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T. Popoviciu, Notes sur les généralisations des fonctions convexes d’ordre supérieur (I), Disquisitiones mathematicae et physicae, 1 (1940), pp. 35-42 (in French).
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Disquisitiones mathematicae et physicae
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[MR0021038, JFM 66.0241.01]
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NOTES ON GENERALIZATIONS OF HIGHER ORDER CONVEX FUNCTIONS (I)
In a series of notes entitled "Notes on Higher Order Convex Functions" we continue the study of higher order functionsnn. In this new series of notes we propose to examine the various classes of functions which generalize the functions of ordernnof a variable.
The order functions (n∣kn \mid k).
- Let us first recall the definition of functions of order n. The functionf=f(x)f=f(x), real, finite, uniform and defined on any linear setEEis said to be of ordernnonEEif its difference divided by ordern+2,[x_(1),x_(2),dots,x_(n+2);f]n+2,\left[x_{1}, x_{2}, \ldots, x_{n+2}; f\right]does not change sign onEE. More precisely we have the following definition
The function f is consex, non-concave, polynomial, non-consex resp. concape of ordernnonEE, following that inequality
(1)
(1)
[x_(1),x_(2),dots,x_(n+2);f] > , >= ,=, <= " resp. " < 0\left[x_{1}, x_{2}, \ldots, x_{n+2}; f\right]>, \geqq,=, \leqq \text { resp. }<0
is serified, whatever the pointsx_(1),x_(2),dots,x_(n+2)epsi Ex_{1}, x_{2}, \ldots, x_{n+2} \varepsilon E.
Convexity and polynomiality of ordernnare special cases of non-concavity of ordernn. Ifffis convex, non-concave, etc. of ordernn, the function-f-fconcave, non-convex, etc., of ordernnand vice versa. We can therefore take as type of order functionnnthe non-concave function of ordernn.
Convexity and polynomiality of ordernnare special cases of non-concavity of ordernn. Ifffis convex, non-concave, etc. of ordernn, the function-f-fconcave, non-convex, etc., of ordernnand vice versa. We can therefore take as type of order functionnnthe non-concave function of ordernn.
In particular, the definition applies ton=-1n=-1and then we have the functions which do not change sign onEE, more precisely the positive, non-negative, identically zero, non-positive resp. negative functions. Forn=0n=0we have monotonic, increasing, non-decreasing, constant, non-increasing resp. decreasing functions
. Finally, forn=1n=1, we have the usual convex, non-concave, linear, non-convex resp. concave functions.
. Finally, forn=1n=1, we have the usual convex, non-concave, linear, non-convex resp. concave functions.
It is clear that the order functionsnnare thus only defined on setsEEhaving at leastn+2n+2points. However, in some statements it is useful to assume that any function defined on less thann+2n+2points is of ordernnand indifferently convex or concave of ordernn.
We will designate by{x_(1),x_(2),dots,x_(m)}\left\{x_{1}, x_{2}, \ldots, x_{m}\right\}an ordered sequence of points, therefore of points such that we havex_(1) < x_(2) < dots < x_(n)x_{1}<x_{2}<\ldots<x_{n}. For the functionffeither convex, non-concave, polynomial, non-convex, or concave of ordernnon the ordered sequence{x_(1),x_(2),dots,x_(m)}(m >= n+2)\left\{x_{1}, x_{2}, \ldots, x_{m}\right\}(m \geq n+2), it is necessary and sufficient that we have
Delta_(n+2)^(i)(f) > , >= ,=, <= resp. < 0,i=1,2,dots,m-n-1,\Delta_{n+2}^{i}(f)>, \geq,=, \leqq \mathrm{resp} .<0, i=1,2, \ldots, m-n-1,
by posing
{:[(2)Delta_(j)^(i)(f)=[x_(i),x_(i+1),dots,x_(i+j);f]","Delta_(0)^(i)(f)=f(x_(i))],[i=1","2","dots","m-j","j=0","1","dots","m-1]:}\begin{gather*}
\Delta_{j}^{i}(f)=\left[x_{i}, x_{i+1}, \ldots, x_{i+j} ; f\right], \Delta_{0}^{i}(f)=f\left(x_{i}\right) \tag{2}\\
i=1,2, \ldots, m-j, j=0,1, \ldots, m-1
\end{gather*}
This property is an immediate consequence of what we can call the theorem of the mean of divided differences. This theorem expresses the property that any divided difference[x_(i_(1)),x_(i_(2)),dots,x_(i_(n+2));f]\left[x_{i_{1}}, x_{i_{2}}, \ldots, x_{i_{n+2}} ; f\right]onn+2n+2points of the ordered sequence{x_(1),x_(2),dots,x_(m)}\left\{x_{1}, x_{2}, \ldots, x_{m}\right\}, is a (generalized) arithmetic mean of the divided differences
Delta_(n+1)^(1)(f),Delta_(n+1)^(2)(f),dots,Delta_(n+1)^(m-n-1)(f).\Delta_{n+1}^{1}(f), \Delta_{n+1}^{2}(f), \ldots, \Delta_{n+1}^{m-n-1}(f) .
