Gromov proved that if $$ f,g:\left[ {a,b} \right] \to R $$ are integrable functions, such that the function $$ t \to \frac{{f\left( t \right)}} {{g\left( t \right)}} $$ is also integrable, and decreasing. Then the function $$ r \to \frac{{\int\limits_a^r {f\left( t \right)dt} }} {{\int\limits_a^r {g\left( t \right)dt} }} $$ is decreasing. I could not proved, and I could not find a proof )=
non-trivial result about integrals due Gromov
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2Have a look at [this](http://www.jstor.org/stable/27642062). – 2011-10-17
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0The short proof by Gromov can be found [here](http://projecteuclid.org/euclid.jdg/1214436699), on p. 11. – 2011-10-17
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0This was an unexpectedly and fun result. :) – 2011-10-17
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0For those puzzled by @Gerben 's second comment, the proof is given on page 42 (page 28 of the pdf) of Cheeger-Gromov-Taylor, *[Finite propagation speed, kernel estimates for functions of the Laplace operator, and the geometry of complete Riemannian manifolds](http://projecteuclid.org/euclid.jdg/1214436699)*, J. Differential Geom. Volume **17** (1), (1982), 15-53. [Here's a screen shot of the relevant passage.](http://i.stack.imgur.com/tm0vj.png) – 2011-10-18
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0@t.b. Thanks, I somehow managed to refer to a non-existing page. – 2011-10-18
3 Answers
Extra assumption 1: $g$ is non-negative or non-negative. (Thanks robjohn)
Extra assumption 2: $f$ and $g$ are absolute continuous (e.g. they are strictly increasing/decreasing). (Thanks Mariano Suárez-Alvarez and t.b.)
Fix $r$. Since $f/g$ is decreasing we have $$\frac{f(x)}{g(x)}\ge\frac{f(r)}{g(r)}$$ for all $a\le x\le r$. Hence (by Extra assumption 1) $$f(x)g(r)\ge f(r)g(x)$$ next we integrate this with respect to $x$ over $[a,r]$ which leads to $$g(r)\int_a^rf(x)dx\ge f(r)\int_a^rg(x)dx$$ (recall $r$ was fixed). But then (by Extra assumption 2) $$\left(\frac{\int_a^rf(x)dx}{\int_a^rg(x)dx}\right)'= \frac{f(r)\int_a^rg(x)dx-g(r)\int_a^rf(x)dx}{\left(\int_a^rg(x)dx\right)^2}\le0.$$
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0You then also need to assume that $g$ is either always non-negative or non-positive so that multiplying by $g(x)\;g(r)$ maintains the inequality. – 2011-10-17
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0@robjohn: Good point. I will add that. – 2011-10-17
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0You seem to assuming that $\int_a^rf$ has a derivative, no? – 2011-10-17
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0@Mariano Suárez-Alvarez♦ : Ha ha.. you have a point there too! I will add that :) – 2011-10-17
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0So, shouldn't we also start to look for a possible counterexample in the case that $f, g$ are not absolutely continuous? Or a full proof. – 2011-10-17
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0@Jonas Teuwen: Yes Jonas, I agree -- I am thinking (but I should go to sleep :) – 2011-10-17
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0Anyway, nice proof. Good night! – 2011-10-17
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0@Jonas Teuwen: Thanks and good night! – 2011-10-17
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0@AD: I have modified my proof due to comments by Mariano Suárez-Alvarez. I think it is free of bad assumptions. – 2011-10-17
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0@AD: You should probably correct that *e.g. they are monotone*. Monotone functions are BV but they can be singular (devil's staircase). – 2011-10-18
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0@t.b.: You are right, strictly increasing/decreasing is sufficient. I will change that, thanks. – 2011-10-18
