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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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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 functionsf(x)f(x)real, of the real variablexx, defined and uniform onmmpoints
(I)
Note presented by MG Tzitzéica, MAR
x. - I consider functionsf(x)f(x)real, of the real variablexx, defined and uniform onmmpoints
(I)
x_(1) < x_(2) < dots < x_(m).x_{1}<x_{2}<\ldots<x_{m} .
I have demonstrated that if the functionf(x)f(x)is of ordernnAndn > xn>xIn general, this function cannot be extended into an interval containing the points (I)^(1){ }^{1}The 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.nntaking the valuesf(x_(i))f(x_{i})at the pointsx_(i)x_{i}It is also evident that this is the case if the function is not convex or concave, without being polynomial of ordernnon (I). On the other hand, a monotonic function and the order functionxxare 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:
- Ifm > 2m>2Andf(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), we can find an increasing polynomialP(x)P(x)such as one hasP(x_(i))=f(x_(i)),i=1,2,dots,mP\left(x_{i}\right)=f\left(x_{i}\right), i=1,2, \ldots, mII .
Ifm > 3m>3Andf(x)f(x)is convex on (I), therefore[x_(i),x_(i+x),x_(i+2);f] > 0\left[x_{i}, x_{i+x}, x_{i+2}; f\right]>0,i=1,2,dots,m-2^(2)i=1,2, \ldots, m-2^{2}), we can find a convex polynomialP(x)P(x)such asl^(')l^{\prime}we haveP(x_(i))=f(x_(i)),i=1,2dots,mP\left(x_{i}\right)=f\left(x_{i}\right), i=1,2 \ldots, 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. LetP(x)P(x)the polynomial we are looking for. The numbers must then
{:(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)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*}
Solent positive, we immediately see that Theotem I returns to
Lemma I. Given mI positive numbersa_(1),a_(2),dots,a_(m-1)a_{1}, a_{2}, \ldots, a_{m-1}, we can find an increasing polynomialP(x)P(x)such as one has
Lemma I. Given mI positive numbersa_(1),a_(2),dots,a_(m-1)a_{1}, a_{2}, \ldots, a_{m-1}, we can find an increasing polynomialP(x)P(x)such as one has
p_(i)=a_(i),i=1,2,dots,m-1,p_{i}=a_{i}, i=1,2, \ldots, m-1,
Let Di be the numbers (2)
Consider the pointM(p_(1),p_(2),dots,p_(mx))M\left(p_{1}, p_{2}, \ldots, p_{mx}\right)coordinatesp_(i)p_{i}in ordinary space amm-I dimensions. At each increasing polynomialP(x)P(x)thus corresponds a pointMM. WhenP(x)P(x)varied,MMdescribes a domain that is obviously convex (since ifP,P_(1)P, P_{1}are two increasing polynomials, the polynonePPtPP(is also increasing). On the other hand, point 2A(a_(1),a_(2),2,3,4,-1:}A\left(a_{1}, a_{2}, 2,3,4,-1\right.of positive coordinatesa_(i)a_{i}also belongs to a convex domain. We conclude that Lemma I will follow from
Consider the pointM(p_(1),p_(2),dots,p_(mx))M\left(p_{1}, p_{2}, \ldots, p_{mx}\right)coordinatesp_(i)p_{i}in ordinary space amm-I dimensions. At each increasing polynomialP(x)P(x)thus corresponds a pointMM. WhenP(x)P(x)varied,MMdescribes a domain that is obviously convex (since ifP,P_(1)P, P_{1}are two increasing polynomials, the polynonePPtPP(is also increasing). On the other hand, point 2A(a_(1),a_(2),2,3,4,-1:}A\left(a_{1}, a_{2}, 2,3,4,-1\right.of positive coordinatesa_(i)a_{i}also belongs to a convex domain. We conclude that Lemma I will follow from
Iemme. I. For ohquie valuent dej=1,2,5j=1,2,5,mm- I and for allepsi > 0\varepsilon>0, we can find a crossing polynomialP(x)P(x)such as one has
To demonstrate either(a,b)(a, b), an interval containing the points (I) and consider the continuous functiong(x)g(x)defined by
We know that we can find a polynomialQ(x)Q(x)growing in(-oo,+oo)(-\infty,+\infty)such that 1 we have 3 )
|*gQ| < epsi^(')," in "(a,b),|\cdot gQ|<\varepsilon^{\prime}, \text { in }(a, b),
Orepsi^(') < min((I)/(4),(epsi)/(4))\varepsilon^{\prime}<\min \left(\frac{I}{4}, \frac{\varepsilon}{4}\right)We can then immediately see that the polynomial
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)}
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","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*}
are positive. The theorem then leads back to
Lemma III. Givenm-2m-2positive numbersa_(1),a_(2),dotsa_(n-2)a_{1}, a_{2}, \ldots a_{n-2}, we can find a convex polynomialP(x)P(x)such as one has
p_(i)=a_(i),i=I,2,dots,m-2p_{i}=a_{i}, i=I, 2, \ldots, m-2
p_(i)p_{i}being the numbers (3).
