On certain inequalities which characterize the convex functions

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

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T. Popoviciu
Institutul de Calcul

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T. Popoviciu, Sur certaines inégalités qui caractérisent les fonctions convexes, An. Şti. Univ. “Al. I. Cuza” Iaşi, Secţ. I a Mat. (N.S.), 11B, (1965), pp. 155-164 (in French).

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1965 b -Popoviciu- An. Sti. Univ. Al. I. Cuza Iasi – Sur certaines inegalites qui characterisent les fonctions convexes

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Analele Stiintifice a Universitatii Al. I. Cuza Iasi

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1965 b -Popoviciu- An. Sti. Univ. Al. I. Cuza Iasi - On certain inequalities which characterize the
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ON CERTAIN INEQUALITIES THAT CHARACTERIZE CONVEX FUNCTIONS

BYTIBERIU POPOVICIUin Cluj

Tribute to MO Mayer on the occasion of his 70th birthday
  1. A function f ( x ) f ( x ) f(x)f(x)f(x), defined on an interval I I III, (of non-zero length) is said to be non-concave, respectively convex if its difference divided by order 2 , [ x 1 , x 2 , x 3 ; f ] x 1 , x 2 , x 3 ; f [x_(1),x_(2),x_(3);f]\left[x_{1}, x_{2}, x_{3}; f\right][x1,x2,x3;f]East ≧ 0 ≧ 0 >= 0\geqq 0≧0, respectively > 0 > 0 > 0>0>0on any group of 3 distinct points x 1 , x 2 , x 3 ∈ I x 1 , x 2 , x 3 ∈ I x_(1),x_(2),x_(3)in Ix_{1}, x_{2}, x_{3} \in Ix1,x2,x3∈IA convex function is, in fact, a special case of a non-concave function.
In what follows, we will study some inequalities that characterize continuous and non-concave functions.
2. Theorem 1. For the continuous function f f fffeither non-concave over the interval I I III, it is necessary and sufficient that the inequality of fensen
(1) 2 f ( x + y 2 ) ≦ f ( x ) + f ( y ) (1) 2 f x + y 2 ≦ f ( x ) + f ( y ) {:(1)2f((x+y)/(2)) <= f(x)+f(y):}\begin{equation*} 2 f\left(\frac{x+y}{2}\right) \leqq f(x)+f(y) \tag{1} \end{equation*}(1)2f(x+y2)≦f(x)+f(y)
be checked for all x , y ∈ I x , y ∈ I x,y in Ix, y \in Ix,y∈I
Inequality (1), for all x , y ∈ I x , y ∈ I x,y in Ix, y \in Ix,y∈I, therefore characterizes continuous and non-concave functions on I I III.
Several proofs of Theorem 1 are known.
The condition is sufficient. Indeed, if inequality (1) is satisfied for all x , y ∈ I x , y ∈ I x,y in Ix, y \in Ix,y∈I, the function f f fff, assumed to be continuous, is non-concave on I I III.
The condition is necessary. This follows immediately from the formula.
[ x , y , x + y 2 ; f ] = 2 ( y − x ) 2 [ f ( x ) + f ( y ) − 2 f ( x + y 2 ) ] ( x ≠ y ) x , y , x + y 2 ; f = 2 ( y − x ) 2 f ( x ) + f ( y ) − 2 f x + y 2 ( x ≠ y ) [x,y,(x+y)/(2);f]=(2)/((yx)^(2))[f(x)+f(y)-2f((x+y)/(2))](x!=y)\left[x, y, \frac{x+y}{2}; f\right]=\frac{2}{(yx)^{2}}\left[f(x)+f(y)-2 f\left(\frac{x+y}{2}\right)\right](x \neq y)[x,y,x+y2;f]=2(y−x)2[f(x)+f(y)−2f(x+y2)](x≠y)
But we will give here another demonstration, which is based on the approximation of continuous functions by polygonal lines inscribed in the representative curve.
The demonstration of the necessity of the condition in Theorem 1 follows from the following properties:
A. Functions λ x + μ , | x − ϱ | λ x + μ , | x − ϱ | lambda x+mu,|x-ϱ|\lambda x+\mu,|x-\varrho|λx+μ,|x−ϱ|, Or λ , μ , ϱ λ , μ , ϱ lambda,mu,ϱ\lambda, \mu, \varrhoλ,μ,ϱare arbitrary constants, satisfying inequality (1) for all x , y ∈ I x , y ∈ I x,y in Ix, y \in Ix,y∈IB.
If a finite number of functions satisfy inequality (1) for all x , y ∈ I x , y ∈ I x,y in Ix, y \in Ix,y∈I, any linear combination of these functions, with non-negative coefficients, enjoys the same property.
C. If the terms of a convergent sequence on I I IIIof functions defined on I I III, satisfy inequality (1) for all x , y ∈ I x , y ∈ I x,y in Ix, y \in Ix,y∈IThe limit function enjoys the same property.
D. Every continuous and non-concave function is the limit of a sequence of functions of the form
(2) λ x + μ + ∑ has = 1 n C has | x − c has | , (2) λ x + μ + ∑ has = 1 n   C has x − c has , {:(2)lambda x+mu+sum_(a=1)^(n)C_(a)|x-c_(a)|",":}\begin{equation*} \lambda x+\mu+\sum_{a=1}^{n} C_{a}\left|x-c_{a}\right|, \tag{2} \end{equation*}(2)λx+μ+∑has=1nChas|x−chas|,
THE C has , α = 1 , 2 , … , n C has , α = 1 , 2 , … , n C_(a),alpha=1,2,dots,nC_{a}, \alpha=1,2, \ldots, nChas,α=1,2,…,n, being non-negative constants and c has , α = 1 , 2 , … , n c has , α = 1 , 2 , … , n c_(a),alpha=1,2,dots,nc_{a}, \alpha=1,2, \ldots, nchas,α=1,2,…,npoints of the interval I I III3.
To complete the previous demonstration, we will examine properties A and D more closely from above. We can omit providing proofs of these properties. B , C B , C B,CB, CB,Cwhich are immediate.
Let us first consider property A.
