On the finite increments formula

Tiberiu Popoviciu
(original), Mathematical Analysis, mean value results, paper

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T. Popoviciu

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T. Popoviciu, Sur la formule des accroissements finis, Mathematica, 23 (1947-1948), pp. 123-126 (in French).

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1948 c -Popoviciu- Mathematica – Sur la formule des accroissements finis (2)

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1948 c -Popoviciu- Mathematica - On the formula for finite increments (2)
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MATHEMATICA

VOLUMUL XXIII1947-1948

PAGE 123-126

TIBERIU POPOVICIU:

ON THE FORMULA FOR FINITE INCREASES

ON THE FORMULA FOR FINITE INCREASES

BY

TIBERIU POPOVICIU

Received on April 27, 1948

    • Either f ( x ) f ( x ) f(x)f(x)f(x)a continuous function in the bounded and closed interval [ has , b ] [ has , b ] [a,b][a, b][has,b]. We will designate by X the abscissa point x x xxxof the real axis and by X ′ X ′ X^(')X^{\prime}X′the coordinate point ( x , f ( x ) ) ( x , f ( x ) ) (x,f(x))(x, f(x))(x,f(x)). Especially HAS , HAS ′ , B , B ′ HAS , HAS ′ , B , B ′ A,A^('),B,B^(')A, A^{\prime}, B, B^{\prime}HAS,HAS′,B,B′are the points ( has , 0 ) , ( has , f ( has ) , ( b , 0 ) , ( b , f ( b ) ) ( has , 0 ) , ( has , f ( has ) , ( b , 0 ) , ( b , f ( b ) ) (a,0),(a,f(a),(b,0),(b,f(b))(a, 0),(a, f(a),(b, 0),(b, f(b))(has,0),(has,f(has),(b,0),(b,f(b)). We will also designate by
[ x 1 , x 2 ; f ] = f ( x 2 ) − f ( x 1 ) x 2 − x 1 x 1 , x 2 ; f = f x 2 − f x 1 x 2 − x 1 [x_(1),x_(2);f]=(f(x_(2))-f(x_(1)))/(x_(2)-x_(1))\left[x_{1}, x_{2}; f\right]=\frac{f\left(x_{2}\right)-f\left(x_{1}\right)}{x_{2}-x_{1}}[x1,x2;f]=f(x2)−f(x1)x2−x1
the divided difference of the function f ( x ) f ( x ) f(x)f(x)f(x)on the points x 1 , x 2 x 1 , x 2 x_(1),x_(2)x_{1}, x_{2}x1,x2and by
[ x 1 , x 2 , x 3 ; f ] = [ x 2 , x 3 ; f ] − [ x 1 , x 2 ; f ] x 3 − x 1 x 1 , x 2 , x 3 ; f = x 2 , x 3 ; f − x 1 , x 2 ; f x 3 − x 1 [x_(1),x_(2),x_(3);f]=([x_(2),x_(3);f]-[x_(1),x_(2);f])/(x_(3)-x_(1))\left[x_{1}, x_{2}, x_{3}; f\right]=\frac{\left[x_{2}, x_{3}; f\right]-\left[x_{1}, x_{2}; f\right]}{x_{3}-x_{1}}[x1,x2,x3;f]=[x2,x3;f]−[x1,x2;f]x3−x1
the divided difference of this same function on the points x 1 , x 2 , x 8 x 1 , x 2 , x 8 x_(1),x_(2),x_(8)x_{1}, x_{2}, x_{8}x1,x2,x8.
The area of ​​the trapezoid X X ′ Y ′ Y ( x ≦ y ) X X ′ Y ′ Y ( x ≦ y ) XX^(')Y^(')Y(x <= y)\mathrm{X} \mathrm{X}^{\prime} \mathrm{Y}^{\prime} \mathrm{Y}(x \leqq y)XX′Y′Y(x≦y)is then, by definition, equal to
φ ( x , y ) = 1 2 [ f ( x ) + f ( y ) ] ( y − x ) φ ( x , y ) = 1 2 [ f ( x ) + f ( y ) ] ( y − x ) varphi(x,y)=(1)/(2)[f(x)+f(y)](yx)\varphi(x, y)=\frac{1}{2}[f(x)+f(y)](yx)φ(x,y)=12[f(x)+f(y)](y−x)