As above, we see that this property results from the following, expressed by the
{:(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*}
are positive. The theorem then leads back to
Lemma III. Givenm-2m-2positive numbersa_(1),a_(2),dotsa_(n-2)a_{1}, a_{2}, \ldots a_{n-2}, we can find a convex polynomialP(x)P(x)such as one has
p_(i)=a_(i),i=I,2,dots,m-2p_{i}=a_{i}, i=I, 2, \ldots, m-2
p_(i)p_{i}being the numbers (3).
As above, we see that this property results from the following, expressed by the
Lemma IV. For each value ofj=1,2,dots,m-2j=1,2, \ldots, m-2and for everythingepsi > 0\varepsilon>0, we can find a convex polynomialP(x)P(x)such as one has
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 .
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)= \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}
In this case we can find a polynomialQ(x)Q(x)convex in-oo,+oo-\infty,+\infty) such that one has
|h-Q| < epsi^(')," dans "(a,b)|h-Q|<\varepsilon^{\prime}, \text { dans }(a, b)
no matter how small the positive numberepsi^(')^(4)\varepsilon^{\prime}{ }^{4}4.
- Without dwelling on the proofs, let us say, in conclusion, that we still have the following properties;
III. Ifm > 2m>2and the functionf(x)f(x)is increasing and convex on (1), we can find a polynomialP(x)P(x)increasing within an interval(a,+oo)(a,+\infty)oita <= x_(1)a \leqq x_{1}and convex in(-oo,+oo)(-\infty,+\infty)such as one hasP(x_(i))=f(x_(i));i=1P\left(x_{i}\right)=f\left(x_{i}\right) ; i=1,2,dots,m2, \ldots, mIV
. Ifm > 2m>2and the functionf(x)f(x)is decreasing and convex on (1), we can find a polynomialP(x)P(x)decreasing within an interval(-oo,b)(-\infty, b)Orb >= x_(m)b \geq x_{m}and convex in(-oo,+oo)(-\infty,+\infty)such as one hasP(x_(i))=f(x_(i))P\left(x_{i}\right)=f\left(x_{i}\right),i=1,2,dots,mi=1,2, \ldots, m.
- Without dwelling on the proofs, let us say, in conclusion, that we still have the following properties;
III. Ifm > 2m>2and the functionf(x)f(x)is increasing and convex on (1), we can find a polynomialP(x)P(x)increasing within an interval(a,+oo)(a,+\infty)oita <= x_(1)a \leqq x_{1}and convex in(-oo,+oo)(-\infty,+\infty)such as one hasP(x_(i))=f(x_(i));i=1P\left(x_{i}\right)=f\left(x_{i}\right) ; i=1,2,dots,m2, \ldots, mIV
. Ifm > 2m>2and the functionf(x)f(x)is decreasing and convex on (1), we can find a polynomialP(x)P(x)decreasing within an interval(-oo,b)(-\infty, b)Orb >= x_(m)b \geq x_{m}and convex in(-oo,+oo)(-\infty,+\infty)such as one hasP(x_(i))=f(x_(i))P\left(x_{i}\right)=f\left(x_{i}\right),i=1,2,dots,mi=1,2, \ldots, m.
Cernăuti, September 28, 2938.
- ^(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}) For the notations, see my previous work. - ^(3){ }^{3}) See Tiberiu Popoviciu, Suy l'approzimation des fonotions convexes d'ordre superiore, in this Bulletin.
- ^(4){ }^{4}) See loc. cit. (previous note).