Every first-degree polynomial λ x + μ λ x + μ lambda x+mu\lambda x+\muλx+μverifies inequality (1), with the sign = = ===, for all x , y ∈ I x , y ∈ I x,y in Ix, y \in Ix,y∈I.
The function | x | | x | |x||x||x|verifies inequality (1) as a consequence of the well-known inequality
(3) | x + y | ≦ | x | + | y | (3) | x + y | ≦ | x | + | y | {:(3)|x+y| <= |x|+|y|:}\begin{equation*} |x+y| \leqq|x|+|y| \tag{3} \end{equation*}(3)|x+y|≦|x|+|y|
between the absolute value of the sum and the sum of the absolute values ​​of the terms.
It immediately follows that the function | x − ϱ | | x − ϱ | |x-ϱ||x-\varrho||x−ϱ|also verifies inequality (1) for all x , y ∈ I x , y ∈ I x,y in Ix, y \in Ix,y∈Iand all constant ϱ ϱ ϱ\varrhoϱ.
Note that the non-concavity of the functions λ x + μ , | x | λ x + μ , | x | lambda x+mu,|x|\lambda x+\mu,|x|λx+μ,|x|(so also of | x − ϱ | | x − ϱ | |x-ϱ||x-\varrho||x−ϱ|) results directly from inequality (1), without the assumption of continuity. Indeed, the function f f fffis non-concave on I I IIIif and only if we have
(4) f ( p x + q y p + q ) ≦ p f ( x ) + q f ( y ) p + q (4) f p x + q y p + q ≦ p f ( x ) + q f ( y ) p + q {:(4)f((px+qy)/(p+q)) <= (pf(x)+qf(y))/(p+q):}\begin{equation*} f\left(\frac{p x+qy}{p+q}\right) \leqq \frac{pf(x)+qf(y)}{p+q} \tag{4} \end{equation*}(4)f(px+qyp+q)≦pf(x)+qf(y)p+q
for everything x , y ∈ I x , y ∈ I x,y in Ix, y \in Ix,y∈Iand for everything p , q > 0 p , q > 0 p,q > 0p, q>0p,q>0.
In the case of a first-degree polynomial P ( x ) = λ x + μ P ( x ) = λ x + μ P(x) = lambda x + muP(x) = λx + μP(x)=λx+μby asking Q ( x ) = λ x + μ ( p + q ) ( p , q > 0 ) Q ( x ) = λ x + μ ( p + q ) ( p , q > 0 ) Q(x)=lambda x+mu(p+q)(p,q > 0)Q(x)=\lambda x+\mu(p+q)(p, q>0)Q(x)=λx+μ(p+q)(p,q>0)inequality 2 Q ( p x + q y 2 ) ≦≦ Q ( p x ) + Q ( q y ) 2 Q p x + q y 2 ≦≦ Q ( p x ) + Q ( q y ) 2Q((px+qy)/(2))≦≦Q(px)+Q(qy)2 Q\left(\frac{p x+qy}{2}\right) \leqq \leqq Q(px)+Q(qy)2Q(px+qy2)≦≦Q(px)+Q(qy)After some simplifications, we deduce P ( p x + q y p + q ) ≦≦ p P ( x ) + q P ( y ) p + q P p x + q y p + q ≦≦ p P ( x ) + q P ( y ) p + q P((px+qy)/(p+q))≦≦(pP(x)+qP(y))/(p+q)P\left(\frac{p x+qy}{p+q}\right) \leqq \leqq \frac{p P(x)+q P(y)}{p+q}P(px+qyp+q)≦≦pP(x)+qP(y)p+q(in these relationships it is the sign = = ===which always takes place).
In the case of the function | x | | x | |x||x||x|, of | p x + q y | ≦ | p x | + | q y | = p | x | + + q | y | ( p , q > 0 ) | p x + q y | ≦ | p x | + | q y | = p | x | + + q | y | ( p , q > 0 ) |px+qy| <= |px|+|qy|=p|x|++q|y|(p,q > 0)|p x+q y| \leqq|p x|+|q y|=p|x|+ +q|y|(p, q>0)|px+qy|≦|px|+|qy|=p|x|++q|y|(p,q>0)it also results | p x + q y p + q | ≦ p | x | + q | y | p + q p x + q y p + q ≦ p | x | + q | y | p + q |(px+qy)/(p+q)| <= (p|x|+q|y|)/(p+q)\left|\frac{p x+q y}{p+q}\right| \leqq \frac{p|x|+q|y|}{p+q}|px+qyp+q|≦p|x|+q|y|p+qWe have a similar property for the function | x − ϱ | | x − ϱ | |x-ϱ||x-\varrho||x−ϱ|4.
To demonstrate property D, it suffices to assume that the function f f fffeither continuous and non-concave on the finite and closed interval [ a , b ] [ a , b ] [a,b][a, b][has,b]So then a = c 0 < c 1 < … < c m − 1 < c m = b a = c 0 < c 1 < … < c m − 1 < c m = b a=c_(0) < c_(1) < dots < c_(m-1) < c_(m)=ba=c_{0}<c_{1}<\ldots<c_{m-1}<c_{m}=bhas=c0<c1<…<cm−1<cm=bthe points that divide the interval [ a , b ] [ a , b ] [a,b][a, b][has,b]in m m mmmequal parts and φ m ( x ) φ m ( x ) varphi_(m)(x)\varphi_{m}(x)φm(x)the function represented by the polygonal line inscribed in the curve y = f ( x ) y = f ( x ) y=f(x)y=f(x)y=f(x)along the peaks ( c a , f ( c a ) ) , α = 0 , 1 , … , m c a , f c a , α = 0 , 1 , … , m (c_(a),f(c_(a))),alpha=0,1,dots,m\left(c_{a}, f\left(c_{a}\right)\right), \alpha=0,1, \ldots, m(chas,f(chas)),α=0,1,…,mSo we have
φ m ( x ) = ∑ a = 0 m C a | x − c a | φ m ( x ) = ∑ a = 0 m   C a x − c a varphi_(m)(x)=sum_(a=0)^(m)C_(a)|x-c_(a)|\varphi_{m}(x)=\sum_{a=0}^{m} C_{a}\left|x-c_{a}\right|φm(x)=∑has=0mChas|x−chas|
which is indeed of the form (2) where the coefficients C a , α = 1 , 2 , … , m − 1 ( m > 1 ) C a , α = 1 , 2 , … , m − 1 ( m > 1 ) C_(a),alpha=1,2,dots,m-1(m > 1)C_{a}, \alpha=1,2, \ldots, m-1 (m>1)Chas,α=1,2,…,m−1(m>1)are non-negative. Indeed, the sequence of slopes
( ∑ a = 0 v C a − ∑ a = v + 1 m C a ) v = 0 m − 1 ∑ a = 0 v   C a − ∑ a = v + 1 m   C a v = 0 m − 1 (sum_(a=0)^(v)C_(a)-sum_(a=v+1)^(m)C_(a))_(v=0)^(m-1)\left(\sum_{a=0}^{v} C_{a}-\sum_{a=v+1}^{m} C_{a}\right)_{v=0}^{m-1}(∑has=0vChas−∑has=v+1mChas)v=0m−1The sides of the polygonal line are non-decreasing.
If ω ( δ ) ω ( δ ) omega(delta)\omega(\delta)ω(δ)is the modulus of oscillation of the function f f fff, We have