Thanks to Mariano Suárez-Alvarez for pointing out a bad assumption I made in my previous attempt
For all $u\le v$, in $[a,b]$ we have
$$
\frac{f(u)}{g(u)}\ge\frac{f(v)}{g(v)}
$$
Assuming that $g$ is either non-negative or non-positive on all of [a,b], we get
$$
f(u)g(v)\ge f(v)g(u)
$$
Let $r\le s$. Then, integrating in $u$ from $a$ to $r$ and then in $v$ from $r$ to $s$, we get
$$
\int_a^rf(u)\mathrm{d}u\;\int_r^sg(v)\mathrm{d}v\ge\int_a^rg(u)\mathrm{d}u\;\int_r^sf(v)\mathrm{d}v
$$
Then we have
$$
\begin{align}
&\frac{\int_a^rf(t)\mathrm{d}t}{\int_a^rg(t)\mathrm{d}t}-\frac{\int_a^sf(t)\mathrm{d}t}{\int_a^sg(t)\mathrm{d}t}\\
&=\frac{\int_a^rf(t)\mathrm{d}t\;\int_a^sg(t)\mathrm{d}t-\int_a^rg(t)\mathrm{d}t\;\int_a^sf(t)\mathrm{d}t}{\int_a^rg(t)\mathrm{d}t\;\int_a^sg(t)\mathrm{d}t}\\
&=\frac{\int_a^rf(t)\mathrm{d}t\;(\int_a^rg(t)\mathrm{d}t+\int_r^sg(t)\mathrm{d}t)-\int_a^rg(t)\mathrm{d}t\;(\int_a^rf(t)\mathrm{d}t+\int_r^sf(t)\mathrm{d}t)}{\int_a^rg(t)\mathrm{d}t\;\int_a^sg(t)\mathrm{d}t}\\
&=\frac{\int_a^rf(t)\mathrm{d}t\;\int_r^sg(t)\mathrm{d}t-\int_a^rg(t)\mathrm{d}t\;\int_r^sf(t)\mathrm{d}t}{\int_a^rg(t)\mathrm{d}t\;\int_a^sg(t)\mathrm{d}t}\\
&\ge0
\end{align}
$$
Update: The requirement that $g$ stay either non-negative or non-positive is reasonable since the result is false for $f(t)=1-t$ and $g(t)=1-t^2$ on $[0,\frac{3}{2}]$. Here is the graph of $\frac{\int_0^x(1-t)\;\mathrm{d}t}{\int_0^x(1-t^2)\;\mathrm{d}t}$:
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2You are making certain assumptions on the functions which are not in the question, no? (Moreover, in (1) you are picking the same point $\xi$ for both $f$ and $g$) – 2011-10-17
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0@Mariano: yes; I stated that I was assuming that $f$ and $g$ are positive. I guess there is no reason for $f$ to be positive, but $g$ needs to be either non-negative on all of $[a,b]$ or non-positive on all of $[a,b]$. – 2011-10-17
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0@Mariano: the same $\xi$ is guaranteed by the [Extended Mean Value Theorem](http://en.wikipedia.org/wiki/Mean_value_theorem#Cauchy.27s_mean_value_theorem). I usually just call this the Mean Value Theorem, but I will amend my answer. – 2011-10-17
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0No need for MVT - as you can see in my post. – 2011-10-17
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2The function $f(x)=\left\{\begin{array}{ll}0&x\in[0,1/2];\\2,&x\in[1/2,1]\end{array}\right.$ never takes the value $1=\int_0^1f$, and its primitive function $F(x)=\int_0^xf$ is not differentiable at all points. – 2011-10-17
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0@Mariano: aha. I am assuming that $f$ and $g$ are continuous, which is not assumed in the question. I defer to AD's answer then. – 2011-10-17
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0@Mariano: I have appended the answer with a proof not using derivatives. Thanks for pointing out my previous error. – 2011-10-17
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0To reduce confusion, I will delete the first part that assumes continuity of $f$ and $g$. – 2011-10-17
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1Very nice :) ${}$ – 2011-10-18
The geometric interpretation of the result is fairly clear if you draw the picture of a particle with velocity vector $(f(t), g(t))$ that at time $t=a$ is at $(0,0) \quad$ (assume $g(t) > 0$ so that the particle moves to the right at all times). Decreasing $f(t)/g(t)$ means the path of the particle is convex, curving downward. This implies the second property if the particle goes through $0$; the slope of the velocity vector when $t>a$ is always less than the slope of the line from the particle to $0$ so that continued motion forces the latter to decrease.