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T. Popoviciu, Sur les diffèrences des fonctions d’une variable réelle, Comptes Rendus des séances de l’Académie des Sciences de Roumanie, 2 (1938) no. 2, pp. 112-114 (in Romanian).
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28. On the differences of functions of a real variable
by TIBERIU POPOVICIU
Presented by Miron Nicolesco, Mc. ASR
(Session of + June 1937).
- We consider real functions of the real variablexx, defined and uniform in a finite and closed interval which can be assumed to be ( 0,1 ). Let us set, as usual,
Delta_(h)^(n)f(x)=sum_(i=o)^(n)(-I)^(ni)((i)/(i))//(x+ih),quad x+ih sub(O,I).\Delta_{h}^{n} f(x)=\sum_{i=o}^{n}(-\mathrm{I})^{ni}\binom{i}{i} /(x+ih), \quad x+ih \subset(\mathrm{O}, \mathrm{I}) .
MA Marchaud demonstrated^(1){ }^{1}) that iff(x)f(x)is bounded and ifDelta_(h)^(n)f(x)\Delta_{h}^{n} f(x)tends uniformly to zero whenh rarr0h \rightarrow 0, the functionf(x)f(x)is continuous in (0,1).
We have the more precise property:
Iff(x)\mathrm{f}(\mathrm{x})is bounded in a subinterval, however small it may be, and ifDelta_(h)^(n)f(x)\Delta_{h}^{n} \mathrm{f}(\mathrm{x})tends uniformly to zero forhrarr0\mathrm{h} \rightarrow 0, the functionf(x)\mathrm{f}(\mathrm{x})is continuous in (0,1).
Iff(x)\mathrm{f}(\mathrm{x})is bounded in a subinterval, however small it may be, and ifDelta_(h)^(n)f(x)\Delta_{h}^{n} \mathrm{f}(\mathrm{x})tends uniformly to zero forhrarr0\mathrm{h} \rightarrow 0, the functionf(x)\mathrm{f}(\mathrm{x})is continuous in (0,1).
It is enough to demonstrate that iff(x)f(x)is bounded in (c;Ic ; \mathrm{I}) it is also in(0.1)(0.1). Either|f(x)| < M,x sub(c,1)|f(x)|<M, x \subset(c, 1)Ando < eta < (Ic)/(n-1)o<\eta<\frac{Ic}{n-1}such as|Delta_(h)^(n)f(x)| < M\left|\Delta_{h}^{n} f(x)\right|<\mathrm{M}For|h| < eta|h|<\eta. We have|f(x)| < 2^(n)M|f(x)|<2^{n} \mathrm{M}Forx >= c-etax \geqslant c-\eta, from which we deduce that
|f(x)| < 2^(ns)M,quad x sub(0,1),quad s > (c)/( eta).|f(x)|<2^{ns} \mathrm{M}, \quad x \subset(0,1), \quad s>\frac{c}{\eta} .
We will now demonstrate that MA Marchaud's theorem extends to measurable functions:
Iff(x)\mathrm{f}(\mathrm{x})is measurable in(O,I)(\mathrm{O}, \mathrm{I})and ifDelta_(h)^(n)f(x)\Delta_{h}^{n} \mathrm{f}(\mathrm{x})tends uniformly to zero forhrarr0\mathrm{h} \rightarrow 0, the functionf(x)\mathrm{f}(\mathrm{x})is continuous in(0,I)(0, \mathrm{I}).
Indeed, iff(x)f(x)is not bounded, nor is it bounded in the interval(0,(I)/(2))\left(0, \frac{I}{2}\right). Either0 < eta < (I)/(2n)0<\eta<\frac{I}{2 n}such as|Delta_(h)^(n)f(x)| < I\left|\Delta_{h}^{n} f(x)\right|<IFor|h| <= eta|h| \leqslant \eta. Given a numberA > 0\mathrm{A}>0, there is axi sub(0,(1)/(2))\xi \subset\left(0, \frac{1}{2}\right)such as|f(xi)| > I+(2^(n)-I)|f(\xi)|>\mathrm{I}+\left(2^{n}-\mathrm{I}\right)A. But we have|Delta_(h)^(n)f(xi)| < I\left|\Delta_{h}^{n} f(\xi)\right|<\mathrm{I}For(nI)/(h)eta < h < eta\frac{n-\mathrm{I}}{h} \eta<h<\eta,
112
therefore|f(xi+ih)| >|f(\xi+ih)|>. A for at least onei,i <= i <= ni, i \leqslant i \leqslant n. We deduce that the set E||f(x)| > A|||f(x)|>\mathrm{A}|is measuring>= (eta )/(n)\geqslant \frac{\eta}{n}. The number A being arbitrary, this is in contradiction with a theorem of ME B ore 1 on measurable functions. The stated property results from this. The case of the equationDelta_(h)^(n)f(x)=O\Delta_{h}^{n} f(x)=\mathrm{O}was studied by MW Sierpinski{:(n=2)^(1))\left.(n=2)^{1}\right)and by ourselves{:(n > 2)^(3))\left.(n>2)^{3}\right).
