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Assume f:[a,b] $\Rightarrow \mathbb{R}$ is continuous and f(x) >= 0 for all x $\in$[a,b]. Prove that if $\int_a^b f dx$ = 0, then f(x) = 0 for all x $\in$ [a,b].

My attempt at a proof a little obvious, but at the same time, pretty wordy. I think it can be done more easily/briefly; please recommend ways I could do that.

"Theorem 5.10" which I employ here states that for a < c

Suppose f:[a,b] $\Rightarrow \mathbb{R}$ is continuous and f(x) >= 0 for all x $\in$[a,b]. By theorem 5.10, interval [a,b] can be cut into some large finite number of subintervals, and the sum of their integrals equal the integral from a to b. No integral of any subinterval of [a,b] can be less than 0, by definition. It follows that none can be greater than 0, since that would result in a positive value for the integral of [a,b]. Therefore, f(x) = 0 for all [a,b].

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  • $\begingroup$ I would start by saying suppose $f$ is positive somewhere, say equal to $p\gt 0$ at $x=c$. Then in some interval (perhaps a half-interval if $c$ is an end-point) of width say $\delta$, we have $f\gt p/2$. So any upper Riemann sum is $\gt (p/2)\delta$. $\endgroup$ – André Nicolas Oct 29 '13 at 5:39
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enter image description hereHint Suppose $f(x)\ne 0\forall x$ so $\exists c$ such that $f(c)>0$ Then by sign preserving property of continuous function there exist $\delta>0$ such that $f(x)>0\forall x\in (c-\delta,c+\delta)$

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  • $\begingroup$ can you explain what you mean by "sign preserving property" please? $\endgroup$ – user102447 Oct 29 '13 at 5:43
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In my humble opinion there is nothing wrong with a wordy proof! I guess if you wanted to write more formally, you would need to remove words such as 'large' -- instead explicitly state what you mean.

That is, you could say for example given the interval $[a,b]$, there exists a finite sequence $(a_{i},b_{i})$ such that $[a,b]=\bigcup\limits_{i=1}^{k}[a_{i},b_{i}]$ with $a_1=a$ and $b_k=b$ and some finite $k$.

If you wanted to be more precise, you could for example state such a partition -- that is explicitly give what $a_i$ and $b_i$ are. This is a matter of taste, and also how important these values would be in the proof. In this case, it really would be overkill since the nature of the partition does not play an important part.

I guess another minor point is that when writing a proof, flow is important. So at times it might be better to explicitly state if you plan on using a proof by contradiction at the start, rather than say it in the middle of the proof.

But for me, what you have said is a nice intuitive expression of mathematics!

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