Abstract
Given a continuous function \(u:[0,\infty)\rightarrow \mathbb{R}\), a family of functionals \(\varphi_{\U{3b1} }:C(I)\rightarrow \mathbb{R},\U{3b1} >0,\) is defined by \(\varphi_{\U{3b1} }(f)=\frac{1}{\U{3b1} }\int \limits_{0}^{\alpha}u(t)f(t/\U{3b1} )dt\). It is proved that the necessary and sufficient conditions for the family (\varphi_{\U{3b1} },\U{3b1} >0\) to satisfy \(\lim \limits_{\U{3b1} \rightarrow \infty}\varphi_{\U{3b1} }(f)=\left(\lim \limits_{\U{3b1} \rightarrow \infty}\frac{1}{\U{3b1} }\int \limits_{0}^{\alpha}u(t)dt\right) \cdot \int \limits_{0}^{1}\) \(f\) are: I. \(\exists\lim \limits_{\U{3b1} \rightarrow \infty}\frac{1}{\alpha}\int \limits_{0}^{\alpha}u(t)dt;\) II.\(\sup \limits_{\alpha>0}\frac{1}{\alpha}\int \limits_{0}^{\alpha}|u(t)|dt<\infty\).
If \(f\in \boldsymbol{C}^{1}(I)\), condition \(I\) alone implies the existence of \(\lim \limits_{\U{3b1} \rightarrow \infty}\varphi_{\alpha}(f)\). \newline A sequence of functionals \((\varphi_{n})_{n\in N}\) is attached to a numerical sequence \((a_{n})_{n\in N}\) which is Cesaro-convergent to \(\alpha\), namely
\[
\varphi_{n}(f)=\frac{1}{n}\sum_{k=1}^{n}a_{k}f(k/n)\text{, }f\text{ Riemann
integrable}
\]Additional conditions are imposed on the sequence \((\alpha_{n})_{n\in N}\) in order to prove that
\[
\lim \limits_{n\rightarrow \infty}\varphi_{n}\left( f\right)
=a\text{\textperiodcentered}\int \limits_{1}^{0}f.
\]
Authors
Mira-Cristiana Anisiu
Tiberiu Popoviciu Institute of Numerical Analysis 37, Republicii st., 3400 Cluj-Napoca, Romania
Valeriu Anisiu
Babes-Bolyai University, Faculty of Mathematics 1, Kogalniceanu st., 3400 Cluj-Napoca, Romania
Keywords
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Paper coordinates
M.-C. Anisiu, V. Anisiu, On the pointwise convergence of a family of functionals on C(I), Proceedings of the Tiberiu Popoviciu Itinerant Seminar of Functional Equations, Approximation and Convexity, Cluj-Napoca, May 21-May 25, 2003, ed. E. Popoviciu, Srima, 2003, 3-13 (pdf file here)
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[1] M.-C. Anisiu, V. Anisiu. Sequences of linear operators related to Cesaro-convergent sequences. Revue d’Analyse Num. Th. Approx. 31 (2) (2002), 139-145.
[2] R. Gologan. On the convergence of some partial sums and a contest problem. Gazeta Matematica seria A, Anul XIX nr. 2 (2001), 70-73.
[3] I. J Maddox, The norm of a linear functional. Amer. Math. Monthly 96 (1989), 434-436.
[4] G. Szasz, L. Geher, I. Kovacs, L. Pinter (editors). Contests in Higher Mathematics. Akademiai Kiado, Budapest 1968.
[5] E. C. Titchmarch. The Theory of Functions. Oxford University Press 1939.
[6] A. Wilanski. Modern Methods in Topological Vector Spaces. McGraw-Hill 1978.
