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Tiberiu Popoviciu

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T. Popoviciu, Sur quelques inegalités entre les fonctions convexes (III), Comptes Rendus de l’Instit. des Sci. de Roumanie, 3 (1939) no. 4, pp. 396-402 (in French).

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1939 c -Popoviciu- Comptes Rendus Seances Inst. Sci. Roum. – Sur quelques inegalites entre les fonctions convexes (III)

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Comptes Rendus de l’Instit. des Sci. de Roumanie

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[Zbl 0021.30305, JFM 65.0215.01]

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1939 c -Popoviciu- Reports of the Sessions of the Roum. Sci. Institute - On some inequalities between functions
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REPORTS

SESSIONS

OF

THE ROMANIAN INSTITUTE OF SCIENCES

FORMER
ACADEMY OF SCIENCES OF ROMANIA
PUBLISHED BY THE EDITORIAL BOARD OF THE INSTITUTE

Published in Bucharest on July 1, 1939

MINUTES OF THE MEETINGS

OF

THE ROMANIAN INSTITUTE OF SCIENCES

FORMER ACADEMY OF SCIENCES OF ROMANIA

SUMMARY

371. ON SOME INEQUALITIES BETWEEN CONVEX FUNCTIONS

(THIRD NOTE)

By TIBERIU POPOVICIU, Mc. ISR
(Session of May 6, 1939)
I. Let us pose, retaining the notations of the two previous notes 1 1 ^(1){ }^{1}1),
B ( φ ) = HAS φ − φ ( HAS ) . B ( φ ) = HAS φ − φ ( HAS ) . B(varphi)=A_(varphi)-varphi(A).\mathrm{B}(\varphi)=\mathrm{A}_{\varphi}-\varphi(\mathrm{A}) .B(φ)=HASφ−φ(HAS).
We have the classical inequality B ( φ ) ≧ 0 B ( φ ) ≧ 0 B(varphi) >= 0\mathrm{B}(\varphi) \geqq 0B(φ)≧0.
We will look for a contrary inequality, so the maximum of B ( φ ) B ( φ ) B(varphi)\mathrm{B}(\varphi)B(φ), when φ φ varphi\varphiφis given and f f fffare functions of ( E has b ) n E has b n (E_(a)^(b))_(n)\left(\mathrm{E}_{a}^{b}\right)_{n}(Ehasb)nfor one n n nnngiven.
In this Note we will examine the case n = 1 n = 1 n=1n=1n=1and we will also say a few words about the case n = 0 n = 0 n=0n=0n=0. The case n > I n > I n > In>In>Iwill be studied in the fifth Note.
Let us remember that φ φ varphi\varphiφis continuous and convex in the closed interval ( 0 , 1 ) ( 0 , 1 ) (0.1)(0.1)(0,1). We can assume that φ ( 0 ) = 0 , φ ( 1 ) = I , 0 ≦ φ ≦ I φ ( 0 ) = 0 , φ ( 1 ) = I , 0 ≦ φ ≦ I varphi(0)=0,varphi(1)=I,0 <= varphi <= I\varphi(0)=0, \varphi(1)=I, 0 \leqq \varphi \leqq Iφ(0)=0,φ(1)=I,0≦φ≦I, without restricting the generality, since B ( φ ) B ( φ ) B(varphi)B(\varphi)B(φ)does not vary when adding to φ φ varphi\varphiφa linear function. The function φ φ varphi\varphiφis then positive and grow-
health for 0 < x ≦ I 0 < x ≦ I 0 < x <= I0<x \leqq \mathrm{I}0<x≦I. The whole ( E has b ) n E has b n (E_(a)^(b))_(n)\left(\mathrm{E}_{a}^{b}\right)_{n}(Ehasb)nis formed by the functions f f fff, continuous and non-concave of order 0 , 1 , … , n 0 , 1 , … , n 0,1,dots,n0.1, \ldots, n0,1,…,nin the interval ( 0,1 ) such that f ( O ) = has ≧ o f ( I ) = b ≦ I f ( O ) = has ≧ o f ( I ) = b ≦ I f(O)=a >= oquad f(I)=b <= If(\mathrm{O})=a \geqq \mathrm{o} \quad f(\mathrm{I})=b \leqq \mathrm{I}f(O)=has≧of(I)=b≦I. The function φ φ varphi\varphiφis a convex function of ( E 0 1 ) 1 E 0 1 1 (E_(0)^(1))_(1)\left(\mathrm{E}_{0}^{1}\right)_{1}(E01)1.
