On some inequalities between convex functions (II)

Tiberiu Popoviciu
(original), Approximation Theory, convexity, Inequalities, Mathematical Analysis, paper

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T. Popoviciu, Sur quelques inegalités entre les fonctions convexes (II), Comptes Rendus de l’Instit. de Sci. de Roum., 2 (1938), pp. 454-458 (in French)

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1938 d -Popoviciu- Comptes Rendus des séances de l’Acad. des Sci. de Roumanie – Sur quelques inegalités entre les fonction convexes (II)

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1938 d -Popoviciu- Reports of the meetings of the Romanian Academy of Sciences - On some inequalities
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113. On SOME INEQUALITIES BETWEEN CONVEX FUNCTIONS

(SECOND NOTE)
By tiberiu popoviciu, Mc. ASR
(Session of March 4, 1938)
I. We have demonstrated in the first Note that the maximum of HAS φ ( f ) HAS φ ( f ) A_(varphi)(f)\mathrm{A}_{\varphi}(f)HASφ(f)In ( E has b ) n E has b n (E_(a)^(b))_(n)\left(\mathrm{E}_{a}^{b}\right)_{n}(Ehasb)n, when HAS ( f ) HAS ( f ) A(f)\mathrm{A}(f)HAS(f)is given, can only be reached by an elementary function of degree n n nnnat most i vertex. More precisely, there always exists an elementary function h ( x ) h ( x ) h(x)h(x)h(x)of ( E has b ) n E has b n (E_(a)^(b))_(n)\left(\mathrm{E}_{a}^{b}\right)_{n}(Ehasb)n, having at most I vertex, such that HAS ( h ) = HAS ( f ) , HAS φ ( h ) > HAS φ ( f ) HAS ( h ) = HAS ( f ) , HAS φ ( h ) > HAS φ ( f ) A(h)=A(f),A_(varphi)(h) > A_(varphi)(f)\mathrm{A}(h)=\mathrm{A}(f), \mathrm{A}_{\varphi}(h)>\mathrm{A}_{\varphi}(f)HAS(h)=HAS(f),HASφ(h)>HASφ(f), equality not being possible if f f fffis not an elementary function at o or I vertex.
In particular, for n = 1 n = 1 n=1n=1n=1, the maximum problem is completely solved since the value of the integral HAS ( f ) HAS ( f ) A(f)\mathrm{A}(f)HAS(f)completely determines the function h h hhh.
We now propose to examine the case n > 1 n > 1 n > 1n>1n>1. It is then necessary to choose among the elementary functions at o or I vertex, the one (or ones) which gives the maximum of HAS φ ( f ) HAS φ ( f ) A_(varphi)(f)A_{\varphi}(f)HASφ(f).
We will use an auxiliary property, which we have already used in the previous note, and which we now state in the following form:
Either φ φ varphi\varphiφa function of the form indicated in the first note. Let then f , f ∗ f , f ∗ f,f^(**)f, f^{*}f,f∗two continuous, differentiable and increasing functions in the closed interval ( o , I o , I o,I\mathrm{o}, \mathrm{I}o,I). In addition, the functions f , f ∗ f , f ∗ f,f^(**)f, f^{*}f,f∗check the properties:
I 0 . f ( o ) = f ∗ ( o ) = has ≥ o , f ( I ) = f ∗ ( I ) = b ≦ I , has ≥ b I 0 . f ( o ) = f ∗ ( o ) = has ≥ o , f ( I ) = f ∗ ( I ) = b ≦ I , has ≥ b I^(0).f(o)=f^(**)(o)=a >= o,f(I)=f^(**)(I)=b <= I,a >= b\mathrm{I}^{0} . f(\mathrm{o})=f^{*}(\mathrm{o})=a \geq \mathrm{o}, f(\mathrm{I})=f^{*}(\mathrm{I})=b \leqq \mathrm{I}, a \geq bI0.f(o)=f∗(o)=has≥o,f(I)=f∗(I)=b≦I,has≥b.
