On some inequalities between convex functions (I)

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T. Popoviciu, Sur quelques inegalités entre les fonction convexes (I), Comptes Rendus des séances de l’Académie des Sciences de Roumanie, 2 (1938), pp. 449-454 (in French).

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1938 e -Popoviciu- Comptes Rendus des séances de l_Acad. des Sci. de Roumanie - Sur quelques inegali
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112. ON SOME INEQUALITIES BETWEEN CONVEX FUNCTIONS

(FIRST NOTE)
By tiberiu popoviciu, Mc. AS r.
(Meeting of March 4, 1938).
  1. Either φ = φ ( x ) φ = φ ( x ) varphi=varphi(x)\varphi=\varphi(x)φ=φ(x)a defined, uniform, continuous and convex function (of order I) in the closed interval ( 0,1 ). We assume that φ ( 0 ) = 0 , φ ( 1 ) = 1 0 φ ( x ) 1 φ ( 0 ) = 0 , φ ( 1 ) = 1 0 φ ( x ) 1 varphi(0)=0,varphi(1)=1quad0 <= varphi(x) <= 1\varphi(0)=0, \varphi(1)=1 \quad 0 \leqq \varphi(x) \leqq 1φ(0)=0,φ(1)=10φ(x)1. It follows that φ ( x ) φ ( x ) varphi(x)\varphi(x)φ(x)is a positive and increasing function for 0 < x I 0 < x I 0 < x <= I0<x \leqq \mathrm{I}0<xI. The right derivative φ ( x ) φ ( x ) varphi^(')(x)\varphi^{\prime}(x)φ(x)exists, is increasing for 0 x < I 0 x < I 0 <= x < I0 \leqq x<\mathrm{I}0x<I, but may not be bounded in ( 0 , I 0 , I 0,I0, \mathrm{I}0,I).
Let us also consider a function f = f ( x ) f = f ( x ) f=f(x)f=f(x)f=f(x), definite, uniform, continuous and non-concave of order 0 , 1 , , n 0 , 1 , , n 0,1,dots,n0,1, \ldots, n0,1,,nin the meantime ( 0 , 1 ) 1 ( 0 , 1 ) 1 (0,1)^(1)(0,1)^{1}(0,1)1). We assume that f ( 0 ) = a 0 , f ( I ) = b I , a < b f ( 0 ) = a 0 , f ( I ) = b I , a < b f(0)=a >= 0,f(I)=b <= I,a < bf(0)=a \geqq 0, f(\mathrm{I})=b \leqq \mathrm{I}, a<bf(0)=has0,f(I)=bI,has<band we will designate by ( E a b ) n E a b n (E_(a)^(b))_(n)\left(\mathrm{E}_{a}^{b}\right)_{n}(Ehasb)nthe set of functions verifying all these properties. We see that ( E a b ) n ( E a b ) n 1 E a b n E a b n 1 (E_(a)^(b))_(n)sub(E_(a)^(b))_(n-1)\left(\mathrm{E}_{a}^{b}\right)_{n} \subset\left(\mathrm{E}_{a}^{b}\right)_{n-1}(Ehasb)n(Ehasb)n1, SO ( E a b ) n ( E a b ) o , n > 0 E a b n E a b o , n > 0 (E_(a)^(b))_(n)sub(E_(a)^(b))_(o),n > 0\left(\mathrm{E}_{a}^{b}\right)_{n} \subset\left(\mathrm{E}_{a}^{b}\right)_{o}, n>0(Ehasb)n(Ehasb)o,n>0. In this work we
we will suppose n > 0 n > 0 n > 0n>0n>0. We will also indicate the results for n = 0 n = 0 n=0n=0n=0, which are almost all known.
The problem we will examine in this work is the following: Given the function φ φ varphi\varphiφ, determine the maximum of A φ ( f ) A φ ( f ) A_(varphi)(f)\mathrm{A}_{\varphi}(f)HASφ(f)when A ( f ) A ( f ) A(f)\mathrm{A}(f)HAS(f)is given and f f fffruns through the whole ( E a b ) n E a b n (E_(a)^(b))_(n)\left(\mathrm{E}_{a}^{b}\right)_{n}(Ehasb)n.