Note that in all our formulas only the values ​​of φ φ varphi\varphiφIn ( has , b ) ( has , b ) (a,b)(a, b)(has,b). It is therefore sufficient to assume that φ φ varphi\varphiφis a function that is continuous and convex in the closed interval ( has , b has , b a,ba, bhas,b). If the right derivative φ d ′ ( has ) φ d ′ ( has ) varphi_(d)^(')(a)\varphi_{d}^{\prime}(a)φd′(has)in has has hashashasand the left derivative φ g ′ ( b ) φ g ′ ( b ) varphi_(g)^(')(b)\varphi_{g}^{\prime}(b)φg′(b)are finished, we can modify the function φ φ varphi\varphiφin the intervals ( o , has ) ( o , has ) (o,a)(\mathrm{o}, a)(o,has), ( b , I b , I b,Ib, \mathrm{I}b,I) so that it becomes, up to a linear additive function, a convex function of ( E 0 1 ) 1 E 0 1 1 (E_(0)^(1))_(1)\left(\mathrm{E}_{0}^{1}\right)_{1}(E01)1.
Let us also note that we have analogous inequalities, but in opposite directions, if φ φ varphi\varphiφinstead of being convex is concave. The inequalities are true, of course, also when φ φ varphi\varphiφis only non-concave (or non-convex). Only the uniqueness of the maxima (or minima) can possibly be faulty.
2. The maximum problem posed above has been solved for n = o n = o n=on=on=oby MK K no pp 1 1 ^(1){ }^{1}1). In this case we have
(I) B ( φ ) ≦ max ( a , b ) [ ( b − x ) φ ( a ) + ( x − a ) φ ( b ) b − a − φ ( x ) ] (I) B ( φ ) ≦ max ( a , b )   ( b − x ) φ ( a ) + ( x − a ) φ ( b ) b − a − φ ( x ) {:(I)B(varphi) <= max_((a,b))[((b-x)varphi(a)+(x-a)varphi(b))/(b-a)-varphi(x)]:}\begin{equation*} \mathrm{B}(\varphi) \leqq \max _{(a, b)}\left[\frac{(b-x) \varphi(a)+(x-a) \varphi(b)}{b-a}-\varphi(x)\right] \tag{I} \end{equation*}(I)B(φ)≦max(has,b)[(b−x)φ(has)+(x−has)φ(b)b−has−φ(x)]
which immediately results from the inequality
( b − A ) φ ( a ) + ( A − a ) φ ( b ) b − a − A φ = I b − a ∫ o 1 [ ( f − a ) φ ( a ) − ( b − a ) φ ( f ) + + ( b − f ) φ ( b ) ] d x ≧ 0 ( b − A ) φ ( a ) + ( A − a ) φ ( b ) b − a − A φ = I b − a ∫ o 1   [ ( f − a ) φ ( a ) − ( b − a ) φ ( f ) + + ( b − f ) φ ( b ) ] d x ≧ 0 {:[((b-A)varphi(a)+(A-a)varphi(b))/(b-a)-A_(varphi)=(I)/(b-a)int_(o)^(1)[(f-a)varphi(a)-(b-a)varphi(f)+],[+(b-f)varphi(b)]dx >= 0]:}\begin{gathered} \frac{(b-\mathrm{A}) \varphi(a)+(\mathrm{A}-a) \varphi(b)}{b-a}-\mathrm{A}_{\varphi}=\frac{\mathrm{I}}{b-a} \int_{o}^{1}[(f-a) \varphi(a)-(b-a) \varphi(f)+ \\ +(b-f) \varphi(b)] d x \geqq 0 \end{gathered}(b−HAS)φ(has)+(HAS−has)φ(b)b−has−HASφ=Ib−has∫o1[(f−has)φ(has)−(b−has)φ(f)++(b−f)φ(b)]dx≧0
The maximum is reached for a single value x 1 x 1 x_(1)x_{1}x1of x x xxx, since the function of the second member of (I) is concave. The maximum of B (%) cannot be reached by a function of ( E a b ) o E a b o (E_(a)^(b))_(o)\left(\mathrm{E}_{a}^{b}\right)_{o}(Ehasb)o, but only by certain limit functions of this family. This is the function
f = { a , 0 ≦ x ≦ x b − x 1 b − a 1 b , b − x 1 b − a < x ≦ 1 f = a ,      0 ≦ x ≦ x b − x 1 b − a 1 b ,      b − x 1 b − a < x ≦ 1 f={[a",",0 <= x <= x(b-x_(1))/(b-a_(1))],[b",",(b-x_(1))/(b-a) < x <= 1]:}f= \begin{cases}a, & 0 \leqq x \leqq x \frac{b-x_{1}}{b-a_{1}} \\ b, & \frac{b-x_{1}}{b-a}<x \leqq 1\end{cases}f={has,0≦x≦xb−x1b−has1b,b−x1b−has<x≦1
It can be noted that x 1 x 1 x_(1)x_{1}x1, never coincides with a a aahasOr b b bbb. This number can be ⟨ o u ⟩ a + b 2 ⟨ o u ⟩ a + b 2 (:ou:)(a+b)/(2)\langle o u\rangle \frac{a+b}{2}⟨ou⟩has+b2. So that we have x 1 ≧ a + b 2 x 1 ≧ a + b 2 x_(1) >= (a+b)/(2)x_{1} \geqq \frac{a+b}{2}x1≧has+b2it is necessary and sufficient that
φ ( b ) − φ ( a ) b − a ≧ φ g ′ ( a + b 2 ) φ ( b ) − φ ( a ) b − a ≧ φ g ′ a + b 2 (varphi(b)-varphi(a))/(b-a) >= varphi_(g)^(')((a+b)/(2))\frac{\varphi(b)-\varphi(a)}{b-a} \geqq \varphi_{g}^{\prime}\left(\frac{a+b}{2}\right)φ(b)−φ(has)b−has≧φg′(has+b2)
