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I am doing a task right now as an excercice for myself.

Let $f,g:[a,b] \to \mathbb{R}$ continuous, $g(x)>0 \forall x \in [a,b]$. Then there is $c \in (a,b)$ such that $\int_{a}^{b} f(x)g(x)dx = f(c)\int_{a}^{b} g(x)$.

There are many good solutions that can be find here. This one for example: Existence of $c$ such that $\int_{a}^{b} f(x)g(x)dx = f(c)\int_{a}^{b} g(x)$

My problem is, that I would like to know, what happens, if g(x) is not always positive but has some negative values as well. Here is the solution from the linked post for $g(x)\geq0$ $$m\le f(x)\le M$$

thus

$$mg(x)\le g(x)f(x)\le Mg(x)$$ since $g(x)>0.$

by integrating

$$m\int_a^bg\le \int_a^bfg\le M\int_a^bg$$

but $\int_a^bg>0$, by division

$$m\le \frac{\int_a^bfg}{\int_a^bg}\le M$$

If we have g<0 for all x, we would get same result right? The relation signs would change twice. When mulitplicating with g, and when dividing by $\int_a^b g$.

But what can we say if only some of g's values are negative?

Is there anyone who could help? I would be very grateful.

Analysis_Mark
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  • If for example the integral of $g$ is zero then there is in general no $c$ as the integral of $fg$ is generally not zero – Conrad Apr 16 '22 at 14:23
  • @Conrad what do you mean? This should also work with g=0, we had to prove this for our task, with $g\geq 0$ – Analysis_Mark Apr 18 '22 at 12:47
  • Yes but if $g\ge 0$ and its integral is zero, then $g=0$ ae and the result is trivial; however if $g$ changes sign, its integral can be zero without $g$ being zero and then the result is false in general; you should look up the general second mean theorem for integrals to see the most general condition under which it holds and the precise statement – Conrad Apr 18 '22 at 12:57
  • What reason can I name that the result is false in general....That there is a possibility that the integral is zero and that we aren't allowed to divide by 0? Would that be enough? Also is my assumption right that the result is also true if $g\leq 0$? – Analysis_Mark Apr 19 '22 at 09:24
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    Just give a counterxample (take $g(x)=x, f(x)=x$ on $[-1,1]$); if $g \le 0$ then $-g\ge 0$ and the result follows by linearity – Conrad Apr 19 '22 at 12:42

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