| f ( x ) − φ m ( x ) | = | [ f ( x ) − f ( c a − 1 ) ] x − c a c a − 1 − c a + + [ f ( x ) − f ( c a ) ] x − c a − 1 c a − c a − 1 | ≦ ω ( x − c a − 1 ) c a − x c a − c a − 1 + + ω ( c a − x ) x − c a − 1 c a − c a − 1 ≦ ω ( b − a m ) f ( x ) − φ m ( x ) = f ( x ) − f c a − 1 x − c a c a − 1 − c a + + f ( x ) − f c a x − c a − 1 c a − c a − 1 ≦ ω x − c a − 1 c a − x c a − c a − 1 + + ω c a − x x − c a − 1 c a − c a − 1 ≦ ω b − a m {:[|f(x)-varphi_(m)(x)|=|[f(x)-f(c_(a-1))](x-c_(a))/(c_(a-1)-c_(a))+:}],[+[f(x)-f(c_(a))](x-c_(a-1))/(c_(a)-c_(a-1))| <= omega(x-c_(a-1))(c_(a)-x)/(c_(a)-c_(a-1))+:}],[+omega(c_(a)-x)(x-c_(a-1))/(c_(a)-c_(a-1)) <= omega((b-a)/(m))]:}\begin{gathered} \left|f(x)-\varphi_{m}(x)\right|=\left\lvert\,\left[f(x)-f\left(c_{a-1}\right)\right] \frac{x-c_{a}}{c_{a-1}-c_{a}}+\right. \\ +\left[f(x)-f\left(c_{a}\right)\right] \frac{x-c_{a-1}}{c_{a}-c_{a-1}} \left\lvert\, \leqq \omega\left(x-c_{a-1}\right) \frac{c_{a}-x}{c_{a}-c_{a-1}}+\right. \\ +\omega\left(c_{a}-x\right) \frac{x-c_{a-1}}{c_{a}-c_{a-1}} \leqq \omega\left(\frac{b-a}{m}\right) \end{gathered}|f(x)−φm(x)|=|[f(x)−f(chas−1)]x−chaschas−1−chas++[f(x)−f(chas)]x−chas−1chas−chas−1|≦ω(x−chas−1)chas−xchas−chas−1++ω(chas−x)x−chas−1chas−chas−1≦ω(b−hasm)
Or x ∈ [ c a − 1 , c a ] , a = 1 , 2 , … , m x ∈ c a − 1 , c a , a = 1 , 2 , … , m x in[c_(a-1),c_(a)],a=1,2,dots,mx \in\left[c_{a-1}, c_{a}\right], a=1,2, \ldots, mx∈[chas−1,chas],has=1,2,…,m, SO
| f ( x ) − φ m ( x ) | ≦ ω ( b − a m ) , pour x ∈ [ a , b ] f ( x ) − φ m ( x ) ≦ ω b − a m ,  pour  x ∈ [ a , b ] |f(x)-varphi_(m)(x)| <= omega((b-a)/(m))," pour "x in[a,b]\left|f(x)-\varphi_{m}(x)\right| \leqq \omega\left(\frac{b-a}{m}\right), \text { pour } x \in[a, b]|f(x)−φm(x)|≦ω(b−hasm), For x∈[has,b]
and the rest ( φ m ( x ) ) m = 1 ∞ φ m ( x ) m = 1 ∞ (varphi_(m)(x))_(m=1)^(oo)\left(\varphi_{m}(x)\right)_{m=1}^{\infty}(φm(x))m=1∞converges (and even uniformly) towards the function f f fff5.
We have shown above the role of inequality (3) in the characterization by inequality (1) of continuous and non-concave functions. We will show how certain generalizations of inequality (3) can be used to obtain other characterizations of the same nature.
Let us consider the inequality of H. Hornich [4],
(5) | x + y | + | y + z | + | z + x | ≦ | x | + | y | + | z | + | x + y + z | (5) | x + y | + | y + z | + | z + x | ≦ | x | + | y | + | z | + | x + y + z | {:(5)|x+y|+|y+z|+|z+x| <= |x|+|y|+|z|+|x+y+z|:}\begin{equation*} |x+y|+|y+z|+|z+x| \leqq|x|+|y|+|z|+|x+y+z| \tag{5} \end{equation*}(5)|x+y|+|y+z|+|z+x|≦|x|+|y|+|z|+|x+y+z|
which is a generalization of inequality (3) and is verified, in particular, for all real numbers x , y , z x , y , z x,y,zx, y, zx,y,zWe then have the
Theorem 2. For the function to continue f f fffeither non-concave over the interval I I IIIIt is necessary and sufficient that inequality
(6) 2 [ f ( x + y 2 ) + f ( y + z 2 ) + f ( z + x 2 ) ] ≦ ≦ f ( x ) + f ( y ) + f ( z ) + 3 f ( x + y + z 3 ) (6) 2 f x + y 2 + f y + z 2 + f z + x 2 ≦ ≦ f ( x ) + f ( y ) + f ( z ) + 3 f x + y + z 3 {:[(6)2[f((x+y)/(2))+f((y+z)/(2))+f((z+x)/(2))] <= ],[quad <= f(x)+f(y)+f(z)+3f((x+y+z)/(3))]:}\begin{align*} & 2\left[f\left(\frac{x+y}{2}\right)+f\left(\frac{y+z}{2}\right)+f\left(\frac{z+x}{2}\right)\right] \leqq \tag{6}\\ & \quad \leqq f(x)+f(y)+f(z)+3 f\left(\frac{x+y+z}{3}\right) \end{align*}(6)2[f(x+y2)+f(y+z2)+f(z+x2)]≦≦f(x)+f(y)+f(z)+3f(x+y+z3)
be checked for all x , y , z ∈ I x , y , z ∈ I x,y,z in Ix, y, z \in Ix,y,z∈IThis theorem
is a special case ( n = 3 , k = 2 n = 3 , k = 2 n=3,k=2n=3, k=2n=3,k=2) of a more general theorem that results from a generalization of inequality (5). Dušan D. Adamovič [1] and Dragomir Z. Dokovič [2] generalized inequality (5), the former for k = 2 k = 2 k=2k=2k=2and the second for k k kkkany, by inequality
(7) ∑ ( k ) | x i 1 + x i 2 + ⋯ + x i k | ≦ ≦ ( n − 2 k − 2 ) [ n − k k − 1 ∑ i = 1 n | x i | + | ∑ i = 1 n x i | ] , k = 2 , 3 , … , n − 1 , n = 3 , 4 , … (7) ∑ ( k )   x i 1 + x i 2 + ⋯ + x i k ≦ ≦ ( n − 2 k − 2 ) n − k k − 1 ∑ i = 1 n   x i + ∑ i = 1 n   x i , k = 2 , 3 , … , n − 1 , n = 3 , 4 , … {:[(7)sum(k)|x_(i_(1))+x_(i_(2))+cdots+x_(i_(k))| <= ],[ <= ((n-2)/(k-2))[(n-k)/(k-1)sum_(i=1)^(n)|x_(i)|+|sum_(i=1)^(n)x_(i)|]","k=2","3","dots","n-1","n=3","4","dots]:}\begin{gather*} \sum^{(k)}\left|x_{i_{1}}+x_{i_{2}}+\cdots+x_{i_{k}}\right| \leqq \tag{7}\\ \leqq\binom{ n-2}{k-2}\left[\frac{n-k}{k-1} \sum_{i=1}^{n}\left|x_{i}\right|+\left|\sum_{i=1}^{n} x_{i}\right|\right], k=2,3, \ldots, n-1, n=3,4, \ldots \end{gather*}(7)∑(k)|xi1+xi2+⋯+xik|≦≦(n−2k−2)[n−kk−1∑i=1n|xi|+|∑i=1nxi|],k=2,3,…,n−1,n=3,4,…
the summons Σ ( k ) Σ ( k ) Sigma^((k))\Sigma^{(k)}Σ(k)being extended to all combinations i 1 , i 2 , … , i k i 1 , i 2 , … , i k i_(1),i_(2),dots,i_(k)i_{1}, i_{2}, \ldots, i_{k}i1,i2,…,ik, k k kkkhas k k kkkclues 1 , 2 , … , n 1 , 2 , … , n 1,2,dots,n1,2, \ldots, n1,2,…,n.
We can then generalize Theorem 2 by
