Remarks on the functional definition of a polynomial of a real variable

Tiberiu Popoviciu
(original), Approximation Theory, functional equations, Mathematical Analysis, paper

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T. Popoviciu, Remarques sur la definition fonctionnelle d’un polynôme d’une variable réelle, Mathematica, 12 (1936), pp. 5-12 (in French).

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1936 f -Popoviciu- Mathematica – Remarques sur la definition fonctionnelle d’un polynôme d’une variable reelle

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1936 f -Popoviciu- Mathematica - Remarks on the functional definition of a polynomial of a varia
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REMARKS ON THE FUNCTIONAL DEFINITION OF A POLYNOMIAL OF A REAL VARIABLE

by

Tiberiu Popoviciu.

Received December 18, 1935.
    • Either f ( x ) f ( x ) f(x)f(x)f(x)a uniform function of the real variable x x xxxand consider its difference in order n n nnn
Δ h n = Δ h n f ( x ) = ∑ i = 0 n ( − 1 ) i ( n i ) f ( x + i h ) Δ h n = Δ h n f ( x ) = ∑ i = 0 n   ( − 1 ) i ( n i ) f ( x + i h ) Delta_(h)^(n)=Delta_(h)^(n)f(x)=sum_(i=0)^(n)(-1)^(i)((n)/(i))f(x+ih)\Delta_{h}^{n}=\Delta_{h}^{n} f(x)=\sum_{i=0}^{n}(-1)^{i}\binom{n}{i} f(x+i h)Δhn=Δhnf(x)=∑i=0n(−1)i(ni)f(x+ih)
When f ( x ) f ( x ) f(x)f(x)f(x)is a polynomial of degree n − 1 ( 1 ) n − 1 1 n-1(^(1))n-1\left({ }^{1}\right)n−1(1)we have Δ h n = 0 Δ h n = 0 Delta_(h)^(n)=0\Delta_{h}^{n}=0Δhn=0whatever x x xxxAnd h h hhh. Conversely we know that if we have Δ h n = 0 Δ h n = 0 Delta_(h)^(n)=0\Delta_{h}^{n}=0Δhn=0for everything x x xxxand for everything h h hhh, the function f ( x ) f ( x ) f(x)f(x)f(x), under fairly general assumptions, is reduced to a polynomial of degree n − 1 ( 2 ) n − 1 2 n-1(^(2))n-1\left({ }^{2}\right)n−1(2).
In his lecture given at the University of Cluj in May 1935 first and in a note of the CR then, Mr. Paul Montel demonstrated the following property ( 3 ) ( 3 ) ^((3)){ }^{(3)}(3):
Any continuous function of a real variable satisfying the functional equations
(1) Δ ω 1 n = 0 , Δ ω 2 n = 0 (1) Δ ω 1 n = 0 , Δ ω 2 n = 0 {:(1)Delta_(omega_(1))^(n)=0","quadDelta_(omega_(2))^(n)=0:}\begin{equation*} \Delta_{\omega_{1}}^{n}=0, \quad \Delta_{\omega_{2}}^{n}=0 \tag{1} \end{equation*}(1)Δω1n=0,Δω2n=0
relating to two periods ω 4 , ω 2 ω 4 , ω 2 omega_(4),omega_(2)\omega_{4}, \omega_{2}ω4,ω2whose ratio is irrational is a polynomial of degree n − 1 n − 1 n-1n-1n−1.
( 1 ) ( 1 ) ^((1)){ }^{(1)}(1)We say that a polynomial is of degree n n nnnwhen its effective degree is at most equal to n n nnn.
(2) See: Th. Anghelutza "On a functional equation characterizing polynomials". Mathematica t. VI (1932), p. 1-7., Tiberiu Popoviciu "On some properties of functions of one and two real variables". Mathematica t. Vlll (1934), p. 1-85, sp. p. 57.
( 3 3 ^(3){ }^{3}3) Paul Montel "On a theorem of Jacobi" CR Acad. Sc. Paris, t. 201 (1935), p. 586.
Mr. Paul Montel also noted that in the case n = 1 n = 1 n=1n=1n=1continuity at a single point is sufficient to be able to affirm that the function reduces to a constant. In the following we will extend this result to the general system (1).
Note that if the function f ( x ) f ( x ) f(x)f(x)f(x)check the equation Δ h n = 0 Δ h n = 0 Delta_(h)^(n)=0\Delta_{h}^{n}=0Δhn=0for one h h hhhgiven it reduces to a polynomial of degree n − 1 n − 1 n-1n-1n−1at equidistant points x , x ± h , x ± 2 h , … x , x ± h , x ± 2 h , … x,x+-h,x+-2h,dotsx, x \pm h, x \pm 2 h, \ldotsx,x±h,x±2h,…and this for any value of x x xxx. It follows that we also have Δ p h n = 0 Δ p h n = 0 Delta_(ph)^(n)=0\Delta_{p h}^{n}=0Δphn=0whatever the integer p p ppp.
2. - Now suppose that the function f ( x ) f ( x ) f(x)f(x)f(x)checks the system (1), ω 1 , ω 2 ω 1 , ω 2 omega_(1),omega_(2)\omega_{1}, \omega_{2}ω1,ω2being two real numbers (the periods) given. We: have
Δ ω 1 n f ( x + p ω 1 + q ω 2 ) = 0 . Δ ω 1 n f x + p ω 1 + q ω 2 = 0 . Delta_(omega_(1))^(n)f(x+pomega_(1)+qomega_(2))=0.\Delta_{\omega_{1}}^{n} f\left(x+p \omega_{1}+q \omega_{2}\right)=0 .Δω1nf(x+pω1+qω2)=0.