The formula for finite increments results from the observation that at least one of the absolute extrema (maximum or minimum) of the sum of the areas of the trapezoids HAS HAS ′ X ′ X , X X ′ B ′ B HAS HAS ′ X ′ X , X X ′ B ′ B AA^(')X^(')X,XX^(')B^(')BAA^{\prime} X^{\prime}HASHAS′X′X,XX′B′Bis necessarily reached at least at one point of the open interval ( has , b ) ( has , b ) (a,b)(a, b)(has,b). In fact, this sum is equal to
F 1 ( x ) = 1 2 [ f ( has ) + f ( x ) ] ( x − has ) + 1 2 [ f ( x ) + f ( b ) ] ( b − x ) F 1 ( x ) = 1 2 [ f ( has ) + f ( x ) ] ( x − has ) + 1 2 [ f ( x ) + f ( b ) ] ( b − x ) F_(1)(x)=(1)/(2)[f(a)+f(x)](xa)+(1)/(2)[f(x)+f(b)](bx)\mathrm{F}_{1}(x)=\frac{1}{2}[f(a)+f(x)](xa)+\frac{1}{2}[f(x)+f(b)](bx)F1(x)=12[f(has)+f(x)](x−has)+12[f(x)+f(b)](b−x)
And F 1 ( x ) F 1 ( x ) F_(1)(x)\mathrm{F}_{1}(x)F1(x)is a continuous function of x x xxxIn [ has , b ] [ has , b ] [a,b][a, b][has,b], taking the same value (1)
F 0 = φ ( has , b ) F 0 = φ ( has , b ) F_(0)=varphi(a,b)\mathrm{F}_{0}=\varphi(a, b)F0=φ(has,b)
at the ends has , b has , b a,ba, bhas,bof this interval. If the derivative f ′ ( x ) f ′ ( x ) f^(')(x)f^{\prime}(x)f′(x)exists in the open interval ( has , b has , b a,ba, bhas,b), we conclude the existence of at least one x ∈ ( has , b ) x ∈ ( has , b ) x in(a,b)x \in(a, b)x∈(has,b)such that we have F 1 ′ ( x ) = 0 F 1 ′ ( x ) = 0 F_(1)^(')(x)=0F_{1}^{\prime}(x)=0F1′(x)=0, SO
f ′ ( x ) = [ has , b ; f ] = f ( b ) − f ( has ) b − has f ′ ( x ) = [ has , b ; f ] = f ( b ) − f ( has ) b − has f^(')(x)=[a,b;f]=(f(b)-f(a))/(ba)f^{\prime}(x)=[a, b ; f]=\frac{f(b)-f(a)}{ba}f′(x)=[has,b;f]=f(b)−f(has)b−has
    • Let us consider n n nnnpoints x 1 , x 2 , … , x n x 1 , x 2 , … , x n x_(1),x_(2),dots,x_(n)x_{1}, x_{2}, \ldots, x_{n}x1,x2,…,xnin the meantime [ has , b ] [ has , b ] [a,b][a, b][has,b]and let's still assume that has ≦ x 1 ≦ x 2 ≦ ⋯ ≦ x n ≦ b has ≦ x 1 ≦ x 2 ≦ ⋯ ≦ x n ≦ b a <= x_(1) <= x_(2) <= cdots <= x_(n) <= ba \leqq x_{1} \leqq x_{2} \leqq \cdots \leqq x_{n} \leqq bhas≦x1≦x2≦⋯≦xn≦b. The sum of the areas of the trapezoids HAS HAS ′ X 1 ′ X 1 , X 1 X 1 ′ X 2 ′ X 2 , … , X n − 1 X n − 1 ′ X n ′ X n , X n X n ′ B ′ B HAS HAS ′ X 1 ′ X 1 , X 1 X 1 ′ X 2 ′ X 2 , … , X n − 1 X n − 1 ′ X n ′ X n , X n X n ′ B ′ B AA^(')X_(1)^(')X_(1),X_(1)X_(1)^(')X_(2)^(')X_(2),dots,X_(n-1)X_(n-1)^(')X_(n)^(')X_(n),X_(n)X_(n)^(')B^(')BA A^{\prime} X_{1}^{\prime} X_{1}, X_{1} X_{1}^{\prime} X_{2}^{\prime} X_{2}, \ldots, X_{n-1} X_{n-1}^{\prime} X_{n}^{\prime} X_{n}, X_{n} X_{n}^{\prime} B^{\prime} BHASHAS′X1′X1,X1X1′X2′X2,…,Xn−1Xn−1′Xn′Xn,XnXn′B′Bis equal to