We asked
A = A ( f ) = 0 1 f d x , A φ = A φ ( f ) = 0 1 φ ( f ) d x A = A ( f ) = 0 1 f d x , A φ = A φ ( f ) = 0 1 φ ( f ) d x A=A(f)=int_(0)^(1)fdx,quadA_(varphi)=A_(varphi)(f)=int_(0)^(1)varphi(f)dx\mathrm{A}=\mathrm{A}(f)=\int_{0}^{1} f d x, \quad \mathrm{~A}_{\varphi}=\mathrm{A}_{\varphi}(f)=\int_{0}^{1} \varphi(f) d xHAS=HAS(f)=01fdx, HASφ=HASφ(f)=01φ(f)dx
We will then make some applications, generalizing certain known results.
2. Let us call an elementary function of degree n n nnnany function whose ( n 1 ) ème ( n 1 ) ème  (n-1)^("ème ")(n-1)^{\text {ème }}(n1)th derivative is a polygonal line. Such a function is therefore of the form
(I) g ( x ) = P ( x ) + i = 1 m c i [ x x i + | x x i | 2 ( I x i ) ] n (I) g ( x ) = P ( x ) + i = 1 m c i x x i + x x i 2 I x i n {:(I)g(x)=P(x)+sum_(i=1)^(m)c_(i)[(x-x_(i)+|x-x_(i)|)/(2(I-x_(i)))]^(n):}\begin{equation*} g(x)=\mathrm{P}(x)+\sum_{i=1}^{m} c_{i}\left[\frac{x-x_{i}+\left|x-x_{i}\right|}{2\left(\mathrm{I}-x_{i}\right)}\right]^{n} \tag{I} \end{equation*}(I)g(x)=P(x)+i=1mci[xxi+|xxi|2(Ixi)]n
P ( x ) = a o + a 1 x + a 2 x 2 + + a n x n 1 , 0 x 1 < x 2 < < x m < I P ( x ) = a o + a 1 x + a 2 x 2 + + a n x n 1 , 0 x 1 < x 2 < < x m < I P(x)=a_(o)+a_(1)x+a_(2)x^(2)+dots+a_(n)quadx^(n-1),0 <= x_(1) < x_(2) < dots < x_(m) < I\mathrm{P}(x)=a_{o}+a_{1} x+a_{2} x^{2}+\ldots+a_{n} \quad x^{n-1}, 0 \leqslant x_{1}<x_{2}<\ldots<x_{m}<\mathrm{I}P(x)=haso+has1x+has2x2++hasnxn1,0x1<x2<<xm<I.
We assume that c i 0 , i = 1 , 2 , , m c i 0 , i = 1 , 2 , , m c_(i)!=0,i=1,2,dots,mc_{i} \neq 0, i=1,2, \ldots, mci0,i=1,2,,mand we will then say that the elementary function g ( x ) g ( x ) g(x)g(x)g(x)is at m m mmmvertices. We will say that the vertices are at the points x 1 , x 2 , , x m x 1 , x 2 , , x m x_(1),x_(2),dots,x_(m)x_{1}, x_{2}, \ldots, x_{m}x1,x2,,xm. If x 1 = 0 x 1 = 0 x_(1)=0x_{1}=0x1=0there is a summit at the point o 2 o 2 o^(2)o^{2}o2). For the elementary function g ( x ) g ( x ) g(x)g(x)g(x)has m m mmmsummits belong to ( E a b ) n E a b n (E_(a)^(b))_(n)\left(\mathrm{E}_{a}^{b}\right)_{n}(Ehasb)nit is necessary and sufficient that one has
a o = a , a r 0 , r = 1 , 2 , , n 1 , c i > 0 , i = 1 , 2 , , m , P ( 1 ) + i = 1 m c i = b a o = a , a r 0 , r = 1 , 2 , , n 1 , c i > 0 , i = 1 , 2 , , m , P ( 1 ) + i = 1 m c i = b a_(o)=a,a_(r) >= 0,r=1,2,dots,n-1,c_(i) > 0,i=1,2,dots,m,P(1)+sum_(i=1)^(m)c_(i)=ba_{o}=a, a_{r} \geqq 0, r=1,2, \ldots, n-1, c_{i}>0, i=1,2, \ldots, m, \mathrm{P}(1)+\sum_{i=1}^{m} c_{i}=bhaso=has,hasr0,r=1,2,,n1,ci>0,i=1,2,,m,P(1)+i=1mci=b.