Let's take for x x xxxAnd q q qqqfixed values. We see that there exists a polynomial P q ( u ) P q ( u ) P_(q)(u)\mathrm{P}_{q}(u)Pq(u)of degree n − 1 n − 1 n-1n-1n−1compared to u u uuusuch that
(2)
f ( x + p ω 1 + q ω 2 ) = P q ( x + p ω 1 + q ω 2 ) p = 0 , ± 1 , ± 2 , … f x + p ω 1 + q ω 2 = P q x + p ω 1 + q ω 2 p = 0 , ± 1 , ± 2 , … {:[f(x+pomega_(1)+qomega_(2))=P_(q)(x+pomega_(1)+qomega_(2))],[p=0","+-1","+-2","dots]:}\begin{gathered} f\left(x+p \omega_{1}+q \omega_{2}\right)=P_{q}\left(x+p \omega_{1}+q \omega_{2}\right) \\ p=0, \pm 1, \pm 2, \ldots \end{gathered}f(x+pω1+qω2)=Pq(x+pω1+qω2)p=0,±1,±2,…
Similarly, assuming x x xxxAnd p p pppfixed we see that there exists a polynomial Q p ( v ) Q p ( v ) Q_(p)(v)Q_{p}(v)Qp(v)of degree n − 1 n − 1 n-1n-1n−1compared to v v vvvsuch that
(3)
f ( x + p ω 1 + q ω 2 ) = Q p ( x + p ω 1 + q ω 2 ) q = 0 , ± 1 , ± 2 , … f x + p ω 1 + q ω 2 = Q p x + p ω 1 + q ω 2 q = 0 , ± 1 , ± 2 , … {:[f(x+pomega_(1)+qomega_(2))=Q_(p)(x+pomega_(1)+qomega_(2))],[q=0","+-1","+-2","dots]:}\begin{gathered} f\left(x+p \omega_{1}+q \omega_{2}\right)=Q_{p}\left(x+p \omega_{1}+q \omega_{2}\right) \\ q=0, \pm 1, \pm 2, \ldots \end{gathered}f(x+pω1+qω2)=Qp(x+pω1+qω2)q=0,±1,±2,…
From (2) and (3) we deduce that we can find a polynomial P ( u , v ) P ( u , v ) P(u,v)\mathrm{P}(u, v)P(u,v); of two variables u u uuuAnd v v vvvof degree n − 1 n − 1 n-1n-1n−1with respect to each of these variables and such that for a x x xxx-given we have
(4) f ( x + p ω 1 + q ω 2 ) = P ( p ω 1 , q ω 2 ) (4) f x + p ω 1 + q ω 2 = P p ω 1 , q ω 2 {:(4)f(x+pomega_(1)+qomega_(2))=P(pomega_(1),qomega_(2)):}\begin{equation*} f\left(x+p \omega_{1}+q \omega_{2}\right)=\mathrm{P}\left(p \omega_{1}, q \omega_{2}\right) \tag{4} \end{equation*}(4)f(x+pω1+qω2)=P(pω1,qω2)
The polynomial P ( u , v ) P ( u , v ) P(u,v)\mathrm{P}(u, v)P(u,v)can be written in the following form
P ( u , v ) = ∑ i = 0 2 n − 2 A 2 n − 2 − i u i P ( u , v ) = ∑ i = 0 2 n − 2   A 2 n − 2 − i u i P(u,v)=sum_(i=0)^(2n-2)A_(2n-2-i)u^(i)\mathrm{P}(u, v)=\sum_{i=0}^{2 n-2} \mathrm{~A}_{2 n-2-i} u^{i}P(u,v)=∑i=02n−2 HAS2n−2−iui
Or A i A i A_(i)\mathbf{A}_{i}HASiare polynomials depending only on u + v u + v u+vu+vu+vand are of degree n − 1 , A 0 , A 1 , … , A n − 1 n − 1 , A 0 , A 1 , … , A n − 1 n-1,A_(0),A_(1),dots,A_(n-1)n-1, \mathrm{~A}_{0}, \mathrm{~A}_{1}, \ldots, \mathrm{~A}_{n-1}n−1, HAS0, HAS1,…, HASn−1being of degree equal to their index.
Let us now suppose that for n − 1 n − 1 n-1n-1n−1values a 1 , a 2 , … , a n − 1 a 1 , a 2 , … , a n − 1 a_(1),a_(2),dots,a_(n-1)a_{1}, a_{2}, \ldots, a_{n-1}has1,has2,…,hasn−1the constant has the polynomial P ( u , v ) P ( u , v ) P(u,v)\mathrm{P}(u, v)P(u,v)remains limited to u + v = a u + v = a u+v=au+v=au+v=has. We immediately see that for these values ​​the coefficients A 0 , A 1 , … A 2 n − 3 A 0 , A 1 , … A 2 n − 3 A_(0),A_(1),dotsA_(2n-3)A_{0}, A_{1}, \ldots A_{2 n-3}HAS0,HAS1,…HAS2n−3must cancel each other out. It follows that
P ( u , v ) = ( u + v − a 1 ) ( u + v − a 2 ) … ( u + v − a n − 1 ) ( b 0 u n − 1 + + b 1 u n − 2 + , … , + b n − 1 ) + A 2 n − 2 P ( u , v ) = u + v − a 1 u + v − a 2 … u + v − a n − 1 b 0 u n − 1 + + b 1 u n − 2 + , … , + b n − 1 + A 2 n − 2 {:[P(u","v)=(u+v-a_(1))(u+v-a_(2))dots(u+v-a_(n-1))(b_(0)u^(n-1)+:}],[{:+b_(1)u^(n-2)+,dots,+b_(n-1))+A_(2n-2)]:}\begin{gathered} \mathrm{P}(u, v)=\left(u+v-a_{1}\right)\left(u+v-a_{2}\right) \ldots\left(u+v-a_{n-1}\right)\left(b_{0} u^{n-1}+\right. \\ \left.+b_{1} u^{n-2}+, \ldots,+b_{n-1}\right)+\mathrm{A}_{2 n-2} \end{gathered}P(u,v)=(u+v−has1)(u+v−has2)…(u+v−hasn−1)(b0un−1++b1un−2+,…,+bn−1)+HAS2n−2
For this polynomial to have the required form, it is necessary that b 0 = b 1 = , … , = b n − 1 = 0 b 0 = b 1 = , … , = b n − 1 = 0 b_(0)=b_(1)=,dots,=b_(n-1)=0b_{0}=b_{1}=, \ldots,=b_{n-1}=0b0=b1=,…,=bn−1=0and we then see that we have the following property:
If the polynomial P ( u , v ) P ( u , v ) P(u,v)\mathrm{P}(u, v)P(u,v)remains limited to n − 1 n − 1 n-1n-1n−1values ​​of u + v u + v u+vu+vu+vit remains bounded for any given value of u + v u + v u+vu+vu+v, so it reduces to a: polynomial of degree n − 1 n − 1 n-1n-1n−1compared to u + v u + v u+vu+vu+v.