or
(2) 1 b − a ∫ a 1 φ g ′ ( x ) d x ≧ φ g ′ ( a + b 2 ) 1 b − a ∫ a 1   φ g ′ ( x ) d x ≧ φ g ′ a + b 2 quad(1)/(b-a)int_(a)^(1)varphi_(g)^(')(x)dx >= varphi_(g)^(')((a+b)/(2))\quad \frac{1}{b-a} \int_{a}^{1} \varphi_{g}^{\prime}(x) d x \geqq \varphi_{g}^{\prime}\left(\frac{a+b}{2}\right)1b−has∫has1φg′(x)dx≧φg′(has+b2),
by designating by φ g ′ φ g ′ varphi_(g)^(')\varphi_{g}^{\prime}φg′the left derivative of φ φ varphi\varphiφ. So that it is so for all the values ​​of a , b a , b a,ba, bhas,bit is necessary and sufficient that φ φ varphi\varphiφbe a function of ( E o 1 ) 2 E o 1 2 (E_(o)^(1))_(2)\left(\mathrm{E}_{o}^{1}\right)_{2}(Eo1)2, therefore non-concave of order 2.
The demonstration of this fact is not difficult. The function φ g ′ φ g ′ varphi_(g)^(')\varphi_{g}^{\prime}φg′is increasing and positive. I say that it must be continuous in the open interval ( O , I O , I O,I\mathrm{O}, \mathrm{I}O,I). So, in fact, x 0 , O < x 0 < I x 0 , O < x 0 < I x_(0),O < x_(0) < Ix_{0}, \mathrm{O}<x_{0}<\mathrm{I}x0,O<x0<I, a point of possible discontinuity. Or also
0 < ε < φ g ′ ( x 0 + 0 ) − φ g ′ ( x 0 − 0 ) 2 0 < ε < φ g ′ x 0 + 0 − φ g ′ x 0 − 0 2 0 < epsi < (varphi_(g)^(')(x_(0)+0)-varphi_(g)^(')(x_(0)-0))/(2)0<\varepsilon<\frac{\varphi_{g}^{\prime}\left(x_{0}+0\right)-\varphi_{g}^{\prime}\left(x_{0}-0\right)}{2}0<ε<φg′(x0+0)−φg′(x0−0)2
We can then find a η > 0 η > 0 eta > 0\eta>0η>0, such that we have 0 < x 0 − η < x 0 + + 2 η < 1 0 < x 0 − η < x 0 + + 2 η < 1 0 < x_(0)-eta < x_(0)++2eta < 10<x_{0}-\eta<x_{0}+ +2 \eta<10<x0−η<x0++2η<1And φ g ′ ( x 0 + 2 η ) < ε + φ g ′ ( x 0 + 0 ) φ g ′ x 0 + 2 η < ε + φ g ′ x 0 + 0 varphi_(g)^(')(x_(0)+2eta) < epsi+varphi_(g)^(')(x_(0)+0)\varphi_{g}^{\prime}\left(x_{0}+2 \eta\right)<\varepsilon+\varphi_{g}^{\prime}\left(x_{0}+0\right)φg′(x0+2η)<ε+φg′(x0+0). We have, by taking a = x 0 − η , b = x 0 + 2 η a = x 0 − η , b = x 0 + 2 η a=x_(0)-eta,b=x_(0)+2etaa=x_{0}-\eta, b=x_{0}+2 \etahas=x0−η,b=x0+2η,
1 3 η ∫ x 0 − η x 0 + 2 ⋅ η φ g ′ ( x ) d x − φ g ′ ( x 0 + η 2 ) < φ g ( x 0 − 0 + 2 [ ε + φ g ′ ( x 0 + 0 ) ] 3 − − φ g ′ ( x 0 + 0 ) = − φ g ′ ( x 0 + 0 ) − φ g ′ ( x 0 − 0 ) 3 + 2 ε 3 < 0 1 3 η ∫ x 0 − η x 0 + 2 ⋅ η   φ g ′ ( x ) d x − φ g ′ x 0 + η 2 < φ g x 0 − 0 + 2 ε + φ g ′ x 0 + 0 3 − − φ g ′ x 0 + 0 = − φ g ′ x 0 + 0 − φ g ′ x 0 − 0 3 + 2 ε 3 < 0 {:[(1)/(3eta)int_(x_(0)-eta)^(x_(0)+2*eta)varphi_(g)^(')(x)dx-varphi_(g)^(')(x_(0)+(eta)/(2)) < (varphi_(g)(x_(0)-0+2[epsi+varphi_(g)^(')(x_(0)+0)])/(3)-],[quad-varphi_(g)^(')(x_(0)+0)=-(varphi_(g)^(')(x_(0)+0)-varphi_(g)^(')(x_(0)-0))/(3)+(2epsi)/(3) < 0]:}\begin{aligned} & \frac{1}{3 \eta} \int_{x_{0}-\eta}^{x_{0}+2 \cdot \eta} \varphi_{g}^{\prime}(x) d x-\varphi_{g}^{\prime}\left(x_{0}+\frac{\eta}{2}\right)<\frac{\varphi_{g}\left(x_{0}-0+2\left[\varepsilon+\varphi_{g}^{\prime}\left(x_{0}+0\right)\right]\right.}{3}- \\ & \quad-\varphi_{g}^{\prime}\left(x_{0}+0\right)=-\frac{\varphi_{g}^{\prime}\left(x_{0}+0\right)-\varphi_{g}^{\prime}\left(x_{0}-0\right)}{3}+\frac{2 \varepsilon}{3}<0 \end{aligned}13η∫x0−ηx0+2⋅ηφg′(x)dx−φg′(x0+η2)<φg(x0−0+2[ε+φg′(x0+0)]3−−φg′(x0+0)=−φg′(x0+0)−φg′(x0−0)3+2ε3<0