In the following we only consider elementary functions belonging to ( E a b ) n E a b n (E_(a)^(b))_(n)\left(\mathrm{E}_{a}^{b}\right)_{n}(Ehasb)n.
According to the previous definition, any polynomial of degree n I n I n-In-InIis an elementary function of degree n n nnnwith o vertices. A polynomial of effective degree n n nnnis an elementary function of degree n n nnnat . I vertex, the vertex being at point 0 .
Among the I-vertex functions we find the following
(2) h λ ( x ) = a + ( b a ) [ x λ + | x λ | 2 ( I λ ) ] n , ( 0 λ < I ) . (2) h λ ( x ) = a + ( b a ) x λ + | x λ | 2 ( I λ ) n , ( 0 λ < I ) . {:(2)h_(lambda)(x)=a+(b-a)[(x-lambda+|x-lambda|)/(2(I-lambda))]^(n)","quad(0 <= lambda < I).:}\begin{equation*} h_{\lambda}(x)=a+(b-a)\left[\frac{x-\lambda+|x-\lambda|}{2(I-\lambda)}\right]^{n}, \quad(0 \leqq \lambda<I) . \tag{2} \end{equation*}(2)hλ(x)=has+(bhas)[xλ+|xλ|2(Iλ)]n,(0λ<I).
In the case n = 1 n = 1 n=1n=1n=1all one-vertex functions are of this form.
Function (1) can be written
g ( x ) = ( b a ) P ( x ) + a [ P ( I ) b ] b a + I b a i = 1 m c i h x i ( x ) , g ( x ) = ( b a ) P ( x ) + a [ P ( I ) b ] b a + I b a i = 1 m c i h x i ( x ) , g(x)=((b-a)P(x)+a[P(I)-b])/(b-a)+(I)/(b-a)sum_(i=1)^(m)c_(i)h_(x_(i))(x),g(x)=\frac{(b-a) \mathrm{P}(x)+a[\mathrm{P}(\mathrm{I})-b]}{b-a}+\frac{\mathrm{I}}{b-a} \sum_{i=1}^{m} c_{i} h_{x_{i}}(x),g(x)=(bhas)P(x)+has[P(I)b]bhas+Ibhasi=1mcihxi(x),
which shows us that it is the (generalized) arithmetic mean of a polynomial of degree n n nnn-I and of m m mmmfunctions of the form (2).
Any function of ( E a b ) n E a b n (E_(a)^(b))_(n)\left(\mathrm{E}_{a}^{b}\right)_{n}(Ehasb)nis the limit of a uniformly convergent sequence of elementary functions of ( E a b ) n E a b n (E_(a)^(b))_(n)\left(\mathrm{E}_{a}^{b}\right)_{n}(Ehasb)n. This property can also be specified in the following way. If f f fffbelongs to ( E a b ) n E a b n (E_(a)^(b))_(n)\left(\mathrm{E}_{a}^{b}\right)_{n}(Ehasb)n, the derivatives f , f , , f ( n 1 ) f , f , , f ( n 1 ) f^('),f^(''),dots,f^((n-1))f^{\prime}, f^{\prime \prime}, \ldots, f^{(n-1)}f,f,,f(n1)exist for o x < I o x < I o <= x < I\mathrm{o} \leqq x<\mathrm{I}ox<I. The function f f fffis then the limit of a uniformly convergent sequence of elementary functions of ( E a b ) n E a b n (E_(a)^(b))_(n)\left(\mathrm{E}_{a}^{b}\right)_{n}(Ehasb)n, these functions all having the same first term P ( x ) = a + x f ( 0 ) + x 2 2 ! f ( 0 ) + + x n 1 ( n 1 ) ! f ( n 1 ) ( 0 ) P ( x ) = a + x f ( 0 ) + x 2 2 ! f ( 0 ) + + x n 1 ( n 1 ) ! f ( n 1 ) ( 0 ) P(x)=a+xf^(')(0)+(x^(2))/(2!)f^('')(0)+dots+(x^(n-1))/((n-1)!)f^((n-1))(0)P(x)=a+x f^{\prime}(0)+\frac{x^{2}}{2!} f^{\prime \prime}(0)+\ldots+\frac{x^{n-1}}{(n-1)!} f^{(n-1)}(0)P(x)=has+xf(0)+x22!f(0)++xn1(n1)!f(n1)(0). Moreover, we can assume that the value of the integral A A AAHASis the same for all these functions.