(2) F n ( M ) = F n ( x 1 , x 2 , … , x n ) = = 1 2 ∑ i = 0 n [ f ( x i ) + f ( x i + 1 ) ] ( x i + 1 − x i ) = ∑ i = 0 n φ ( x i , x i + 1 ) (2) F n ( M ) = F n x 1 , x 2 , … , x n = = 1 2 ∑ i = 0 n   f x i + f x i + 1 x i + 1 − x i = ∑ i = 0 n   φ x i , x i + 1 {:[(2)F_(n)(M)=F_(n)(x_(1),x_(2),dots,x_(n))=],[=(1)/(2)sum_(i=0)^(n)[f(x_(i))+f(x_(i+1))](x_(i+1)-x_(i))=sum_(i=0)^(n)varphi(x_(i),x_(i+1))]:}\begin{align*} \mathrm{F}_{n}(\mathrm{M}) & =\mathrm{F}_{n}\left(x_{1}, x_{2}, \ldots, x_{n}\right)= \tag{2}\\ & =\frac{1}{2} \sum_{i=0}^{n}\left[f\left(x_{i}\right)+f\left(x_{i+1}\right)\right]\left(x_{i+1}-x_{i}\right)=\sum_{i=0}^{n} \varphi\left(x_{i}, x_{i+1}\right) \end{align*}(2)Fn(M)=Fn(x1,x2,…,xn)==12∑i=0n[f(xi)+f(xi+1)](xi+1−xi)=∑i=0nφ(xi,xi+1)
by posing x 0 = a , x n + 1 = b x 0 = a , x n + 1 = b x_(0)=a,x_(n+1)=bx_{0}=a, x_{n+1}=bx0=has,xn+1=band designating by M the point with coordinates ( x 1 , x 2 , … , x n ) x 1 , x 2 , … , x n (x_(1),x_(2),dots,x_(n))\left(x_{1}, x_{2}, \ldots, x_{n}\right)(x1,x2,…,xn)in ordinary space at n n nnndimensions. F n ( x 1 , x 2 , … , x n ) F n x 1 , x 2 , … , x n F_(n)(x_(1),x_(2),dots,x_(n))\mathrm{F}_{n}\left(x_{1}, x_{2}, \ldots, x_{n}\right)Fn(x1,x2,…,xn)is then a continuous function of x 1 , x 2 , … , x n x 1 , x 2 , … , x n x_(1),x_(2),dots,x_(n)x_{1}, x_{2}, \ldots, x_{n}x1,x2,…,xnin a limited and closed area D D DDD. This domain D D DDDis formed by a simplex M 0 M 1 … M n M 0 M 1 … M n M_(0)M_(1)dotsM_(n)M_{0} M_{1} \ldots M_{n}M0M1…Mnwhose vertices are the points
M i ( a , a , … , a ⏟ i , b , b , … , b ⏟ n − i ) , i = 0 , 1 , … , n ( M 0 ( b , b , … , b ) , M n ( a , a , … , a ) ) . M i ( a , a , … , a ⏟ i , b , b , … , b ⏟ n − i ) , i = 0 , 1 , … , n M 0 ( b , b , … , b ) , M n ( a , a , … , a ) . {:[M_(i)ubrace((a,a,dots,aubrace)_(i)","ubrace(b,b,dots,bubrace)_(n-i))","quad i=0","1","dots","n],[(M_(0)(b,b,dots,b),quadM_(n)(a,a,dots,a)).]:}\begin{aligned} & \mathrm{M}_{i} \underbrace{(a, a, \ldots, a}_{i},\underbrace{b, b, \ldots, b}_{n-i}), \quad i=0,1, \ldots, n \\ &\left(\mathrm{M}_{0}(b, b, \ldots, b), \quad \mathrm{M}_{n}(a, a, \ldots, a)\right) . \end{aligned}Mi(has,has,…,has⏟i,b,b,…,b⏟n−i),i=0,1,…,n(M0(b,b,…,b),Mn(has,has,…,has)).
The interior points M ( x 1 , x 2 , … , x n ) M x 1 , x 2 , … , x n M(x_(1),x_(2),dots,x_(n))M\left(x_{1}, x_{2}, \ldots, x_{n}\right)M(x1,x2,…,xn)of D D DDDare then characterized by inequalities a < x 1 < x 2 < ⋯ < x n < b a < x 1 < x 2 < ⋯ < x n < b a < x_(1) < x_(2) < cdots < x_(n) < ba<x_{1}<x_{2}<\cdots<x_{n}<bhas<x1<x2<⋯<xn<b. Point M belongs to the face D i D i D_(i)D_{i}Diopposite the summit M i M i M_(i)M_{i}Miof D D DDDif x i = x i + 1 x i = x i + 1 x_(i)=x_(i+1)x_{i}=x_{i+1}xi=xi+1and this point M M MMMis an interior point of this face if a < x 1 < ⋯ < x i = x i + 1 < x i + 2 << ⋯ < x n < b a < x 1 < ⋯ < x i = x i + 1 < x i + 2 << ⋯ < x n < b a < x_(1) < cdots < x_(i)=x_(i+1) < x_(i+2)<<cdots < x_(n) < ba<x_{1}<\cdots<x_{i}=x_{i+1}<x_{i+2}< <\cdots<x_{n}<bhas<x1<⋯<xi=xi+1<xi+2<<⋯<xn<b.