2 0 2 0 2^(0)2^{0}20. We have HAS ( f ∗ ) = HAS ( f ) HAS f ∗ = HAS ( f ) A(f^(**))=A(f)\mathrm{A}\left(f^{*}\right)=\mathrm{A}(f)HAS(f∗)=HAS(f).
3. There is a number x 0 , 0 < x 0 < 1 x 0 , 0 < x 0 < 1 x_(0),0 < x_(0) < 1x_{0}, 0<x_{0}<1x0,0<x0<1, such that we have f ∗ ( x ) ⋚ f ( x ) f ∗ ( x ) ⋚ f ( x ) f^(**)(x)⋚f(x)f^{*}(x) \lesseqgtr f(x)f∗(x)⋚f(x), depending on whether x ⋚ x 0 , 0 < x < 1 x ⋚ x 0 , 0 < x < 1 x⋚x_(0),0 < x < 1x \lesseqgtr x_{0}, 0<x<1x⋚x0,0<x<1.
More generally we can assume that f* is first constantly equal to a and then is increasing.
We then have the inequality HAS φ ( f ∗ ) > HAS φ ( f ) 1 HAS φ f ∗ > HAS φ ( f ) 1 A_(varphi)(f^(**)) > A_(varphi)(f)^(1)\mathrm{A}_{\varphi}\left(f^{*}\right)>\mathrm{A}_{\varphi}(f){ }^{1}HASφ(f∗)>HASφ(f)1).
The demonstration is simple. According to the representation of φ φ varphi\varphiφby a Stieltjes integral, it suffices to demonstrate the property for the functions φ Λ . 2 φ Λ . 2 varphi_(Lambda).^(2)\varphi_{\Lambda} .{ }^{2}φΛ.2). This amounts to demonstrating that
G ( λ ) = ∫ i ∗ 1 f ∗ d x − ∫ t 1 f d x + λ ( t ∗ − t ) > 0 , has < λ < b G ( λ ) = ∫ i ∗ 1   f ∗ d x − ∫ t 1   f d x + λ t ∗ − t > 0 , has < λ < b G(lambda)=int_(i^(**))^(1)f^(**)dx-int_(t)^(1)fdx+lambda(t^(**)-t) > 0,quad a < lambda < b\mathrm{G}(\lambda)=\int_{i^{*}}^{1} f^{*} d x-\int_{t}^{1} f d x+\lambda\left(t^{*}-t\right)>0, \quad a<\lambda<bG(λ)=∫i∗1f∗dx−∫t1fdx+λ(t∗−t)>0,has<λ<b
Or λ = f ( l ) = f ∗ ( t ∗ ) λ = f ( l ) = f ∗ t ∗ lambda=f(l)=f^(**)(t^(**))\lambda=f(l)=f^{*}\left(t^{*}\right)λ=f(L)=f∗(t∗). Gold, t , t ∗ t , t ∗ t,t^(**)t, t^{*}t,t∗and the first member of this inequality are functions derivable from λ λ lambda\lambdaλ. We have
d G d λ = − f ∗ ( t ∗ ) d t ∗ d λ + f ( t ) d t d λ + λ ( d t ∗ d λ − d t d λ ) + ( t ∗ − t ) = t ∗ − t d G d λ = − f ∗ t ∗ d t ∗ d λ + f ( t ) d t d λ + λ d t ∗ d λ − d t d λ + t ∗ − t = t ∗ − t (dG)/(d lambda)=-f^(**)(t^(**))(dt^(**))/(d lambda)+f(t)(dt)/(d lambda)+lambda((dt^(**))/(d lambda)-(dt)/(d lambda))+(t^(**)-t)=t^(**)-t\frac{d \mathrm{G}}{d \lambda}=-f^{*}\left(t^{*}\right) \frac{d t^{*}}{d \lambda}+f(t) \frac{d t}{d \lambda}+\lambda\left(\frac{d t^{*}}{d \lambda}-\frac{d t}{d \lambda}\right)+\left(t^{*}-t\right)=t^{*}-tdGdλ=−f∗(t∗)dt∗dλ+f(t)dtdλ+λ(dt∗dλ−dtdλ)+(t∗−t)=t∗−t
Ownership follows immediately from the fact that t ∗ ⋛ t t ∗ ⋛ t t^(**)⋛tt^{*} \gtreqless tt∗⋛tfollowing that λ ⋚ f ( x 0 ) λ ⋚ f x 0 lambda⋚f(x_(0))\lambda \lesseqgtr f\left(x_{0}\right)λ⋚f(x0).