These considerations apply, in particular, to the function φ φ varphi\varphiφ, which belongs to ( E 0 1 ) 1 E 0 1 1 (E_(0)^(1))_(1)\left(\mathrm{E}_{0}^{1}\right)_{1}(E01)1, and serve to establish the following assertions with complete rigor.
3. Let φ φ varphi\varphiφgiven and g ( x ) g ( x ) g(x)g(x)g(x)an elementary function of ( E a b ) n E a b n (E_(a)^(b))_(n)\left(\mathrm{E}_{a}^{b}\right)_{n}(Ehasb)nhaving m > 1 m > 1 m > 1m>1m>1vertices and given by formula (I). Let us construct the function.
g ( x ) = P ( x ) + i = 1 m 2 c i [ x x i + | x x i | 2 ( I x i ) ] n + c [ x μ + | x μ | 2 ( I μ ) ] n g ( x ) = P ( x ) + i = 1 m 2 c i x x i + x x i 2 I x i n + c x μ + | x μ | 2 ( I μ ) n g^(**)(x)=P(x)+sum_(i=1)^(m-2)c_(i)[(x-x_(i)+|x-x_(i)|)/(2(I-x_(i)))]^(n)+c[(x-mu+|x-mu|)/(2(I-mu))]^(n)g^{*}(x)=\mathrm{P}(x)+\sum_{i=1}^{m-2} c_{i}\left[\frac{x-x_{i}+\left|x-x_{i}\right|}{2\left(\mathrm{I}-x_{i}\right)}\right]^{n}+c\left[\frac{x-\mu+|x-\mu|}{2(\mathrm{I}-\mu)}\right]^{n}g(x)=P(x)+i=1m2ci[xxi+|xxi|2(Ixi)]n+c[xμ+|xμ|2(Iμ)]n
and determine the constants c , μ c , μ c,muc, \muc,μso that we have g ( I ) = b g ( I ) = b g^(**)(I)=bg^{*}(\mathrm{I})=bg(I)=b, A ( g ) = A ( g ) A g = A ( g ) A(g^(**))=A(g)\mathrm{A}\left(g^{*}\right)=\mathrm{A}(g)HAS(g)=HAS(g). We find that c = c m 1 + c m c = c m 1 + c m c=c_(m-1)+c_(m)c=c_{m-1}+c_{m}c=cm1+cmAnd μ = c m 1 x m 1 + c m x m c m 1 + c m μ = c m 1 x m 1 + c m x m c m 1 + c m mu=(c_(m-1)x_(m-1)+c_(m)x_(m))/(c_(m-1)+c_(m))\mu=\frac{c_{m-1} x_{m-1}+c_{m} x_{m}}{c_{m-1}+c_{m}}μ=cm1xm1+cmxmcm1+cm, SO x m 1 < μ < x m , c > 0 x m 1 < μ < x m , c > 0 x_(m-1) < mu < x_(m),c > 0x_{m-1}<\mu<x_{m}, c>0xm1<μ<xm,c>0And g g g^(**)g^{*}gis indeed an elementary function of degree n n nnnhas m 1 m 1 m-1m-1m1summits belonging to ( E a b ) n E a b n (E_(a)^(b))_(n)\left(\mathrm{E}_{a}^{b}\right)_{n}(Ehasb)n.