From formula (2) it immediately follows that if M ( x 1 , x 2 , … , x n ) M x 1 , x 2 , … , x n M(x_(1),x_(2),dots,x_(n))M\left(x_{1}, x_{2}, \ldots, x_{n}\right)M(x1,x2,…,xn)is on the face D i D i D_(i)D_{i}Diwe have
F n ( M ) = F n − 1 ( x 1 , x 2 , … , x i , x i + 2 , x i + 3 , … , x n ) F n ( M ) = F n − 1 x 1 , x 2 , … , x i , x i + 2 , x i + 3 , … , x n F_(n)(M)=F_(n-1)(x_(1),x_(2),dots,x_(i),x_(i+2),x_(i+3),dots,x_(n))\mathrm{F}_{n}(\mathrm{M})=\mathrm{F}_{n-1}\left(x_{1}, x_{2}, \ldots, x_{i}, x_{i+2}, x_{i+3}, \ldots, x_{n}\right)Fn(M)=Fn−1(x1,x2,…,xi,xi+2,xi+3,…,xn)
In particular we have
(3)
F n ( a , x 1 , x 2 , … , x n − 1 ) = F n ( x 1 , x 1 , x 2 , … , x n − 1 ) = = F n ( x 1 , x 2 , x 2 , x 3 , … , x n − 1 ) = … = F ( x 1 , x 2 , … , x n − 2 , x n − 1 , x n − 1 ) = = F n ( x 1 , x 2 , … , x n − 1 , b ) = F n − 1 ( x 1 , x 2 , … , x n − 1 ) , F n a , x 1 , x 2 , … , x n − 1 = F n x 1 , x 1 , x 2 , … , x n − 1 = = F n x 1 , x 2 , x 2 , x 3 , … , x n − 1 = … = F x 1 , x 2 , … , x n − 2 , x n − 1 , x n − 1 = = F n x 1 , x 2 , … , x n − 1 , b = F n − 1 x 1 , x 2 , … , x n − 1 , {:[F_(n)(a,x_(1),x_(2),dots,x_(n-1))=F_(n)(x_(1),x_(1),x_(2),dots,x_(n-1))=],[=F_(n)(x_(1),x_(2),x_(2),x_(3),dots,x_(n-1))=dots=F(x_(1),x_(2),dots,x_(n-2),x_(n-1),x_(n-1))=],[=F_(n)(x_(1),x_(2),dots,x_(n-1),b)=F_(n-1)(x_(1),x_(2),dots,x_(n-1))","]:}\begin{aligned} & \mathrm{F}_{n}\left(a, x_{1}, x_{2}, \ldots, x_{n-1}\right)=\mathrm{F}_{n}\left(x_{1}, x_{1}, x_{2}, \ldots, x_{n-1}\right)= \\ = & \mathrm{F}_{n}\left(x_{1}, x_{2}, x_{2}, x_{3}, \ldots, x_{n-1}\right)=\ldots=\mathrm{F}\left(x_{1}, x_{2}, \ldots, x_{n-2}, x_{n-1}, x_{n-1}\right)= \\ = & \mathrm{F}_{n}\left(x_{1}, x_{2}, \ldots, x_{n-1}, b\right)=\mathrm{F}_{n-1}\left(x_{1}, x_{2}, \ldots, x_{n-1}\right), \end{aligned}Fn(has,x1,x2,…,xn−1)=Fn(x1,x1,x2,…,xn−1)==Fn(x1,x2,x2,x3,…,xn−1)=…=F(x1,x2,…,xn−2,xn−1,xn−1)==Fn(x1,x2,…,xn−1,b)=Fn−1(x1,x2,…,xn−1),
which shows us that F n ( M ) F n ( M ) F_(n)(M)F_{n}(M)Fn(M)takes the same values ​​on the faces D i D i D_(i)D_{i}Di' i = 0 , 1 , … , n i = 0 , 1 , … , n i=0,1,dots,ni=0,1, \ldots, ni=0,1,…,nThe geometric interpretation of this fact is, moreover, very simple.
Finally we can also note that
(4)
F n ( M i ) = F 0 , i = 0 , 1 , … , n , F n M i = F 0 , i = 0 , 1 , … , n , F_(n)(M_(i))=F_(0),quad i=0,1,dots,n,\mathrm{F}_{n}\left(\mathrm{M}_{i}\right)=\mathrm{F}_{0}, \quad i=0,1, \ldots, n,Fn(Mi)=F0,i=0,1,…,n,
F 0 F 0 F_(0)\mathrm{F}_{0}F0being defined by (1).
3. - This being so, we can demonstrate the
Theorem 1. - If the function f ( x ) f ( x ) f(x)f(x)f(x)is continuous in the closed interval [ a , b ] [ a , b ] [a,b][a, b][has,b]and is differentiable in the open interval ( a , b ) ( a , b ) (a,b)(a, b)(has,b), the function
F n ( x 1 , x 2 , … , x n ) F n x 1 , x 2 , … , x n F_(n)(x_(1),x_(2),dots,x_(n))\mathrm{F}_{n}\left(x_{1}, x_{2}, \ldots, x_{n}\right)Fn(x1,x2,…,xn)reaches at least one of its absolute extrema at at least one interior point of the domain D .