2. Now consider the function of ( E n b ) n E n b n (E_(n)^(b))_(n)\left(\mathrm{E}_{n}^{b}\right)_{n}(Enb)n,
h ( x ) = a + a 1 x + a 2 x 2 + … a k x k + ( b − a − ∑ i = 1 k a i ) [ x − μ + | x − μ | 2 ( I − μ ) ] n h ( x ) = a + a 1 x + a 2 x 2 + … a k x k + b − a − ∑ i = 1 k   a i x − μ + | x − μ | 2 ( I − μ ) n h(x)=a+a_(1)x+a_(2)x^(2)+dotsa_(k)x^(k)+(b-a-sum_(i=1)^(k)a_(i))[(x-mu+|x-mu|)/(2(I-mu))]^(n)h(x)=a+a_{1} x+a_{2} x^{2}+\ldots a_{k} x^{k}+\left(b-a-\sum_{i=1}^{k} a_{i}\right)\left[\frac{x-\mu+|x-\mu|}{2(\mathrm{I}-\mu)}\right]^{n}h(x)=has+has1x+has2x2+…haskxk+(b−has−∑i=1khasi)[x−μ+|x−μ|2(I−μ)]n
Or, 0 < μ < I , a i ≧ 0 , i = 1 , 2 , … , k − I , a k > 0 , k ≦ n − I , b − a − ∑ i = 1 k a i > 0 0 < μ < I , a i ≧ 0 , i = 1 , 2 , … , k − I , a k > 0 , k ≦ n − I , b − a − ∑ i = 1 k   a i > 0 0 < mu < I,a_(i) >= 0,quad i=1,2,dots,k-I,quada_(k) > 0,k <= n-I,b-a-sum_(i=1)^(k)a_(i) > 00<\mu<\mathrm{I}, a_{i} \geqq 0, \quad i=1,2, \ldots, k-\mathrm{I}, \quad a_{k}>0, k \leqq n-\mathrm{I}, b-a-\sum_{i=1}^{k} a_{i}>00<μ<I,hasi≧0,i=1,2,…,k−I,hask>0,k≦n−I,b−has−∑i=1khasi>0.
  1. This is only a special case of an inequality which, for the finite case, has been given by Messrs. GH 'Hardy, JE Littlewood, G. Pólya, yoir: ,,Some simple inequalities satisfied by convex /unctions" Messenger of. math. 58 (1929) p. 145-152. See also J. Karamata ,On an inequality relating to convex functions', Publ. Math. Univ. Belgrade t. i (1932) p. 145-148. We will return to this inequality in another work.
  2. I ask the reader to refer to the first note for the assumptions made on the functions and for the notations.
We have
A ( h ) = a + a 1 2 + a 2 3 + … + a k k + 1 + ( b − a − ∑ i = 1 k a l ) ( 1 − μ ) n + 1 . A ( h ) = a + a 1 2 + a 2 3 + … + a k k + 1 + b − a − ∑ i = 1 k   a l ( 1 − μ ) n + 1 . A(h)=a+(a_(1))/(2)+(a_(2))/(3)+dots+(a_(k))/(k+1)+((b-a-sum_(i=1)^(k)a_(l))(1-mu))/(n+1).\mathrm{A}(h)=a+\frac{a_{1}}{2}+\frac{a_{2}}{3}+\ldots+\frac{a_{k}}{k+1}+\frac{\left(b-a-\sum_{i=1}^{k} a_{l}\right)(1-\mu)}{n+1} .HAS(h)=has+has12+has23+…+haskk+1+(b−has−∑i=1khasL)(1−μ)n+1.