There is a number μ , μ < μ < I μ , μ < μ < I mu^('),mu < mu^(') < I\mu^{\prime}, \mu<\mu^{\prime}<Iμ,μ<μ<Isuch that we have
(3) g ( x ) g ( x ) g ( x ) g ( x ) quadg^(**)(x) <= g(x)\quad g^{*}(x) \leqq g(x)g(x)g(x)following that x μ x μ x <= mu^(')x \leqq \mu^{\prime}xμ, For x m 1 < x < I 4 x m 1 < x < I 4 x_(m-1) < x < I^(4)x_{m-1}<x<I^{4}xm1<x<I4)
Consider the function
φ λ ( x ) = x λ + | x λ | 2 , 0 λ < 1 φ λ ( x ) = x λ + | x λ | 2 , 0 λ < 1 varphi_(lambda)(x)=(x-lambda+|x-lambda|)/(2),quad0 <= lambda < 1\varphi_{\lambda}(x)=\frac{x-\lambda+|x-\lambda|}{2}, \quad 0 \leqq \lambda<1φλ(x)=xλ+|xλ|2,0λ<1
If g ( x m 1 ) < λ < b g x m 1 < λ < b g(x_(m-1)) < lambda < bg\left(x_{m-1}\right)<\lambda<bg(xm1)<λ<b, we have
A φ λ ( g ) A φ λ ( g ) = i 1 g d x t 1 g d x + λ ( t t ) A φ λ g A φ λ ( g ) = i 1 g d x t 1 g d x + λ t t Avarphi_(lambda)(g^(**))-Avarphi_(lambda)(g)=int_(i^(**))^(1)g^(**)dx-int_(t)^(1)gdx+lambda(t^(**)-t)\mathrm{A} \varphi_{\lambda}\left(g^{*}\right)-\mathrm{A} \varphi_{\lambda}(g)=\int_{i^{*}}^{1} g^{*} d x-\int_{t}^{1} g d x+\lambda\left(t^{*}-t\right)HASφλ(g)HASφλ(g)=i1gdxt1gdx+λ(tt)
Or λ = g ( t ) = g ( t ) λ = g ( t ) = g t lambda=g(t)=g^(**)(t^(**))\lambda=g(t)=g^{*}\left(t^{*}\right)λ=g(t)=g(t)and property (3) shows us that this function of λ λ lambda\lambdaλcancels out for λ = g ( x m 1 ) λ = g x m 1 lambda=g(x_(m-1))\lambda=g\left(x_{m-1}\right)λ=g(xm1)And λ = b λ = b lambda=b\lambda=bλ=b, is increasing for g ( x m 1 ) < λ < g ( μ ) g x m 1 < λ < g μ g(x_(m-1)) < lambda < g(mu^('))g\left(x_{m-1}\right)<\lambda<g\left(\mu^{\prime}\right)g(xm1)<λ<g(μ)and decreasing for g ( μ ) < λ < b g μ < λ < b g(mu^(')) < lambda < bg\left(\mu^{\prime}\right)<\lambda<bg(μ)<λ<b. It therefore follows that
(4) A φ λ ( g ) > A φ λ ( g ) , g ( x m 1 ) < λ < b (4) A φ λ g > A φ λ ( g ) , g x m 1 < λ < b {:(4)Avarphi_(lambda)(g^(**)) > Avarphi_(lambda)(g)","quad g(x_(m-1)) < lambda < b:}\begin{equation*} \mathrm{A} \varphi_{\lambda}\left(g^{*}\right)>\mathrm{A} \varphi_{\lambda}(g), \quad g\left(x_{m-1}\right)<\lambda<b \tag{4} \end{equation*}(4)HASφλ(g)>HASφλ(g),g(xm1)<λ<b
  1. We now propose to demonstrate that A φ ( g ) >> A φ ( g ) A φ g >> A φ ( g ) A_(varphi)(g^(**))>>A_(varphi)(g)\mathrm{A}_{\varphi}\left(g^{*}\right)> >\mathrm{A}_{\varphi}(g)HASφ(g)>>HASφ(g). The inequality with the sign >=\geqqresults from (4), but it is a matter of proving that equality is not possible.