- We will now generalize the order functionsnnby introducing the order functions (n∣kn \mid k).
Eithere={x_(1),x_(2),dots,x_(m)}e=\left\{x_{1}, x_{2}, \ldots, x_{m}\right\}a finite subset (a sequence) ofEE.
Definition 1. We will say that the sequence (with the notations (2))
Definition 1. We will say that the sequence (with the notations (2))
{:(3)Delta_(n+1)^(1)(f)","Delta_(n+1)^(2)(f)","dots","Delta_(n+1)^(m-n-1)(f)",":}\begin{equation*}
\Delta_{n+1}^{1}(f), \Delta_{n+1}^{2}(f), \ldots, \Delta_{n+1}^{m-n-1}(f), \tag{3}
\end{equation*}
is the sequeld_(n+1)d_{n+1}corresponding to the followingee, or simply the continuationd_(n+1)d_{n+1}ofee. In particular, the followingd_(0)d_{0}ofeeEast
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)
Let us now introduce the following definition.
Definition 2. We will say that the functionffis of order (n∣kn \mid k) onEEif the maximum number of variations of the sequencesd_(n+1)d_{n+1}, of all finite sequences e ofEE, is equal tokk.
Definition 2. We will say that the functionffis of order (n∣kn \mid k) onEEif the maximum number of variations of the sequencesd_(n+1)d_{n+1}, of all finite sequences e ofEE, is equal tokk.
The numberkkis equal to 0 or a natural number. The definition therefore requires thatkkis finished, so that the number of variations of the sequencesd_(n+n)d_{n+n}is bounded. It is clear that there then exists at least one sequenceeewhose
continuationd_(n+1)d_{n+1}presents exactlykkvariations. The numbernncan take the values-1,0,1,2,dots-1,0,1,2, \ldots
continuationd_(n+1)d_{n+1}presents exactlykkvariations. The numbernncan take the values-1,0,1,2,dots-1,0,1,2, \ldots
The order functions (n∣0n \mid 0) coincide with the order functionsnn.
To simplify the language, we will say that the order (n∣kn \mid k) is less than, at most equal, equal, at least equal resp. greater than the order (n∣k^(')n \mid k^{\prime}) depending on whetherk < , <= ,=, >=k<, \leqslant,=, \geqslantresp.> k^(')>k^{\prime}.
To simplify the language, we will say that the order (n∣kn \mid k) is less than, at most equal, equal, at least equal resp. greater than the order (n∣k^(')n \mid k^{\prime}) depending on whetherk < , <= ,=, >=k<, \leqslant,=, \geqslantresp.> k^(')>k^{\prime}.
It is clear that ifffis of order (n∣kn \mid k) onEE, it is at the most orderly(n∣k)(n \mid k)Above allE_(1)sub EE_{1} \subset E. Ifffis in order(n∣k)(n \mid k), the functioncfc f, Orccis a non-zero constant and is also of order (n∣kn \mid k). The same is true of the functionf+Pf+P, OrPPis any polynomial of degreenn.
3. The order functions (n∣kn \mid k) are therefore characterized by a property of sequencesd_(n+1)d_{n+1}finite subsets ofEE. We must therefore first study more closely the structure of these sequences.d_(n+1)d_{n+1}.
3. The order functions (n∣kn \mid k) are therefore characterized by a property of sequencesd_(n+1)d_{n+1}finite subsets ofEE. We must therefore first study more closely the structure of these sequences.d_(n+1)d_{n+1}.