The function F n ( M ) F n ( M ) F_(n)(M)\mathrm{F}_{n}(\mathrm{M})Fn(M)takes the value F 0 F 0 F_(0)\mathrm{F}_{0}F0. If this function reduces to a constant the theorem is demonstrated. Otherwise, and to clarify the ideas, let us suppose that F n ( M ) F n ( M ) F_(n)(M)F_{n}(M)Fn(M)take values > F 0 > F 0 > F_(0)>F_{0}>F0. It is then sufficient to demonstrate that the maximum of F n ( M ) F n ( M ) F_(n)(M)F_{n}(M)Fn(M)is reached at an interior point of D.
We can now prove the theorem by complete induction on n n nnn. Suppose the theorem is true for the function F n − 1 ( x 1 , x 2 , … , x n − 1 ) F n − 1 x 1 , x 2 , … , x n − 1 F_(n-1)(x_(1),x_(2),dots,x_(n-1))\mathrm{F}_{n-1}\left(x_{1}, x_{2}, \ldots, x_{n-1}\right)Fn−1(x1,x2,…,xn−1). We will demonstrate that the maximum of F n ( M ) F n ( M ) F_(n)(M)\mathrm{F}_{n}(\mathrm{M})Fn(M)can only be reached on the faces D i D i D_(i)D_{i}Diof the domain D D DDDOtherwise this maximum would be reached at an interior point of the face. D i D i D_(i)D_{i}Di, as a result of equalities (3), (4) and as a result of the hypothesis that the theorem is verified for the function F n − 1 F n − 1 F_(n-1)\mathrm{F}_{n-1}Fn−1. Either ( x 1 , x 2 , … , x i , x i , x i + 1 , … , x n − 1 ) x 1 , x 2 , … , x i , x i , x i + 1 , … , x n − 1 (x_(1),x_(2),dots,x_(i),x_(i),x_(i+1),dots,x_(n-1))\left(x_{1}, x_{2}, \ldots, x_{i}, x_{i}, x_{i+1}, \ldots, x_{n-1}\right)(x1,x2,…,xi,xi,xi+1,…,xn−1)an interior point of D i D i D_(i)D_{i}Diwhere the maximum is reached. We have a < x 1 << x 2 < … < x n − 1 < b a < x 1 << x 2 < … < x n − 1 < b a < x_(1)<<x_(2) < dots < x_(n-1) < ba<x_{1}< <x_{2}<\ldots<x_{n-1}<bhas<x1<<x2<…<xn−1<band the fact that inside D F n ( M ) D F n ( M ) DF_(n)(M)\mathrm{D} \mathrm{F}_{n}(\mathrm{M})DFn(M)takes smaller values ​​it results that
F n ( x 1 , x 2 , … , x i , t , x i + 1 , … , x n − 1 ) < F n − 1 ( x 1 , x 2 , … , x n − 1 ) , x i < t < x i + 1 F n x 1 , x 2 , … , x i , t , x i + 1 , … , x n − 1 < F n − 1 x 1 , x 2 , … , x n − 1 , x i < t < x i + 1 F_(n)(x_(1),x_(2),dots,x_(i),t,x_(i+1),dots,x_(n-1)) < F_(n-1)(x_(1),x_(2),dots,x_(n-1)),x_(i) < t < x_(i+1)\mathrm{F}_{n}\left(x_{1}, x_{2}, \ldots, x_{i}, t, x_{i+1}, \ldots, x_{n-1}\right)<\mathrm{F}_{n-1}\left(x_{1}, x_{2}, \ldots, x_{n-1}\right), x_{i}<t<x_{i+1}Fn(x1,x2,…,xi,t,xi+1,…,xn−1)<Fn−1(x1,x2,…,xn−1),xi<t<xi+1,
i = 0 , 1 , … , n − 1 i = 0 , 1 , … , n − 1 i=0,1,dots,n-1i=0,1, \ldots, n-1i=0,1,…,n−1
Or
φ ( x i , t ) + φ ( t , x i + 1 ) < φ ( x i , x i + 1 ) , x i < t < x i + 1 , i = 0 , 1 , … , 1 − n φ x i , t + φ t , x i + 1 < φ x i , x i + 1 , x i < t < x i + 1 , i = 0 , 1 , … , 1 − n varphi(x_(i),t)+varphi(t,x_(i+1)) < varphi(x_(i),x_(i+1)),x_(i) < t < x_(i+1),quad i=0,1,dots,1-n\varphi\left(x_{i}, t\right)+\varphi\left(t, x_{i+1}\right)<\varphi\left(x_{i}, x_{i+1}\right), x_{i}<t<x_{i+1}, \quad i=0,1, \ldots, 1-nφ(xi,t)+φ(t,xi+1)<φ(xi,xi+1),xi<t<xi+1,i=0,1,…,1−n. A simple calculation gives us