On the pointwise convergence of a family of functionals on C(I)\mathcal{C}(I)
Abstract. Given a continuous function u:[0,oo)rarrRu:[0, \infty) \rightarrow \mathbb{R}, a family of functionals varphi_(alpha):C(I)rarrR,alpha > 0\varphi_{\alpha}: \mathcal{C}(I) \rightarrow \mathbb{R}, \alpha>0, is defined by varphi_(alpha)(f)=(1)/(alpha)int_(0)^(alpha)u(t)f(t//alpha)dt\varphi_{\alpha}(f)=\frac{1}{\alpha} \int_{0}^{\alpha} u(t) f(t / \alpha) \mathrm{d} t. It is proved that the necessary and sufficient conditions for the family varphi_(alpha),alpha > 0\varphi_{\alpha}, \alpha>0 to satisfy lim_(alpha rarr oo)varphi_(alpha)(f)=(lim_(alpha rarr oo)(1)/(alpha)int_(0)^(alpha)u(t)dt)*int_(0)^(1)f\lim _{\alpha \rightarrow \infty} \varphi_{\alpha}(f)=\left(\lim _{\alpha \rightarrow \infty} \frac{1}{\alpha} \int_{0}^{\alpha} u(t) \mathrm{d} t\right) \cdot \int_{0}^{1} f are:
I. EElim_(alpha rarr oo)(1)/(alpha)int_(0)^(alpha)u(t)dt;quad\exists \lim _{\alpha \rightarrow \infty} \frac{1}{\alpha} \int_{0}^{\alpha} u(t) \mathrm{d} t ; \quad II. s u p_(alpha > 0)(1)/(alpha)int_(0)^(alpha)|u(t)|dt < oo\sup _{\alpha>0} \frac{1}{\alpha} \int_{0}^{\alpha}|u(t)| \mathrm{d} t<\infty.
I. EElim_(alpha rarr oo)(1)/(alpha)int_(0)^(alpha)u(t)dt;quad\exists \lim _{\alpha \rightarrow \infty} \frac{1}{\alpha} \int_{0}^{\alpha} u(t) \mathrm{d} t ; \quad II. s u p_(alpha > 0)(1)/(alpha)int_(0)^(alpha)|u(t)|dt < oo\sup _{\alpha>0} \frac{1}{\alpha} \int_{0}^{\alpha}|u(t)| \mathrm{d} t<\infty.
If f inC^(1)(I)f \in \mathcal{C}^{1}(I), condition I alone implies the existence of lim_(alpha rarr oo)varphi_(alpha)(f)\lim _{\alpha \rightarrow \infty} \varphi_{\alpha}(f).
A sequence of functionals (varphi_(n))_(n inN)\left(\varphi_{n}\right)_{n \in \mathbb{N}} is attached to a numerical sequence (a_(n))_(n inN)\left(a_{n}\right)_{n \in \mathbb{N}} which is Cesàro-convergent to aa, namely
A sequence of functionals (varphi_(n))_(n inN)\left(\varphi_{n}\right)_{n \in \mathbb{N}} is attached to a numerical sequence (a_(n))_(n inN)\left(a_{n}\right)_{n \in \mathbb{N}} which is Cesàro-convergent to aa, namely
varphi_(n)(f)=(1)/(n)sum_(k=1)^(n)a_(k)f(k//n),f" Riemann integrable. "\varphi_{n}(f)=\frac{1}{n} \sum_{k=1}^{n} a_{k} f(k / n), f \text { Riemann integrable. }
Additional conditions are imposed on the sequence (a_(n))_(n inN)\left(a_{n}\right)_{n \in \mathbb{N}} in order to prove that
lim_(n rarr oo)varphi_(n)(f)=a*int_(0)^(1)f\lim _{n \rightarrow \infty} \varphi_{n}(f)=a \cdot \int_{0}^{1} f
MSC 2000: 47B38, 26E60
1 Introduction
For the interval I=[0,1]I=[0,1] and a Banach space F!={0}F \neq\{0\}, let us denote by B(I,F)\mathcal{B}(I, F) the Banach space of bounded functions f:I rarr Ff: I \rightarrow F endowed with the sup norm. The subspace of B(I,F)\mathcal{B}(I, F) of regular functions (which admit side limits at each t in It \in I ) will be denoted by R(I,F)\mathcal{R}(I, F); the Banach space of continuous functions C(I,F)\mathcal{C}(I, F) is a subspace of R(I,F)\mathcal{R}(I, F).
