delta$, then $|f(x)-f(y) | < \epsilon$. Choose a partition $\pi$ with $\text{mesh } \pi < \delta$. Since $f \geq 0$, we have $L(f,\pi) = 0$, and hence on every subinterval $[x,y]$ of $\pi$, we have some $\xi$ such that $f(\xi) = 0$. Uniform continuity shows that $f(t) < \epsilon$ for all $t \in [x,y]$. Hence $f(t) < \epsilon$ for all $t \in [a,b]$. Since $\epsilon>0$ was arbitrary, we are finished.
copper.hat
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Great proof. One (little) thing I can comment on is that the subintervals of $\pi$ don't have to be of the form $[x,y]$ but can also have the form $(x,y]$ or $(x,y)$ etc. But this is no problem since the length of the intervals do not change by removing or adding single points. – Jan 09 '13 at 18:28
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@André: A partition is of the form $\pi=(x_0=a,x_1,...,x_n=b)$ and the intervals formed from the partition are $[x_i, x_{i+1}]$, all of which are closed. – copper.hat Dec 17 '14 at 16:24