Now let the function
h ∗ ( x ) = a + a 1 x + a 2 x 2 + … + a k − 1 x k − 1 + ( a k − ε ) x k + + ( b − a − ∑ i = 1 k a i + ε ) [ x − μ ∗ + | x − μ ∗ | 2 ( 1 − μ ∗ ) ] n h ∗ ( x ) = a + a 1 x + a 2 x 2 + … + a k − 1 x k − 1 + a k − ε x k + + b − a − ∑ i = 1 k   a i + ε x − μ ∗ + x − μ ∗ 2 1 − μ ∗ n {:[h^(**)(x)=a+a_(1)x+a_(2)x^(2)+dots+a_(k-1)x^(k-1)+(a_(k)-epsi)x^(k)+],[+(b-a-sum_(i=1)^(k)a_(i)+epsi)[(x-mu^(**)+|x-mu^(**)|)/(2(1-mu^(**)))]^(n)]:}\begin{aligned} h^{*}(x) & =a+a_{1} x+a_{2} x^{2}+\ldots+a_{k-1} x^{k-1}+\left(a_{k}-\varepsilon\right) x^{k}+ \\ & +\left(b-a-\sum_{i=1}^{k} a_{i}+\varepsilon\right)\left[\frac{x-\mu^{*}+\left|x-\mu^{*}\right|}{2\left(1-\mu^{*}\right)}\right]^{n} \end{aligned}h∗(x)=has+has1x+has2x2+…+hask−1xk−1+(hask−ε)xk++(b−has−∑i=1khasi+ε)[x−μ∗+|x−μ∗|2(1−μ∗)]n
Or μ ∗ = μ − ε n − k + ( k + 1 ) μ ( k + 1 ) ( b − a − ∑ i = 1 k a i + ε ) μ ∗ = μ − ε n − k + ( k + 1 ) μ ( k + 1 ) b − a − ∑ i = 1 k   a i + ε mu^(**)=mu-epsi(n-k+(k+1)mu)/((k+1)(b-a-sum_(i=1)^(k)a_(i)+epsi))\mu^{*}=\mu-\varepsilon \frac{n-k+(k+1) \mu}{(k+1)\left(b-a-\sum_{i=1}^{k} a_{i}+\varepsilon\right)}μ∗=μ−εn−k+(k+1)μ(k+1)(b−has−∑i=1khasi+ε)And ε ε epsi\varepsilonεa positive number such that a k − ε > 0 , b − a − Σ k a i + ε > 0 , 0 < μ ∗ < 1 a k − ε > 0 , b − a − Σ k a i + ε > 0 , 0 < μ ∗ < 1 a_(k)-epsi > 0,b-a-Sigma^(k)a_(i)+epsi > 0,0 < mu^(**) < 1a_{k}-\varepsilon>0, b-a-\Sigma^{k} a_{i}+\varepsilon>0,0<\mu^{*}<1hask−ε>0,b−has−Σkhasi+ε>0,0<μ∗<1, which is quite feasible.
The function h ∗ ∗ h ∗ ∗ h^(**^(**))h^{\stackrel{*}{*}}h∗∗is still an elementary function with at most r vertices belonging ( E a b ) n E a b n (E_(a)^(b))_(n)\left(\mathrm{E}_{a}^{b}\right)_{n}(Ehasb)n. We have μ ∗ < μ 0 μ ∗ < μ 0 mu^(**) < mu_(0)\mu^{*}<\mu_{0}μ∗<μ0And A ( h ∗ ) = A ( h ) A h ∗ = A ( h ) A(h^(**))=A(h)\mathrm{A}\left(h^{*}\right)=\mathrm{A}(h)HAS(h∗)=HAS(h). We also check that the difference h ∗ − h h ∗ − h h^(**)-hh^{*}-hh∗−his zero for x = 0 , I x = 0 , I x=0,Ix=0, \mathrm{I}x=0,I, is negative in the neighborhood of o and can only be zero at a single point outside o and i. I, the property of the previous No. applies and we have
A φ ( h ∗ ) > A φ ( h ) . A φ h ∗ > A φ ( h ) . A_(varphi)(h^(**)) > A_(varphi)(h).\mathrm{A}_{\varphi}\left(h^{*}\right)>\mathrm{A}_{\varphi}(h) .HASφ(h∗)>HASφ(h).