Given a sequence of real numbers (a_(n))_(n inN)\left(a_{n}\right)_{n \in \mathbb{N}}, a sequence of operators varphi_(n):R(I,F)rarr F,n inN\varphi_{n}: \mathcal{R}(I, F) \rightarrow F, n \in \mathbb{N},
{:(1.1)varphi_(n)(f)=(1)/(n)sum_(k=1)^(n)a_(k)f((k)/(n)):}\begin{equation*}
\varphi_{n}(f)=\frac{1}{n} \sum_{k=1}^{n} a_{k} f\left(\frac{k}{n}\right) \tag{1.1}
\end{equation*}
can be generated. It was proved in [1] that:
AA. The operators varphi_(n)\varphi_{n} are linear and continuous.
BB. If the numeric sequence satisfies the conditions:
B_(1)*(a_(n))_(n inN)B_{1} \cdot\left(a_{n}\right)_{n \in \mathbb{N}} is Cesàro-convergent to a(lim_(n rarr oo)(a_(1)+dots+a_(n))/(n)=a)a\left(\lim _{n \rightarrow \infty} \frac{a_{1}+\ldots+a_{n}}{n}=a\right);
B_(2)B_{2}. the sequence ((|a_(1)|+dots+|a_(n)|)/(n))_(n inN)\left(\frac{\left|a_{1}\right|+\ldots+\left|a_{n}\right|}{n}\right)_{n \in \mathbb{N}} is bounded, then the sequence (varphi_(n)(f))_(n inN)\left(\varphi_{n}(f)\right)_{n \in \mathbb{N}} is convergent and
AA. The operators varphi_(n)\varphi_{n} are linear and continuous.
BB. If the numeric sequence satisfies the conditions:
B_(1)*(a_(n))_(n inN)B_{1} \cdot\left(a_{n}\right)_{n \in \mathbb{N}} is Cesàro-convergent to a(lim_(n rarr oo)(a_(1)+dots+a_(n))/(n)=a)a\left(\lim _{n \rightarrow \infty} \frac{a_{1}+\ldots+a_{n}}{n}=a\right);
B_(2)B_{2}. the sequence ((|a_(1)|+dots+|a_(n)|)/(n))_(n inN)\left(\frac{\left|a_{1}\right|+\ldots+\left|a_{n}\right|}{n}\right)_{n \in \mathbb{N}} is bounded, then the sequence (varphi_(n)(f))_(n inN)\left(\varphi_{n}(f)\right)_{n \in \mathbb{N}} is convergent and
{:(1.2)lim_(n rarr oo)varphi_(n)(f)=a*int_(0)^(1)f:}\begin{equation*}
\lim _{n \rightarrow \infty} \varphi_{n}(f)=a \cdot \int_{0}^{1} f \tag{1.2}
\end{equation*}
CC. If lim_(n rarr oo)varphi_(n)(f)\lim _{n \rightarrow \infty} \varphi_{n}(f) exists for every f inC(I,F)subeR(I,F)f \in \mathcal{C}(I, F) \subseteq \mathcal{R}(I, F), the conditions B_(1)B_{1} and B_(2)B_{2} from above are also necessary.
DD. If f inC^(1)(I,F)f \in \mathcal{C}^{1}(I, F) (i.e. ff is continuous with a continuous derivative), the result in BB holds even if condition B_(2)B_{2} is omitted.
DD. If f inC^(1)(I,F)f \in \mathcal{C}^{1}(I, F) (i.e. ff is continuous with a continuous derivative), the result in BB holds even if condition B_(2)B_{2} is omitted.
The aim of this paper is to provide continuous variants of these results for families of functionals (for the sake of simplicity we consider F=RF=\mathbb{R} ).
Let u:[0,oo)rarrRu:[0, \infty) \rightarrow \mathbb{R} be a continuous function. We define a family of
functionals associated to uu, namely varphi_(alpha):C(I)rarrR,alpha > 0\varphi_{\alpha}: \mathcal{C}(I) \rightarrow \mathbb{R}, \alpha>0, given by
functionals associated to uu, namely varphi_(alpha):C(I)rarrR,alpha > 0\varphi_{\alpha}: \mathcal{C}(I) \rightarrow \mathbb{R}, \alpha>0, given by
{:(1.3)varphi_(alpha)(f)=(1)/(alpha)int_(0)^(alpha)u(t)f((t)/( alpha))dt","f inC(I).:}\begin{equation*}
\varphi_{\alpha}(f)=\frac{1}{\alpha} \int_{0}^{\alpha} u(t) f\left(\frac{t}{\alpha}\right) \mathrm{d} t, f \in \mathcal{C}(I) . \tag{1.3}
\end{equation*}
Proposition 1.1 For each alpha > 0\alpha>0, the functional varphi_(alpha)\varphi_{\alpha} is linear and continuous, and its norm is given by
{:(1.4)||varphi_(alpha)||=(1)/(alpha)int_(0)^(alpha)|u(t)|dt:}\begin{equation*}
\left\|\varphi_{\alpha}\right\|=\frac{1}{\alpha} \int_{0}^{\alpha}|u(t)| \mathrm{d} t \tag{1.4}
\end{equation*}
Proof. This result is classical; see for a simple proof [3]. It also holds if C(I)\mathcal{C}(I) is replaced with the Banach space R(I)\mathcal{R}(I) of regular functions.