From a more general theorem of MI Schoenberg^(1){ }^{1}
), Lemma 1 results . Ifgamma_(i),mu_(i),i=1,2,dots,m-1\gamma_{i}, \mu_{i}, i=1,2, \ldots, m-1are non-negative numbers, the number of variations of the sequence
(4)lambda_(1)c_(1)+mu_(1)c_(2),lambda_(2)c_(2)+mu_(2)c_(3),dots,lambda_(m-1)c_(m-1)+mu_(m-1)c_(m)\lambda_{1} c_{1}+\mu_{1} c_{2}, \lambda_{2} c_{2}+\mu_{2} c_{3}, \ldots, \lambda_{m-1} c_{m-1}+\mu_{m-1} c_{m}
is at most equal to the number of bets in the sequence
(5)
), Lemma 1 results . Ifgamma_(i),mu_(i),i=1,2,dots,m-1\gamma_{i}, \mu_{i}, i=1,2, \ldots, m-1are non-negative numbers, the number of variations of the sequence
(4)lambda_(1)c_(1)+mu_(1)c_(2),lambda_(2)c_(2)+mu_(2)c_(3),dots,lambda_(m-1)c_(m-1)+mu_(m-1)c_(m)\lambda_{1} c_{1}+\mu_{1} c_{2}, \lambda_{2} c_{2}+\mu_{2} c_{3}, \ldots, \lambda_{m-1} c_{m-1}+\mu_{m-1} c_{m}
is at most equal to the number of bets in the sequence
(5)
c_(1),c_(2),dots,c_(m)c_{1}, c_{2}, \ldots, c_{m}
We complete this property with
Lemma 2. If the sequences (4) and (5) have the same numberkkof variations, the first non-zero terms in these sequences have the same sign and the last non-zero terms also have the same sign.
Lemma 2. If the sequences (4) and (5) have the same numberkkof variations, the first non-zero terms in these sequences have the same sign and the last non-zero terms also have the same sign.
In a sequence (5) the first non-zero term is the termc_(s)c_{s}such asc_(1)=c_(2)=dots=c_(s-1)=0,c_(s)!=0c_{1}=c_{2}=\ldots=c_{s-1}=0, c_{s} \neq 0and the last non-zero term is defined in an analogous way. The proof of Lemma 2 is easily done by induction on the numberkk. It is of course assumed that, ifk=0k=0, none of the sequences (4), (5) is identically zero.
Now consider a finite sequencee={x_(1),x_(2),dots,x_(m)}e=\left\{x_{1}, x_{2}, \ldots, x_{m}\right\}ofEE. We have the
Theorem 1. The number of variations of the sequenced_(n+1)d_{n+1}of a partial sequence of e is at most equal to the number of pariations of the sequenced_(n+1)d_{n+1}of e.
It is obviously sufficient to demonstrate the property for partial sequencese_(l)={x_(1),x_(2),dots,x_(j-1),x_(j+1)dots,x_(m)},l <= j <= me_{l}=\left\{x_{1}, x_{2}, \ldots, x_{j-1}, x_{j+1} \ldots, x_{m}\right\}, l \leqq j \leqq mofee. Let (3) be the sequenced_(n+1)d_{n+1}ofeeand
(6)
(6)
Delta_(n+1)^(**1)(f),Delta_(n+1)^(**2)(f),dots,Delta_(n+1)^(**m-n-2)(f)\Delta_{n+1}^{* 1}(f), \Delta_{n+1}^{* 2}(f), \ldots, \Delta_{n+1}^{* m-n-2}(f)
the sequeld_(n+1)d_{n+1}ofe_(j)e_{j}. We have
{:(7){[Delta_(n+1)^(^(**)i),(f)=Delta_(n+1)^(i)(f)","],[Delta_(n+1)^(^(**)i)(f)=((x_(j)-x_(i))Delta_(n+1)^(i)(f)+(x_(i+n+2)-x_(j))Delta_(n+1)^(i+1)(f))/(x_(i+n+2)-x_(i)),],[,i=j-n-1","j-n","dots","j-1],[Delta_(n+1)^(^(**)i)(f)=Delta_(n+1)^(i+1)(f)",",i=j","j+1","dots","m-n-2]:}:}\begin{cases}\Delta_{n+1}^{{ }^{*} i} & (f)=\Delta_{n+1}^{i}(f), \tag{7}\\ \Delta_{n+1}^{{ }^{*} i}(f)=\frac{\left(x_{j}-x_{i}\right) \Delta_{n+1}^{i}(f)+\left(x_{i+n+2}-x_{j}\right) \Delta_{n+1}^{i+1}(f)}{x_{i+n+2}-x_{i}} & \\ & i=j-n-1, j-n, \ldots, j-1 \\ \Delta_{n+1}^{{ }^{*} i}(f)=\Delta_{n+1}^{i+1}(f), & i=j, j+1, \ldots, m-n-2\end{cases}
and the property results from Lemma 1. In formulas (7) the first two groups are to be deleted ifj=1j=1. The first group is to be deleted and in the secondiivaries from 1 toj-1j-1if1 < j <= n+21<j \leq n+2. The same applies to the last two groups ifi >= m-n-1i \geqq m-n-1.
