I've known few days ago the known as Steffensen's inequality, see the article *Steffensen's inequality* from Wolfram MathWorld and the cited bibliography. It seems that there are applications (I don't know what are) in research of this inequality or generalizations. I wondered if one can to elaborate some exercise at undergradute level concerning this didactic unit, and involving some major theorem of convergence in real analysis for special functions or sequences of functions.



> 
> **Question.** Can you show an example in the contex of real analysis as illustration of Steffensen's inequality (thus provide your result and what were the functions or sequences of functions that you've used; also if you want what convergence theorems, or other theorems from real analysis at undergraduted level, you need to combine in the proof of yourself inequality)? **Many thanks.**
> 
> 
> 


Now I add my example of such exercise that I evoke.


**Example.** I believe that using the dominated convergence theorem, particuar values of Dirichlet series and the evaluation of $\int x^a\log\left(\frac{1}{x}-1\right)dx$, one can to state the inequality $$\frac{(\zeta(3)-1)\log\left(\zeta(3)-1\right)}{\zeta(3)}-\log\zeta(3)\leq -\sum\_{m=1}^\infty\frac{\mu(m)}{m^3}H\_{m-1}$$
$$\qquad\qquad \qquad\qquad \qquad\qquad \qquad\qquad \leq \log\zeta(3)+\left(\frac{1}{\zeta(3)}-1\right)\log (\zeta(3)-1),$$
where $\mu(m)$ denotes the Möbius function and $H\_a$ the harmonic numbers. My functions were thus $g(x)=\sum\_{m=1}^\infty\frac{\mu(m)}{m^2}x^{m-1}$ defined as $g:[0,1]\to[0,1]$, and taking $f:[0,1]\to\mathbb{R}$ as $f(x)=\log\left(\frac{1}{x}-1\right)$ the inverse function from one of the answers (see the comment below the answer that I evoke) of the question 200180 from MSE (look at if you want [1]).


**Remarks.** I know the asymptotic of $H\_n$, thus one can to calculate a better approximation of previous one. Thus I feel that my example isn't the best example that one can to wait from this nice inequality. I add that my example is in the context of real analysis because the Möbius function is also an important function of real analysis.


References:
-----------


[1] [*Is there a bijective map from* $(0,1)$ *to* $\mathbb{R}$?](https://math.stackexchange.com/questions/200180/is-there-a-bijective-map-from-0-1-to-mathbbr) , Mathematics Stack Exchange (Sep 21, 2012).



Examples of Steffensen's inequality at undergraduated level studies
As pointed out by [Mitrinovich](https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=8&cad=rja&uact=8&ved=2ahUKEwiE45zCrsHkAhVLXKwKHYyPDVcQFjAHegQIABAB&url=https%3A%2F%2Fwww.jstor.org%2Fstable%2Fpdf%2F43667360.pdf&usg=AOvVaw0-KYTTRVu5OtsosK7su5Z2), Steffensen himself used his inequality to derive, in particular, the following generalized version of Jensen's inequality with possibly negative weights: 
$$f\Big(\frac{\sum\_1^n c\_kx\_k}{\sum\_1^n c\_k}\Big)\le\frac{\sum\_1^n c\_kf(x\_k)}{\sum\_1^n c\_k}, 
$$
where $f$ is a convex function, $x\_1\le\dots\le x\_n$, $0\le\sum\_m^n c\_k\le\sum\_1^n c\_k$ for $m=1,\dots,n$, and $\sum\_1^n c\_k>0$. 


Steffensen also used his inequality to derive lower and upper bounds on the cosine and sine transforms of functions of a certain class; see again [Mitrinovich, top of page 4](https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=8&cad=rja&uact=8&ved=2ahUKEwiE45zCrsHkAhVLXKwKHYyPDVcQFjAHegQIABAB&url=https%3A%2F%2Fwww.jstor.org%2Fstable%2Fpdf%2F43667360.pdf&usg=AOvVaw0-KYTTRVu5OtsosK7su5Z2). 




---


By the way, here is a very short and no-ingenuity-required proof of Steffensen's inequality 
$$\int\_{b-h}^b f(t)\,dt\le\int\_a^b f(t)g(t)\,dt\le\int\_a^{a+h} f(t)\,dt, \tag{1}
$$
where $a<b$, $f$ is nonincreasing, $0\le g\le1$, and $h:=h(b):=\int\_a^b g(t)\,dt$: 


By approximation, without loss of generality $f$ and $g$ are continuous. Let 
$$d(u):=\int\_a^u f(t)g(t)\,dt-\int\_a^{a+h(u)} f(t)\,dt. 
$$ 
Then $d(a)=0$ and 
$$d'(u)=f(u)g(u)-f(a+h(u))h'(u)
=[f(u)-f(a+h(u))]g(u)\le0
$$ 
for $u\in[a,b]$, because (i) $g\le1$ and hence $a+h(u)\le u$; (ii) $f$ is nonincreasing, and (iii) $0\le g$. So, $d(b)\le0$; that is, we have the second inequality in (1). The first inequality can be proved quite similarly or, as pointed out in [Mitrinovich, page 2](https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=8&cad=rja&uact=8&ved=2ahUKEwiE45zCrsHkAhVLXKwKHYyPDVcQFjAHegQIABAB&url=https%3A%2F%2Fwww.jstor.org%2Fstable%2Fpdf%2F43667360.pdf&usg=AOvVaw0-KYTTRVu5OtsosK7su5Z2), it can be easily deduced from the second inequality in (1). 




---


Here is another short proof, requiring (almost) no approximation, but only the observation that each of the three terms of Steffensen's inequalities is linear in $f$: Without loss of generality, $a=0$ and the nonincreasing function $f$ is right-continuous, with $f(b)=0$. Then $f=-\int\_{(0,b]}df(u)f\_u$, where $f\_u:=1\_{[0,u)}$, so that $f$ is a mixture of the "basic" nonincreasing functions $f\_u$ with nonnegative "coefficients". So, it is enough to check Steffensen's inequalities with $f\_u$ in place of $f$, where $u\in(0,b]$. Then (say) the second one of the two inequalities becomes $\int\_0^u g\le\min(h,u)$ for $u\in(0,b]$, where $h:=\int\_0^b g$, as before. Finally, the latter inequality holds because (i) $\int\_0^u g\le h$, since $g\ge0$ and $u\in(0,b]$ and (ii) $\int\_0^u g\le u$, since $g\le1$ and $u>0$. 