2 Main results for families of functionals
As mentioned in BB and CC in the introduction, in the discrete case the conditions B_(1)B_{1} and B_(2)B_{2} are necessary and sufficient for the sequence (1.1) to converge, the limit being given by (1.2). We can prove a similar result for the continuous case.
Theorem 2.1 Let there be given f inC(I)f \in \mathcal{C}(I) and u inC([0,oo))u \in \mathcal{C}([0, \infty)). If the function uu satisfies the conditions:
I. EElim_(alpha rarr oo)(1)/(alpha)int_(0)^(alpha)u(t)dt\exists \lim _{\alpha \rightarrow \infty} \frac{1}{\alpha} \int_{0}^{\alpha} u(t) \mathrm{d} t;
II. s u p_(alpha > 0)(1)/(alpha)int_(0)^(alpha)|u(t)|dt < oo\sup _{\alpha>0} \frac{1}{\alpha} \int_{0}^{\alpha}|u(t)| \mathrm{d} t<\infty,
there exists the limit lim_(alpha rarr oo)varphi_(alpha)(f)\lim _{\alpha \rightarrow \infty} \varphi_{\alpha}(f) and
I. EElim_(alpha rarr oo)(1)/(alpha)int_(0)^(alpha)u(t)dt\exists \lim _{\alpha \rightarrow \infty} \frac{1}{\alpha} \int_{0}^{\alpha} u(t) \mathrm{d} t;
II. s u p_(alpha > 0)(1)/(alpha)int_(0)^(alpha)|u(t)|dt < oo\sup _{\alpha>0} \frac{1}{\alpha} \int_{0}^{\alpha}|u(t)| \mathrm{d} t<\infty,
there exists the limit lim_(alpha rarr oo)varphi_(alpha)(f)\lim _{\alpha \rightarrow \infty} \varphi_{\alpha}(f) and
{:(2.1)lim_(alpha rarr oo)varphi_(alpha)(f)=(lim_(alpha rarr oo)(1)/(alpha)int_(0)^(alpha)u(t)dt)*int_(0)^(1)f:}\begin{equation*}
\lim _{\alpha \rightarrow \infty} \varphi_{\alpha}(f)=\left(\lim _{\alpha \rightarrow \infty} \frac{1}{\alpha} \int_{0}^{\alpha} u(t) \mathrm{d} t\right) \cdot \int_{0}^{1} f \tag{2.1}
\end{equation*}
The conditions I and II are also necessary in order to have (2.1) for each f inC(I)f \in \mathcal{C}(I).
Proof. Let us suppose that conditions I and II hold. We can prove (2.1) even for the more general case of a regular function ff. To this end it suffices to prove (2.1) for f in Ef \in E with E subeR(I)E \subseteq \mathcal{R}(I) and bar(sp)E=R(I)\overline{\mathrm{sp}} E=\mathcal{R}(I) (condition II allows then to obtain (2.1) for any f inR(I)f \in \mathcal{R}(I) ). As a set EE we choose the set of characteristic functions chi_([a,b]),[a,b]sube I\chi_{[a, b]},[a, b] \subseteq I.