Consider the non-decreasing and non-concave function,
Φ ε ( x ) = { φ ( x ) , dans ( 0 , I ε ) φ ( I ε ) + ( x I + ε ) φ ( I ε ) , dans ( I ε , I ) , Φ ε ( x ) = φ ( x ) ,       dans  ( 0 , I ε ) φ ( I ε ) + ( x I + ε ) φ ( I ε ) ,       dans  ( I ε , I ) , Phi_(epsi)(x)={[varphi(x)","," dans "(0","I-epsi)],[varphi(I-epsi)+(x-I+epsi)varphi(I-epsi)","," dans "(I-epsi","I)","]:}\Phi_{\varepsilon}(x)=\left\{\begin{array}{lr} \varphi(x), & \text { dans }(0, \mathrm{I}-\varepsilon) \\ \varphi(\mathrm{I}-\varepsilon)+(x-\mathrm{I}+\varepsilon) \varphi(\mathrm{I}-\varepsilon), & \text { dans }(\mathrm{I}-\varepsilon, \mathrm{I}), \end{array}\right.Φε(x)={φ(x), In (0,Iε)φ(Iε)+(xI+ε)φ(Iε), In (Iε,I),
Or ε ε epsi\varepsilonεis a fairly small positive number and φ φ varphi^(')\varphi^{\prime}φis the right derivative of φ φ varphi\varphiφ. We have
(5) Φ z ( x ) = 0 1 φ t ( x ) d Φ ε ( t ) + α x (5) Φ z ( x ) = 0 1 φ t ( x ) d Φ ε ( t ) + α x {:(5)Phi_(z)(x)=int_(0)^(1)varphi_(t)(x)dPhi_(epsi)^(')(t)+alpha x:}\begin{equation*} \Phi_{z}(x)=\int_{0}^{1} \varphi_{t}(x) d \Phi_{\varepsilon}^{\prime}(t)+\alpha x \tag{5} \end{equation*}(5)Φz(x)=01φt(x)dΦε(t)+αx
the integral being Stieltjes and α α alpha\alphaαa constant of no importance to us 5 5 ^(5){ }^{5}5).
It can be easily demonstrated that
A Φ Σ ( g ) A Φ ε ( g ) = 0 1 { A φ l ( g ) A ψ t ( g ) } d Φ ε ( t ) A Φ Σ g A Φ ε ( g ) = 0 1 A φ l g A ψ t ( g ) d Φ ε ( t ) A_(Phi Sigma)(g^(**))-A_(Phi epsi)(g)=int_(0)^(1){(A)varphi_(l)(g^(**))-Apsi_(t)(g)}dPhi_(epsi)^(')(t)\mathrm{A}_{\Phi \Sigma}\left(g^{*}\right)-\mathrm{A}_{\Phi \varepsilon}(g)=\int_{0}^{1}\left\{\mathrm{~A} \varphi_{l}\left(g^{*}\right)-\mathrm{A} \psi_{t}(g)\right\} d \Phi_{\varepsilon}^{\prime}(t)HASΦΣ(g)HASΦε(g)=01{ HASφL(g)HASψt(g)}dΦε(t)
  1. This formula is analogous to that given for non-convex functions by Messrs. W. Blascke and G. Pick, see: „Distanzschätzungen im Funktionenraum II". Math. Ann. Bd. 77 (1916) p. 277-300. The introduction of the function Φ ε Φ ε Phi_(epsi)\boldsymbol{\Phi}_{\varepsilon}Φεis only necessary if b = 1 b = 1 b=1b=1b=1. It may happen in fact that the right derivative of φ ( x ) φ ( x ) varphi(x)\varphi(x)φ(x)is not of bounded variation.
    which results from the fact that the integral of S t i e 1 t j e s S t i e 1 t j e s Stie1tjesS t i e 1 t j e sStie1tIesof the second member exists.