which is in contradiction with (2). The function φ g ′ φ g ′ varphi_(g)^(')\varphi_{g}^{\prime}φg′must therefore be continuous 1 1 ^(1){ }^{1}1). We also see that the function φ g ′ + α x φ g ′ + α x varphi_(g)^(')+alpha x\varphi_{g}^{\prime}+\alpha xφg′+αxstill verifies property (2), regardless α α alpha\alphaα. It is easily demonstrated on (2) that φ g ′ + α x φ g ′ + α x varphi_(g)^(')+alpha x\varphi_{g}^{\prime}+\alpha xφg′+αxcannot reach its maximum in an interval ( a , b ) ( a , b ) (a,b)(a, b)(has,b)that at the ends a a aahasOr b b bbb, unless it is constant in ( a , b a , b a,ba, bhas,b). According to a remark by MS S aks 2 2 ^(2){ }^{2}2), the function φ g ′ φ g ′ varphi_(g)^(')\varphi_{g}^{\prime}φg′is non-concave of order I I III, SO φ φ varphi\varphiφis non-concave of order 2.
  1. It follows that φ φ varphi\varphiφhas a continuous derivative.
  2. S. Saks „O funckjach wypuklych i podharmoniczhych” Mathesis Polska, 6, 43-64 (1931).
We demonstrate in the same way that to have x 1 ≦ a + b 2 x 1 ≦ a + b 2 x_(1) <= (a+b)/(2)x_{1} \leqq \frac{a+b}{2}x1≦has+b2, for all values ​​of a a aahasAnd b b bbb, it is necessary and sufficient that φ φ varphi\varphiφbe non-convex of order 2.
For φ = log ⁡ x , a > 0 φ = log ⁡ x , a > 0 varphi=log x,a > 0\varphi=\log x, a>0φ=log⁡x,has>0, inequality (1) gives us
A G ≦ c − I c c − 1 − log ⁡ c ( c − 1 ) log ⁡ c ⋅ log ⁡ c = I − J e ⋅ log ⁡ I c ( I c ) 1 1 − c , c = a b A G ≦ c − I c c − 1 − log ⁡ c ( c − 1 ) log ⁡ c ⋅ log ⁡ c = I − J e ⋅ log ⁡ I c I c 1 1 − c , c = a b (A)/(G) <= (c-I)/(c^((c-1-log c)/((c-1)log c))*log c)=(I-J)/(e*log((I)/(c)))((I)/(c))^((1)/(1-c)),quad c=(a)/(b)\frac{\mathrm{A}}{\mathrm{G}} \leqq \frac{c-\mathrm{I}}{c^{\frac{c-1-\log c}{(c-1) \log c}} \cdot \log c}=\frac{\mathrm{I}-\mathrm{J}}{e \cdot \log \frac{\mathrm{I}}{c}}\left(\frac{\mathrm{I}}{c}\right)^{\frac{1}{1-c}}, \quad c=\frac{a}{b}HASG≦c−Icc−1−log⁡c(c−1)log⁡c⋅log⁡c=I−Ie⋅log⁡Ic(Ic)11−c,c=hasb
Or G = exp . ∫ 0 1 log ⁡ f d x G = exp . ∫ 0 1   log ⁡ f d x G=exp.int_(0)^(1)log fdx\mathrm{G}=\exp . \int_{0}^{1} \log f d xG=exp.∫01log⁡fdxis the geometric mean of f f fff. We have demonstrated this inequality in a previous work 1 1 ^(1){ }^{1}1).
3. Let us now move on to the study of the case n = 1 n = 1 n=1n=1n=1. Formula (6) and the results of the first note allow us to state the following property
If φ φ varphi\varphiφis a function of the form indicated and f f fffa function of ( E a b ) 1 E a b 1 (E_(a)^(b))_(1)\left(\mathrm{E}_{a}^{b}\right)_{1}(Ehasb)1, we have inequality
(3) A ψ − φ ( A ) 幺 ( a , a + b 2 ) max a [ φ ( a ) + 2 ( x − a ) ( b − a ) 2 ∫ a b [ φ ( t ) − φ ( a ) ] d t − φ ( x ) ] A ψ − φ ( A ) 幺 a , a + b 2 max a   φ ( a ) + 2 ( x − a ) ( b − a ) 2 ∫ a b   [ φ ( t ) − φ ( a ) ] d t − φ ( x ) A_(psi)-varphi(A)幺_((a,(a+b)/(2)))max_(a)[varphi(a)+(2(x-a))/((b-a)^(2))int_(a)^(b)[varphi(t)-varphi(a)]dt-varphi(x)]\mathrm{A}_{\psi}-\varphi(\mathrm{A}) \underset{\left(a, \frac{a+b}{2}\right)}{幺} \max _{a}\left[\varphi(a)+\frac{2(x-a)}{(b-a)^{2}} \int_{a}^{b}[\varphi(t)-\varphi(a)] d t-\varphi(x)\right]HASψ−φ(HAS)幺(has,has+b2)maxhas[φ(has)+2(x−has)(b−has)2∫hasb[φ(t)−φ(has)]dt−φ(x)]
equality being possible only for a single value x 2 x 2 x_(2)x_{2}x2of x x xxxand for function
(4) f = a + ( b − a ) x − λ + | x − λ | 2 ( 1 − λ ) , λ = b + a − 2 x 2 b − a f = a + ( b − a ) x − λ + | x − λ | 2 ( 1 − λ ) , λ = b + a − 2 x 2 b − a f=a+(b-a)(x-lambda+|x-lambda|)/(2(1-lambda)),lambda=(b+a-2x_(2))/(b-a)f=a+(b-a) \frac{x-\lambda+|x-\lambda|}{2(1-\lambda)}, \lambda=\frac{b+a-2 x_{2}}{b-a}f=has+(b−has)x−λ+|x−λ|2(1−λ),λ=b+has−2x2b−hasonly.