{:(7){[Delta_(n+1)^(^(**)i),(f)=Delta_(n+1)^(i)(f)","],[Delta_(n+1)^(^(**)i)(f)=((x_(j)-x_(i))Delta_(n+1)^(i)(f)+(x_(i+n+2)-x_(j))Delta_(n+1)^(i+1)(f))/(x_(i+n+2)-x_(i)),],[,i=j-n-1","j-n","dots","j-1],[Delta_(n+1)^(^(**)i)(f)=Delta_(n+1)^(i+1)(f)",",i=j","j+1","dots","m-n-2]:}:}\begin{cases}\Delta_{n+1}^{{ }^{*} i} & (f)=\Delta_{n+1}^{i}(f), \tag{7}\\ \Delta_{n+1}^{{ }^{*} i}(f)=\frac{\left(x_{j}-x_{i}\right) \Delta_{n+1}^{i}(f)+\left(x_{i+n+2}-x_{j}\right) \Delta_{n+1}^{i+1}(f)}{x_{i+n+2}-x_{i}} & \\ & i=j-n-1, j-n, \ldots, j-1 \\ \Delta_{n+1}^{{ }^{*} i}(f)=\Delta_{n+1}^{i+1}(f), & i=j, j+1, \ldots, m-n-2\end{cases}
and the property results from Lemma 1. In formulas (7) the first two groups are to be deleted ifj=1j=1. The first group is to be deleted and in the secondiivaries from 1 toj-1j-1if1 < j <= n+21<j \leq n+2. The same applies to the last two groups ifi >= m-n-1i \geqq m-n-1.
It also follows that if the followingd_(n+1)d_{n+1}of a partial sequence ofeehas exactly as many variations as the sequence (3), the first non-zero terms in both sequencesd_(n+1)d_{n+1}have the same sign and the last non-zero terms also have the same sign.
4. Consider a functionffset toEEand an integern >= -1n \geq-1.
4. Consider a functionffset toEEand an integern >= -1n \geq-1.
Definition 3. We will say that the finite sequencee={x_(1),x_(2),dots,x_(m)}e=\left\{x_{1}, x_{2}, \ldots, x_{m}\right\}ofEEis irreducible if we cannot remove any point of e without reducing the number of variations in its sequenced_(n+1)d_{n+1}.
Otherwise we will say that the sequence e is reducible.
It is clear that reducibility depends onnn, of the function///and the numberkkvariations of the suited_(n+1)d_{n+1}ofee. We will see that ifnnAndkkare given, the number of terms of an irreducible sequence has a maximum independent of the functionff.
It is clear that reducibility depends onnn, of the function///and the numberkkvariations of the suited_(n+1)d_{n+1}ofee. We will see that ifnnAndkkare given, the number of terms of an irreducible sequence has a maximum independent of the functionff.
In the casek=0k=0, so thateeto be irreducible it is necessary and sufficient that it be formed byn+2n+2points. In the demonstrations which follow we will assumek > 0k>0. The results fork=0k=0can be easily deduced from this.
So thateeis irreducible it is necessary and sufficient, according to Theorem 1, that the sequencesd_(n+1)d_{n+1}partial sequencese_(j)={x_(1),x_(2),dots,x_(i-1),x_(j+1),dots:}e_{j}=\left\{x_{1}, x_{2}, \ldots, x_{i-1}, x_{j+1}, \ldots\right.,{:x_(m)},j=1,2,dots,m\left.x_{m}\right\}, j=1,2, \ldots, mall present fewer variations than the followingd_(n+1)d_{n+1}ofee. So that the continuationeebe reducible it is necessary and it sulfites that the continuationd_(n+1)d_{n+1}of at least one of the suitesc_(j)c_{j}has exactly as many variations as the sequenced_(n+1)d_{n+1}ofee.