It immediately follows that A φ ( h ∗ ) A φ h ∗ A_(varphi)(h^(**))\mathrm{A}_{\varphi}\left(h^{*}\right)HASφ(h∗)grows when ε ε epsi\varepsilonεgrows. Now, ε ε epsi\varepsilonεcan grow until one of the equalities ε = a k , μ ∗ = 0 ε = a k , μ ∗ = 0 epsi=a_(k),mu^(**)=0\varepsilon=a_{k}, \mu^{*}=0ε=hask,μ∗=0takes place, therefore
When the function φ φ varphi\varphiφAnd 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)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 if f f fffis either a function h λ ( x ) h λ ( x ) h_(lambda)(x)h_{\lambda}(x)hλ(x), or a polynomial of degree at most equal to n n nnn.
Regardless of f f fffIn ( E a b ) n E a b n (E_(a)^(b))_(n)\left(\mathrm{E}_{a}^{b}\right)_{n}(Ehasb)nwe have a ≦ A ( f ) ≦ a + b 2 a ≦ A ( f ) ≦ a + b 2 a <= A(f) <= (a+b)/(2)a \leqq \mathrm{~A}(f) \leqq \frac{a+b}{2}has≦ HAS(f)≦has+b2. For a function h λ ( x ) h λ ( x ) h_(lambda)(x)h_{\lambda}(x)hλ(x)we have
A ( h λ ) = a + ( b − a ) ( r − λ ) n + I A h λ = a + ( b − a ) ( r − λ ) n + I A(h_(lambda))=a+((b-a)(r-lambda))/(n+I)\mathrm{A}\left(h_{\lambda}\right)=a+\frac{(b-a)(\mathrm{r}-\lambda)}{n+\mathrm{I}}HAS(hλ)=has+(b−has)(r−λ)n+I
SO a ≦ A ( h λ ) ≦ n a + b n + I a ≦ A h λ ≦ n a + b n + I a <= A(h_(lambda)) <= (na+b)/(n+I)a \leqq \mathrm{~A}\left(h_{\lambda}\right) \leqq \frac{n a+b}{n+\mathrm{I}}has≦ HAS(hλ)≦nhas+bn+I. It follows that a function h λ h λ h_(lambda)h_{\lambda}hλcan only be maximizing if a ≦ A ( f ) ≦ n a + b n + I a ≦ A ( f ) ≦ n a + b n + I a <= A(f) <= (na+b)/(n+I)a \leqq \mathrm{~A}(f) \leqq \frac{n a+b}{n+\mathrm{I}}has≦ HAS(f)≦nhas+bn+I. It is clear that if A ( f ) > n a + b n + I A ( f ) > n a + b n + I A(f) > (na+b)/(n+I)\mathrm{A}(f)>\frac{n a+b}{n+\mathrm{I}}HAS(f)>nhas+bn+IThe maximizing function (or functions) can only be a polynomial.
3. It remains to find the maximizing polynomials of ( E a b ) n E a b n (E_(a)^(b))_(n)\left(\mathrm{E}_{a}^{b}\right)_{n}(Ehasb)n. Either
P ( x ) = a + a 1 x + a 2 x 2 + … a n x n , P ( x ) = a + a 1 x + a 2 x 2 + … a n x n , P(x)=a+a_(1)x+a_(2)x^(2)+dotsa_(n)x^(n),P(x)=a+a_{1} x+a_{2} x^{2}+\ldots \mathrm{a}_{n} x^{n},P(x)=has+has1x+has2x2+…hasnxn,
a i ≥ 0 , i = I , 2 … , n , ∑ i = 1 n a i = b − a a i ≥ 0 , i = I , 2 … , n , ∑ i = 1 n   a i = b − a a_(i) >= 0,i=I,2dots,n,sum_(i=1)^(n)a_(i)=b-aa_{i} \geq 0, i=\mathrm{I}, 2 \ldots, n, \sum_{i=1}^{n} a_{i}=b-ahasi≥0,i=I,2…,n,∑i=1nhasi=b−has, a polynomial of ( E a b ) n E a b n (E_(a)^(b))^(n)\left(\mathrm{E}_{a}^{b}\right)^{n}(Ehasb)n.