The number x 2 x 2 x_(2)x_{2}x2never coincides with a a aahasand maybe > o u < > o u < > ou <>o u<>ou<that 2 a + b 3 2 a + b 3 (2a+b)/(3)\frac{2 a+b}{3}2has+b3. To have x 2 ≧ 2 a + b 3 x 2 ≧ 2 a + b 3 x_(2) >= (2a+b)/(3)x_{2} \geqq \frac{2 a+b}{3}x2≧2has+b3it is necessary and sufficient that one has 2 ( b − a ) 2 ∫ a b [ φ ( t ) − φ ( a ) ] d t − φ g ′ ( 2 a + b 3 ) = 2 ( b − a ) 2 ∫ a b ( b − t ) φ g ′ ( t ) d t − 2 ( b − a ) 2 ∫ a b   [ φ ( t ) − φ ( a ) ] d t − φ g ′ 2 a + b 3 = 2 ( b − a ) 2 ∫ a b   ( b − t ) φ g ′ ( t ) d t − (2)/((b-a)^(2))int_(a)^(b)[varphi(t)-varphi(a)]dt-varphi_(g)^(')((2a+b)/(3))=(2)/((b-a)^(2))int_(a)^(b)(b-t)varphi_(g)^(')(t)dt-\frac{2}{(b-a)^{2}} \int_{a}^{b}[\varphi(t)-\varphi(a)] d t-\varphi_{g}^{\prime}\left(\frac{2 a+b}{3}\right)=\frac{2}{(b-a)^{2}} \int_{a}^{b}(b-t) \varphi_{g}^{\prime}(t) d t-2(b−has)2∫hasb[φ(t)−φ(has)]dt−φg′(2has+b3)=2(b−has)2∫hasb(b−t)φg′(t)dt−
− φ g ′ ( 2 a + b 3 ) ≧ 0 . − φ g ′ 2 a + b 3 ≧ 0 . -varphi_(g)^(')((2a+b)/(3)) >= 0.-\varphi_{g}^{\prime}\left(\frac{2 a+b}{3}\right) \geqq 0 .−φg′(2has+b3)≧0.
  1. Tiberiu Popovici „Asupra mediilor aritmetice şi geometrice" Gazeta Matematică, 40, 155-160 (1934). M. Knop also reported this inequality, as an application, with a slight error in the second member.
We demonstrate exactly as above for inequality (2) that:
So that x 2 x 2 x_(2)x_{2}x2either ≧ 2 a + b 3 ≧ 2 a + b 3 >= (2a+b)/(3)\geqq \frac{2 a+b}{3}≧2has+b3, whatever a and b b bbb, it is necessary and sufficient that the function φ φ varphi\varphiφis non-concave of order 2, therefore a function of ( E 0 1 ) 2 E 0 1 2 (E_(0)^(1))_(2)\left(\mathrm{E}_{0}^{1}\right)_{2}(E01)2.
So that x 2 x 2 x_(2)x_{2}x2either ≦ 2 a + b 3 ≦ 2 a + b 3 <= (2a+b)/(3)\leqq \frac{2 a+b}{3}≦2has+b3, whatever a a aahasAnd b b bbb, it is necessary and sufficient that the function φ φ varphi\varphiφbe non-convex of order 2.
The number x 2 x 2 x_(2)x_{2}x2may also coincide with a + b 2 a + b 2 (a+b)/(2)\frac{a+b}{2}has+b2. We then have the following statement.
If φ φ varphi\varphiφis a function of the form indicated, f f fffa function of ( E a b ) 1 E a b 1 (E_(a)^(b))_(1)\left(\mathrm{E}_{a}^{b}\right)_{1}(Ehasb)1and if
(5) 2 ( b − a ) 2 ∫ a b [ φ ( t ) − φ ( a ) ] d t − φ g ′ ( a + b 2 ) ≧ 0 (5) 2 ( b − a ) 2 ∫ a b   [ φ ( t ) − φ ( a ) ] d t − φ g ′ a + b 2 ≧ 0 {:(5)(2)/((b-a)^(2))int_(a)^(b)[varphi(t)-varphi(a)]dt-varphi_(g)^(')((a+b)/(2)) >= 0:}\begin{equation*} \frac{2}{(b-a)^{2}} \int_{a}^{b}[\varphi(t)-\varphi(a)] d t-\varphi_{g}^{\prime}\left(\frac{a+b}{2}\right) \geqq 0 \tag{5} \end{equation*}(5)2(b−has)2∫hasb[φ(t)−φ(has)]dt−φg′(has+b2)≧0
we have inequality
A φ − φ ( A ) ≦ I b − a ∫ a b φ ( t ) d t − φ ( a + b 2 ) A φ − φ ( A ) ≦ I b − a ∫ a b   φ ( t ) d t − φ a + b 2 A_(varphi)-varphi(A) <= (I)/(b-a)int_(a)^(b)varphi(t)dt-varphi((a+b)/(2))\mathrm{A}_{\varphi}-\varphi(\mathrm{A}) \leqq \frac{\mathrm{I}}{b-a} \int_{a}^{b} \varphi(t) d t-\varphi\left(\frac{a+b}{2}\right)HASφ−φ(HAS)≦Ib−has∫hasbφ(t)dt−φ(has+b2)
equality being possible only for the function f = a + ( b − a ) x f = a + ( b − a ) x f=a+(b-a)xf=a+(b-a) xf=has+(b−has)x.
But, it should be noted that this cannot happen for all values ​​of a and b. The property results from the above if φ φ varphi\varphiφis not non-concave of order 2. If φ φ varphi\varphiφis non-concave of order 2 the derivative φ ′ φ ′ varphi^(')\varphi^{\prime}φ′exists, is continuous, increasing and non-concave of order I. Let ξ , 0 < ξ < 1 ξ , 0 < ξ < 1 xi,0 < xi < 1\xi, 0<\xi<1ξ,0<ξ<1a point where the second derivative φ ′ ′ ( ξ ) φ ′ ′ ( ξ ) varphi^('')(xi)\varphi^{\prime \prime}(\xi)φ′′(ξ)exists and is > 0 > 0 > 0>0>0. There is obviously such a point. 1 1 ^(1){ }^{1}1). So then ε ε epsi\varepsilonεa positive number and < 2 φ ′ ′ ( ξ ) 3 < 2 φ ′ ′ ( ξ ) 3 < (2varphi^('')(xi))/(3)<\frac{2 \varphi^{\prime \prime}(\xi)}{3}<2φ′′(ξ)3And η > 0 η > 0 eta > 0\eta>0η>0such as 0 < ξ − η < ξ + η < 1 0 < ξ − η < ξ + η < 1 0 < xi-eta < xi+eta < 10<\xi-\eta<\xi+\eta<10<ξ−η<ξ+η<1and that