Let (3) always be the sequenced_(n+1)d_{n+1}ofeeand (6) the followingd_(n+1)d_{n+1}, ofe_(j)e_{j}. Formulas (7) show us that if the partial sequence
Delta_(n+1)^(j-n-1)(f),Delta_(n+1)^(j-n)(f),dots,Delta_(n+1)^(j)(f)\Delta_{n+1}^{j-n-1}(f), \Delta_{n+1}^{j-n}(f), \ldots, \Delta_{n+1}^{j}(f)
does not present any variations, the number of variations of the sequences (3),
(6) is the same. It immediately follows that if the partial sequence of (3).
(8)quadDelta_(n+1)^(alpha-n)(f),Delta_(n+1)^(alpha-n+1)(f),dots,Delta_(n+1)^(beta)(f),a < beta\quad \Delta_{n+1}^{\alpha-n}(f), \Delta_{n+1}^{\alpha-n+1}(f), \ldots, \Delta_{n+1}^{\beta}(f), a<\beta,
does not present any variations, the followingd_(n+1)d_{n+1}, from the following{x_(1),x_(2),dots:}\left\{x_{1}, x_{2}, \ldots\right.,{:x_(alpha),xbeta_(+1),dots,x_(m)}\left.x_{\alpha}, x \beta_{+1}, \ldots, x_{m}\right\}presents exactly as many variations as the sequence (3). We can see easily how the property must be modified if
(6) is the same. It immediately follows that if the partial sequence of (3).
(8)quadDelta_(n+1)^(alpha-n)(f),Delta_(n+1)^(alpha-n+1)(f),dots,Delta_(n+1)^(beta)(f),a < beta\quad \Delta_{n+1}^{\alpha-n}(f), \Delta_{n+1}^{\alpha-n+1}(f), \ldots, \Delta_{n+1}^{\beta}(f), a<\beta,
does not present any variations, the followingd_(n+1)d_{n+1}, from the following{x_(1),x_(2),dots:}\left\{x_{1}, x_{2}, \ldots\right.,{:x_(alpha),xbeta_(+1),dots,x_(m)}\left.x_{\alpha}, x \beta_{+1}, \ldots, x_{m}\right\}presents exactly as many variations as the sequence (3). We can see easily how the property must be modified if
a-n < 1" ou "beta > m-n-1a-n<1 \text { ou } \beta>m-n-1
We deduce that we can always find ae^(***) <= ee^{\star} \leqq ewhose continuationd_(n+1)d_{n+1}has as many variations as the sequence (3) and in which the first two terms are non-zero and of opposite signs and the last two terms are also non-zero and of opposite signs. This results in:
Theorem 2. If (3) is the followingd_(n+1)d_{n+1}of an irreducible sequence e, we have
Delta_(n+1)^(1)(f)*Delta_(n+1)^(2)(f) < 0,Delta_(n+1)^(m-n-2)(f)*Delta_(n+1)^(m-n-1)(f) < 0.\Delta_{n+1}^{1}(f) \cdot \Delta_{n+1}^{2}(f)<0, \Delta_{n+1}^{m-n-2}(f) \cdot \Delta_{n+1}^{m-n-1}(f)<0 .
Theorem 3. Whenk=0,1k=0,1, any irreducible sequence ean+k+2n+k+2points.
Especially ifk=1k=1, the sequeld_(n+1)d_{n+1}of aeeirreducible is formed by two non-zero terms with opposite signs.
We also see, easily:
Theorem 4. Whenn=-1,0n=-1,0, any irreducible sequence ean+k+2n+k+2points.
Theorem 4. Whenn=-1,0n=-1,0, any irreducible sequence ean+k+2n+k+2points.
It is clear that the terms of the sequenced_(n+1)d_{n+1}ofeeare then all different from zero and alternately positive and negative.