3. - Let's return to our problem. If the periods ω ˙ 1 , ω ˙ 2 ω ˙ 1 , ω ˙ 2 omega^(˙)_(1),omega^(˙)_(2)\dot{\omega}_{1}, \dot{\omega}_{2}ω˙1,ω˙2her U U U\mathfrak{U}Udependent there are two whole numbers p p pppAnd q q qqqsuch as ω 4 = p ω ω 4 = p ω omega_(4)=p omega\omega_{4}=p \omegaω4=pω, ω 2 = q ω ω 2 = q ω omega_(2)=q omega\omega_{2}=q \omegaω2=qωand we see immediately that the function f ( x ) f ( x ) f(x)f(x)f(x)check the equation Δ ω n = 0 Δ ω n = 0 Delta_(omega)^(n)=0\Delta_{\omega}^{n}=0Δωn=0. In this case the polynomials P q , Q p P q , Q p P_(q),Q_(p)P_{q}, Q_{p}Pq,Qpcoincide at an infinite number of points and are therefore identical. We leave aside: this case.
So let's assume that the periods ω 4 , ω 2 ω 4 , ω 2 omega_(4),omega_(2)\omega_{4}, \omega_{2}ω4,ω2are independent ( 4 ) 4 (^(4))\left({ }^{4}\right)(4). All points p ω 4 + q ω 2 , p , q p ω 4 + q ω 2 , p , q pomega_(4)+qomega_(2),p,qp \omega_{4}+q \omega_{2}, p, qpω4+qω2,p,qbeing integers, is then everywhere dense on the real axis.
Now suppose that the function f ( x ) f ( x ) f(x)f(x)f(x)be bounded at a point x ′ x ′ x^(')x^{\prime}x′, which amounts to saying that it is bounded in an interval containing the point x ′ x ′ x^(')x^{\prime}x′. There then exist two infinite and non-bozed sequences of integers
p 1 , p 2 , … , p ν , … q 1 , q 2 , … , q ν , … p 1 , p 2 , … , p ν , … q 1 , q 2 , … , q ν , … {:[p_(1)","p_(2)","dots","p_(nu)","dots],[q_(1)","q_(2)","dots","q_(nu)","dots]:}\begin{aligned} & p_{1}, p_{2}, \ldots, p_{\nu}, \ldots \\ & q_{1}, q_{2}, \ldots, q_{\nu}, \ldots \end{aligned}p1,p2,…,pν,…q1,q2,…,qν,…
such as
p ν ω 1 + q ν ω 2 → x ′ − x ν → ∞ . p ν ω 1 + q ν ω 2 → x ′ − x ν → ∞ . {:[p_(nu)omega_(1)+q_(nu)omega_(2) rarrx^(')-x],[nu rarr oo.]:}\begin{aligned} p_{\nu} \omega_{1}+q_{\nu} \omega_{2} & \rightarrow x^{\prime}-x \\ \nu & \rightarrow \infty . \end{aligned}pνω1+qνω2→x′−xν→∞.
From (4) then follows that the polynomial P ( u , v ) P ( u , v ) P(u,v)\mathrm{P}(u, v)P(u,v)must be limited for u + v = x ′ − x u + v = x ′ − x u+v=x^(')-xu+v=x^{\prime}-xu+v=x′−x. The function being bounded at a point it is bounded at n − 1 n − 1 n-1n-1n−1points. It follows that the polynomial P ( u , v ) P ( u , v ) P(u,v)\mathrm{P}(u, v)P(u,v)reduces to a polynomial of degree n − 1 n − 1 n-1n-1n−1compared to u + v u + v u+vu+vu+vTonsion f ( x ) f ( x ) f(x)f(x)f(x)coincides: so on the points x + p ω 1 + q ω 2 , p , q = 0 . ± 1 , ± 2 , … x + p ω 1 + q ω 2 , p , q = 0 . ± 1 , ± 2 , … x+pomega_(1)+qomega_(2),p,q=0.+-1,+-2,dotsx+p \omega_{1}+q \omega_{2}, p, q=0 . \pm 1, \pm 2, \ldotsx+pω1+qω2,p,q=0.±1,±2,…with a polynomial of degree n − 1 n − 1 n-1n-1n−1In particular, we have
Δ p ω 1 + q ω 2 n = 0 Δ p ω 1 + q ω 2 n = 0 Delta_(p_(omega_(1))+qomega_(2))^(n)=0\Delta_{p_{\omega_{1}}+q \omega_{2}}^{n}=0Δpω1+qω2n=0
regardless of the integers p p pppAnd q q qqq. So we have the following property:
If the periods ω 1 , ω 2 ω 1 , ω 2 omega_(1),omega_(2)\omega_{1}, \omega_{2}ω1,ω2are independent, if the function f ( x ) f ( x ) f(x)f(x)f(x)
(4) Numbers ω 1 , ω 2 ω 1 , ω 2 omega_(1),omega_(2)\omega_{1}, \omega_{2}ω1,ω2are independent when equality p ω 1 + q ω 2 = 0 p ω 1 + q ω 2 = 0 pomega_(1)+qomega_(2)=0\boldsymbol{p} \boldsymbol{\omega}_{1}+\boldsymbol{q} \boldsymbol{\omega}_{2}=0pω1+qω2=0cannot be satisfied in whole numbers p p p\boldsymbol{p}p, And q q q\boldsymbol{q}q, that if p = q = 0 p = q = 0 p=q=0p=q=0p=q=0. It is therefore necessary and sufficient that the ratio of the numbers ω 1 , ω 2 ω 1 , ω 2 omega_(1),omega_(2)\omega_{1}, \omega_{2}ω1,ω2be irrational.
