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This is the statement:

Let $f \colon [a,b] \to \mathbb{R}$ be continuous, g monotonous and positive such that $g'$ is continious on [a,b]. Prove that there exists a $\xi \in [a,b]$ s.t:

$\int_{a}^b f(x)g(x) dx = g(a)\int_a^{\xi}f(x)dx + g(b) \int_{\xi}^b f(x)dx$

So far I've done this:

Let $\int_{a}^b f(x)g(x) dx = \int_{a}^{\xi} f(x)g(x) dx + \int_{\xi}^b f(x)g(x) dx $ for $\xi \in [a,b]$. From FTC:

$$\int_{a}^x g'(t) = g(x) - g(a)$$

and

$$\int_{x}^b g'(x) = g(b)-g(x)$$

So the previous equation becomes:

$$\int_{a}^{\xi} f(x)\int_{a}^{x}g'(x) dx + g(a)\int_{a}^{\xi} f(x) dx + g(b)\int_{\xi}^b f(x) dx -\int_{\xi}^b f(x) dx\int_{x}^b g'(x) dx$$

Is there a way to prove $\int_{a}^{\xi} f(x)\int_{a}^{x}g'(x) dx-\int_{\xi}^b f(x) dx\int_{x}^b g'(x) dx = 0$?

Thanks!

Eduardo V. Kuri
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  • This is true only given that $g$ is monotonous and $f$ is Riemann integrable. – RRL May 09 '22 at 17:53
  • Please see my edit: the original post had a basic error. I have provided counterexample now for when we forget the assumption of positivity – FShrike May 09 '22 at 18:07

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$\newcommand{\d}{\,\mathrm{d}}$Some of the conditions are superfluous. We only need $g$ monotone as all monotone functions are integrable ($g$ differentiable is unnecessary) and we do not need $f$ continuous, only integrable. $g$ and $f$ always positive (or always negative) is necessary so that the multiplicative inequalities hold (! my original post forgot this).

Consider the function: $$I:[a,b]\to\Bbb R,\,I(x)=g(a)\cdot\int_a^xf(t)\d t+g(b)\cdot\int_x^bf(t)\d t$$By assumption we have that $I$ is a continuous function; by the monotonicity assumption and that fact that $I(a)=g(b)\int_a^bf(x)\d x$, $I(b)=g(a)\int_a^bf(x)\d x$, either: $$I(a)\le\int_a^bf(x)g(x)\d x\le I(b)$$If $g$ is monotone nonincreasing, or the same holds but with reversed inequalities.

We conclude by the intermediate value theorem that $I$ attains the intermediate value of the integral at some $\xi\in[a,b]$. Note that similar constructions of $I$ will work to produce similar looking variants of the same problem.

FShrike
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  • Given $g(b) \leq g(x) \leq g(a)$ for $x \in [a,b]$, how do you show that $g(b)\int_a^b f(x) dx \leq \int_a^b f(x)g(x) dx \leq g(a)\int_a^b f(x) dx$? – RRL May 09 '22 at 17:52
  • @RRL Standard integral properties. It depends on how you define the integral but these inequalities are well known – FShrike May 09 '22 at 17:54
  • I know it is easy to show if $f$ is nonnegative (or nonpositive) but what if that is not the case? – RRL May 09 '22 at 17:59
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    @RRL Oh my! I have made a very basic error here, thank you for pointing it out. Post edited to emphasise the assumption $g$ always positive / always negative (as per the OP) – FShrike May 09 '22 at 18:02
  • Also where did you use continuity of $f$? This "second mean value" theorem can be proved only with $f$ Riemann integrable and the intermediate step is more involved as I show here. – RRL May 09 '22 at 18:03
  • Otherwise your proof is fine under more restrictive assumptions. – RRL May 09 '22 at 18:05
  • Under the corrections ($f$ and $g$ always of the same sign) my proof works fine: continuity of $f$ guarantees the continuity of $I(x)$, the inequalities and intermediate value theorem do the rest – FShrike May 09 '22 at 18:09
  • $F(x) = \int_a^x f(t) dt$ is continuous without continuity of $f$. – RRL May 09 '22 at 18:11
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    Absolutely continuous, even! I am quite forgetful today. You should leave your linked post as a comment to the OP as it is educational and more general than my solution @RRL – FShrike May 09 '22 at 18:14
  • Just had time to check the comment section. Thanks for the feedback! Noted – Eduardo V. Kuri May 18 '22 at 21:48