x i + 1 ) − φ ( x i , t ) − φ ( t , x i + 1 ) = = 1 2 ( x i + 1 − x i ) ( t − x i ) ( x i + 1 − t ) [ x i , t , x i + 1 ; f ] = = 1 2 ( x i + 1 − x i ) ( t − x i ) { [ x i , x i + 1 ; f ] − [ x i , t ; f ] } = = 1 2 ( x i + 1 − x i ) ( t − x i + 1 ) { [ x i , x i + 1 ; f ] − [ x i + 1 , t ; f ] } x i + 1 − φ x i , t − φ t , x i + 1 = = 1 2 x i + 1 − x i t − x i x i + 1 − t x i , t , x i + 1 ; f = = 1 2 x i + 1 − x i t − x i x i , x i + 1 ; f − x i , t ; f = = 1 2 x i + 1 − x i t − x i + 1 x i , x i + 1 ; f − x i + 1 , t ; f {:[{:x_(i+1))-varphi(x_(i),t)-varphi(t,x_(i+1))=],[=(1)/(2)(x_(i+1)-x_(i))(t-x_(i))(x_(i+1)-t)[x_(i),t,x_(i+1);f]=],[=(1)/(2)(x_(i+1)-x_(i))(t-x_(i)){[x_(i),x_(i+1);f]-[x_(i),t;f]}=],[=(1)/(2)(x_(i+1)-x_(i))(t-x_(i+1)){[x_(i),x_(i+1);f]-[x_(i+1),t;f]}]:}\begin{aligned} & \left.x_{i+1}\right)-\varphi\left(x_{i}, t\right)-\varphi\left(t, x_{i+1}\right)= \\ & =\frac{1}{2}\left(x_{i+1}-x_{i}\right)\left(t-x_{i}\right)\left(x_{i+1}-t\right)\left[x_{i}, t, x_{i+1} ; f\right]= \\ & =\frac{1}{2}\left(x_{i+1}-x_{i}\right)\left(t-x_{i}\right)\left\{\left[x_{i}, x_{i+1} ; f\right]-\left[x_{i}, t ; f\right]\right\}= \\ & =\frac{1}{2}\left(x_{i+1}-x_{i}\right)\left(t-x_{i+1}\right)\left\{\left[x_{i}, x_{i+1} ; f\right]-\left[x_{i+1}, t ; f\right]\right\} \end{aligned}xi+1)−φ(xi,t)−φ(t,xi+1)==12(xi+1−xi)(t−xi)(xi+1−t)[xi,t,xi+1;f]==12(xi+1−xi)(t−xi){[xi,xi+1;f]−[xi,t;f]}==12(xi+1−xi)(t−xi+1){[xi,xi+1;f]−[xi+1,t;f]}
and inequality (5) allows us to write
[ x i , t ; f ] < [ x i , x i + 1 ; f ] < [ x i + 1 , t ; f ] , x i < t < x i + 1 , i = 0 , 1 , … , n − 1 x i , t ; f < x i , x i + 1 ; f < x i + 1 , t ; f , x i < t < x i + 1 , i = 0 , 1 , … , n − 1 [x_(i),t;f] < [x_(i),x_(i+1);f] < [x_(i+1),t;f],quadx_(i) < t < x_(i+1),quad i=0,1,dots,n-1\left[x_{i}, t ; f\right]<\left[x_{i}, x_{i+1} ; f\right]<\left[x_{i+1}, t ; f\right], \quad x_{i}<t<x_{i+1}, \quad i=0,1, \ldots, n-1[xi,t;f]<[xi,xi+1;f]<[xi+1,t;f],xi<t<xi+1,i=0,1,…,n−1.
Making it tender t t ttttowards x i x i x_(i)x_{i}xithen towards x i + 1 x i + 1 x_(i+1)x_{i+1}xi+1, we deduce
from where
f ′ ( x i ) ≦ [ x i , x i + 1 ; f ] , i = 1 , 2 , … , n − 1 [ x i , x i + 1 ; f ] ≦ f ′ ( x i + 1 ) , i = 0 , 1 , … , n − 2 f ′ x i ≦ x i , x i + 1 ; f ,      i = 1 , 2 , … , n − 1 x i , x i + 1 ; f ≦ f ′ x i + 1 ,      i = 0 , 1 , … , n − 2 {:[f^(')(x_(i)) <= [x_(i),x_(i+1);f]",",i=1","2","dots","n-1],[[x_(i),x_(i+1);f] <= f^(')(x_(i+1))",",i=0","1","dots","n-2]:}\begin{array}{ll} f^{\prime}\left(x_{i}\right) \leqq\left[x_{i}, x_{i+1} ; f\right], & i=1,2, \ldots, n-1 \\ {\left[x_{i}, x_{i+1} ; f\right] \leqq f^{\prime}\left(x_{i+1}\right),} & i=0,1, \ldots, n-2 \end{array}f′(xi)≦[xi,xi+1;f],i=1,2,…,n−1[xi,xi+1;f]≦f′(xi+1),i=0,1,…,n−2