φ ′ ( ξ ) − φ ′ ( ξ − η ) η > φ ′ ′ ( ξ ) − ε , φ ′ ( ξ + η ) − φ ′ ( ξ ) η < φ ′ ′ ( ξ ) + ε . φ ′ ( ξ ) − φ ′ ( ξ − η ) η > φ ′ ′ ( ξ ) − ε , φ ′ ( ξ + η ) − φ ′ ( ξ ) η < φ ′ ′ ( ξ ) + ε . (varphi^(')(xi)-varphi^(')(xi-eta))/(eta) > varphi^('')(xi)-epsiquad,quad(varphi^(')(xi+eta)-varphi^(')(xi))/(eta) < varphi^('')(xi)+epsi.\frac{\varphi^{\prime}(\xi)-\varphi^{\prime}(\xi-\eta)}{\eta}>\varphi^{\prime \prime}(\xi)-\varepsilon \quad, \quad \frac{\varphi^{\prime}(\xi+\eta)-\varphi^{\prime}(\xi)}{\eta}<\varphi^{\prime \prime}(\xi)+\varepsilon .φ′(ξ)−φ′(ξ−η)η>φ′′(ξ)−ε,φ′(ξ+η)−φ′(ξ)η<φ′′(ξ)+ε.
If we take a = ξ − η , b = ξ + η a = ξ − η , b = ξ + η a=xi-eta,b=xi+etaa=\xi-\eta, b=\xi+\etahas=ξ−η,b=ξ+ηand if we take into account the fact that φ ′ φ ′ varphi^(')\varphi^{\prime}φ′remains, in the interval ( ξ − η , ξ + η ξ − η , ξ + η xi-eta,xi+eta\xi-\eta, \xi+\etaξ−η,ξ+η), not above the seg-
straight lines joining the points of the curve y = φ ′ ( x ) y = φ ′ ( x ) y=varphi^(')(x)y=\varphi^{\prime}(x)y=φ′(x)For x = ξ − η x = ξ − η x=xi-etax=\xi-\etax=ξ−η, ξ , ξ + η ξ , ξ + η xi,xi+eta\xi, \xi+\etaξ,ξ+η, we find that
I 2 η 2 ∫ ξ − η ξ + η ( ξ + η − t ) φ ′ ( t ) d t − φ ′ ( ξ ) ≦ η I 2 [ φ ′ ( ξ + η ) − φ ′ ( ξ ) η − − 5 φ ′ ( ξ ) − φ ′ ( ξ − η ) η ] < η I 2 [ − 4 φ ′ ′ ( ξ ) + 6 ε ] < 0 I 2 η 2 ∫ ξ − η ξ + η   ( ξ + η − t ) φ ′ ( t ) d t − φ ′ ( ξ ) ≦ η I 2 φ ′ ( ξ + η ) − φ ′ ( ξ ) η − − 5 φ ′ ( ξ ) − φ ′ ( ξ − η ) η < η I 2 − 4 φ ′ ′ ( ξ ) + 6 ε < 0 {:[(I)/(2eta^(2))int_(xi-eta)^(xi+eta)(xi+eta-t)varphi^(')(t)dt-varphi^(')(xi) <= (eta)/(I2)[(varphi^(')(xi+eta)-varphi^(')(xi))/(eta)-:}],[{:-(5varphi^(')(xi)-varphi^(')(xi-eta))/(eta)] < (eta)/(I2)[-4varphi^('')(xi)+6epsi] < 0]:}\begin{aligned} \frac{\mathrm{I}}{2 \eta^{2}} \int_{\xi-\eta}^{\xi+\eta}(\xi+\eta-t) \varphi^{\prime}(t) d t-\varphi^{\prime}(\xi) \leqq \frac{\eta}{\mathrm{I} 2}\left[\frac{\varphi^{\prime}(\xi+\eta)-\varphi^{\prime}(\xi)}{\eta}-\right. \\ \left.-\frac{5 \varphi^{\prime}(\xi)-\varphi^{\prime}(\xi-\eta)}{\eta}\right]<\frac{\eta}{\mathrm{I} 2}\left[-4 \varphi^{\prime \prime}(\xi)+6 \varepsilon\right]<0 \end{aligned}I2η2∫ξ−ηξ+η(ξ+η−t)φ′(t)dt−φ′(ξ)≦ηI2[φ′(ξ+η)−φ′(ξ)η−−5φ′(ξ)−φ′(ξ−η)η]<ηI2[−4φ′′(ξ)+6ε]<0
which demonstrates the property.
We do not insist on this question. In general, it can be demonstrated that if for a a a aahasgiven the number b b bbbis close enough to a a aahaswe necessarily have x 2 < a + b 2 x 2 < a + b 2 x_(2) < (a+b)/(2)x_{2}<\frac{a+b}{2}x2<has+b2.
4. Let us make some applications of the previous formulas.
I 0 . φ = x p , p > I , a = 0 , b = I I 0 . φ = x p , p > I , a = 0 , b = I I^(0).varphi=x^(p),p > I,a=0,b=I\mathrm{I}^{0} . \varphi=x^{p}, p>\mathrm{I}, a=0, b=\mathrm{I}I0.φ=xp,p>I,has=0,b=I. Inequality (5) is written 2 p ≧ p ( p + I ) 2 p ≧ p ( p + I ) 2^(p) >= p(p+I)2^{p} \geqq p(p+\mathrm{I})2p≧p(p+I)and we can state the following property.
If f f fffis a function of ( F 0 1 ) 1 F 0 1 1 (F_(0)^(1))_(1)\left(\mathrm{F}_{0}^{1}\right)_{1}(F01)1we have inequality
