On the extension of monotone functions and convex functions defined on a finite number of points

Tiberiu Popoviciu
(original), Approximation Theory, convexity, Mathematical Analysis, monotonicity, not-at-ICTP, paper

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T. Popoviciu, Sur le prolongement des fonctions monotones et des fonctions convexes definies sur un nombre fini de points, Bull. de le Sect. sci de l’Acad. Roum., 20 (1938) no. 7, pp. 196-199 (in French).

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Bulletin Mathematique de la Societe des Sciences Mathematiques de Roumanie

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Romanian Society of Mathematical Sciences

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[Zbl 0021.11702, JFM 64.1025.02]

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1938 a -Popoviciu- Bull. Sect. Sci. Acad. Roum. - On the continuation of monotonic functions and of
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ON THE EXTENSION OF MONOTONE AND CONVEX FUNCTIONS DEFINED ON A FINITE NUMBER OF POINTS

TIBERIU POPOVICIU
Note presented by MG Tzitzéica, MAR
x. - I consider functions f ( x ) f ( x ) f(x)f(x)f(x)real, of the real variable x x xxx, defined and uniform on m m mmmpoints
(I)
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.
I have demonstrated that if the function f ( x ) f ( x ) f(x)f(x)f(x)is of order n n nnnAnd n > x n > x n > xn>xn>xIn general, this function cannot be extended into an interval containing the points (I) 1 1 ^(1){ }^{1}1The extension is, a fortiori, impossible, in general, along the entire real axis ( − ∞ , + ∞ − ∞ , + ∞ -oo,+oo-\infty,+\infty−∞,+∞). It follows that there is no polynomial of order in general. n n nnntaking the values f ( x i ) f x i f(x_(i))f(x_{i})f(xi)at the points x i x i x_(i)x_{i}xiIt is also evident that this is the case if the function is not convex or concave, without being polynomial of order n n nnnon (I). On the other hand, a monotonic function and the order function x x xxxare always and everywhere. extendable.
In what follows, I propose to demonstrate that if the function is strictly monotonic (i.e., increasing or decreasing) or if it is convex or concave of order r, the extension is always possible by a polynomial. In other words, I will prove the following theorems:
  1. If m > 2 m > 2 m > 2m>2m>2And f ( x 1 ) < f ( x 2 ) < … < f ( x m ) f x 1 < f x 2 < … < f x m f(x_(1)) < f(x_(2)) < dots < f(x_(m))f\left(x_{1}\right)<f\left(x_{2}\right)<\ldots<f\left(x_{m}\right)f(x1)<f(x2)<…<f(xm), we can find an increasing polynomial P ( x ) P ( x ) P(x)P(x)P(x)such as one has P ( x i ) = f ( x i ) , i = 1 , 2 , … , m P x i = f x i , i = 1 , 2 , … , m P(x_(i))=f(x_(i)),i=1,2,dots,mP\left(x_{i}\right)=f\left(x_{i}\right), i=1,2, \ldots, mP(xi)=f(xi),i=1,2,…,mII .