Check the functional equations
Δ ω 1 n = 0 , Δ ω 2 n = 0 Δ ω 1 n = 0 , Δ ω 2 n = 0 Delta_(omega_(1))^(n)=0,quadDelta_(omega_(2))^(n)=0\Delta_{\omega_{1}}^{n}=0, \quad \Delta_{\omega_{2}}^{n}=0Δω1n=0,Δω2n=0
and is bounded at a point, we also have
Δ p 1 + q ω 2 n = 0 Δ p 1 + q ω 2 n = 0 Delta_(p_(1)+qomega_(2))^(n)=0\Delta_{p_{1}+q \omega_{2}}^{n}=0Δp1+qω2n=0
for any value of x x xxxand whatever the integers p p pppAnd q q qqq.
If the function is not bounded at any point the property may not be true. For example the function which is equal to p q p q pqp qpqat a point of the form p ω 1 + q ω 2 p ω 1 + q ω 2 pomega_(1)+qomega_(2)p \omega_{1}+q \omega_{2}pω1+qω2and is zero everywhere else, check the system carefully
Δ ω 1 n = 0 , Δ ω 2 n = 0 ( n > 1 ) Δ ω 1 n = 0 , Δ ω 2 n = 0 ( n > 1 ) Delta_(omega_(1))^(n)=0,quadDelta_(omega_(2))^(n)=0quad(n > 1)\Delta_{\omega_{1}}^{n}=0, \quad \Delta_{\omega_{2}}^{n}=0 \quad(n>1)Δω1n=0,Δω2n=0(n>1)
but we generally have
Δ p ω 1 + q ω 2 n ≠ 0 Δ p ω 1 + q ω 2 n ≠ 0 Delta_(pomega_(1)+qomega_(2))^(n)!=0\Delta_{p \omega_{1}+q \omega_{2}}^{n} \neq 0Δpω1+qω2n≠0
    • We are in the conditions of the previous case if f ( x ) f ( x ) f(x)f(x)f(x)is continuous at a point x ′ x ′ x^(')x^{\prime}x′. In this case P ( u , v ) P ( u , v ) P(u,v)\mathrm{P}(u, v)P(u,v)depends only on u + v u + v u+vu+vu+vand all these polynomials must coincide at the point x ′ x ′ x^(')x^{\prime}x′. These polynomials being of degree n − 1 n − 1 n-1n-1n−1We can state the following property:
If the periods ω 1 , ω 2 ω 1 , ω 2 omega_(1),omega_(2)\omega_{1}, \omega_{2}ω1,ω2are independent, if the function f ( x ) f ( x ) f(x)f(x)f(x)check the functional equations
Δ ω 1 n = 0 , Δ ω 2 n = 0 Δ ω 1 n = 0 , Δ ω 2 n = 0 Delta_(omega_(1))^(n)=0,quadDelta_(omega_(2))^(n)=0\Delta_{\omega_{1}}^{n}=0, \quad \Delta_{\omega_{2}}^{n}=0Δω1n=0,Δω2n=0
and is continuous at n points, it reduces to a polynomial of degree n-1.
The property may not be true if f ( x ) f ( x ) f(x)f(x)f(x)is supposed to be continuous in less than n n nnnpoints. For example, the function that takes the values ​​of x x xxxin every respect of the form p ω ˙ 4 + q ω 2 p ω ˙ 4 + q ω 2 pomega^(˙)_(4)+qomega_(2)p \dot{\omega}_{4}+q \omega_{2}pω˙4+qω2and is zero everywhere else is well continuous for x = 0 x = 0 x=0x=0x=0, check the equations Δ ω 1 n = 0 Δ ω 1 n = 0 Delta_(omega_(1))^(n)=0\Delta_{\omega_{1}}^{n}=0Δω1n=0, Δ ω 3 n = 0 , n > 1 Δ ω 3 n = 0 , n > 1 Delta_(omega_(3))^(n)=0,quad n > 1\Delta_{\omega_{3}}^{n}=0, \quad n>1Δω3n=0,n>1and yet does not reduce to a polynomial. We can also give other examples of this nature.
5. - Let us consider the more general equation
(5)
∑ i = 0 n a i f ( x + i h ) = 0 ∑ i = 0 n   a i f ( x + i h ) = 0 sum_(i=0)^(n)a_(i)f(x+ih)=0\sum_{i=0}^{n} a_{i} f(x+i h)=0∑i=0nhasif(x+ih)=0
where the a i a i a_(i)a_{i}hasiare constants. We can assume a 0 ≠ 0 , a n ≠ 0 a 0 ≠ 0 , a n ≠ 0 a_(0)!=0,a_(n)!=0a_{0} \neq 0, a_{n} \neq 0has0≠0,hasn≠0and that the first member of the characteristic equation
F ( z ) = a 0 + a 1 z + , … , a n z n = 0 F ( z ) = a 0 + a 1 z + , … , a n z n = 0 F(z)=a_(0)+a_(1)z+,dots,a_(n)z^(n)=0F(z)=a_{0}+a_{1} z+, \ldots, a_{n} z^{n}=0F(z)=has0+has1z+,…,hasnzn=0
does not reduce me to a polynomial with respect to an integer power > 1 > 1 > 1>1>1of z z zzz.