For f=chi_([a,b])f=\chi_{[a, b]} we have
{:[varphi_(alpha)(f)=int_(0)^(1)u(alpha t)f(t)dt=int_(a)^(b)u(alpha t)dt=(1)/(alpha)int_(alpha a)^(alpha b)u(t)dt],[=b(1)/(alpha b)int_(0)^(alpha b)u(t)dt-a(1)/(alpha a)int_(0)^(alpha a)u(t)dt]:}\begin{aligned}
\varphi_{\alpha}(f) & =\int_{0}^{1} u(\alpha t) f(t) \mathrm{d} t=\int_{a}^{b} u(\alpha t) \mathrm{d} t=\frac{1}{\alpha} \int_{\alpha a}^{\alpha b} u(t) \mathrm{d} t \\
& =b \frac{1}{\alpha b} \int_{0}^{\alpha b} u(t) \mathrm{d} t-a \frac{1}{\alpha a} \int_{0}^{\alpha a} u(t) \mathrm{d} t
\end{aligned}
It follows
{:[lim_(alpha rarr oo)varphi_(alpha)(f)=(b-a)lim_(alpha rarr oo)(1)/(alpha)int_(0)^(alpha)u(t)dt],[=(lim_(alpha rarr oo)(1)/(alpha)int_(0)^(alpha)u(t)dt)*int_(0)^(1)f]:}\begin{aligned}
\lim _{\alpha \rightarrow \infty} \varphi_{\alpha}(f) & =(b-a) \lim _{\alpha \rightarrow \infty} \frac{1}{\alpha} \int_{0}^{\alpha} u(t) \mathrm{d} t \\
& =\left(\lim _{\alpha \rightarrow \infty} \frac{1}{\alpha} \int_{0}^{\alpha} u(t) \mathrm{d} t\right) \cdot \int_{0}^{1} f
\end{aligned}
Let us suppose now that (2.1) takes place for each f inC(I)f \in \mathcal{C}(I). For f(x)=1,AA x in If(x)=1, \forall x \in I we obtain condition I. To prove II, we apply the Banach-Steinhaus principle for sequences alpha_(n)rarr oo\alpha_{n} \rightarrow \infty, because it cannot be used directly for generalized sequences. (In fact, if XX is an infinite dimensional Banach space, then its dual X^(**)X^{*} is sequentially closed and dense in (X^(#):}\left(X^{\#}\right., weak {:^(**))\left.^{*}\right), see [6, p. 138].) It follows that s u p||varphi_(alpha)|| < oo\sup \left\|\varphi_{\alpha}\right\|<\infty, and using the expression of ||varphi_(alpha)||\left\|\varphi_{\alpha}\right\| given in (1.4) we obtain II.
Remark 2.1 If the function uu is periodic with u(t+T)=u(t)u(t+T)=u(t) for each t > 0t>0, then
lim_(alpha rarr oo)(1)/(alpha)int_(0)^(alpha)u(t)dt=(1)/(T)int_(0)^(T)u(t)dt\lim _{\alpha \rightarrow \infty} \frac{1}{\alpha} \int_{0}^{\alpha} u(t) \mathrm{d} t=\frac{1}{T} \int_{0}^{T} u(t) \mathrm{d} t
and condition II automatically holds because of the boundedness of uu. In
this special case we obtain
this special case we obtain
{:(2.2)lim_(alpha rarr oo)varphi_(alpha)(f)=(1)/(T)int_(0)^(T)u(t)dt*int_(0)^(1)f:}\begin{equation*}
\lim _{\alpha \rightarrow \infty} \varphi_{\alpha}(f)=\frac{1}{T} \int_{0}^{T} u(t) \mathrm{d} t \cdot \int_{0}^{1} f \tag{2.2}
\end{equation*}
which is a result due to L. Fejér, see [4, p. 114].
Remark 2.2 Condition II does not follow from condition I. Indeed, let u inC([0,oo))u \in \mathcal{C}([0, \infty)) be given by
u(x)={[2sqrt(n+1)(x-2n)","x in[2n","2n+1//2)],[-2sqrt(n+1)(x-2n-1)","x in[2n+1//2","2n+3//2)],[2sqrt(n+1)(x-2n-2)","x in[2n+3//2","2n+2)]:}u(x)=\left\{\begin{array}{l}
2 \sqrt{n+1}(x-2 n), x \in[2 n, 2 n+1 / 2) \\
-2 \sqrt{n+1}(x-2 n-1), x \in[2 n+1 / 2,2 n+3 / 2) \\
2 \sqrt{n+1}(x-2 n-2), x \in[2 n+3 / 2,2 n+2)
\end{array}\right.