    If m > 3 m > 3 m > 3m>3m>3And f ( x ) f ( x ) f(x)f(x)f(x)is convex on (I), therefore [ x i , x i + x , x i + 2 ; f ] > 0 x i , x i + x , x i + 2 ; f > 0 [x_(i),x_(i+x),x_(i+2);f] > 0\left[x_{i}, x_{i+x}, x_{i+2}; f\right]>0[xi,xi+x,xi+2;f]>0, i = 1 , 2 , … , m − 2 2 i = 1 , 2 , … , m − 2 2 i=1,2,dots,m-2^(2)i=1,2, \ldots, m-2^{2}i=1,2,…,m−22), we can find a convex polynomial P ( x ) P ( x ) P(x)P(x)P(x)such as L ′ L ′ l^(')l^{\prime}L′we have P ( x i ) = f ( x i ) , i = 1 , 2 … , m P x i = f x i , i = 1 , 2 … , m P(x_(i))=f(x_(i)),i=1,2dots,mP\left(x_{i}\right)=f\left(x_{i}\right), i=1,2 \ldots, mP(xi)=f(xi),i=1,2…,m.
In these two theorems and in what follows, we are dealing with increasing polynomials and convex polynomials on the entire real axis ( − ∞ , + ∞ − ∞ , + ∞ -oo,+oo-\infty,+\infty−∞,+∞), unless, of course, we say otherwise.
2. Let's prove Theorem I. Let P ( x ) P ( x ) P(x)P(x)P(x)the polynomial we are looking for. The numbers must then
(2) P C 0 = P ( x i + 1 ) − P ( x i ) ; i = 1 , 2 , … , m − 1 (2) P C 0 = P x i + 1 − P x i ; i = 1 , 2 , … , m − 1 {:(2)PC_(0)=P(x_(i)+1)-P(x_(i))quad;quad i=1","2","dots","m-1:}\begin{equation*} P C_{0}=P\left(x_{i}+1\right)-P\left(x_{i}\right) \quad ; \quad i=1,2, \ldots, m-1 \tag{2} \end{equation*}(2)PC0=P(xi+1)−P(xi);i=1,2,…,m−1
Solent positive, we immediately see that Theotem I returns to
Lemma I. Given mI positive numbers has 1 , has 2 , … , has m − 1 has 1 , has 2 , … , has m − 1 a_(1),a_(2),dots,a_(m-1)a_{1}, a_{2}, \ldots, a_{m-1}has1,has2,…,hasm−1, we can find an increasing polynomial P ( x ) P ( x ) P(x)P(x)P(x)such as one has
p i = has i , i = 1 , 2 , … , m − 1 , p i = has i , i = 1 , 2 , … , m − 1 , p_(i)=a_(i),i=1,2,dots,m-1,p_{i}=a_{i}, i=1,2, \ldots, m-1,pi=hasi,i=1,2,…,m−1,
Let Di be the numbers (2)
Consider the point M ( p 1 , p 2 , … , p m − x ) M p 1 , p 2 , … , p m − x M(p_(1),p_(2),dots,p_(mx))M\left(p_{1}, p_{2}, \ldots, p_{mx}\right)M(p1,p2,…,pm−x)coordinates p i p i p_(i)p_{i}piin ordinary space a m m mmm-I dimensions. At each increasing polynomial P ( x ) P ( x ) P(x)P(x)P(x)thus corresponds a point M M MMM. When P ( x ) P ( x ) P(x)P(x)P(x)varied, M M MMMdescribes a domain that is obviously convex (since if P , P 1 P , P 1 P,P_(1)P, P_{1}P,P1are two increasing polynomials, the polynone P P PPPt P P PPP(is also increasing). On the other hand, point 2 HAS ( has 1 , has 2 , 2 , 3 , 4 , − 1 HAS has 1 , has 2 , 2 , 3 , 4 , − 1 A(a_(1),a_(2),2,3,4,-1:}A\left(a_{1}, a_{2}, 2,3,4,-1\right.HAS(has1,has2,2,3,4,−1of positive coordinates has i has i a_(i)a_{i}hasialso belongs to a convex domain. We conclude that Lemma I will follow from