From this formula and the definition of Φ ε Φ ε Phi_(epsi)\Phi_{\varepsilon}Φεit follows that A Φ ( g ) A Φ g A_(Phi)(g^(**))\mathrm{A}_{\Phi}\left(g^{*}\right)HASΦ(g)-- A Φ ε ( g ) A Φ ε ( g ) APhi_(epsi)(g)\mathrm{A} \Phi_{\varepsilon}(g)HASΦε(g)is positive and does not decrease when ε ε epsi\varepsilonεdecreases. We have therefore
lim 3 0 [ A Φ ε ( g ) A Φ ε ( g ) ] = A φ ( g ) A φ ( g ) > 0 lim 3 0 A Φ ε g A Φ ε ( g ) = A φ g A φ ( g ) > 0 lim_(3rarr0)[A_(Phi epsi)(g^(**))-A_(Phi epsi)(g)]=A_(varphi)(g^(**))-A_(varphi)(g) > 0\lim _{3 \rightarrow 0}\left[\mathrm{~A}_{\Phi \varepsilon}\left(g^{*}\right)-\mathrm{A}_{\Phi \varepsilon}(g)\right]=\mathrm{A}_{\varphi}\left(g^{*}\right)-\mathrm{A}_{\varphi}(g)>0lim30[ HASΦε(g)HASΦε(g)]=HASφ(g)HASφ(g)>0
  1. Now consider a function f f fffof ( E a b ) n E a b n (E_(a)^(b))_(n)\left(\mathrm{E}_{a}^{b}\right)_{n}(Ehasb)n. Let the elementary function of degree n n nnnat the summit
    h ( x ) = a + x f ( o ) + x 2 2 ! f ( o ) + + x n 1 ( n I ) ! f ( n 1 ) ( o ) + d [ x ρ + | x ρ | 2 ( I ρ ) ] n h ( x ) = a + x f ( o ) + x 2 2 ! f ( o ) + + x n 1 ( n I ) ! f ( n 1 ) ( o ) + d x ρ + | x ρ | 2 ( I ρ ) n h(x)=a+xf^(')(o)+(x^(2))/(2!)f^('')(o)+dots+(x^(n-1))/((n-I)!)f^((n-1))(o)+d[(x-rho+|x-rho|)/(2(I-rho))]^(n)h(x)=a+x f^{\prime}(\mathrm{o})+\frac{x^{2}}{2!} f^{\prime \prime}(\mathrm{o})+\ldots+\frac{x^{n-1}}{(n-\mathrm{I})!} f^{(n-1)}(\mathrm{o})+d\left[\frac{x-\rho+|x-\rho|}{2(\mathrm{I}-\rho)}\right]^{n}h(x)=has+xf(o)+x22!f(o)++xn1(nI)!f(n1)(o)+d[xρ+|xρ|2(Iρ)]n
    Or d , ρ d , ρ d,rhod, \rhod,ρare completely determined by the conditions h ( I ) = b h ( I ) = b h(I)=bh(I)=bh(I)=b, A ( h ) = A ( f ) . h ( x ) A ( h ) = A ( f ) . h ( x ) A(h)=A(f).h(x)\mathrm{A}(h)=\mathrm{A}(f) . h(x)HAS(h)=HAS(f).h(x)belongs to ( E a b ) n E a b n (E_(a)^(b))_(n)\left(\mathrm{E}_{a}^{b}\right)_{n}(Ehasb)nand we have A φ ( h ) A φ ( f ) A φ ( h ) A φ ( f ) A_(varphi)(h) >= A_(varphi)(f)\mathrm{A}_{\varphi}(h) \geqq \mathrm{A}_{\varphi}(f)HASφ(h)HASφ(f)We propose to demonstrate that equality is only possible if h = f h = f h=fh=fh=f.
Let us take formula (5); we have
A Φ s ( h ) A Φ s ( f ) = 0 1 { A φ t ( h ) A φ t ( f ) } d Φ ε ( t ) A Φ s ( h ) A Φ s ( f ) = 0 1 A φ t ( h ) A φ t ( f ) d Φ ε ( t ) A_(Phi s)(h)-A_(Phi s)(f)=int_(0)^(1){A_(varphi t)(h)-A_(varphi t)(f)}dPhi_(epsi)^(')(t)\mathrm{A}_{\Phi s}(h)-\mathrm{A}_{\Phi s}(f)=\int_{0}^{1}\left\{\mathrm{~A}_{\varphi t}(h)-\mathrm{A}_{\varphi t}(f)\right\} d \Phi_{\varepsilon}^{\prime}(t)HASΦs(h)HASΦs(f)=01{ HASφt(h)HASφt(f)}dΦε(t)
The function A φ i ( h ) A φ t ( f ) A φ i ( h ) A φ t ( f ) Avarphi_(i)(h)-Avarphi_(t)(f)\mathrm{A} \varphi_{i}(h)-\mathrm{A} \varphi_{t}(f)HASφi(h)HASφt(f)is continuous in t t tttand is 0 0 >= 0\geqq 00.