2. Consider the more general expression^(3){ }^{3}).
(1)
therefore|f(xi+ih)| >|f(\xi+ih)|>. A for at least onei,i <= i <= ni, i \leqslant i \leqslant n. We deduce that the set E||f(x)| > A|||f(x)|>\mathrm{A}|is measuring>= (eta )/(n)\geqslant \frac{\eta}{n}. The number A being arbitrary, this is in contradiction with a theorem of ME B ore 1 on measurable functions. The stated property results from this. The case of the equationDelta_(h)^(n)f(x)=O\Delta_{h}^{n} f(x)=\mathrm{O}was studied by MW Sierpinski{:(n=2)^(1))\left.(n=2)^{1}\right)and by ourselves{:(n > 2)^(3))\left.(n>2)^{3}\right).
2. Consider the more general expression^(3){ }^{3}).
(1)
Delta_(h)^((n))f(x)=sum_(i=o)^(n)a_(i)f(x+ih)\Delta_{h}^{(n)} f(x)=\sum_{i=o}^{n} a_{i} f(x+ih)
THEa_(i)a_{i}being given constants. This expression is of order k ifsum_(i=o)^(n)a_(i)i^(j)=0,j=0,I,dots,kI,sum_(i=o)^(n)a_(i)i^(k)!=0\sum_{i=o}^{n} a_{i} i^{j}=0, j=0, \mathrm{I}, \ldots, k-\mathrm{I}, \sum_{i=o}^{n} a_{i} i^{k} \neq 0, so if the polynomialF(x)=sum_(i=o)^(n)a_(i)x^(i)\mathrm{F}(x)=\sum_{i=o}^{n} a_{i} x^{i}has the root I of orderkkof multiplicity. In this case we can determine a positive integerppand two polynomialsvarphi(x),psi(x)\varphi(x), \psi(x), so that we have^(4){ }^{4}) :
varphi(x)F(x)+psi(lambda)F(x^(P))-=(I-lambda^(P))^(k).\varphi(x) \mathrm{F}(x)+\psi(\lambda) \mathrm{F}\left(x^{\mathrm{P}}\right) \equiv\left(\mathrm{I}-\lambda^{\mathrm{P}}\right)^{k} .
It follows that we have a relation of the form
{:(2)Delta_(ph)^(k)f(x)=sum_(i=o)^(pn-h-i)alpha_(i)Delta_(h)^((n))f(x+ih)+sum_(i=o)^(n-k-i)beta_(i)Delta_(ph)^((v))f(x+ih)",":}\begin{equation*}
\Delta_{p h}^{k} f(x)=\sum_{i=o}^{p n-h-i} \alpha_{i} \Delta_{h}^{(n)} f(x+i h)+\sum_{i=o}^{n-k-i} \beta_{i} \Delta_{p h}^{(v)} f(x+i h), \tag{2}
\end{equation*}
THEalpha_(i),beta_(i)\alpha_{i}, \beta_{i}being constants independent ofx,hx, hand functionf(x)f(x).
We see that the properties of No. I remain true for expression (1). In particular, ifsum_(i=o)^(n)a_(i)=0\sum_{i=o}^{n} a_{i}=0and if (1) tends uniformly to zero, the functionf(x)f(x)is zero identically in (0,I0, \mathrm{I}). If
- W. Sierpinski: On measurable convex functions. Fund. Math. t. I (1920), pp. 125-129). M.W. Sierpinski also proved this theorem without using Zermelo's axiom, see: On the functional equationf(x+y)=f(x)+f(y)f(x+y)=f(x)+f(y)id. p. 116-122.
- T. Popoviciu: On some properties of functions of one or two real variables. Mathematica vol. VIII (1934) p. I— 85.
- T. Pcpoviciu: On certain functional equations defining polynomials". Mathematica t. X, (1934), p. 197-211.
- We will give the demonstration in a work which will appear in vol. XIV of the journal "Mathematica".
sum_(i=o)^(n)a_(i)=0,sum_(i=o)^(n)a_(i)i=0\sum_{i=o}^{n} a_{i}=0, \sum_{i=o}^{n} a_{i} i=0and if (1) tends to zero for allxx, the function is continuous in(0,1)(0,1). - MA Marchaud also demonstrated that iff(x)f(x)is bounded and if the ratioh^(-h)Delta_(h)^(h)f(x)h^{-h} \Delta_{h}^{h} f(x)tends uniformly towards a determined functiong(x)g(x), the functionf(x)f(x)has a derivativek^(ème)k^{e ̀ m e}continues which is (obviously) equal tog(x)g(x). Let us demonstrate that this theorem is still true for measurable functions, so
Iff(x)\mathrm{f}(\mathrm{x})is measurable and ifh^(-k)Delta_(h)^(k)f(x)\mathrm{h}^{-k} \Delta_{h}^{k} \mathrm{f}(\mathrm{x})tends uniformly to a functiong(x)\mathrm{g}(\mathrm{x}), the derivativek^("ème ")k^{\text {ème }}èoff(x)\mathrm{f}(\mathrm{x})exists, is continuous and we havef^((k))(x)=g(x)\mathrm{f}^{(k)}(\mathrm{x})=\mathrm{g}(\mathrm{x})In(0,I)(0, \mathrm{I}).