Iemme. I. For ohquie valuent de j = 1 , 2 , 5 j = 1 , 2 , 5 j=1,2,5j=1,2,5j=1,2,5, m m mmm- I and for all ε > 0 ε > 0 epsi > 0\varepsilon>0ε>0, we can find a crossing polynomial P ( x ) P ( x ) P(x)P(x)P(x)such as one has
To demonstrate either ( has , b ) ( has , b ) (a,b)(a, b)(has,b), an interval containing the points (I) and consider the continuous function g ( x ) g ( x ) g(x)g(x)g(x)defined by
We know that we can find a polynomial Q ( x ) Q ( x ) Q(x)Q(x)Q(x)growing in ( − ∞ , + ∞ ) ( − ∞ , + ∞ ) (-oo,+oo)(-\infty,+\infty)(−∞,+∞)such that 1 we have 3 )
| ⋅ g − Q | < ε ′ , In ( has , b ) , | ⋅ g − Q | < ε ′ ,  In  ( has , b ) , |*gQ| < epsi^(')," in "(a,b),|\cdot gQ|<\varepsilon^{\prime}, \text { in }(a, b),|⋅g−Q|<ε′, In (has,b),
Or ε ′ < min ( I 4 , ε 4 ) ε ′ < min I 4 , ε 4 epsi^(') < min((I)/(4),(epsi)/(4))\varepsilon^{\prime}<\min \left(\frac{I}{4}, \frac{\varepsilon}{4}\right)ε′<min(I4,ε4)We can then immediately see that the polynomial
P ( x ) = Q ( x ) Q ( x j + 1 ) − Q ( x j ) P ( x ) = Q ( x ) Q x j + 1 − Q x j P(x)=(Q(x))/(Q(x_(j+1))-Q(x_(j)))P(x)=\frac{Q(x)}{Q\left(x_{j+1}\right)-Q\left(x_{j}\right)}P(x)=Q(x)Q(xj+1)−Q(xj)
check lemma II.
3. - Let's move on to the proof of Theorem II. In this case, the numbers must
(3) p i = [ x i , x i + 1 , x i + 2 ; f ] , i = 1 , 2 , … , m − 2 (3) p i = x i , x i + 1 , x i + 2 ; f , i = 1 , 2 , … , m − 2 {:(3)p_(i)=[x_(i),x_(i+1),x_(i+2);f]","i=1","2","dots","m-2:}\begin{equation*} p_{i}=\left[x_{i}, x_{i+1}, x_{i+2} ; f\right], i=1,2, \ldots, m-2 \tag{3} \end{equation*}(3)pi=[xi,xi+1,xi+2;f],i=1,2,…,m−2
are positive. The theorem then leads back to
Lemma III. Given m − 2 m − 2 m-2m-2m−2positive numbers a 1 , a 2 , … a n − 2 a 1 , a 2 , … a n − 2 a_(1),a_(2),dotsa_(n-2)a_{1}, a_{2}, \ldots a_{n-2}has1,has2,…hasn−2, we can find a convex polynomial P ( x ) P ( x ) P(x)P(x)P(x)such as one has
p i = a i , i = I , 2 , … , m − 2 p i = a i , i = I , 2 , … , m − 2 p_(i)=a_(i),i=I,2,dots,m-2p_{i}=a_{i}, i=I, 2, \ldots, m-2pi=hasi,i=I,2,…,m−2
p i p i p_(i)p_{i}pibeing the numbers (3).
As above, we see that this property results from the following, expressed by the
Lemma IV. For each value of j = 1 , 2 , … , m − 2 j = 1 , 2 , … , m − 2 j=1,2,dots,m-2j=1,2, \ldots, m-2j=1,2,…,m−2and for everything ε > 0 ε > 0 epsi > 0\varepsilon>0ε>0, we can find a convex polynomial P ( x ) P ( x ) P(x)P(x)P(x)such as one has
p j = I , p i < ε , i = I , 2 , … , j − I , j + I , … , m − 2 . p j = I , p i < ε , i = I , 2 , … , j − I , j + I , … , m − 2 . p_(j)=I,p_(i) < epsi,i=I,2,dots,j-I,j+I,dots,m-2.p_{j}=\mathrm{I}, p_{i}<\varepsilon, i=\mathrm{I}, 2, \ldots, j-\mathrm{I}, j+\mathrm{I}, \ldots, m-2 .pj=I,pi<ε,i=I,2,…,j−I,j+I,…,m−2.