or else
[ a , x 1 ; f ] ≦ [ x 1 , x 2 , ; f ] ≦ [ x 2 , x 3 ; f ] ≦ … ≦ [ x n − 1 , b ; f ] a , x 1 ; f ≦ x 1 , x 2 , ; f ≦ x 2 , x 3 ; f ≦ … ≦ x n − 1 , b ; f [a,x_(1);f] <= [x_(1),x_(2),;f] <= [x_(2),x_(3);f] <= dots <= [x_(n-1),b;f]\left[a, x_{1} ; f\right] \leqq\left[x_{1}, x_{2}, ; f\right] \leqq\left[x_{2}, x_{3} ; f\right] \leqq \ldots \leqq\left[x_{n-1}, b ; f\right][has,x1;f]≦[x1,x2,;f]≦[x2,x3;f]≦…≦[xn−1,b;f]
(7)
[ x i , x i + 1 , x i + 2 ; f ] ≧ 0 , i == 0 , 1 , … , n − 2 . x i , x i + 1 , x i + 2 ; f ≧ 0 , i == 0 , 1 , … , n − 2 . [x_(i),x_(i+1),x_(i+2);f] >= 0,quad i==0,1,dots,n-2.\left[x_{i}, x_{i+1}, x_{i+2} ; f\right] \geqq 0, \quad i==0,1, \ldots, n-2 .[xi,xi+1,xi+2;f]≧0,i==0,1,…,n−2.
The function f ( x ) f ( x ) f(x)f(x)f(x)is therefore non-concave (of order 1) on the points a a aahas, x 1 , … , x n − 1 , b x 1 , … , x n − 1 , b x_(1),dots,x_(n-1),bx_{1}, \ldots, x_{n-1}, bx1,…,xn−1,b. We then know that we also have
(8)
[ a , x i , b ; f ] ≧ 0 , i = 1 , 2 , … , n − 1 a , x i , b ; f ≧ 0 , i = 1 , 2 , … , n − 1 [a,x_(i),b;f] >= 0,quad i=1,2,dots,n-1\left[a, x_{i}, b ; f\right] \geqq 0, \quad i=1,2, \ldots, n-1[has,xi,b;f]≧0,i=1,2,…,n−1
But, I say that these inequalities are in contradiction with
(9) F n − 1 ( x 1 , x 2 , … , x n − 1 ) > F 0 (9) F n − 1 x 1 , x 2 , … , x n − 1 > F 0 {:(9)F_(n-1)(x_(1),x_(2),dots,x_(n-1)) > F_(0):}\begin{equation*} \mathrm{F}_{n-1}\left(x_{1}, x_{2}, \ldots, x_{n-1}\right)>\mathrm{F}_{0} \tag{9} \end{equation*}(9)Fn−1(x1,x2,…,xn−1)>F0
which results from the hypothesis made on the points x 1 , x 2 , … , x n − 1 x 1 , x 2 , … , x n − 1 x_(1),x_(2),dots,x_(n-1)x_{1}, x_{2}, \ldots, x_{n-1}x1,x2,…,xn−1. Indeed, a simple calculation gives us
∑ i = 1 n − 1 [ φ ( a , x i ) + φ ¯ ( x i , b ) − φ ( a , b ) ] = ( b − a ) [ F n − 1 ( x 1 , x 2 , … , x n − 1 ) − F 0 ] ∑ i = 1 n − 1   φ a , x i + φ ¯ x i , b − φ ( a , b ) = ( b − a ) F n − 1 x 1 , x 2 , … , x n − 1 − F 0 sum_(i=1)^(n-1)[varphi(a,x_(i))+( bar(varphi))(x_(i),b)-varphi(a,b)]=(b-a)[F_(n-1)(x_(1),x_(2),dots,x_(n-1))-F_(0)]\sum_{i=1}^{n-1}\left[\varphi\left(a, x_{i}\right)+\bar{\varphi}\left(x_{i}, b\right)-\varphi(a, b)\right]=(b-a)\left[\mathrm{F}_{n-1}\left(x_{1}, x_{2}, \ldots, x_{n-1}\right)-\mathrm{F}_{0}\right]∑i=1n−1[φ(has,xi)+φ¯(xi,b)−φ(has,b)]=(b−has)[Fn−1(x1,x2,…,xn−1)−F0]
and, taking into account (6),
− 1 2 ∑ i = 1 n − 1 ( x i − a ) ( b − x i ) [ a , x i , b ; f ] = F n − 1 ( x 1 , x 2 , … , x n − 1 ) − F 0 − 1 2 ∑ i = 1 n − 1   x i − a b − x i a , x i , b ; f = F n − 1 x 1 , x 2 , … , x n − 1 − F 0 -(1)/(2)sum_(i=1)^(n-1)(x_(i)-a)(b-x_(i))[a,x_(i),b;f]=F_(n-1)(x_(1),x_(2),dots,x_(n-1))-F_(0)-\frac{1}{2} \sum_{i=1}^{n-1}\left(x_{i}-a\right)\left(b-x_{i}\right)\left[a, x_{i}, b ; f\right]=\mathrm{F}_{n-1}\left(x_{1}, x_{2}, \ldots, x_{n-1}\right)-\mathrm{F}_{0}−12∑i=1n−1(xi−has)(b−xi)[has,xi,b;f]=Fn−1(x1,x2,…,xn−1)−F0