Theorem 3. If n n nnnis a natural number ≧ 3 ≧ 3 >= 3\geqq 3≧3And k k kkka natural number that satisfies the inequalities 2 ≦ k ≦ n − 1 2 ≦ k ≦ n − 1 2 <= k <= n-12 \leqq k \leqq n-12≦k≦n−1, so that the function continues f f fffeither non-concave over the interval I I IIIIt is necessary and sufficient that inequality
k ∑ ( k ) f ( x i 1 + x i 2 + … + x i k k ) ≦ k ∑ ( k ) f x i 1 + x i 2 + … + x i k k ≦ k sum^((k))f((x_(i_(1))+x_(i_(2))+dots+x_(i_(k)))/(k)) <=k \sum{ }^{(k)} f\left(\frac{x_{i_{1}}+x_{i_{2}}+\ldots+x_{i_{k}}}{k}\right) \leqqk∑(k)f(xi1+xi2+…+xikk)≦
(8) ≦ ( n − 2 k − 2 ) [ n − k k − 1 ∑ i = 1 n f ( x i ) + n f ( x 1 + x 2 + … + x n n ) ] (8) ≦ ( n − 2 k − 2 ) n − k k − 1 ∑ i = 1 n   f x i + n f x 1 + x 2 + … + x n n {:(8) <= ((n-2)/(k-2))[(n-k)/(k-1)sum_(i=1)^(n)f(x_(i))+nf((x_(1)+x_(2)+dots+x_(n))/(n))]:}\begin{equation*} \leqq\binom{ n-2}{k-2}\left[\frac{n-k}{k-1} \sum_{i=1}^{n} f\left(x_{i}\right)+n f\left(\frac{x_{1}+x_{2}+\ldots+x_{n}}{n}\right)\right] \tag{8} \end{equation*}(8)≦(n−2k−2)[n−kk−1∑i=1nf(xi)+nf(x1+x2+…+xnn)]
be checked for all x 1 , x 2 , … , x n ∈ I x 1 , x 2 , … , x n ∈ I x_(1),x_(2),dots,x_(n)in Ix_{1}, x_{2}, \ldots, x_{n} \in Ix1,x2,…,xn∈I.
For n = 3 , k = 2 n = 3 , k = 2 n=3,k=2n=3, k=2n=3,k=2We find ourselves back in Theorem 2.
Inequality (8), and in particular inequality (6), therefore characterize continuous and non-concave functions.
6. The proof of Theorem 3 can be done in the same way as that of Theorem 1.
The condition is sufficient, therefore, if inequality (8) is satisfied for all x 1 , x 2 , … , x n ∈ I x 1 , x 2 , … , x n ∈ I x_(1),x_(2),dots,x_(n)in Ix_{1}, x_{2}, \ldots, x_{n} \in Ix1,x2,…,xn∈I, the continuous function f f fffis non-concave on I I IIIIndeed, let us set in (8), first x 1 = x , x 2 = x 3 = … = x n = n y − x n − 1 x 1 = x , x 2 = x 3 = … = x n = n y − x n − 1 x_(1)=x,x_(2)=x_(3)=dots=x_(n)=(ny-x)/(n-1)x_{1}=x, x_{2}=x_{3}=\ldots=x_{n}=\frac{n y-x}{n-1}x1=x,x2=x3=…=xn=ny−xn−1, Next x 1 = y , x 2 = x 3 = … = x n = n x − y n − 1 x 1 = y , x 2 = x 3 = … = x n = n x − y n − 1 x_(1)=y,x_(2)=x_(3)=dots=x_(n)=(nx-y)/(n-1)x_{1}=y, x_{2}=x_{3}=\ldots=x_{n}=\frac{n x-y}{n-1}x1=y,x2=x3=…=xn=nx−yn−1and adding the inequalities thus obtained term by term. We deduce
(9) f ( x + y 2 + y − x 2 δ ) + f ( x + y 2 − y − x 2 δ ) ≦ f ( x ) + f ( y ) (9) f x + y 2 + y − x 2 δ + f x + y 2 − y − x 2 δ ≦ f ( x ) + f ( y ) {:(9)f((x+y)/(2)+(y-x)/(2delta))+f((x+y)/(2)-(y-x)/(2delta)) <= f(x)+f(y):}\begin{equation*} f\left(\frac{x+y}{2}+\frac{y-x}{2 \delta}\right)+f\left(\frac{x+y}{2}-\frac{y-x}{2 \delta}\right) \leqq f(x)+f(y) \tag{9} \end{equation*}(9)f(x+y2+y−x2δ)+f(x+y2−y−x2δ)≦f(x)+f(y)
Or δ = k ( n − 1 ) ( n + 1 ) k − 2 n ≧ ( n − 1 ) 2 ( n − 1 ) 2 − 2 > 1 δ = k ( n − 1 ) ( n + 1 ) k − 2 n ≧ ( n − 1 ) 2 ( n − 1 ) 2 − 2 > 1 delta=(k(n-1))/((n+1)k-2n) >= ((n-1)^(2))/((n-1)^(2)-2) > 1\delta=\frac{k(n-1)}{(n+1) k-2 n} \geqq \frac{(n-1)^{2}}{(n-1)^{2}-2}>1δ=k(n−1)(n+1)k−2n≧(n−1)2(n−1)2−2>1As a result, the points x + y 2 + + y − x 2 δ , x + y 2 − y − x 2 δ x + y 2 + + y − x 2 δ , x + y 2 − y − x 2 δ (x+y)/(2)++(y-x)/(2delta),(x+y)/(2)-(y-x)/(2delta)\frac{x+y}{2}+ +\frac{y-x}{2 \delta}, \frac{x+y}{2}-\frac{y-x}{2 \delta}x+y2++y−x2δ,x+y2−y−x2δare always between x x xxxAnd y y yyyWe can then repeat inequality (9) by taking these points for x x xxxAnd y y yyyContinuing in this way, we find
(10) f ( x + y 2 + y − x 2 δ m ) + f ( x + y 2 − y − x 2 δ m ) ≦ f ( x ) + f ( y ) , m = 1 , 2 , … (10) f x + y 2 + y − x 2 δ m + f x + y 2 − y − x 2 δ m ≦ f ( x ) + f ( y ) , m = 1 , 2 , … {:(10)f((x+y)/(2)+(y-x)/(2delta^(m)))+f((x+y)/(2)-(y-x)/(2delta^(m))) <= f(x)+f(y)","m=1","2","dots:}\begin{equation*} f\left(\frac{x+y}{2}+\frac{y-x}{2 \delta^{m}}\right)+f\left(\frac{x+y}{2}-\frac{y-x}{2 \delta^{m}}\right) \leqq f(x)+f(y), m=1,2, \ldots \tag{10} \end{equation*}(10)f(x+y2+y−x2δm)+f(x+y2−y−x2δm)≦f(x)+f(y),m=1,2,…
Taking into account δ > 1 δ > 1 delta > 1\delta>1δ>1and the continuity of the function f f fff, by doing m → ∞ m → ∞ m rarr oom \rightarrow \inftym→∞, we deduce from (10) Jensen's inequality (1).
The condition is necessary, therefore any non-concave and continuous function on I I IIIverifies inequality (8) for all x 1 , x 2 , … , x n ∈ I x 1 , x 2 , … , x n ∈ I x_(1),x_(2),dots,x_(n)in Ix_{1}, x_{2}, \ldots, x_{n} \in Ix1,x2,…,xn∈IIndeed, the properties A , B , C , D A , B , C , D A,B,C,DA, B, C, DHAS,B,C,Dof no. 2 are verified for inequality (8).
7. Consider the inequality
(11) ∑ a = 1 m p a f ( z a ) ≧ 0 (11) ∑ a = 1 m   p a f z a ≧ 0 {:(11)sum_(a=1)^(m)p_(a)f(z_(a)) >= 0:}\begin{equation*} \sum_{a=1}^{m} p_{a} f\left(z_{a}\right) \geqq 0 \tag{11} \end{equation*}(11)∑has=1mphasf(zhas)≧0
Or p a ≠ 0 , α = 1 , 2 , … , m p a ≠ 0 , α = 1 , 2 , … , m p_(a)!=0,alpha=1,2,dots,mp_{a} \neq 0, \alpha=1,2, \ldots, mphas≠0,α=1,2,…,mAnd z 1 < z 2 < … < z m z 1 < z 2 < … < z m z_(1) < z_(2) < dots < z_(m)z_{1}<z_{2}<\ldots<z_{m}z1<z2<…<zmare m m mmmdistinct points of I I IIIIf inequality (11) is satisfied for any function f f fffnon-concave on I I III, We have m > 2 m > 2 m > 2m>2m>2And