We say that equation (5) is of degree n. If 1 is a root of order k k kkkof multiplicity of the characteristic equation we say that equation (5) is of order k k kkk. If F ( 1 ) ≠ 0 F ( 1 ) ≠ 0 F(1)!=0\mathrm{F}(1) \neq 0F(1)≠0the equation is of order 0. An equation of order and degree n n nnnis necessarily of the form
Δ h n = 0 Δ h n = 0 Delta_(h)^(n)=0\Delta_{h}^{n}=0Δhn=0
For an order equation k k kkkwe have
F ( 1 ) = F ′ ( 1 ) = , … , = F ( k − 1 ) ( 1 ) = 0 , F ( k ) ( 1 ) ≠ 0 F ( 1 ) = F ′ ( 1 ) = , … , = F ( k − 1 ) ( 1 ) = 0 , F ( k ) ( 1 ) ≠ 0 F(1)=F^(')(1)=,dots,=F^((k-1))(1)=0,F^((k))(1)!=0F(1)=F^{\prime}(1)=, \ldots,=F^{(k-1)}(1)=0, F^{(k)}(1) \neq 0F(1)=F′(1)=,…,=F(k−1)(1)=0,F(k)(1)≠0
If equation (5) remains verified, with the same constants a i a i a_(i)a_{i}hasi, when replacing h h hhhby m h , m = 2 , 3 , … m h , m = 2 , 3 , … mh,m=2,3,dotsm h, m=2,3, \ldotsmh,m=2,3,…, we can find an integer r r r\boldsymbol{r}rsuch as one has
(6) Δ r h n = − 0 (6) Δ r h n = − 0 {:(6)Delta_(rh)^(n)=-0:}\begin{equation*} \Delta_{r h}^{n}=-0 \tag{6} \end{equation*}(6)Δrhn=−0
From the properties of equation (5) it follows that it is sufficient to take for m m mmmonly a finite number of values. These values, among which r r rrrstill exists, are indicated by the way in which equation (6) is obtained and which we have specified in another work ( 5 5 ^(5){ }^{5}5These values ​​can be chosen. m m mmmin an infinite number of ways: In the following when we write an equation (5) it is implied that this equation remains verified also when we replace h h hhhby m h m h mhm hmhwhere the positive integer m m mmmtakes a number of values ​​such that reduction to the form (6) is possible.
The number h h hhhbeing fixed, we immediately verify that any polynomial solution of equation] (5) is necessarily an arbitrary polynomial of degree k − 1 k − 1 k-1k-1k−1(if k = 0 k = 0 k=0k=0k=0any solution of this nature is identically zero).
This property also takes place without the restriction imposed on the substitution of h h hhhby m h m h mhm hmhIn other words, it takes place by taking for m m mmmthe only value 1.
6. As a generalization of the problem studied we propose to examine the system
(7) ∑ i = 0 n 1 a i f ( x + i ω 1 ) = 0 , ∑ i = 0 n 2 b i f ( x + i ω 2 ) = 0 (7) ∑ i = 0 n 1   a i f x + i ω 1 = 0 , ∑ i = 0 n 2   b i f x + i ω 2 = 0 {:(7)sum_(i=0)^(n_(1))a_(i)f(x+iomega_(1))=0","quadsum_(i=0)^(n_(2))b_(i)f(x+iomega_(2))=0:}\begin{equation*} \sum_{i=0}^{n_{1}} a_{i} f\left(x+i \omega_{1}\right)=0, \quad \sum_{i=0}^{n_{2}} b_{i} f\left(x+i \omega_{2}\right)=0 \tag{7} \end{equation*}(7)∑i=0n1hasif(x+iω1)=0,∑i=0n2bif(x+iω2)=0
(5) Tib. Popoviciu "On certain functional equations defining polynomials", Mathematica t. X (1935), pp. 197-211. We ask the reader to refer to this work.
formed by equations of degree n 1 , n 2 n 1 , n 2 n_(1),n_(2)n_{1}, n_{2}n1,n2and order k 1 , k 2 k 1 , k 2 k_(1),k_(2)k_{1}, k_{2}k1,k2respect ⊤ ⊤ _(TT){ }_{\top}⊤tively. The periods, that is to say the numbers ω 1 , ω 2 ω 1 , ω 2 omega_(1),omega_(2)\omega_{1}, \omega_{2}ω1,ω2, are assumed to be independent. The number n = max . ( n 1 , n 2 ) n = max . n 1 , n 2 n=max.(n_(1),n_(2))n=\max .\left(n_{1}, n_{2}\right)n=max.(n1,n2)will be called the degree of the system "(7) and the number k = min . ( k 1 , k 2 ) k = min . k 1 , k 2 k=min.(k_(1),k_(2))k=\min .\left(k_{1}, k_{2}\right)k=min.(k1,k2)the order of this system. To fix the ideas we suppose that k = k 1 ≤ k 2 k = k 1 ≤ k 2 k=k_(1) <= k_(2)k=k_{1} \leq k_{2}k=k1≤k2.
