I would like to prove that the function $$ F(x) = \frac{1}{a^2}\int_0^a f(s) s \mathrm{d}s - \frac{1}{x^2}\int_0^x f(s) s \mathrm{d}s $$ is non-positive for $x \in [0, a] $, given that the function $f$ is continuous, positive and non-increasing over the same interval.

I think I managed as I show afterwards: I still would like to see how it could be done more concisely/elegantly/effectively.

My first approach: firstly I compute the derivative of $F$, $$ F ^\prime (x) = \Big[- \frac{1}{x^2} f(x) x + \frac{2}{x^3} \int_0^x f(s) s \mathrm{d}s \Big] $$

As $f(x)$ is non-increasing, the following inequality stands

$$ \frac{2}{x} \int_0^xf(s) s \mathrm{d}s \geq 2f(x) x$$ and then $$ F ^\prime(x) = [- \frac{1}{x^2} f(x) x + \frac{2}{x^3} \int_0^x f(s) s \mathrm{d}s ] \geq \frac{1}{x^2} [-f(x) x +2f(x) x] = \frac{1}{x^2} f(x) > 0$$

I compute then $F(0)$. As $$ \lim_{x\to 0} \frac{1}{x^2} \int_0^x f(s) s \mathrm{d}s = \frac{1}{2}f(0)$$ it follows that $$F(0) = \frac{1}{a^2}\int_0^a f(s) s \mathrm{d}s - \frac{1}{2}f(0)$$ which is non-positive (it is actually nihil when $f$ is constant) As $F(0) \leq 0$, $F(a) = 0$ and $F$ is non-decreasing, I conclude it must be non-positive, a desired.

The second approach I thought about is consists in arguing that the expression $$\frac{x^2} {a^2} \leq \frac { \int_0^x f(s) s \mathrm{d}s } {\int_0^a f(s) s \mathrm{d}s }$$ stands. One notices that the equality stands if $f(x)$ is constant. As $f(x)$ is non increasing, the second term must be larger than in the case of costant $f(x)$, which proves the claim.

As said, I would like to know how the same claim can be proved in more efficient/elegant/coincise fashion.


This inequality is a direct result of Cauchy's mean value theorem.

Note that $F(x) \leq 0$ can be equivalently written as $$\frac{\int_{0}^x f(s)s \,ds}{\int_0^x s\,ds} \geq \frac{\int_{0}^a f(s)s \,ds}{\int_0^a s\,ds}. $$ From Cauchy's mean value theorem, we can show that $\exists \,\xi \in (0, x), \eta \in (x, a)$ such that $$ f(\xi) = \frac{\int_{0}^x f(s)s \,ds}{\int_0^x s\,ds}, \,f(\eta) = \frac{\int_{x}^a f(s)s \,ds}{\int_x^a s\,ds}. $$ Since $f(x)$ is non-increasing, we have $f(\xi) \geq f(\eta)$, i.e., $$ \frac{\int_{0}^x f(s)s \,ds}{\int_0^x s\,ds} \geq \frac{\int_{x}^a f(s)s \,ds}{\int_x^a s\,ds}. $$ As a result, $$ \frac{\int_{0}^x f(s)s \,ds}{\int_0^x s\,ds} \geq \frac{\int_{0}^x f(s)s \,ds + \int_{x}^a f(s)s \,ds}{\int_0^x s\,ds + \int_x^a s\,ds} = \frac{\int_{0}^a f(s)s \,ds}{\int_0^a s\,ds}, $$ where we haved use the following fact for ratios:

Suppose $a, b, c, d > 0$, then $$\frac{a}{b} \geq \frac{c}{d} \implies \frac{a}{b} \geq \frac{a + c}{b + d} .$$

In general, when $f(x), g(x)$ are positive and continuous, and $f(x)$ is non-increasing, we have $$ \frac{\int_{0}^x f(s)g(s) \,ds}{\int_0^x g(s)\,ds} \geq \frac{\int_{0}^a f(s)g(s) \,ds}{\int_0^a g(s)\,ds}. $$

  • $\begingroup$ Thanks a lot. I thought about your approach but then I have also to say, what I did is shorter. Thank you a lot anyhow, much appreciated $\endgroup$ – An aedonist Jul 23 '18 at 8:47
  • $\begingroup$ @Anaedonist Thanks. I have fixed some typos, and please see the latest version. IMHO, although this method seems wordy, it reveals some intrinsic properties of this problem, and can be generalized :) $\endgroup$ – Xiangxiang Xu Jul 23 '18 at 13:00

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