(n inN)(n \in \mathbb{N}). For this function we have
s u p_(alpha > 0)(1)/(alpha)int_(0)^(alpha)|u(t)|dt >= s u p_(n inN^(**))(1)/(2n)int_(0)^(2n)|u(t)|dt=s u p_(n inN^(**))(1)/(2n)sum_(k=1)^(n)sqrtk=oo\sup _{\alpha>0} \frac{1}{\alpha} \int_{0}^{\alpha}|u(t)| \mathrm{d} t \geq \sup _{n \in \mathbb{N}^{*}} \frac{1}{2 n} \int_{0}^{2 n}|u(t)| \mathrm{d} t=\sup _{n \in \mathbb{N}^{*}} \frac{1}{2 n} \sum_{k=1}^{n} \sqrt{k}=\infty
For 2n <= alpha < 2n+22 n \leq \alpha<2 n+2 it follows (1)/(alpha)int_(0)^(alpha)u(t)dt=(1)/(alpha)int_(2n)^(alpha)u(t)dt <= (sqrt(n+1))/(2alpha) <= (sqrt(n+1))/(4n)\frac{1}{\alpha} \int_{0}^{\alpha} u(t) \mathrm{d} t=\frac{1}{\alpha} \int_{2 n}^{\alpha} u(t) \mathrm{d} t \leq \frac{\sqrt{n+1}}{2 \alpha} \leq \frac{\sqrt{n+1}}{4 n} and lim_(alpha rarr oo)(1)/(alpha)int_(0)^(alpha)u(t)dt=0\lim _{\alpha \rightarrow \infty} \frac{1}{\alpha} \int_{0}^{\alpha} u(t) \mathrm{d} t=0.
If we consider only functions in C^(1)(I)\mathcal{C}^{1}(I), we obtain the corresponding property as in DD mentioned in the introduction for sequences of operators.
Theorem 2.2 Let there be given a function f inC^(1)(I)f \in \mathcal{C}^{1}(I) and u inC([0,oo))u \in \mathcal{C}([0, \infty)). If the function uu satisfies condition I from Theorem 2.1 (EElim_(alpha rarr oo)(1)/(alpha)int_(0)^(alpha)u(t)dt=a)\left(\exists \lim _{\alpha \rightarrow \infty} \frac{1}{\alpha} \int_{0}^{\alpha} u(t) \mathrm{d} t=a\right), then there exists lim_(alpha rarr oo)varphi_(alpha)(f)\lim _{\alpha \rightarrow \infty} \varphi_{\alpha}(f) and its value is given by (2.1).
Proof. Let UU be an antiderivative of uu with U(0)=0U(0)=0 (i.e. U(x)={:int_(int)^(x)u(t)dt)U(x)= \left.\int_{\int}^{x} u(t) \mathrm{d} t\right). We have for f inC^(1)(I)f \in \mathcal{C}^{1}(I)
0
0
{:[varphi_(alpha)(f)=(1)/(alpha)int_(0)^(alpha)U^(')(t)f((t)/( alpha))dt],[=(1)/(alpha)U(alpha)f(1)-(1)/(alpha^(2))int_(0)^(alpha)U(t)f^(')((t)/( alpha))dt],[=(1)/(alpha)U(alpha)f(1)-(1)/(alpha)int_(0)^(1)U(alpha t)f^(')(t)dt]:}\begin{aligned}
\varphi_{\alpha}(f) & =\frac{1}{\alpha} \int_{0}^{\alpha} U^{\prime}(t) f\left(\frac{t}{\alpha}\right) \mathrm{d} t \\
& =\frac{1}{\alpha} U(\alpha) f(1)-\frac{1}{\alpha^{2}} \int_{0}^{\alpha} U(t) f^{\prime}\left(\frac{t}{\alpha}\right) \mathrm{d} t \\
& =\frac{1}{\alpha} U(\alpha) f(1)-\frac{1}{\alpha} \int_{0}^{1} U(\alpha t) f^{\prime}(t) \mathrm{d} t
\end{aligned}
But (1)/(alpha)U(alpha t)=(1)/(alpha)int_(0)^(alpha t)u(s)ds\frac{1}{\alpha} U(\alpha t)=\frac{1}{\alpha} \int_{0}^{\alpha t} u(s) \mathrm{d} s and we get lim_(alpha rarr oo)(1)/(alpha)U(alpha t)=ta\lim _{\alpha \rightarrow \infty} \frac{1}{\alpha} U(\alpha t)=t a. Using the theorem of dominated convergence we get
lim_(alpha rarr oo)varphi_(alpha)(f)=af(1)-aint_(0)^(1)tf^(')(t)dt=a*int_(0)^(1)f.\lim _{\alpha \rightarrow \infty} \varphi_{\alpha}(f)=a f(1)-a \int_{0}^{1} t f^{\prime}(t) \mathrm{d} t=a \cdot \int_{0}^{1} f .