We immediately see that we can find two integers r 1 , r 2 r 1 , r 2 r_(1),r_(2)r_{1}, r_{2}r1,r2: such as
(8) Δ ω 1 ′ n = 0 , Δ ω 2 ′ n = 0 (8) Δ ω 1 ′ n = 0 , Δ ω 2 ′ n = 0 {:(8)Delta_(omega_(1)^('))^(n)=0","quadDelta_(omega_(2)^('))^(n)=0:}\begin{equation*} \Delta_{\omega_{1}^{\prime}}^{n}=0, \quad \Delta_{\omega_{2}^{\prime}}^{n}=0 \tag{8} \end{equation*}(8)Δω1′n=0,Δω2′n=0
by posing ω 1 ′ = r 1 ω 1 , ω 2 ′ = r 2 ω 2 ω 1 ′ = r 1 ω 1 , ω 2 ′ = r 2 ω 2 omega_(1)^(')=r_(1)omega_(1),omega_(2)^(')=r_(2)omega_(2)\omega_{1}^{\prime}=r_{1} \omega_{1}, \omega_{2}^{\prime}=r_{2} \omega_{2}ω1′=r1ω1,ω2′=r2ω2. The numbers ω 1 ′ , ω 2 ′ ω 1 ′ , ω 2 ′ omega_(1)^('),omega_(2)^(')\omega_{1}^{\prime}, \omega_{2}^{\prime}ω1′,ω2′are still independent and we can therefore state the following property:
If the periods ω 1 , ω 2 ω 1 , ω 2 omega_(1),omega_(2)\omega_{1}, \omega_{2}ω1,ω2are independent, if the function f ( x ) f ( x ) f(x)f(x)f(x)checks the system of functional equations
∑ i = 0 n 1 a i f ( x + i ω 1 ) = 0 , ∑ i = 0 n 2 b i f ( x + i ω 2 ) = 0 ∑ i = 0 n 1   a i f x + i ω 1 = 0 , ∑ i = 0 n 2   b i f x + i ω 2 = 0 sum_(i=0)^(n_(1))a_(i)f(x+iomega_(1))=0,quadsum_(i=0)^(n_(2))b_(i)f(x+iomega_(2))=0\sum_{i=0}^{n_{1}} a_{i} f\left(x+i \omega_{1}\right)=0, \quad \sum_{i=0}^{n_{2}} b_{i} f\left(x+i \omega_{2}\right)=0∑i=0n1hasif(x+iω1)=0,∑i=0n2bif(x+iω2)=0
of degree n n nnn, of order k k kkkand is continuous in n n nnnpoints, it reduces to a polynomial of degree k − 1 k − 1 k-1k-1k−1.
One may wonder whether one could not reduce the number of points of continuity of the function such that the conclusions remain the same. In particular, the continuity in k k kkkpoints alone is enough to conclude that the function reduces to a polynomial?'
We will demonstrate that this is indeed the case.
7. - Equations (8) show us that if we give to x x xxxa fixed value and if the function f ( x ) f ( x ) f(x)f(x)f(x)is assumed to be bounded at a point. It reduces to the points x + p ω 1 ′ + q ω 2 ′ , p , q = 0 , ± 1 , ± 2 , … x + p ω 1 ′ + q ω 2 ′ , p , q = 0 , ± 1 , ± 2 , … x+pomega_(1)^(')+qomega_(2)^('),p,q=0,+-1,+-2,dotsx+p \omega_{1}^{\prime}+q \omega_{2}^{\prime}, p, q=0, \pm 1, \pm 2, \ldotsx+pω1′+qω2′,p,q=0,±1,±2,…. to a polynomial of degree n − 1 n − 1 n-1n-1n−1. In particular, for a x x xxxgiven, the function reduces to the points x + s ω ˙ 1 + p ω 1 ′ + q ω 2 ′ , p , q = 0 , ± 1 , ± 2 , … x + s ω ˙ 1 + p ω 1 ′ + q ω 2 ′ , p , q = 0 , ± 1 , ± 2 , … x+somega^(˙)_(1)+pomega_(1)^(')+qomega_(2)^('),p,q=0,+-1,+-2,dotsx+s \dot{\omega}_{1}+p \omega_{1}^{\prime}+q \omega_{2}^{\prime}, p, q=0, \pm 1, \pm 2, \ldotsx+sω˙1+pω1′+qω2′,p,q=0,±1,±2,…to a polynomial R s ( x ) R s ( x ) R_(s)(x)\mathrm{R}_{s}(x)Rs(x)of degree n − 1 n − 1 n-1n-1n−1. We have thus r 1 r 1 r_(1)r_{1}r1polynomials R s ( x ) R s ( x ) R_(s)(x)\mathrm{R}_{s}(x)Rs(x)which can be distinct, but we always have R s ( x ) = R s ′ ( x ) R s ( x ) = R s ′ ( x ) R_(s)(x)=R_(s^('))(x)\mathrm{R}_{s}(x)=\mathrm{R}_{s^{\prime}}(x)Rs(x)=Rs′(x)s. s = s ′ ( mod . r 1 ) s = s ′ mod . r 1 s=s^(')(mod.r_(1))s=s^{\prime}\left(\bmod . r_{1}\right)s=s′(mod.r1).
The functions f ( x ) f ( x ) f(x)f(x)f(x)verifies, by hypothesis, the relationship
∑ i = 0 n 1 a i f ( x + i ω 1 ′ ) = 0 ∑ i = 0 n 1   a i f x + i ω 1 ′ = 0 sum_(i=0)^(n_(1))a_(i)f(x+iomega_(1)^('))=0\sum_{i=0}^{n_{1}} a_{i} f\left(x+i \omega_{1}^{\prime}\right)=0∑i=0n1hasif(x+iω1′)=0
we therefore have, for each s s sss,
(9) ∑ i = 0 n 1 a i R s ( x + i ω 1 ′ ) = 0 (9) ∑ i = 0 n 1   a i R s x + i ω 1 ′ = 0 {:(9)sum_(i=0)^(n_(1))a_(i)R_(s)(x+iomega_(1)^('))=0:}\begin{equation*} \sum_{i=0}^{n_{1}} a_{i} R_{s}\left(x+i \omega_{1}^{\prime}\right)=0 \tag{9} \end{equation*}(9)∑i=0n1hasiRs(x+iω1′)=0
for an infinity of values ​​of x x xxx. It immediately follows that the equalities (9) are verified identically in x x xxx. The polynomial R s ( x ) R s ( x ) R_(s)(x)\mathrm{R}_{s}(x)Rs(x)therefore satisfies an equation of the form (5) and of order k k kkk. It follows that: the polynomials R s ( x ) R s ( x ) R_(s)(x)\mathrm{R}_{s}(x)Rs(x)are of degree k − 1 k − 1 k-1k-1k−1.