Indeed, for everythingh,h^(-k)Delta_(h)^(k)f(x)h, h^{-k} \Delta_{h}^{k} f(x)is a measurable function, sog(x)g(x)is also. Note thatDelta_(2h)^(h)f(x)=sum_(i=o)^(h)((k)/(i))Delta_(h)^(k)f(x+ih)\Delta_{2 h}^{h} f(x)=\sum_{i=o}^{h}\binom{k}{i} \Delta_{h}^{k} f(x+i h), SO
{:[|(I)/(2^(k))sum_(i=o)^(k)((k)/(i))g(x+ih)-g(x)| <= ],[ <= |(Delta_(2h)^(k)f(x))/((2h)^(k))-g(x)|+(I)/(2^(k))sum_(i=o)^(h)((k)/(i))|(Delta_(h)^(h)f(x+ih))/(h^(k))-g(x+ih)|]:}\begin{aligned}
& \left|\frac{\mathrm{I}}{2^{k}} \sum_{i=o}^{k}\binom{k}{i} g(x+i h)-g(x)\right| \leqslant \\
& \leqslant\left|\frac{\Delta_{2 h}^{k} f(x)}{(2 h)^{k}}-g(x)\right|+\frac{\mathrm{I}}{2^{k}} \sum_{i=o}^{h}\binom{k}{i}\left|\frac{\Delta_{h}^{h} f(x+i h)}{h^{k}}-g(x+i h)\right|
\end{aligned}
The expression of orderII,
(I)/(2^(h))sum_(i=o)^(k)((k)/(i))g(x+ih)-g(x),\frac{\mathrm{I}}{2^{h}} \sum_{i=o}^{k}\binom{k}{i} g(x+i h)-g(x),
therefore tends (moreover uniformly) towards zero. It follows thatg(x)g(x)is continuous and, therefore, bounded in (0,I0, \mathrm{I}). SO,Delta_(h)^(h)f(x)\Delta_{h}^{h} f(x)tends uniformly towards zero, sof(x)f(x)is also continuous. The stated property results from this.
If we posef(x)=x^(h)f(x)=x^{h}in (2), we find
(I)/(p^(k))sum_(i=o)^(pn-k-1)alpha_(i)+sum_(i=o)^(n-k-1)beta_(i)=(k!)/(sum_(i=o)^(k)a_(i)i^(k))\frac{\mathrm{I}}{p^{k}} \sum_{i=o}^{p n-k-1} \alpha_{i}+\sum_{i=o}^{n-k-1} \beta_{i}=\frac{k!}{\sum_{i=o}^{k} a_{i} i^{k}}
and we deduce the following property:
Iff(x)\mathrm{f}(\mathrm{x})is measurable or iff(x)\mathrm{f}(x)is bounded and ifh^(-k)Delta_(h)^((n))f(x)\mathrm{h}^{-k} \Delta_{h}^{(n)} \mathrm{f}(\mathrm{x})tends uniformly to a functiong(x)\mathrm{g}(\mathrm{x}), the functionf(x)\mathrm{f}(\mathrm{x})has a derivativek^("ème ")\mathrm{k}^{\text {ème }}ècontinue and we have
Iff(x)\mathrm{f}(\mathrm{x})is measurable or iff(x)\mathrm{f}(x)is bounded and ifh^(-k)Delta_(h)^((n))f(x)\mathrm{h}^{-k} \Delta_{h}^{(n)} \mathrm{f}(\mathrm{x})tends uniformly to a functiong(x)\mathrm{g}(\mathrm{x}), the functionf(x)\mathrm{f}(\mathrm{x})has a derivativek^("ème ")\mathrm{k}^{\text {ème }}ècontinue and we have
f^((k))(x)=(k!)/(sum_(i=o)^(n)a_(i)i^(k))g(x),quad x sub(0,I).f^{(k)}(x)=\frac{k!}{\sum_{i=o}^{n} a_{i} i^{k}} g(x), \quad x \subset(0, \mathrm{I}) .
- A. Marchaud: On the derivatives and on the differences of functions of real variables. Journ, de Math. t. 6 (1927), p. 337-425.