Let's examine the casen >= 1n \geqq 1. Suppose that in formulas (7) we havealpha > n,m-n-1 >= beta >= alpha+2\alpha>n, m-n-1 \geq \beta \geq \alpha+2and that the partial sequence (8) presents a variation. We can then find agamma\gammasuch asalpha-n < gamma <= beta\alpha-n<\gamma \leqq \beta,Delta_(n+1)^(gamma)(f)!=0\Delta_{n+1}^{\gamma}(f) \neq 0and that the consequences
{:[(9)Delta_(n+1)^(alpha-n)(f)","Delta_(n+1)^(alpha-n+1)(f)","dots","Delta_(n+1)^(gamma-1)(f)],[(10)Delta_(n+1)^(gamma)(f)","Delta_(n+1)^(gamma+1)(f)","dots","Delta_(n+1)^(beta)(f)]:}\begin{gather*}
\Delta_{n+1}^{\alpha-n}(f), \Delta_{n+1}^{\alpha-n+1}(f), \ldots, \Delta_{n+1}^{\gamma-1}(f) \tag{9}\\
\Delta_{n+1}^{\gamma}(f), \Delta_{n+1}^{\gamma+1}(f), \ldots, \Delta_{n+1}^{\beta}(f) \tag{10}
\end{gather*}
do not present any variations. In (9) there is, moreover, at least one non-zero term of opposite sign withDelta_(n+1)^(gamma)(f)\Delta_{n+1}^{\gamma}(f). The previous results show us that we can assume that the sequences (9), (10) each have at mostn+1n+1terms. We can, in fact, return to this case by removing a certain number of pointsx_(i)x_{i}without changing the number of variations in the suited_(n+1)d_{n+1}ofee. It also follows that the sequences (9) and (10) each have at least two terms. If now the sequence (9) has at mostnnterms, by removing the periodx_(a+1)x_{a+1}we do not reduce the number of variations of the sequence (8) and consequently we do not reduce the number
of variations of the sequence (3). If the sequence (9) hasn+1n+1terms we arrive at the same result by removing the pointx_(a+2)x_{a+2}. We deduce that if the sequence (3) presentskkvariations, we can find ae^(***) <= ee^{\star} \leqq ewhose continuationd_(n+1)d_{n+1}presentskkvariations and which is formed by at most[(k)/(2)]n+k+1\left[\frac{k}{2}\right] n+k+1terms^(1){ }^{1})
of variations of the sequence (3). If the sequence (9) hasn+1n+1terms we arrive at the same result by removing the pointx_(a+2)x_{a+2}. We deduce that if the sequence (3) presentskkvariations, we can find ae^(***) <= ee^{\star} \leqq ewhose continuationd_(n+1)d_{n+1}presentskkvariations and which is formed by at most[(k)/(2)]n+k+1\left[\frac{k}{2}\right] n+k+1terms^(1){ }^{1})
This results in the
Theorem 5. Ifn >= 0n \geqq 0, any irreducible sequence, whose sequenced_(n+1)d_{n+1}presentskkpariations, at most[(k+6)/(2)]n+k+2\left[\frac{k+6}{2}\right] n+k+2points.
We can notice that by removing the pointx_(alpha+1)x_{\alpha+1}, we do not reduce the number of variations of the sequence (3) ifbeta=alpha+1\beta=\alpha+1AndDelta_(n+1)^(gamma-1)(f)=0\Delta_{n+1}^{\gamma-1}(f)=0. Applying this property to the casen=1n=1, we can easily see that ifn=-0,1n=-0,1and if the sequence (3) presentskkvariations, we can find ae^(***) <= ee^{\star} \leqslant ewhose continuationd_(n+1)d_{n+1}presentskkvariations and has all its non-zero terms. Done
Theorem 6. Whenn=-1,0,1n=-1,0,1, all the terms of the sequenced_(n+1)d_{n+1}of an irreducible sequence e are different from zero.
We can easily see that in the suitesd_(n+1)d_{n+1}, of all irreducible partial sequences ofee, whose sequelsd_(n+1)d_{n+1}have as many variations as the sequenced_(n+1)d_{n+1}ofee, the first terms have the same sign and the last terms also have the same sign.
It is obvious, moreover, that ifeeis reducible, we can find ae^(***) < ee^{\star}<eirreducible whose continuationd_(n+1)d_{n+1}has exactly as many variations as the sequenced_(n+1)d_{n+1}ofee.