The proof is done as above for Lemma II, by considering the function
h ( x ) = { 0 , dans ( a , x i + 1 ) x − x j + x , dans ( x i + x , b ) h ( x ) = 0 ,       dans  a , x i + 1 x − x j + x ,       dans  x i + x , b h(x)={[0","," dans "(a,x_(i+1))],[x-x_(j+x)","," dans "(x_(i+x),b)]:}h(x)= \begin{cases}0, & \text { dans }\left(a, x_{i+1}\right) \\ x-x_{j+x}, & \text { dans }\left(x_{i+x}, b\right)\end{cases}h(x)={0, In (has,xi+1)x−xj+x, In (xi+x,b)
In this case we can find a polynomial Q ( x ) Q ( x ) Q(x)Q(x)Q(x)convex in − ∞ , + ∞ − ∞ , + ∞ -oo,+oo-\infty,+\infty−∞,+∞) such that one has
| h − Q | < ε ′ , dans ( a , b ) | h − Q | < ε ′ ,  dans  ( a , b ) |h-Q| < epsi^(')," dans "(a,b)|h-Q|<\varepsilon^{\prime}, \text { dans }(a, b)|h−Q|<ε′, In (has,b)
no matter how small the positive number ε ′ 4 ε ′ 4 epsi^(')^(4)\varepsilon^{\prime}{ }^{4}ε′44.
- Without dwelling on the proofs, let us say, in conclusion, that we still have the following properties;
III. If m > 2 m > 2 m > 2m>2m>2and the function f ( x ) f ( x ) f(x)f(x)f(x)is increasing and convex on (1), we can find a polynomial P ( x ) P ( x ) P(x)P(x)P(x)increasing within an interval ( a , + ∞ ) ( a , + ∞ ) (a,+oo)(a,+\infty)(has,+∞)oit a ≦ x 1 a ≦ x 1 a <= x_(1)a \leqq x_{1}has≦x1and convex in ( − ∞ , + ∞ ) ( − ∞ , + ∞ ) (-oo,+oo)(-\infty,+\infty)(−∞,+∞)such as one has P ( x i ) = f ( x i ) ; i = 1 P x i = f x i ; i = 1 P(x_(i))=f(x_(i));i=1P\left(x_{i}\right)=f\left(x_{i}\right) ; i=1P(xi)=f(xi);i=1, 2 , … , m 2 , … , m 2,dots,m2, \ldots, m2,…,mIV
. If m > 2 m > 2 m > 2m>2m>2and the function f ( x ) f ( x ) f(x)f(x)f(x)is decreasing and convex on (1), we can find a polynomial P ( x ) P ( x ) P(x)P(x)P(x)decreasing within an interval ( − ∞ , b ) ( − ∞ , b ) (-oo,b)(-\infty, b)(−∞,b)Or b ≥ x m b ≥ x m b >= x_(m)b \geq x_{m}b≥xmand convex in ( − ∞ , + ∞ ) ( − ∞ , + ∞ ) (-oo,+oo)(-\infty,+\infty)(−∞,+∞)such as one has P ( x i ) = f ( x i ) P x i = f x i P(x_(i))=f(x_(i))P\left(x_{i}\right)=f\left(x_{i}\right)P(xi)=f(xi), i = 1 , 2 , … , m i = 1 , 2 , … , m i=1,2,dots,mi=1,2, \ldots, mi=1,2,…,m.
Cernăuti, September 28, 2938.
  1. 1 1 ^(1){ }^{1}1) See: Tiberiu Popoviciu, On the continuation of higher order convex functions, “Bull. Math. Soc. Roumaine des Sc.”, vol. 36 (1934), pp. 75-108.
    2 2 ^(2){ }^{2}2) For the notations, see my previous work.
  2. 3 3 ^(3){ }^{3}3) See Tiberiu Popoviciu, Suy l'approzimation des fonotions convexes d'ordre superiore, in this Bulletin.
  3. 4 4 ^(4){ }^{4}4) See loc. cit. (previous note).
1938

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