Suppose that the given value of A = A ( f ) A = A ( f ) A=A(f)\mathrm{A}=\mathrm{A}(f)HAS=HAS(f)checks the inequalities
and that A ( P ) = A A ( P ) = A A(P)=A\mathrm{A}(\mathrm{P})=\mathrm{A}HAS(P)=HAS. The polynomial of ( E a b ) n E a b n (E_(a)^(b))_(n)\left(\mathrm{E}_{a}^{b}\right)_{n}(Ehasb)n,
(2) P ∗ ( x ) = a + α x i 1 + β x j (2) P ∗ ( x ) = a + α x i 1 + β x j {:(2)P^(**)(x)=a+alphax^(i)^(1)+betax^(j):}\begin{equation*} \mathrm{P}^{*}(x)=a+\alpha x^{i}{ }^{1}+\beta x^{j} \tag{2} \end{equation*}(2)P∗(x)=has+αxi1+βxI
is then completely determined by the condition A ( P ∗ ) = A ( P ) A P ∗ = A ( P ) A(P^(**))=A(P)\mathrm{A}\left(\mathrm{P}^{*}\right)=\mathrm{A}(\mathrm{P})HAS(P∗)=HAS(P). We have
α = j ( j + r ) ( A − j a + b j + I ) , β = j ( j + r ) [ ( j − I ) a + b j − A ] . α = j ( j + r ) A − j a + b j + I , β = j ( j + r ) ( j − I ) a + b j − A . alpha=j(j+r)(A-(ja+b)/(j+I)),quad beta=j(j+r)[((j-I)a+b)/(j)-A].\alpha=j(j+\mathrm{r})\left(\mathrm{A}-\frac{j a+b}{j+\mathrm{I}}\right), \quad \beta=j(j+\mathrm{r})\left[\frac{(j-\mathrm{I}) a+b}{j}-\mathrm{A}\right] .α=I(I+r)(HAS−Ihas+bI+I),β=I(I+r)[(I−I)has+bI−HAS].
But, the polynomial P ∗ − P P ∗ − P P^(**)-P\mathrm{P}^{*}-\mathrm{P}P∗−Pcan only be cancelled at a single point different from o o oooAnd I I IIIand, moreover, this polynomial is negative in the neighborhood of 0. It follows that A φ ( P ∗ ) > A φ ( P ) A φ P ∗ > A φ ( P ) A_(varphi)(P^(**)) > A_(varphi)(P)\mathrm{A}_{\varphi}\left(\mathrm{P}^{*}\right)>\mathrm{A}_{\varphi}(\mathrm{P})HASφ(P∗)>HASφ(P).
So finally if we have inequality (1), the (unique) maximizing function is the polynomial (2).
4. We can now state our final result:
If φ φ varphi\varphiφis a definite, non-decreasing, non-concave (or non-convex) function in the term interval ( O , I ) ( O , I ) (O,I)(\mathrm{O}, \mathrm{I})(O,I)and if f f fffis a function of ( E a b ) n E a b n (E_(a)^(b))_(n)\left(\mathrm{E}_{a}^{b}\right)_{n}(Ehasb)n, we have the inequality
(3) ∫ o 1 φ ( f ) d x ≦ ( o u ≥ ) ∫ o 1 φ { a + j ( j + 1 ) [ ( A − j a + b j + 1 ) + + ( ( j − 1 ) a + b j − A ) x ] x j } 1 d x si j a + b j + I ≦ A ≦ ( j − 1 ) a + b j , 2 ≦ j ≦ n (3) ∫ o 1   φ ( f ) d x ≦ ( o u ≥ ) ∫ o 1   φ a + j ( j + 1 ) A − j a + b j + 1 + + ( j − 1 ) a + b j − A x x j 1 d x  si  j a + b j + I ≦ A ≦ ( j − 1 ) a + b j , 2 ≦ j ≦ n {:[(3)int_(o)^(1)varphi(f)dx <= (ou >= )int_(o)^(1)varphi{a+j(j+1)[(A-(ja+b)/(j+1))+:}],[+(((j-1)a+b)/(j)-A)x]x^(j)}^(1)dx],[" si "(ja+b)/(j+I) <= A <= ((j-1)a+b)/(j)","quad2 <= j <= n]:}\begin{gather*} \int_{o}^{1} \varphi(f) d x \leqq(o u \geq) \int_{o}^{1} \varphi\left\{a+j(j+1)\left[\left(\mathrm{A}-\frac{j a+b}{j+1}\right)+\right.