(12) ∑ a = 1 m p a f ( z a ) = ∑ a = 1 m − 2 q a [ z a , z a + 1 , z a + 2 ; f ] (12) ∑ a = 1 m   p a f z a = ∑ a = 1 m − 2   q a z a , z a + 1 , z a + 2 ; f {:(12)sum_(a=1)^(m)p_(a)f(z_(a))=sum_(a=1)^(m-2)q_(a)[z_(a),z_(a+1),z_(a+2);f]:}\begin{equation*} \sum_{a=1}^{m} p_{a} f\left(z_{a}\right)=\sum_{a=1}^{m-2} q_{a}\left[z_{a}, z_{a+1}, z_{a+2} ; f\right] \tag{12} \end{equation*}(12)∑has=1mphasf(zhas)=∑has=1m−2qhas[zhas,zhas+1,zhas+2;f]
with q a ⩾ 0 , a = 1 , 2 , … , m − 2 q a ⩾ 0 , a = 1 , 2 , … , m − 2 q_(a) >= 0,a=1,2,dots,m-2q_{a} \geqslant 0, a=1,2, \ldots, m-2qhas⩾0,has=1,2,…,m−2[6]. In particular, q 1 = ( z 1 − z 2 ) ( z 1 − z 3 ) p 1 > 0 q 1 = z 1 − z 2 z 1 − z 3 p 1 > 0 q_(1)=(z_(1)-z_(2))(z_(1)-z_(3))p_(1) > 0q_{1}=\left(z_{1}-z_{2}\right)\left(z_{1}-z_{3}\right) p_{1}>0q1=(z1−z2)(z1−z3)p1>0, q m − 2 = ( z m − z m − 1 ) ( z m − z m − 2 ) p m > 0 q m − 2 = z m − z m − 1 z m − z m − 2 p m > 0 q_(m-2)=(z_(m)-z_(m-1))(z_(m)-z_(m-2))p_(m) > 0q_{m-2}=\left(z_{m}-z_{m-1}\right)\left(z_{m}-z_{m-2}\right) p_{m}>0qm−2=(zm−zm−1)(zm−zm−2)pm>0and (12), by setting f ( x ) = x 2 f ( x ) = x 2 f(x)=x^(2)f(x)=x^{2}f(x)=x2It follows that
∑ a = 1 m p a z a 2 = ∑ a = 1 m − 2 q a > 0 ∑ a = 1 m   p a z a 2 = ∑ a = 1 m − 2   q a > 0 sum_(a=1)^(m)p_(a)z_(a)^(2)=sum_(a=1)^(m-2)q_(a) > 0\sum_{a=1}^{m} p_{a} z_{a}^{2}=\sum_{a=1}^{m-2} q_{a}>0∑has=1mphaszhas2=∑has=1m−2qhas>0
We deduce that for any function f f fffconvex on I I IIIstrict inequality
(13) ∑ a = 1 m p a f ( z a ) > 0 (13) ∑ a = 1 m   p a f z a > 0 {:(13)sum_(a=1)^(m)p_(a)f(z_(a)) > 0:}\begin{equation*} \sum_{a=1}^{m} p_{a} f\left(z_{a}\right)>0 \tag{13} \end{equation*}(13)∑has=1mphasf(zhas)>0
is verified.
From this analysis results
Lemma 1. If inequality (11), where p a , α = 1 , 2 , … , m p a , α = 1 , 2 , … , m p_(a),alpha=1,2,dots,mp_{a}, \alpha=1,2, \ldots, mphas,α=1,2,…,m, are constants (any constants, not necessarily different from zero) and z a z a z_(a)z_{a}zhas, α = 1 , 2 , … , m α = 1 , 2 , … , m alpha=1,2,dots,m\alpha=1,2, \ldots, mα=1,2,…,mpoints (distinct or not) of the interval I I III, is checked for every function f f fffnon-concave on I and if ∑ α = 1 m p α z α 2 > 0 ∑ α = 1 m   p α z α 2 > 0 sum_(alpha=1)^(m)p_(alpha)z_(alpha)^(2) > 0\sum_{\alpha=1}^{m} p_{\alpha} z_{\alpha}^{2}>0∑α=1mpαzα2>0, the strict inequality (13) is verified for any function f f fffconvex on I I III.
Let's return to inequality (8). For f ( x ) = x 2 f ( x ) = x 2 f(x)=x^(2)f(x)=x^{2}f(x)=x2the difference between the second and first members is equal to
( n − 2 k − 2 ) n − k n k Σ ( 2 ) ( x l 1 − x l 2 ) 2 ( n − 2 k − 2 ) n − k n k Σ ( 2 ) x l 1 − x l 2 2 ((n-2)/(k-2))(n-k)/(nk)Sigma^((2))(x_(l_(1))-x_(l_(2)))^(2)\binom{n-2}{k-2} \frac{n-k}{n k} \Sigma^{(2)}\left(x_{l_{1}}-x_{l_{2}}\right)^{2}(n−2k−2)n−knkΣ(2)(xL1−xL2)2
Applying Lemma 1, we deduce
Consequence 1. If n n nnnis a natural number ≧ 3 , k ≧ 3 , k >= 3,k\geqq 3, k≧3,ka natural number that satisfies the inequalities 2 ≤ k ≤ n − 1 2 ≤ k ≤ n − 1 2 <= k <= n-12 \leq k \leq n-12≤k≤n−1and if x 1 , x 2 , … , x n ∈ I x 1 , x 2 , … , x n ∈ I x_(1),x_(2),dots,x_(n)in Ix_{1}, x_{2}, \ldots, x_{n} \in Ix1,x2,…,xn∈Iany function f f fffcontinuous and convex on I ¯ ― I ¯ ¯ bar(bar(I))\overline{\bar{I}}I¯—verifies inequality (8), equality holding if and only if 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=…=xn
8. It follows that the functional equation obtained by equating the two members of relation (8) has as its general continuous solution the polynomials of degree 1 .
Indeed, for such a solution it is necessary that the functions f f fffAnd − f − f -f-f−fare both non-concave, which occurs if and only if f f fffis a polynomial of degree 1.
Regarding the functional equation
Σ ( k ) f ( x i 1 + x i 2 + … + x i k ) = (14) = ( n − 2 k − 2 ) [ n − k k − 1 ∑ i = 1 n f ( x i ) + f ( x 1 + x 2 + … + x n ) ] Σ ( k ) f x i 1 + x i 2 + … + x i k = (14) = ( n − 2 k − 2 ) n − k k − 1 ∑ i = 1 n   f x i + f x 1 + x 2 + … + x n {:[Sigma^((k))f(x_(i_(1))+x_(i_(2))+dots+x_(i_(k)))=],[(14)=((n-2)/(k-2))[(n-k)/(k-1)sum_(i=1)^(n)f(x_(i))+f(x_(1)+x_(2)+dots+x_(n))]]:}\begin{gather*} \Sigma^{(k)} f\left(x_{i_{1}}+x_{i_{2}}+\ldots+x_{i_{k}}\right)= \\ =\binom{n-2}{k-2}\left[\frac{n-k}{k-1} \sum_{i=1}^{n} f\left(x_{i}\right)+f\left(x_{1}+x_{2}+\ldots+x_{n}\right)\right] \tag{14} \end{gather*}Σ(k)f(xi1+xi2+…+xik)=(14)=(n−2k−2)[n−kk−1∑i=1nf(xi)+f(x1+x2+…+xn)]
analogous to the Cauchy equation f ( x + y ) = f ( x ) + f ( y ) f ( x + y ) = f ( x ) + f ( y ) f(x+y)=f(x)+f(y)f(x+y)=f(x)+f(y)f(x+y)=f(x)+f(y)We will demonstrate
that its general continuous solution is a polynomial of degree 2 which vanishes for x = 0 x = 0 x=0x=0x=0, therefore a function of the form λ x 2 + μ x λ x 2 + μ x lambdax^(2)+mu x\lambda x^{2}+\mu xλx2+μx.