3 A new result for sequences of operators
In connection with the result mentioned in BB in the introduction for the sequence of operators varphi_(n):R(I,F)rarr F,n inN\varphi_{n}: \mathcal{R}(I, F) \rightarrow F, n \in \mathbb{N}, given by (1.1), an open question was formulated in [1]: Is the conclusion in BB true if F=RF=\mathbb{R} for ff Riemann integrable (instead of regular)?
We shall prove that this result holds if the sequence (a_(n))_(n inN)\left(a_{n}\right)_{n \in \mathbb{N}} is bounded from above or below, or if ((|a_(1)|+cdots+|a_(n)|)/(n))_(n inN)\left(\frac{\left|a_{1}\right|+\cdots+\left|a_{n}\right|}{n}\right)_{n \in \mathbb{N}} is convergent.
Theorem 3.1 Let there be given a Riemann integrable function f:I rarrRf: I \rightarrow \mathbb{R} and a sequence (a_(n))_(n inN)\left(a_{n}\right)_{n \in \mathbb{N}} of real numbers satisfying the conditions:
- lim_(n rarr oo)(a_(1)+dots+a_(n))/(n)=a\lim _{n \rightarrow \infty} \frac{a_{1}+\ldots+a_{n}}{n}=a;
- the sequence ((|a_(1)|+dots+|a_(n)|)/(n))_(n inN)\left(\frac{\left|a_{1}\right|+\ldots+\left|a_{n}\right|}{n}\right)_{n \in \mathbb{N}} is bounded;
- the sequence (a_(n))_(n inN)\left(a_{n}\right)_{n \in \mathbb{N}} is bounded from above or from below, or ((|a_(1)|+dots+|a_(n)|)/(n))_(n inN)\left(\frac{\left|a_{1}\right|+\ldots+\left|a_{n}\right|}{n}\right)_{n \in \mathbb{N}} is convergent.
Then the sequence (varphi_(n)(f))_(n inN)\left(\varphi_{n}(f)\right)_{n \in \mathbb{N}} given by (1.1) is convergent to a*int_(0)^(1)fa \cdot \int_{0}^{1} f.
Proof. (i) Let us consider a sequence (a_(n))_(n inN)\left(a_{n}\right)_{n \in \mathbb{N}} with a_(n) >= 0a_{n} \geq 0, which satisfies condition 1 (hence also condition 2). Given epsi > 0\varepsilon>0, for the Riemann integrable function ff there exist the continuous functions u,vu, v : I rarrRI \rightarrow \mathbb{R} such that
{:(3.1)u <= f <= v" and "int(v-u) < epsi.:}\begin{equation*}
u \leq f \leq v \text { and } \int(v-u)<\varepsilon . \tag{3.1}
\end{equation*}
Then the functionals varphi_(n)\varphi_{n} given by (1.1) will satisfy
{:(3.2)varphi_(n)(u) <= varphi_(n)(f) <= varphi_(n)(v):}\begin{equation*}
\varphi_{n}(u) \leq \varphi_{n}(f) \leq \varphi_{n}(v) \tag{3.2}
\end{equation*}
From the result in [1] mentioned at BB in the introduction we have lim_(n rarr oo)varphi_(n)(v)=a*int_(0)^(1)v\lim _{n \rightarrow \infty} \varphi_{n}(v)=a \cdot \int_{0}^{1} v, hence there exists n_(1)inNn_{1} \in \mathbb{N} so that for any n >= n_(1)n \geq n_{1}, varphi_(n)(v) < a*int_(0)^(1)v+epsi\varphi_{n}(v)<a \cdot \int_{0}^{1} v+\varepsilon. Condition (3.1) implies that int_(0)^(1)v < int_(0)^(1)f+epsi\int_{0}^{1} v<\int_{0}^{1} f+\varepsilon, hence