\right. \tag{3}\\ \left.\left.+\left(\frac{(j-1) a+b}{j}-\mathrm{A}\right) x\right] x^{j}\right\}^{1} d x \\ \text { si } \frac{j a+b}{j+\mathrm{I}} \leqq \mathrm{~A} \leqq \frac{(j-1) a+b}{j}, \quad 2 \leqq j \leqq n \end{gather*}(3)∫o1φ(f)dx≦(ou≥)∫o1φ{has+I(I+1)[(HAS−Ihas+bI+1)++((I−1)has+bI−HAS)x]xI}1dx if Ihas+bI+I≦ HAS≦(I−1)has+bI,2≦I≦n
and inequality
(4) ∫ o 1 φ ( f ) d x ≦ ( o u ≧ ) b + n a − ( n + r ) A b − a φ ( a ) + + ( n + 1 ) ( A − a ) n ( b − a ) b − a n ∫ a b φ ( x ) d x ( x − a ) n − 1 n (4) ∫ o 1   φ ( f ) d x ≦ ( o u ≧ ) b + n a − ( n + r ) A b − a φ ( a ) + + ( n + 1 ) ( A − a ) n ( b − a ) b − a n ∫ a b   φ ( x ) d x ( x − a ) n − 1 n {:[(4)int_(o)^(1)varphi(f)dx <= (ou >= )(b+na-(n+r)A)/(b-a)varphi(a)+],[+((n+1)(A-a))/(n(b-a)root(n)(b-a))int_(a)^(b)(varphi(x)dx)/(root(n)((x-a)^(n-1)))]:}\begin{align*} & \int_{o}^{1} \varphi(f) d x \leqq(o u \geqq) \frac{b+n a-(n+\mathrm{r}) \mathrm{A}}{b-a} \varphi(a)+ \tag{4}\\ & +\frac{(n+1)(\mathrm{A}-a)}{n(b-a) \sqrt[n]{b-a}} \int_{a}^{b} \frac{\varphi(x) d x}{\sqrt[n]{(x-a)^{n-1}}} \end{align*}(4)∫o1φ(f)dx≦(ou≧)b+nhas−(n+r)HASb−hasφ(has)++(n+1)(HAS−has)n(b−has)b−hasn∫hasbφ(x)dx(x−has)n−1n
if a ≦ A ≦ n a + b n + I a ≦ A ≦ n a + b n + I a <= A <= (na+b)/(n+I)a \leqq \mathrm{~A} \leqq \frac{n a+b}{n+\mathrm{I}}has≦ HAS≦nhas+bn+I.
If in addition the function φ φ varphi\varphiφis convex (or concave), the equality in (3) is only possible if
f ( x ) = a + j ( j + I ) [ ( A − j a + b j + I ) + ( ( j − I ) a + b j − A ) x ] x j f ( x ) = a + j ( j + I ) A − j a + b j + I + ( j − I ) a + b j − A x x j f(x)=a+j(j+I)[(A-(ja+b)/(j+I))+(((j-I)a+b)/(j)-A)x]x^(j)f(x)=a+j(j+\mathrm{I})\left[\left(\mathrm{A}-\frac{j a+b}{j+\mathrm{I}}\right)+\left(\frac{(j-\mathrm{I}) a+b}{j}-\mathrm{A}\right) x\right] x^{j}f(x)=has+I(I+I)[(HAS−Ihas+bI+I)+((I−I)has+bI−HAS)x]xI
and the equality in (4) that if
f ( x ) = a + ( b − a ) [ x − λ + ∣ x − λ 2 ( I − λ ) ] n , λ = b + n a − ( n + 1 ) A b − a . f ( x ) = a + ( b − a ) x − λ + ∣ x − λ 2 ( I − λ ) n , λ = b + n a − ( n + 1 ) A b − a . f(x)=a+(b-a)[(x-lambda+∣x-lambda)/(2(I-lambda))]^(n),quad lambda=(b+na-(n+1)A)/(b-a).f(x)=a+(b-a)\left[\frac{x-\lambda+\mid x-\lambda}{2(\mathrm{I}-\lambda)}\right]^{n}, \quad \lambda=\frac{b+n a-(n+1) \mathrm{A}}{b-a} .f(x)=has+(b−has)[x−λ+∣x−λ2(I−λ)]n,λ=b+nhas−(n+1)HASb−has.