4a. The maximum number of points in an irreducible sequence can actually be reached. It is sufficient to consider the casen > 0n>0. Let's continuee={1,2,dots,m},m=[(k+2)/(2)]n+k+2e=\{1,2, \ldots, m\}, m=\left[\frac{k+2}{2}\right] n+k+2and choose the functionffso that we have
{:[Delta_(n+1)^(i(n+2)+1)(f)=1","quad i=0","1","dots","[(k)/(2)]],[Delta_(n+1)^(i(n+2))(f)=Delta_(n+1)^(j(n+2)+2)quad(f)=-n-2],[i=1","2","dots","[(k)/(2)]","quad i=0","1","dots","[(k-1)/(2)]],[Delta_(n+1)^(i)(f) <= 0","i!=0","1","2","(mod n+2)]:}\begin{array}{r}
\Delta_{n+1}^{i(n+2)+1}(f)=1, \quad i=0,1, \ldots,\left[\frac{k}{2}\right] \\
\Delta_{n+1}^{i(n+2)}(f)=\Delta_{n+1}^{j(n+2)+2} \quad(f)=-n-2 \\
i=1,2, \ldots,\left[\frac{k}{2}\right], \quad i=0,1, \ldots,\left[\frac{k-1}{2}\right] \\
\Delta_{n+1}^{i}(f) \leqq 0, i \neq 0,1,2,(\bmod n+2)
\end{array}
The sequeleeis then irreducible. This example shows us that ifn > 1n>1the sequeld_(n+1)d_{n+1}of a sequeleeirreducible can actually have zero terms.
5. Let us return to the order functions (n∣kn \mid k). Let us first introduce Definition 4. We will say that the finite sequence e ofE^(')E^{\prime}is a maximizing sequence, for the functionffof order (n∣kn \mid k), if its continuationd_(n+1)d_{n+1}presents exactlykkbets.
5. Let us return to the order functions (n∣kn \mid k). Let us first introduce Definition 4. We will say that the finite sequence e ofE^(')E^{\prime}is a maximizing sequence, for the functionffof order (n∣kn \mid k), if its continuationd_(n+1)d_{n+1}presents exactlykkbets.
There are obviously maximizing sequences. Any finite sequence ofEEwhich contains a maximizing partial sequence is still maximizing. We immediately deduce that if the sequenced_(n+1)d_{n+1}ofeepresentskkvariations, the function is of order (n∣kn \mid k) onee. It also follows that in the sequelsd_(n+1)d_{n+1}maximizing sequences, the first non-zero terms have the same sign and the last non-zero terms also have the same sign. The sign of the last non-zero term (or the first non-zero term) is therefore characteristic for the function.
Definition 5. We will say that the function f of order (n∣k\mid k) is of order(n∣k)^(+)(n \mid k)^{+}or order (n∣kn \mid k)^(-){ }^{-}following that the last non-zero term of the sequenced_(n+1)d_{n+1}of a maximizing sequence ofEEis positive or negative.
Order functions(n∣0)^(+)(n \mid 0)^{+}are the non-concave functions of ordernnand the order functions (n∣0n \mid 0)- non-convex functions of ordernn. Polynomial functions of ordernnthus escape this definition, but we can agree that such a function is indifferently of order(n∣0)^(+)(n \mid 0)^{+}or order(n∣0)^(-)(n \mid 0)^{-}.
Ifffis in order(n∣k)^(+)(n \mid k)^{+}order manager(n∣k)^(-)(n \mid k)^{-}, it is the same forcfc f, Orccis a positive constant and of the functionf+.Pf+. P, OrPPis a polynomial of degreenn. The function -ffis of order (n∣kn \mid k)- resp. order(n∣k)^(+)(n \mid k)^{+}.
If the function///is not in order<= (n∣k)\leqq(n \mid k)onEE, we can find a subset ofEEon whichffeither of order (n∣k+1n \mid k+1). We also see, easily, that we can find a finite subset ofEEon whichffeither of order(n∣k)^(+)(n \mid k)^{+}and a finite subassembly ofEEon whichffeither of order(n∣k)^(-)(n \mid k)^{-}.
6. Let us recall the well-known property, expressed by Lomme 3. If the sequence
6. Let us recall the well-known property, expressed by Lomme 3. If the sequence
{:(11)c_(2)-c_(1)","c_(3)-c_(2)","dots","c_(m)-c_(m-1):}\begin{equation*}
c_{2}-c_{1}, c_{3}-c_{2}, \ldots, c_{m}-c_{m-1} \tag{11}
\end{equation*}
presentskkvariations, the sequel
{:(12)c_(1)","c_(2)","dots","c_(m):}\begin{equation*}
c_{1}, c_{2}, \ldots, c_{m} \tag{12}
\end{equation*}
present at mostk+1k+1variations.