For n = 3 , k = 2 n = 3 , k = 2 n=3,k=2n=3, k=2n=3,k=2we have the functional equation
(15) f ( x + y ) + f ( y + z ) + f ( z + x ) = f ( x ) + f ( y ) + f ( z ) + f ( x + y + z ) f ( x + y ) + f ( y + z ) + f ( z + x ) = f ( x ) + f ( y ) + f ( z ) + f ( x + y + z ) f(x+y)+f(y+z)+f(z+x)=f(x)+f(y)+f(z)+f(x+y+z)f(x+y)+f(y+z)+f(z+x)=f(x)+f(y)+f(z)+f(x+y+z)f(x+y)+f(y+z)+f(z+x)=f(x)+f(y)+f(z)+f(x+y+z).
Let us therefore consider equation (14) which is verified regardless 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,…,xnreal numbers. First, note that any polynomial of the form γ x 2 + μ x γ x 2 + μ x gammax^(2)+mu x\gamma x^{2}+\mu xγx2+μxcheck this equation.
So now f f fffa solution to equation (14). If we set x 1 = x 2 = … = x r = 0 x 1 = x 2 = … = x r = 0 x_(1)=x_(2)=dots=x_(r)=0x_{1}=x_{2}=\ldots=x_{r}=0x1=x2=…=xr=0, we deduce f ( 0 ) = 0 f ( 0 ) = 0 f(0)=0f(0)=0f(0)=0If we then set (for n > 3 n > 3 n > 3n>3n>3), x 1 = x , x 2 = y , x 3 = z , x 4 = x 5 = … = x 2 = 0 x 1 = x , x 2 = y , x 3 = z , x 4 = x 5 = … = x 2 = 0 x_(1)=x,x_(2)=y,x_(3)=z,x_(4)=x_(5)=dots=x^(2)=0x_{1}=x, x_{2}=y, x_{3}=z, x_{4}=x_{5}=\ldots=x^{2}=0x1=x,x2=y,x3=z,x4=x5=…=x2=0, we find that the function f f fffverifies equation (15). It therefore suffices to show that every continuous solution of this equation is of the form γ x 2 + μ x γ x 2 + μ x gammax^(2)+mu x\gamma x^{2}+\mu xγx2+μxThis is the result of research by M M MMM. Fréchet on the functional characterization of polynomials [3]. We will give the proof. It suffices to demonstrate that the solution f f fffof equation (15) is a polynomial of degree 2. If we set Δ h 2 f ( x ) = f ( x + 2 h ) − 2 f ( x + h ) + f ( x ) Δ h 2 f ( x ) = f ( x + 2 h ) − 2 f ( x + h ) + f ( x ) Delta_(h)^(2)f(x)=f(x+2h)-2f(x+h)+f(x)\Delta_{h}^{2} f(x)=f(x+2 h)-2 f(x+h)+f(x)Δh2f(x)=f(x+2h)−2f(x+h)+f(x)and if we ask y = z = h y = z = h y=z=hy=z=hy=z=hin (15) we find ( f ( 0 ) = 0 ) , Δ h 2 f ( x ) = Δ h 2 f ( 0 ) ( f ( 0 ) = 0 ) , Δ h 2 f ( x ) = Δ h 2 f ( 0 ) (f(0)=0),Delta_(h)^(2)f(x)=Delta_(h)^(2)f(0)(f(0)=0), \Delta_{h}^{2} f(x)=\Delta_{h}^{2} f(0)(f(0)=0),Δh2f(x)=Δh2f(0), therefore also Δ h 2 f ( x + h ) = Δ h 2 f ( 0 ) Δ h 2 f ( x + h ) = Δ h 2 f ( 0 ) Delta_(h)^(2)f(x+h)=Delta_(h)^(2)f(0)\Delta_{h}^{2} f(x+h)=\Delta_{h}^{2} f(0)Δh2f(x+h)=Δh2f(0)for everything x x xxxAnd h h hhhWe deduce from this Δ h 3 f ( x ) = Δ h 2 f ( x + h ) − Δ h 2 f ( x ) = 0 Δ h 3 f ( x ) = Δ h 2 f ( x + h ) − Δ h 2 f ( x ) = 0 Delta_(h)^(3)f(x)=Delta_(h)^(2)f(x+h)-Delta_(h)^(2)f(x)=0\Delta_{h}^{3} f(x)=\Delta_{h}^{2} f(x+h)-\Delta_{h}^{2} f(x)=0Δh3f(x)=Δh2f(x+h)−Δh2f(x)=0for everything x x xxxAnd h h hhh, from which the property follows immediately.
9. One may ask whether it is possible to extend the preceding results to inequality
(16) 2 ∑ f ( x 1 + x 2 2 ) + 4 f ( x 1 + x 2 + x 3 + x 4 4 ) ≦ ∑ f ( x 1 ) + 3 ∑ f ( x 1 + x 2 + x 3 3 ) 2 ∑ f x 1 + x 2 2 + 4 f x 1 + x 2 + x 3 + x 4 4 ≦ ∑ f x 1 + 3 ∑ f x 1 + x 2 + x 3 3 2sum f((x_(1)+x_(2))/(2))+4f((x_(1)+x_(2)+x_(3)+x_(4))/(4)) <= sum f(x_(1))+3sum f((x_(1)+x_(2)+x_(3))/(3))2 \sum f\left(\frac{x_{1}+x_{2}}{2}\right)+4 f\left(\frac{x_{1}+x_{2}+x_{3}+x_{4}}{4}\right) \leqq \sum f\left(x_{1}\right)+3 \sum f\left(\frac{x_{1}+x_{2}+x_{3}}{3}\right)2∑f(x1+x22)+4f(x1+x2+x3+x44)≦∑f(x1)+3∑f(x1+x2+x33)
extended to 4 values x 1 , x 2 , x 3 , x 4 x 1 , x 2 , x 3 , x 4 x_(1),x_(2),x_(3),x_(4)x_{1}, x_{2}, x_{3}, x_{4}x1,x2,x3,x4of the argument (the summonses having obvious meanings).
Any function that satisfies the functional inequality (16) is indeed non-concave, because taking for x 1 , x 2 , x 3 , x 4 x 1 , x 2 , x 3 , x 4 x_(1),x_(2),x_(3),x_(4)x_{1}, x_{2}, x_{3}, x_{4}x1,x2,x3,x4the points 2 x − y , 2 x − y 2 x − y , 2 x − y 2x-y,2x-y2 x-y, 2 x-y2x−y,2x−y, 2 y − x , 2 y − x 2 y − x , 2 y − x 2y-x,2y-x2 y-x, 2 y-x2y−x,2y−xFrom this we deduce Jensen's inequality (1). But the converse is not true, because the non-concave function | x − ϱ | | x − ϱ | |x-ϱ||x-\varrho||x−ϱ|, if ϱ ϱ ϱ\varrhoϱis within the interval I I IIIdoes not satisfy inequality (16), the inequality analogous to that of H. Hornich
(17) Σ | x 1 + x 2 | + | x 1 + x 2 + x 3 + x 4 | ≦ Σ | x 1 | + Σ | x 1 + x 2 + x 3 | (17) Σ x 1 + x 2 + x 1 + x 2 + x 3 + x 4 ≦ Σ x 1 + Σ x 1 + x 2 + x 3 {:(17)Sigma|x_(1)+x_(2)|+|x_(1)+x_(2)+x_(3)+x_(4)| <= Sigma|x_(1)|+Sigma|x_(1)+x_(2)+x_(3)|:}\begin{equation*} \Sigma\left|x_{1}+x_{2}\right|+\left|x_{1}+x_{2}+x_{3}+x_{4}\right| \leqq \Sigma\left|x_{1}\right|+\Sigma\left|x_{1}+x_{2}+x_{3}\right| \tag{17} \end{equation*}(17)Σ|x1+x2|+|x1+x2+x3+x4|≦Σ|x1|+Σ|x1+x2+x3|
not being true in general. We will provide the demonstration of this fact later.
With regard to the functional equation obtained by equating the two members of relation (16), its general continuous solution is still any polynomial of degree 1.