Likewise, for a x x xxxgiven, the function reduces to the points. x + s ω 2 + p ω 1 ′ + q ω 2 ′ , p , q = 0 , ± 1 , ± 2 , … x + s ω 2 + p ω 1 ′ + q ω 2 ′ , p , q = 0 , ± 1 , ± 2 , … x+somega_(2)+pomega_(1)^(')+qomega_(2)^('),p,q=0,+-1,+-2,dotsx+s \omega_{2}+p \omega_{1}^{\prime}+q \omega_{2}^{\prime}, p, q=0, \pm 1, \pm 2, \ldotsx+sω2+pω1′+qω2′,p,q=0,±1,±2,…to a polynomial S s ( x ) S s ( x ) S_(s)(x)S_{s}(x)Ss(x)of degree n − 1 n − 1 n-1n-1n−1. We have thus r 2 r 2 r_(2)r_{2}r2polynomials S s ( x ) S s ( x ) S_(s)(x)S_{s}(x)Ss(x)which can be distinct, but we always have S s ( x ) = S s ′ ( x ) S s ( x ) = S s ′ ( x ) S_(s)(x)=S_(s^('))(x)\mathrm{S}_{s}(x)=\mathrm{S}_{s^{\prime}}(x)Ss(x)=Ss′(x)if s = s ′ ( mod . r 2 ) s = s ′ mod . r 2 s=s^(')(mod.r_(2))s=s^{\prime}\left(\bmod . r_{2}\right)s=s′(mod.r2). Now, we also have, for each s s sss,
∑ i = 0 n 1 a i S s ( x + i ω 1 ′ ) = 0 ∑ i = 0 n 1   a i S s x + i ω 1 ′ = 0 sum_(i=0)^(n_(1))a_(i)S_(s)(x+iomega_(1)^('))=0\sum_{i=0}^{n_{1}} a_{i} \mathrm{~S}_{s}\left(x+i \omega_{1}^{\prime}\right)=0∑i=0n1hasi Ss(x+iω1′)=0
for an infinity of values ​​of x x xxx. It follows that the polynomials S s ( x ) S s ( x ) S_(s)(x)\mathrm{S}_{s}(x)Ss(x)are also of a certain degree k − 1 k − 1 k-1k-1k−1.
The property stated in № 6 therefore remains true if we assume that the function f ( x ) f ( x ) f(x)f(x)f(x)is continuous only in l l llLpoints. In this case, in fact, all the polynomials R s ( x ) , S s ( x ) R s ( x ) , S s ( x ) R_(s)(x),S_(s)(x)\mathrm{R}_{s}(x), \mathrm{S}_{s}(x)Rs(x),Ss(x)coincide.
In particular, we can state the following proposition:
If the periods ω 1 , ω 2 ω 1 , ω 2 omega_(1),omega_(2)\omega_{1}, \omega_{2}ω1,ω2are independent, if the function f ( x ) f ( x ) f(x)f(x)f(x)checks: functional equations
∑ i = 0 n 1 a 1 f ( x + i m ω 1 ) = 0 , ∑ i = 0 n 2 b i f ( x + i m ω 2 ) = 0 m = 1 , 2 , … ∑ i = 0 n 1   a 1 f x + i m ω 1 = 0 , ∑ i = 0 n 2   b i f x + i m ω 2 = 0 m = 1 , 2 , … {:[sum_(i=0)^(n_(1))a_(1)f(x+imomega_(1))=0","quadsum_(i=0)^(n_(2))b_(i)f(x+imomega_(2))=0],[m=1","quad2","dots]:}\begin{gathered} \sum_{i=0}^{n_{1}} a_{1} f\left(x+i m \omega_{1}\right)=0, \quad \sum_{i=0}^{n_{2}} b_{i} f\left(x+i m \omega_{2}\right)=0 \\ m=1, \quad 2, \ldots \end{gathered}∑i=0n1has1f(x+imω1)=0,∑i=0n2bif(x+imω2)=0m=1,2,…
of order 7 and is continuous in k k kkkpoints, it reduces to a polynomial of degree k − 1 k − 1 k-1k-1k−1.
8. - The restriction relating to the substitution of h h hhhby m h m h mhm hmhin: (5) is essential.
