Proofs of Schweitzer's Inequality

Preliminaries

P. Schweitzer proved the following inequality in a 1914 paper:

Diaz-Metcalf Inequality

We already have two proofs on earlier pages:

  1. Sitaru-Schweitzer Inequality
  2. Applications of Schweitzer's inequality

The latter is an extract from D. A. Mitrinovic's book that was brought to my attention by N. N. Taleb. The book contains a wealth of information. Below I borrow from the book two inequalities (Diaz-Metcalf and Grüss) with proofs that are used to deliver two additional proofs of Schweitzer's inequality. The derivation of Schweitzer's inequality from Grüss' inequality is by N. N. Taleb.

Reference

D. S. Mitrinovic, Analytic Inequalities, Springer 1970

Diaz-Metcalf Inequality (1963)

If $a_k\,(\ne 0)\,$ and $b_k,\,$ $k\in\overline{1,n},\,$ are real numbers and if

(1)

$\displaystyle m\le\frac{b_k}{a_k}\le M,\,\text{for}\,k\in\overline{1,n},$

then

(2)

$\displaystyle \sum_{k=1}^{n}b_k^2+mM\sum_{k=1}^{n}a_k^2\le (m+M)\sum_{k=1}^{n}a_kb_k.$

Equality holds iff in each of the $n\,$ inequalities in (1) at least one equality sign holds, i.e., $b_k=ma_k\,$ or $b_k=Ma_k\,$ (where the equation may vary with $k).$

Proof

By virtue of (1), $\displaystyle 0\le\left(\frac{b_k}{a_k}-m\right)\left(M-\frac{b_k}{a_k}\right)\,$ is obvious. Summing up

(3)

$\displaystyle 0\le\sum_{k=1}^{n}\left(\frac{b_k}{a_k}-m\right)\left(M-\frac{b_k}{a_k}\right),$

i.e.,

$\displaystyle 0\le -\sum_{k=1}^{n}[b_k^2-(m+M)a_kb_k+mMa_k^2]$

which is (2).

In inequality (3), or equivalently, in inequality (2), equality holds iff each of the summands $(b_k-ma_k)(Ma_k-b_k)\,$ is equal to zero, which proves the statement.

Grüss' Inequality (1935)

Let $f\,$ and $g\,$ be two functions defined and integrable over $(a,b).\,$ Let

$\phi\le f(x)\le\Phi,\,\gamma\le g(x)\le \Gamma$

for all $x\in (a,b),\,$ where $\phi,\Phi,\gamma,\Gamma\,$ are fixed real constants. Then

(4)

$\displaystyle \left|\frac{1}{b-a}\int_{a}^{b}f(x)g(x)dx-\frac{1}{(b-a)^2}\int_{a}^{b}f(x)dx\int_{a}^{b}g(x)dx\right|\\ \qquad\qquad\le\frac{1}{4}(\Phi-\phi)(\Gamma-\gamma).$

Proof

First, by making the substitution $\displaystyle x=\frac{t-a}{b-a}\,$ the problem is reduced to the special case $a=0,\,b=1.\,$ Denote $F=\displaystyle \int_{0}^{1}f(x)dx,\,$ $G=\displaystyle \int_{0}^{1}g(x)dx,\,$ and consider

$\displaystyle D(f,g)=\int_{0}^{1}f(x)g(x)dx-FG.$

Then (4) reads

(5)

$\displaystyle D(f,g)\le\frac{1}{4}(\Phi-\phi)(\Gamma-\gamma).$

Note that

(6)

$D(f,f)=\displaystyle \int_0^1f^2(x)dx-\left(\int_0^1f(x)dx\right)^2\ge 0,\,$

by the Cauchy-Schwarz inequality. On the other hand,

$\displaystyle D(f,f)=(\Phi-F)(F-\phi)-\int_0^1[\Phi-f(x)][f(x)-\phi]dx,$

which implies

(7)

$\displaystyle D(f,f)\le (\Phi-F)(F-\phi).$

One can easily verify that

$\displaystyle D(f,g)\le \int_0^1(f(x)-F)(g(x)-G)dx.$

Using the Cauchy-Schwarz inequality inequality again, we get

$\displaystyle \begin{align} D(f,g)^2 &\le \int_0^1[f(x)-F]^2dx\int_0^1[g(x)-G]^2dx\\ &=D(f,f)D(g,g). \end{align}$

According to (6) and (7), we infer that

(8)

$\displaystyle D(f,g)^2\le (\Phi-F)(F-\phi)(\Gamma-G)(G-\gamma).$

Since

$4(\Phi-F)(F-\phi)\le (\Phi-\phi)^2,\\ 4(\Gamma-G)(G-\gamma)\le (\Gamma-\gamma)^2,$

we conclude that (8) implies (5).

Taking $f(x)=g(x)=sgn (2x-1),\,$ the constant $\displaystyle \frac{1}{4}\,$ in (5) is seen to be the best possible.

Schweitzer's from Diaz-Metcalf

Letting $b_k^2=x_k,\,$ $\displaystyle a_k^2=\frac{1}{x_k}\,$ and assuming $m\le x_k\le M,\,$ so $\displaystyle \sqrt{m}\le b_k=\frac{1}{a_k}\le \sqrt{M},\,$ such that $\displaystyle m\le\frac{b_k}{a_k}\le M.\,$ The Diaz-Metcalf inequality then takes the form

$\displaystyle \sum_{k=1}^{n}x_k+mM\sum_{k=1}^{n}\frac{1}{x_k}\le (m+M)n,$

from which, employing the AM-GM inequality and subsequently squaring, we get

$\displaystyle \begin{align} &4mM\left(\sum_{k=1}^{n}x_k\right)\left(\sum_{k=1}^{n}\frac{1}{x_k}\right)\\ &\qquad\le\left(\sum_{k=1}^{n}x_k\right)^2+ \left(mM\sum_{k=1}^{n}\frac{1}{x_k}\right)^2+2mM\left(\sum_{k=1}^{n}x_k\right)\left(\sum_{k=1}^{n}\frac{1}{x_k}\right)\\ &\qquad\le (m+M)^2n^2. \end{align}$

Omitting the middle part gives Schweitzer's inequality.

$\displaystyle 4mM\left(\sum_{k=1}^{n}x_k\right)\left(\sum_{k=1}^{n}\frac{1}{x_k}\right)\le (m+M)^2n^2.$

Schweitzer's from Grüss

We shall prove Schweitzer's inequality in integral form. Let in Grüss' inequality, $m\le f(x)\le M\,$ and $g(x)=\displaystyle \frac{1}{f(x)}.\,$ The inequality becomes

$\displaystyle \left|\frac{1}{b-a}\int_a^bdx-\frac{1}{(b-a)^2}\int_a^nf(x)dx\int_a^b\frac{dx}{f(x)}\right|\\ \qquad\le\displaystyle \frac{1}{4}(M-m)\left(\frac{1}{m}-\frac{1}{M}\right)=\frac{(M-m)^2}{4mM}.$

This in particular means that

$\displaystyle \frac{1}{(b-a)^2}\int_a^nf(x)dx\int_a^b\frac{dx}{f(x)}\le\frac{(M-m)^2}{4mM}+1=\frac{(M+m)^2}{4mM}. $

 

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