∫ 0 1 f p d x − ( ∫ 0 1 f d x ) p ≦ { ( p − I ) [ 2 p ( p + I ) ] p p − 1 si I < p ≦ p ′ I p + I − I 2 p si p ≧ p 1 ∫ 0 1   f p d x − ∫ 0 1   f d x p ≦ ( p − I ) 2 p ( p + I ) p p − 1       si       I < p ≦ p ′ I p + I − I 2 p       si       p ≧ p 1 int_(0)^(1)f^(p)dx-(int_(0)^(1)fdx)^(p) <= {[(p-I)[(2)/(p(p+I))]^((p)/(p-1))," si ",I < p <= p^(')],[(I)/(p+I)-(I)/(2^(p))," si ",p >= p_(1)]:}\int_{0}^{1} f^{p} d x-\left(\int_{0}^{1} f d x\right)^{p} \leqq\left\{\begin{array}{lll}(p-\mathrm{I})\left[\frac{2}{p(p+\mathrm{I})}\right]^{\frac{p}{p-1}} & \text { si } & \mathrm{I}<p \leqq p^{\prime} \\ \frac{\mathrm{I}}{p+\mathrm{I}}-\frac{\mathrm{I}}{2^{p}} & \text { si } & p \geqq p_{1}\end{array}\right.∫01fpdx−(∫01fdx)p≦{(p−I)[2p(p+I)]pp−1 if I<p≦p′Ip+I−I2p if p≧p1
Or p 1 p 1 p_(1)p_{1}p1is the root, between 4.79 and 4.8, of the equation 2 p = p ( p + 1 ) 2 p = p ( p + 1 ) 2^(p)=p(p+1)2^{p}=p(p+1)2p=p(p+1).
In the first case equality is only possible if
f = x − λ + | x − λ | 2 ( I − λ ) , λ = I − 2 [ 2 p ( p + I ) ] 1 p − 1 f = x − λ + | x − λ | 2 ( I − λ ) , λ = I − 2 2 p ( p + I ) 1 p − 1 f=(x-lambda+|x-lambda|)/(2(I-lambda))quad,quad lambda=I-2[(2)/(p(p+I))]^((1)/(p-1))f=\frac{x-\lambda+|x-\lambda|}{2(\mathrm{I}-\lambda)} \quad, \quad \lambda=\mathrm{I}-2\left[\frac{2}{p(p+\mathrm{I})}\right]^{\frac{1}{p-1}}f=x−λ+|x−λ|2(I−λ),λ=I−2[2p(p+I)]1p−1
in the second case only if f = x f = x f=xf=xf=x.
2 0 . φ = x p , 0 < p < I , a = 0 , b = I 2 0 . φ = x p , 0 < p < I , a = 0 , b = I 2^(0).varphi=x^(p),0 < p < I,a=0,b=I2^{0} . \varphi=x^{p}, 0<p<\mathrm{I}, a=0, b=\mathrm{I}20.φ=xp,0<p<I,has=0,b=I. The function φ φ varphi\varphiφis concave of order 1 and convex of order 2. We deduce that
If f f fffis a function of ( E 0 1 ) 1 E 0 1 1 (E_(0)^(1))_(1)\left(\mathrm{E}_{0}^{1}\right)_{1}(E01)1, we have the inequality
( ∫ 0 1 f d x ) p − ∫ 0 1 f p d x ≦ ( I − p ) [ p ( p + I ) 2 ] p 1 − p ∫ 0 1   f d x p − ∫ 0 1   f p d x ≦ ( I − p ) p ( p + I ) 2 p 1 − p (int_(0)^(1)fdx)^(p)-int_(0)^(1)f^(p)dx <= (I-p)[(p(p+I))/(2)]^((p)/(1-p))\left(\int_{0}^{1} f d x\right)^{p}-\int_{0}^{1} f^{p} d x \leqq(\mathrm{I}-p)\left[\frac{p(p+\mathrm{I})}{2}\right]^{\frac{p}{1-p}}(∫01fdx)p−∫01fpdx≦(I−p)[p(p+I)2]p1−pif 0 < p < I 0 < p < I quad0 < p < I\quad 0<p<\mathrm{I}0<p<I
equality being possible only for the function
f = x − λ + | x − λ | 2 ( 1 − λ ) , λ = 1 − 2 [ p ( p + 1 ) 2 ] 1 1 − p f = x − λ + | x − λ | 2 ( 1 − λ ) , λ = 1 − 2 p ( p + 1 ) 2 1 1 − p f=(x-lambda+|x-lambda|)/(2(1-lambda)),quad lambda=1-2[(p(p+1))/(2)]^((1)/(1-p))f=\frac{x-\lambda+|x-\lambda|}{2(1-\lambda)}, \quad \lambda=1-2\left[\frac{p(p+1)}{2}\right]^{\frac{1}{1-p}}f=x−λ+|x−λ|2(1−λ),λ=1−2[p(p+1)2]11−p
We have, in particular,
∫ 0 1 f 2 d x − ( ∫ 0 1 f d x ) 2 ≦ 1 9 , ∫ 0 1 f d x − ∫ 0 1 f d x ≦ 3 16 ∫ 0 1   f 2 d x − ∫ 0 1   f d x 2 ≦ 1 9 , ∫ 0 1   f d x − ∫ 0 1   f d x ≦ 3 16 int_(0)^(1)f^(2)dx-(int_(0)^(1)fdx)^(2) <= (1)/(9),quadsqrt(int_(0)^(1)fdx)-int_(0)^(1)sqrtfdx <= (3)/(16)\int_{0}^{1} f^{2} d x-\left(\int_{0}^{1} f d x\right)^{2} \leqq \frac{1}{9}, \quad \sqrt{\int_{0}^{1} f d x}-\int_{0}^{1} \sqrt{f} d x \leqq \frac{3}{16}∫01f2dx−(∫01fdx)2≦19,∫01fdx−∫01fdx≦316
equality being possible only for functions
f = 3 x − 1 + | 3 x − 1 | 4 , f = 32 x − 23 + | 32 x − 23 | 18 f = 3 x − 1 + | 3 x − 1 | 4 , f = 32 x − 23 + | 32 x − 23 | 18 f=(3x-1+|3x-1|)/(4)quad,quad f=(32 x-23+|32 x-23|)/(18)f=\frac{3 x-1+|3 x-1|}{4} \quad, \quad f=\frac{32 x-23+|32 x-23|}{18}f=3x−1+|3x−1|4,f=32x−23+|32x−23|18
respectively．The first inequality is known and we will take it up again－ in the fifth note．