10. Let us consider the inequality

(18) ∑ k = 1 n ( − 1 ) k − 1 ∑ ( k ) | x i 1 + x i 2 + … + x i k | ≧ 0 (18) ∑ k = 1 n   ( − 1 ) k − 1 ∑ ( k )   x i 1 + x i 2 + … + x i k ≧ 0 {:(18)sum_(k=1)^(n)(-1)^(k-1)sum(k)|x_(i_(1))+x_(i_(2))+dots+x_(i_(k))| >= 0:}\begin{equation*} \sum_{k=1}^{n}(-1)^{k-1} \sum^{(k)}\left|x_{i_{1}}+x_{i_{2}}+\ldots+x_{i_{k}}\right| \geqq 0 \tag{18} \end{equation*}(18)∑k=1n(−1)k−1∑(k)|xi1+xi2+…+xik|≧0
of which (5) and (17) are special cases, for n = 3 n = 3 n=3n=3n=3And n = 4 n = 4 n=4n=4n=4respectively. Dragomir Ž. Dokovič, in his cited work [2], showed that inequality (17) is not true in general. Similarly, we can demonstrate that for n > 3 n > 3 n > 3n>3n>3In any case, inequality (18) is not true in general. For this, it suffices to take x 1 = x 2 = … = x n − 1 = 1 x 1 = x 2 = … = x n − 1 = 1 x_(1)=x_(2)=dots=x_(n-1)=1x_{1}=x_{2}=\ldots=x_{n-1}=1x1=x2=…=xn−1=1, x n = − 2 x n = − 2 x_(n)=-2x_{n}=-2xn=−2(or any non-zero values ​​proportional to this sequence of numbers). We then have
Σ ( k ) | x i 1 + x i 2 + … + x i k | = ( n − 1 k ) k + ( n − 1 k − 1 ) | k − 3 | k = 1 , 2 , … , n [ ( n − 1 n ) = 0 ] Σ ( k ) x i 1 + x i 2 + … + x i k = ( n − 1 k ) k + ( n − 1 k − 1 ) | k − 3 | k = 1 , 2 , … , n ( n − 1 n ) = 0 {:[Sigma^((k))|x_(i_(1))+x_(i_(2))+dots+x_(i_(k))|=((n-1)/(k))k+((n-1)/(k-1))|k-3|],[k=1","2","dots","n[((n-1)/(n))=0]]:}\begin{gathered} \Sigma^{(k)}\left|x_{i_{1}}+x_{i_{2}}+\ldots+x_{i_{k}}\right|=\binom{n-1}{k} k+\binom{n-1}{k-1}|k-3| \\ k=1,2, \ldots, n\left[\binom{n-1}{n}=0\right] \end{gathered}Σ(k)|xi1+xi2+…+xik|=(n−1k)k+(n−1k−1)|k−3|k=1,2,…,n[(n−1n)=0]
and if n > 3 n > 3 n > 3n>3n>3, the first member of (18) becomes
∑ k = 1 n ( − 1 ) k − 1 [ ( n − 1 k ) k + ( n − 1 k − 1 ) | k − 3 | ] = ∑ k = 1 n ( − 1 ) k − 1 ( n − 1 k − 1 ) | k − 3 | = = 3 − n + ∑ k = 3 n ( − 1 ) k − 1 ( n − 1 k − 1 ) ( k − 3 ) = 2 ( 3 − n ) < 0 ∑ k = 1 n   ( − 1 ) k − 1 ( n − 1 k ) k + ( n − 1 k − 1 ) | k − 3 | = ∑ k = 1 n   ( − 1 ) k − 1 ( n − 1 k − 1 ) | k − 3 | = = 3 − n + ∑ k = 3 n   ( − 1 ) k − 1 ( n − 1 k − 1 ) ( k − 3 ) = 2 ( 3 − n ) < 0 {:[sum_(k=1)^(n)(-1)^(k-1)[((n-1)/(k))k+((n-1)/(k-1))|k-3|]=sum_(k=1)^(n)(-1)^(k-1)((n-1)/(k-1))|k-3|=],[=3-n+sum_(k=3)^(n)(-1)^(k-1)((n-1)/(k-1))(k-3)=2(3-n) < 0]:}\begin{gathered} \sum_{k=1}^{n}(-1)^{k-1}\left[\binom{n-1}{k} k+\binom{n-1}{k-1}|k-3|\right]=\sum_{k=1}^{n}(-1)^{k-1}\binom{n-1}{k-1}|k-3|= \\ =3-n+\sum_{k=3}^{n}(-1)^{k-1}\binom{n-1}{k-1}(k-3)=2(3-n)<0 \end{gathered}∑k=1n(−1)k−1[(n−1k)k+(n−1k−1)|k−3|]=∑k=1n(−1)k−1(n−1k−1)|k−3|==3−n+∑k=3n(−1)k−1(n−1k−1)(k−3)=2(3−n)<0
In this way, we also answered negatively a problem I posed previously [7].
11. The functional equation
∑ k = 1 n ( − 1 ) k − 1 ∑ ( k ) f ( x i 1 + x i 2 + … + x i k ) = 0 ∑ k = 1 n   ( − 1 ) k − 1 ∑ ( k )   f x i 1 + x i 2 + … + x i k = 0 sum_(k=1)^(n)(-1)^(k-1)sum(k)f(x_(i_(1))+x_(i_(2))+dots+x_(i_(k)))=0\sum_{k=1}^{n}(-1)^{k-1} \sum^{(k)} f\left(x_{i_{1}}+x_{i_{2}}+\ldots+x_{i_{k}}\right)=0∑k=1n(−1)k−1∑(k)f(xi1+xi2+…+xik)=0
a, according to M. Fréchet [3], as a general continuous solution any polynomial of degree n − 1 n − 1 n-1n-1n−1which cancels out for x = 0 x = 0 x=0x=0x=0This property can be demonstrated as in the particular case n = 3 n = 3 n=3n=3n=3of equation (15) by first noting that any polynomial of the form a 0 x n − 1 + a 1 x n − 2 + … + a n − 2 x a 0 x n − 1 + a 1 x n − 2 + … + a n − 2 x a_(0)x^(n-1)+a_(1)x^(n-2)+dots+a_(n-2)xa_{0} x^{n-1} +a_{1} x^{n-2}+\ldots+a_{n-2} xhas0xn−1+has1xn−2+…+hasn−2xchecks the equation and that every solution also satisfies the functional equation Δ h n f ( x ) = ∑ i = 0 n ( − 1 ) n − i ( n i ) f ( x + i h ) = 0 Δ h n f ( x ) = ∑ i = 0 n   ( − 1 ) n − i ( n i ) f ( x + i h ) = 0 Delta_(h)^(n)f(x)=sum_(i=0)^(n)(-1)^(n-i)((n)/(i))f(x+ih)=0\Delta_{h}^{n} f(x)=\sum_{i=0}^{n}(-1)^{n-i}\binom{n}{i} f(x+i h)=0Δhnf(x)=∑i=0n(−1)n−i(ni)f(x+ih)=0whose continuous solution is well known.