If we have A = j a + b j + I A = j a + b j + I A=(ja+b)/(j+I)\mathrm{A}=\frac{j a+b}{j+\mathrm{I}}HAS=Ihas+bI+I, formula (3) or (4) can also be written
∫ o 1 φ ( f ) d x ≦ ( o u ≧ ) 1 j b − a j ∫ n b φ ( x ) d x ( x − a ) j − 1 j ∫ o 1   φ ( f ) d x ≦ ( o u ≧ ) 1 j b − a j ∫ n b   φ ( x ) d x ( x − a ) j − 1 j int_(o)^(1)varphi(f)dx <= (ou >= )(1)/(jroot(j)(b-a))int_(n)^(b)(varphi(x)dx)/(root(j)((x-a)^(j-1)))\int_{o}^{1} \varphi(f) d x \leqq(o u \geqq) \frac{1}{j \sqrt[j]{b-a}} \int_{n}^{b} \frac{\varphi(x) d x}{\sqrt[j]{(x-a)^{j-1}}}∫o1φ(f)dx≦(ou≧)1Ib−hasI∫nbφ(x)dx(x−has)I−1I
  1. In the case n = 0 , f n = 0 , f n=0,fn=0, fn=0,fis a continuous and non-decreasing function. The maximum of A φ ( f ) A φ ( f ) A_(varphi)(f)\mathrm{A}_{\varphi}(f)HASφ(f)is not affected by any function f f fff. We have
max A φ ( f ) = ( b − A ) φ ( a ) + ( A − a ) φ ( b ) b − a max A φ ( f ) = ( b − A ) φ ( a ) + ( A − a ) φ ( b ) b − a maxA_(varphi)(f)=((b-A)varphi(a)+(A-a)varphi(b))/(b-a)\max \mathrm{A}_{\varphi}(f)=\frac{(b-\mathrm{A}) \varphi(a)+(\mathrm{A}-a) \varphi(b)}{b-a}maxHASφ(f)=(b−HAS)φ(has)+(HAS−has)φ(b)b−has
but equality cannot take place if φ φ varphi\varphiφis convex and f f fffcontinuous. The maximum is, in this case, reached by limit functions of ( E a b ) 0 E a b 0 (E_(a)^(b))_(0)\left(\mathrm{E}_{a}^{b}\right)_{0}(Ehasb)0. Such a maximizing function is, for example, the function
f ( x ) = { a , 0 ≦ x ≦ b − A b − a b , b − A b − a ≦ x ≦ I f ( x ) = a ,      0 ≦ x ≦ b − A b − a b ,      b − A b − a ≦ x ≦ I f(x)={[a",",0 <= x <= (b-A)/(b-a)],[b",",(b-A)/(b-a) <= x <= I]:}f(x)= \begin{cases}a, & 0 \leqq x \leqq \frac{b-\mathrm{A}}{b-a} \\ b, & \frac{b-\mathrm{A}}{b-a} \leqq x \leqq \mathrm{I}\end{cases}f(x)={has,0≦x≦b−HASb−hasb,b−HASb−has≦x≦I
If φ φ varphi\varphiφis a concave function we have the same property for the minimum of the integral A φ ( f ) A φ ( f ) A_(varphi)(f)\mathrm{A}_{\varphi}(f)HASφ(f).
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