3 0 3 0 3^(0)3^{0}30．We can also consider φ = x p , p < 0 , a > 0 φ = x p , p < 0 , a > 0 varphi=x^(p),p < 0,a > 0\varphi=x^{p}, p<0, a>0φ=xp,p<0,has>0.In this case φ φ varphi\varphiφis convex of order I I IIIand concave of order 2. Let, in particular, be the case p = − 1 p = − 1 p=-1p=-1p=−1．Let us designate by H = 1 / ∫ 0 1 d x t H = 1 / ∫ 0 1   d x t H=1//int_(0)^(1)(dx)/(t)H=1 / \int_{0}^{1} \frac{d x}{t}H=1/∫01dxtthe harmonic mean of f f fff．We deduce the following property:
If f f fffis a function of ( E a b ) 1 , a > 0 , A E a b 1 , a > 0 , A (E_(a)^(b))_(1),a > 0,A\left(\mathrm{E}_{a}^{b}\right)_{1}, a>0, \mathrm{~A}(Ehasb)1,has>0, HASand H the arithmetic mean and the harmonic mean of f f fff,we have inequality
I H − I A ≦ I b | I c − 2 I − c I − c + c log ⁡ c | 2 , c = a b , I H − I A ≦ I b I c − 2 I − c I − c + c log ⁡ c 2 , c = a b , (I)/(H)-(I)/((A)) <= (I)/(b)|(I)/(sqrtc)-(sqrt2)/(I-c)sqrt(I-c+c log c)|^(2)quad,quad c=(a)/(b),\frac{\mathrm{I}}{\mathrm{H}}-\frac{\mathrm{I}}{\mathrm{~A}} \leqq \frac{\mathrm{I}}{b}\left|\frac{\mathrm{I}}{\sqrt{c}}-\frac{\sqrt{2}}{\mathrm{I}-c} \sqrt{\mathrm{I}-c+c \log c}\right|^{2} \quad, \quad c=\frac{a}{b},IH−I HAS≦Ib|Ic−2I−cI−c+clog⁡c|2,c=hasb,
equality being possible only for a single function of the form (4) where
λ = I + c I − c − 2 c I − c + c log ⁡ c , c = a b λ = I + c I − c − 2 c I − c + c log ⁡ c , c = a b lambda=(I+c)/(I-c)-sqrt((2c)/(I-c+c log c)),quad c=(a)/(b)\lambda=\frac{\mathrm{I}+c}{\mathrm{I}-c}-\sqrt{\frac{2 c}{\mathrm{I}-c+c \log c}}, \quad c=\frac{a}{b}λ=I+cI−c−2cI−c+clog⁡c,c=hasb
4 0 4 0 4^(0)4^{0}40．Let us consider again the case φ = log ⁡ x , a > 0 φ = log ⁡ x , a > 0 varphi=log x,a > 0\varphi=\log x, a>0φ=log⁡x,has>0．This function is concave of order 1 and convex of order 2 ．We deduce the following property．
If f f fffis a function of ( E a b ) 1 , a > 0 , A E a b 1 , a > 0 , A (E_(a)^(b))_(1),a > 0,A\left(\mathrm{E}_{a}^{b}\right)_{1}, a>0, \mathrm{~A}(Ehasb)1,has>0, HASand G the arithmetic mean and the geometric mean of the function f f fff,we have inequality
A G ≦ 2 c ( c − I − log ⁡ c ) ( c − I ) 2 ⋅ e 1 + c 1 − c ( I c ) 2 c ( c − 1 ) 2 , c = a b A G ≦ 2 c ( c − I − log ⁡ c ) ( c − I ) 2 ⋅ e 1 + c 1 − c I c 2 c ( c − 1 ) 2 , c = a b (A)/(G) <= (2c(c-I-log c))/((c-I)^(2)*e^((1+c)/(1-c)))((I)/(c))^((2c)/((c-1)^(2)))quad,quad c=(a)/(b)\frac{\mathrm{A}}{\mathrm{G}} \leqq \frac{2 c(c-\mathrm{I}-\log c)}{(c-\mathrm{I})^{2} \cdot e^{\frac{1+c}{1-c}}}\left(\frac{\mathrm{I}}{c}\right)^{\frac{2 c}{(c-1)^{2}}} \quad, \quad c=\frac{a}{b}HASG≦2c(c−I−log⁡c)(c−I)2⋅e1+c1−c(Ic)2c(c−1)2,c=hasb
equality being possible only for a single function of the form (4) where
λ = I + c I − c − I − c c − I − log ⁡ c , c = a b λ = I + c I − c − I − c c − I − log ⁡ c , c = a b lambda=(I+c)/(I-c)-(I-c)/(c-I-log c)quad,quad c=(a)/(b)\lambda=\frac{\mathrm{I}+c}{\mathrm{I}-c}-\frac{\mathrm{I}-c}{c-\mathrm{I}-\log c} \quad, \quad c=\frac{a}{b}λ=I+cI−c−I−cc−I−log⁡c,c=hasb
For Firm which

dtuan if is dollow：4046

  1. Memorial of Mathematical Sciences. Fasc. XLIV, p. 12. Sequences of functions in general. Real domain, by ML Léau,
    I. I ask the reader to refer to the two preceding notes for the hypotheses and notations. See this CR, 2, 449-454, 454-458 (1938).
  2. JK Knopp,,Über die maximalen Abstände und Verhältnisse verschiedener Mittelwerte'' Math. Zeitschrift, 39, 768-776 (1935). The assumptions made about φ φ varphi\varphiφby MK Knopp are a little more restrictive (the existence of the second derivative φ ′ ′ φ ′ ′ varphi^('')\varphi^{\prime \prime}φ′′).
    1. φ ′ ′ φ ′ ′ varphi^('')\varphi^{\prime \prime}φ′′exists, except perhaps on a set that is at most countable.

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