In the equation under consideration, the variables 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,…,xnare chosen in every possible way with the sole restriction that all
sums x i 1 + x i 2 + … + x i k x i 1 + x i 2 + … + x i k x_(i_(1))+x_(i_(2))+dots+x_(i_(k))x_{i_{1}}+x_{i_{2}}+\ldots+x_{i_{k}}xi1+xi2+…+xik(for all k k kkk) remain within an interval of the real axis that contains the origin. This interval can, in particular, be reduced to the real axis itself.
12. The functional equation
∑ k = 1 n ( − 1 ) k − 1 k ∑ ( k ) f ( x i 1 + x i 2 + ⋯ + x i k k ) = 0 ∑ k = 1 n   ( − 1 ) k − 1 k ∑ ( k )   f x i 1 + x i 2 + ⋯ + x i k k = 0 sum_(k=1)^(n)(-1)^(k-1)ksum(k)f((x_(i_(1))+x_(i_(2))+cdots+x_(i_(k)))/(k))=0\sum_{k=1}^{n}(-1)^{k-1} k \sum^{(k)} f\left(\frac{x_{i_{1}}+x_{i_{2}}+\cdots+x_{i_{k}}}{k}\right)=0∑k=1n(−1)k−1k∑(k)f(xi1+xi2+⋯+xikk)=0
checked for everything x 1 , x 2 , … , x n ∈ I x 1 , x 2 , … , x n ∈ I x_(1),x_(2),dots,x_(n)in Ix_{1}, x_{2}, \ldots, x_{n} \in Ix1,x2,…,xn∈I, has as its general continuous solution any polynomial of degree 1.
This result can be obtained independently of the study of functions for which the left-hand side does not change sign.
Indeed, every polynomial of degree 1 satisfies the equation, but the polynomial x 2 x 2 x^(2)x^{2}x2is not a solution since
∑ k = 1 n ( − k ) k − 1 1 k ∑ ( k ) ( x i 1 + x i 2 + … + x i k ) 2 = 1 n ( n − 1 ) ∑ ( 2 ) ( x i 1 − x i 2 ) 2 ∑ k = 1 n   ( − k ) k − 1 1 k ∑ ( k )   x i 1 + x i 2 + … + x i k 2 = 1 n ( n − 1 ) ∑ ( 2 )   x i 1 − x i 2 2 sum_(k=1)^(n)(-k)^(k-1)(1)/(k)sum(k)(x_(i_(1))+x_(i_(2))+dots+x_(i_(k)))^(2)=(1)/(n(n-1))sum(2)(x_(i_(1))-x_(i_(2)))^(2)\sum_{k=1}^{n}(-k)^{k-1} \frac{1}{k} \sum^{(k)}\left(x_{i_{1}}+x_{i_{2}}+\ldots+x_{i_{k}}\right)^{2}=\frac{1}{n(n-1)} \sum^{(2)}\left(x_{i_{1}}-x_{i_{2}}\right)^{2}∑k=1n(−k)k−11k∑(k)(xi1+xi2+…+xik)2=1n(n−1)∑(2)(xi1−xi2)2
is, in general, different from 0. I I IIIis of non-zero length).
By setting x i = x + ( i − 1 ) n ! h , i = 1 , 2 , … , n x i = x + ( i − 1 ) n ! h , i = 1 , 2 , … , n x_(i)=x+(i-1)n!h,i=1,2,dots,nx_{i}=x+(i-1) n!h, i=1,2, \ldots, nxi=x+(i−1)n!h,i=1,2,…,n, we find that any solution f f fffof the equation in question also satisfies an equation of the form ∑ a = 0 ( n − 1 ) n ! a a f ( x + α h ) = 0 ∑ a = 0 ( n − 1 ) n !   a a f ( x + α h ) = 0 sum_(a=0)^((n-1)n!)a_(a)f(x+alpha h)=0\sum_{a=0}^{(n-1) n!} a_{a} f(x+\alpha h)=0∑has=0(n−1)n!hashasf(x+αh)=0, Or x , x + ( n − 1 ) n ! h ∈ I x , x + ( n − 1 ) n ! h ∈ I x,x+(n-1)n!h in Ix, x+(n-1) n!h \in Ix,x+(n−1)n!h∈I, THE a a a a a_(a)a_{a}hashasare constants and a 0 ≠ 0 , a ( n − 1 ) n ! ≠ 0 a 0 ≠ 0 , a ( n − 1 ) n ! ≠ 0 a_(0)!=0,a_((n-1)n!)!=0a_{0} \neq 0, a_{(n-1) n!} \neq 0has0≠0,has(n−1)n!≠0We know [5] that any continuous solution of such an equation is a polynomial.

BIBLIOGRAPHY

  1. Adamovič, Dušan, D. - Generalization of a Hlawka identity and the corresponding inequality. Matem. Vesnik 1 (16) (1964), 39-43.
  2. Dokovič, Dragomir, Ż. - Generalizations of Hlawka's inequality. Glasnik Mat. Fiz. i Astr., 18 (1963), 169-175.
  3. Fréchet, M. - A functional definition of polynomials. New Ann. of Math. (4) 9 (1909), 145-162.
    h, H. - Eine Ungleichung für Vektorlängen. Math. Zeitschrift, 48 (1942), 268-274.
  4. Popoviciu, T.-On certain functional equations defining polynomials. Mathematica, 10 (1934), 197-211.
  5. „ - Notes on higher order convex functions. III. Mathematica, 16, (1940), 7. "--P. 139, Colloquium Mathematicum III, 2 (1955), 172.

ASUPRA UNOR INEQUALITĂTI CE CHARACTERIZEAZĂ FUNCTIILE CONVEX

Summary

It is demonstrated that inequalities (1), (6), (8) characterize the continuous neconcave function, iar (16) is the sufficient condition, it is necessary, de neconcavitate.

О НЕКОТОРЫХ НЕРАВЕНСТВАХ ХАРАКТЕРИЗУЮЩИХ

ВЫПУКЛЫЕ ФУНКЦИИ

Краткое содержание

This work requires the following (1), (6), (8) characteristics. Unprecedented, no functions and this (13) will be sent to you не необходимым условием выпуклости.
1965

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