which proves our assertion.
This contradiction demonstrates Theorem 1 since for n = 1 n = 1 n=1n=1n=1the property results easily.
The example of the function
f ( x ) = { ( x − a ) 2 , x ∈ [ a , a + b 2 ] ( x − b ) 2 , x ∈ [ a + b 2 , b ] f ( x ) = ( x − a ) 2 ,      x ∈ a , a + b 2 ( x − b ) 2 ,      x ∈ a + b 2 , b f(x)={[(x-a)^(2)",",x in[a,(a+b)/(2)]],[(x-b)^(2)",",x in[(a+b)/(2),b]]:}f(x)= \begin{cases}(x-a)^{2}, & x \in\left[a, \frac{a+b}{2}\right] \\ (x-b)^{2}, & x \in\left[\frac{a+b}{2}, b\right]\end{cases}f(x)={(x−has)2,x∈[has,has+b2](x−b)2,x∈[has+b2,b]
shows us that the property expressed by Theorem 1 is no longer true in general if the function f f fffis not differentiable at every point in the interval ( a , b ) ( a , b ) (a,b)(a, b)(has,b).
4. - We have
∂ F n ( x 1 , x 2 , … , x n ) ∂ x i = 1 2 [ f ′ ( x i ) ( x i + 1 − x i − 1 ) − f ( x i + 1 ) + f ( x i − 1 ) ] i = 1 , 2 , … , n ∂ F n x 1 , x 2 , … , x n ∂ x i = 1 2 f ′ x i x i + 1 − x i − 1 − f x i + 1 + f x i − 1 i = 1 , 2 , … , n {:[(delF_(n)(x_(1),x_(2),dots,x_(n)))/(delx_(i))=(1)/(2)[f^(')(x_(i))(x_(i+1)-x_(i-1))-f(x_(i+1))+f(x_(i-1))]],[i=1","2","dots","n]:}\begin{gathered} \frac{\partial F_{n}\left(x_{1}, x_{2}, \ldots, x_{n}\right)}{\partial x_{i}}=\frac{1}{2}\left[f^{\prime}\left(x_{i}\right)\left(x_{i+1}-x_{i-1}\right)-f\left(x_{i+1}\right)+f\left(x_{i-1}\right)\right] \\ i=1,2, \ldots, n \end{gathered}∂Fn(x1,x2,…,xn)∂xi=12[f′(xi)(xi+1−xi−1)−f(xi+1)+f(xi−1)]i=1,2,…,n
and we deduce
Theorem 2. - If the function f ( x ) f ( x ) f(x)f(x)f(x)is continuous in the closed interval [ a , b ] [ a , b ] [a,b][a, b][has,b]and is differentiable in the open interval ( a , b ) ( a , b ) (a,b)(a, b)(has,b)we can find n n nnndistinct points x 1 , x 2 , … , x n , x 1 < x 2 < … < x n x 1 , x 2 , … , x n , x 1 < x 2 < … < x n x_(1),x_(2),dots,x_(n),x_(1) < x_(2) < dots < x_(n)x_{1}, x_{2}, \ldots, x_{n}, x_{1}<x_{2}<\ldots<x_{n}x1,x2,…,xn,x1<x2<…<xnof ( a , b ) ( a , b ) (a,b)(a, b)(has,b)such that we have
f ′ ( x i ) = [ x i − 1 , x i + 1 ; f ] = f ( x i + 1 ) − f ( x i − 1 ) x i + 1 − x i − 1 i = 1 , 2 , … , n ( x 0 = a , x n + 1 = b ) f ′ x i = x i − 1 , x i + 1 ; f = f x i + 1 − f x i − 1 x i + 1 − x i − 1 i = 1 , 2 , … , n x 0 = a , x n + 1 = b {:[f^(')(x_(i))=[x_(i-1),x_(i+1);f]=(f(x_(i+1))-f(x_(i-1)))/(x_(i+1)-x_(i-1))],[i=1","2","dots","n quad(x_(0)=a,x_(n+1)=b)]:}\begin{gathered} f^{\prime}\left(x_{i}\right)=\left[x_{i-1}, x_{i+1} ; f\right]=\frac{f\left(x_{i+1}\right)-f\left(x_{i-1}\right)}{x_{i+1}-x_{i-1}} \\ i=1,2, \ldots, n \quad\left(x_{0}=a, x_{n+1}=b\right) \end{gathered}f′(xi)=[xi−1,xi+1;f]=f(xi+1)−f(xi−1)xi+1−xi−1i=1,2,…,n(x0=has,xn+1=b)
The geometric interpretation is simple.
1947-1948

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