Consider the system
(10) f ( x ) + f ( x + ω 1 ) − f ( x + 2 ω 1 ) − f ( x + 3 ω 4 ) = 0 f ( x ) + f ( x + ω 2 ) − f ( x + 2 ω 2 ) − f ( x + 3 ω 2 ) = 0 (10) f ( x ) + f x + ω 1 − f x + 2 ω 1 − f x + 3 ω 4 = 0 f ( x ) + f x + ω 2 − f x + 2 ω 2 − f x + 3 ω 2 = 0 {:[(10)f(x)+f(x+omega_(1))-f(x+2omega_(1))-f(x+3omega_(4))=0],[f(x)+f(x+omega_(2))-f(x+2omega_(2))-f(x+3omega_(2))=0]:}\begin{align*} & f(x)+f\left(x+\omega_{1}\right)-f\left(x+2 \omega_{1}\right)-f\left(x+3 \omega_{4}\right)=0 \tag{10}\\ & f(x)+f\left(x+\omega_{2}\right)-f\left(x+2 \omega_{2}\right)-f\left(x+3 \omega_{2}\right)=0 \end{align*}(10)f(x)+f(x+ω1)−f(x+2ω1)−f(x+3ω4)=0f(x)+f(x+ω2)−f(x+2ω2)−f(x+3ω2)=0
which is of degree 3 and order 1. Let us define the function as follows: It is equal to x x xxxat a point of the form p ω 1 + q ω 2 p ω 1 + q ω 2 pomega_(1)+qomega_(2)p \omega_{1}+q \omega_{2}pω1+qω2Or p , q p , q p,qp, qp,qare of the same parity, is equal to − x − x -x-x−xat a point of the form p ω 1 + q ω 2 p ω 1 + q ω 2 pomega_(1)+qomega_(2)p \omega_{1}+q \omega_{2}pω1+qω2Or p , q p , q p,qp, qp,qare of different parity and is zero everywhere else.
This function is continuous at the point x = 0 x = 0 x=0x=0x=0, but it does not reduce to a polynomial and is not a solution of the system (10) in our sense. On the contrary, this function verifies the two equations (10) alone.-
This same example shows us that among the values ​​of m m mmmit is necessary that r appears. In this case in fact we can take r 1 = r 2 = 2 r 1 = r 2 = 2 r_(1)=r_(2)=2r_{1}=r_{2}=2r1=r2=2and we see that the system (8) is a consequence of the two equations (10).
There are special cases where we can greatly reduce the number of values ​​to take for m m mmm.
Note that f ( x ) f ( x ) f(x)f(x)f(x)checks the degree equation n n nnnwhere instead of h h hhhWe have r h r h rhr hrhand "the characteristic equation is G ( z ) = 0 G ( z ) = 0 G(z)=0\mathrm{G}(z)=0G(z)=0, with
G ( z r ) = F ( z ) F ( α z ) , … , F ( α r − 1 z ) G z r = F ( z ) F ( α z ) , … , F α r − 1 z G(z^(r))=F(z)F(alpha z),dots,F(alpha^(r-1)z)\mathrm{G}\left(z^{r}\right)=\mathrm{F}(z) \mathrm{F}(\alpha z), \ldots, \mathrm{F}\left(\alpha^{r-1} z\right)G(zr)=F(z)F(αz),…,F(αr−1z)
a being a primitive root of order r r rrrof unity.
It follows that if all the roots of the characteristic equation F ( z ) = 0 F ( z ) = 0 F(z)=0\mathrm{F}(z)=0F(z)=0are roots of order r r rrrof the unit equation (6) is a consequence of the single equation (5).
We can therefore state the following proposition:
The function f ( x ) f ( x ) f(x)f(x)f(x), verifying the three functional equations
∑ i = 0 n 1 a i f ( x + i ω 1 ) = 0 , ∑ i = 0 n 1 a i f ( x + i r ω 1 ) = 0 , ∑ i = 0 n 2 b i f ( x + i ω 2 ) ) ∑ i = 0 n 1   a i f x + i ω 1 = 0 , ∑ i = 0 n 1   a i f x + i r ω 1 = 0 , ∑ i = 0 n 2   b i f x + i ω 2 {:sum_(i=0)^(n_(1))a_(i)f(x+iomega_(1))=0,quadsum_(i=0)^(n_(1))a_(i)f(x+iromega_(1))=0,quadsum_(i=0)^(n_(2))b_(i)f(x+iomega_(2)))\left.\sum_{i=0}^{n_{1}} a_{i} f\left(x+i \omega_{1}\right)=0, \quad \sum_{i=0}^{n_{1}} a_{i} f\left(x+i r \omega_{1}\right)=0, \quad \sum_{i=0}^{n_{2}} b_{i} f\left(x+i \omega_{2}\right)\right)∑i=0n1hasif(x+iω1)=0,∑i=0n1hasif(x+irω1)=0,∑i=0n2bif(x+iω2))
  • of order k ( k = k 1 ≤ k 2 ) k k = k 1 ≤ k 2 k(k=k_(1) <= k_(2))k\left(k=k_{1} \leq k_{2}\right)k(k=k1≤k2), reduces to a polynomial of degree k − 1 k − 1 k-1k-1k−1if
  1. The periods ω 1 , ω 2 ω 1 , ω 2 omega_(1),omega_(2)\omega_{1}, \omega_{2}ω1,ω2are independent.
    2 0 2 0 2^(0)2^{0}20. All roots of the characteristic equation a 0 + a 1 z + … a 0 + a 1 z + … a_(0)+a_(1)z+dotsa_{0}+a_{1} z+\ldotshas0+has1z+…. … . + a n 1 z n 1 = 0 … . + a n 1 z n 1 = 0 dots.+a_(n_(1))z^(n_(1))=0\ldots .+a_{n_{1}} z^{n_{1}}=0….+hasn1zn1=0are roots of order r r rrrof unity and all the roots of the equation b 0 + b 1 z + … + b n 2 z n 2 = 0 b 0 + b 1 z + … + b n 2 z n 2 = 0 b_(0)+b_(1)z+dots+b_(n_(2))z^(n_(2))=0b_{0}+b_{1} z+\ldots+b_{n_{2}} z^{n_{2}}=0b0+b1z+…+bn2zn2=0are roots of a certain order of unity.
  2. The function is continuous in k k kkkpoints.
If we consider the system formed by the first and third equations (with m = 1 m = 1 m=1m=1m=1only) the property remains true if
1 0 1 0 1^(0)1^{0}10And 2 0 2 0 2^(0)2^{0}20are verified
3 0 3 0 3^(0)3^{0}30The function is continuous in n n nnnpoints, n being the degree of the system.
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