It is easily demonstrated that if f f fffdoes not reduce to an elementary function of degree n n nnnat o or I vertex, this function is not identically zero. We therefore have A Φ ε ^ ( h ) A Φ ε ^ ( f ) > 0 A Φ ε ^ ( h ) A Φ ε ^ ( f ) > 0 A_(Phi widehat(epsi))(h)-A_(Phi widehat(epsi))(f) > 0\mathrm{A}_{\Phi \widehat{\varepsilon}}(h)-\mathrm{A}_{\Phi \widehat{\varepsilon}}(f)>0HASΦε^(h)HASΦε^(f)>0, at least for ε ε epsi\varepsilonεquite small. The requested property results as above.
So finally:
If the function φ φ varphi\varphiφand the number A ( f ) A ( f ) A(f)\mathrm{A}(f)HAS(f)are given, the maximum of A φ ( f ) A φ ( f ) A_(varphi)(f)\mathrm{A}_{\varphi}(f)HASφ(f)In ( E a b ) n E a b n (E_(a)^(b))_(n)\left(\mathrm{E}_{a}^{b}\right)_{n}(Ehasb)ncan only be achieved for an elementary function of degree n n nnnat most I summit.
In the case n = 1 n = 1 n=1n=1n=1the function h ( x ) h ( x ) h(x)h(x)h(x)is completely determined by the value of the integral A . We therefore have the following property:
If φ φ varphi\varphiφis given and f is a continuous, non-decreasing, non-concave function in ( O , I ) ( O , I ) (O,I)(\mathrm{O}, \mathrm{I})(O,I), we have
(6) 0 1 φ ( f ) d x b + a 2 A b a φ ( a ) + 2 ( A a ) ( b a ) 2 a b φ ( x ) d x , A = 0 1 f d x (6) 0 1 φ ( f ) d x b + a 2 A b a φ ( a ) + 2 ( A a ) ( b a ) 2 a b φ ( x ) d x , A = 0 1 f d x {:(6)int_(0)^(1)varphi(f)dx <= (b+a-2(A))/(b-a)varphi(a)+(2((A)-a))/((b-a)^(2))int_(a)^(b)varphi(x)dx","quadA=int_(0)^(1)fdx:}\begin{equation*} \int_{0}^{1} \varphi(f) d x \leqq \frac{b+a-2 \mathrm{~A}}{b-a} \varphi(a)+\frac{2(\mathrm{~A}-a)}{(b-a)^{2}} \int_{a}^{b} \varphi(x) d x, \quad \mathrm{~A}=\int_{0}^{1} f d x \tag{6} \end{equation*}(6)01φ(f)dxb+has2 HASbhasφ(has)+2( HAShas)(bhas)2hasbφ(x)dx, HAS=01fdx
equality being possible only for the function
t = a + ( b a ) x λ + | x λ | 2 ( I λ ) , λ = b + a 2 A b a t = a + ( b a ) x λ + | x λ | 2 ( I λ ) , λ = b + a 2 A b a t=a+(b-a)(x-lambda+|x-lambda|)/(2(I-lambda)),lambda=(b+a-2(A))/(b-a)t=a+(b-a) \frac{x-\lambda+|x-\lambda|}{2(\mathrm{I}-\lambda)}, \lambda=\frac{b+a-2 \mathrm{~A}}{b-a}t=has+(bhas)xλ+|xλ|2(Iλ),λ=b+has2 HASbhas

    1. We also say that such a function is ( n + 1 n + 1 n+1n+1n+1)-times monotonous.
    1. We have already considered such functions in a previous work, see: Tiberiu Popoviciu, „On the extension of convex functions of higher order". Bull. Math. Soc. Roum. Sc., t. 36 (1934), pp. 75-108.
    1. In the case m = 2 m = 2 m=2m=2m=2, we obviously have
    2. The position of the point μ μ mu^(')\mu^{\prime}μin the meantime ( μ , r μ , r mu,r\mu, \mathrm{r}μ,r) depends on the values ​​of c m 1 , c m c m 1 , c m c_(m-1),c_(m)c_{m-1}, c_{m}cm1,cm. In particular, if n = 1 n = 1 n=1n=1n=1we always have μ < μ < x m μ < μ < x m mu < mu^(') < x_(m)\mu<\mu^{\prime}<x_{m}μ<μ<xm.
1938

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