We can complete this lemma with the following
Lemma 4. If the sequence (11) presentsk( >= 0)k(\geq 0)sariations and the continuation (12) exactlyk+1k+1pariations, the last non-zero terms in the two sequences have the same sign.
Lemma 4. If the sequence (11) presentsk( >= 0)k(\geq 0)sariations and the continuation (12) exactlyk+1k+1pariations, the last non-zero terms in the two sequences have the same sign.
Consider the functionf(x_(i))=c_(i),i=1,2,dots,mf\left(x_{i}\right)=c_{i}, i=1,2, \ldots, mon the wholee={x_(1),x_(2),dots,x_(m)}e=\left\{x_{1}, x_{2}, \ldots, x_{m}\right\}. Ifm=k+2m=k+2and the following (12) presentsk+1k+1variations, the sequence (11) necessarily presentskkvariations and the lemma is easily proven. Ifm > k+2m>k+2, the function being of order(-1∣k+1)(-1 \mid k+1)onee, we can find a partial sequencee^(**)={x_(1)^(**),x_(2)^(**)dots,x_(k+2)^(**)}e^{*}=\left\{x_{1}^{*}, x_{2}^{*} \ldots, x_{k+2}^{*}\right\}ofk+2k+2. terms ofee, maximizing and irreducible.ffis also of order(0∣k)(0 \mid k)and in relation to this ordereeAnde^(**)e^{*}are maximizing. Lemma 4 follows immediately.
Given the equalities
Delta_(n+1)^(i)(f)=(Delta_(n)^(i+1)(f)-Delta_(n)^(i)(f))/(x_(n+i+1)-x_(i)),quad i=1,2,dots,m-n-1\Delta_{n+1}^{i}(f)=\frac{\Delta_{n}^{i+1}(f)-\Delta_{n}^{i}(f)}{x_{n+i+1}-x_{i}}, \quad i=1,2, \ldots, m-n-1
relating to ae={x_(1),x_(2),dots,x_(m)}e=\left\{x_{1}, x_{2}, \ldots, x_{m}\right\}, Lemma 3 gives us
Theorem 7. Any function of order (n∣kn \mid k),n >= 0n \geq 0, is at most of order(n-1∣k+1)(n-1 \mid k+1). In general, any function of order (n∣kn \mid k) is at most of order(n-1∣k+i)(n-1 \mid k+i), regardless ofi=1,2,dots,n+1i=1,2, \ldots, n+1.
Theorem 7. Any function of order (n∣kn \mid k),n >= 0n \geq 0, is at most of order(n-1∣k+1)(n-1 \mid k+1). In general, any function of order (n∣kn \mid k) is at most of order(n-1∣k+i)(n-1 \mid k+i), regardless ofi=1,2,dots,n+1i=1,2, \ldots, n+1.
From Lemma 4 the following more complete property results.
Theorem 8. If a function of order (n∣kn \mid k)^(+){ }^{+}is of order (n-1∣k+1n-1 \mid k+1), it is necessarily of order(n-1∣k+1)^(+)(n-1 \mid k+1)^{+}. In general, if the function is of order (n-i∣k+in-i \mid k+i), it is necessarily of order (n-i∣k+i)^(+)n-i \mid k+i)^{+}, regardless ofi=1,2,dots,n+1i=1,2, \ldots, n+1.
Theorem 8. If a function of order (n∣kn \mid k)^(+){ }^{+}is of order (n-1∣k+1n-1 \mid k+1), it is necessarily of order(n-1∣k+1)^(+)(n-1 \mid k+1)^{+}. In general, if the function is of order (n-i∣k+in-i \mid k+i), it is necessarily of order (n-i∣k+i)^(+)n-i \mid k+i)^{+}, regardless ofi=1,2,dots,n+1i=1,2, \ldots, n+1.
In the following notes we will examine some properties of order functions (n∣kn \mid k).
Manuscript received on May 2, 1940.
- ^(1){ }^{1}) I. Schoenberg: Uber variationsvernindernde lineare Transformationen, Math, Zeitschrift, 32, 321-328, (1930).
- ^(1){ }^{1}) We designate, as usual, by[a][a]the largest integer<= a\leqq a.