{:(3.3)varphi_(n)(v) < a*int_(0)^(1)f+epsi(a+1):}\begin{equation*}
\varphi_{n}(v)<a \cdot \int_{0}^{1} f+\varepsilon(a+1) \tag{3.3}
\end{equation*}
Similarly, there exists n_(2)inNn_{2} \in \mathbb{N} so that for any n >= n_(2)n \geq n_{2},
{:(3.4)a*int_(0)^(1)f-epsi(a+1) < varphi_(n)(u):}\begin{equation*}
a \cdot \int_{0}^{1} f-\varepsilon(a+1)<\varphi_{n}(u) \tag{3.4}
\end{equation*}
From (3.2), (3.3) and (3.4) we obtain
a*int_(0)^(1)f-epsi(a+1) < varphi_(n)(f) < a*int_(0)^(1)f+epsi(a+1)a \cdot \int_{0}^{1} f-\varepsilon(a+1)<\varphi_{n}(f)<a \cdot \int_{0}^{1} f+\varepsilon(a+1)
hence |varphi_(n)(f)-a*int_(0)^(1)f| <= epsi(a+1)\left|\varphi_{n}(f)-a \cdot \int_{0}^{1} f\right| \leq \varepsilon(a+1) for n >= max{n_(1),n_(2)}n \geq \max \left\{n_{1}, n_{2}\right\} and the conclusion holds.
(ii) Let us consider (a_(n))_(n inN)\left(a_{n}\right)_{n \in \mathbb{N}} which satisfies the conditions 1 and 2 and is bounded from below, i. e. a_(n) >= -alphaa_{n} \geq-\alpha. The sequence (b_(n))_(n inN)\left(b_{n}\right)_{n \in \mathbb{N}}, b_(n)=a_(n)+alphab_{n}=a_{n}+\alpha is Cesàro-convergent to a+alphaa+\alpha and has b_(n) >= 0b_{n} \geq 0; applying the result proved in (i) it follows that
(ii) Let us consider (a_(n))_(n inN)\left(a_{n}\right)_{n \in \mathbb{N}} which satisfies the conditions 1 and 2 and is bounded from below, i. e. a_(n) >= -alphaa_{n} \geq-\alpha. The sequence (b_(n))_(n inN)\left(b_{n}\right)_{n \in \mathbb{N}}, b_(n)=a_(n)+alphab_{n}=a_{n}+\alpha is Cesàro-convergent to a+alphaa+\alpha and has b_(n) >= 0b_{n} \geq 0; applying the result proved in (i) it follows that
lim_(n rarr oo)(1)/(n)sum_(k=1)^(n)(a_(n)+alpha)f((k)/(n))=(a+alpha)*int_(0)^(1)f\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n}\left(a_{n}+\alpha\right) f\left(\frac{k}{n}\right)=(a+\alpha) \cdot \int_{0}^{1} f
hence lim_(n rarr oo)varphi_(n)(f)=a*int_(0)^(1)f\lim _{n \rightarrow \infty} \varphi_{n}(f)=a \cdot \int_{0}^{1} f.
(iii) If the sequence (a_(n))_(n inN)\left(a_{n}\right)_{n \in \mathbb{N}} satisfies 1 and 2 , and is bounded from above ( a_(n) <= alphaa_{n} \leq \alpha ), the sequence (c_(n))_(n inN),c_(n)=alpha-a_(n)\left(c_{n}\right)_{n \in \mathbb{N}}, c_{n}=\alpha-a_{n} is Cesàro-convergent to alpha-a\alpha-a and has c_(n) >= 0c_{n} \geq 0. Applying again the result proved in (i) we obtain the conclusion.
(iv) Let us consider the sequence (a_(n))_(n inN)\left(a_{n}\right)_{n \in \mathbb{N}} with lim_(n rarr oo)(a_(1)+dots+a_(n))/(n)=a\lim _{n \rightarrow \infty} \frac{a_{1}+\ldots+a_{n}}{n}=a and lim_(n rarr oo)(|a_(1)|+dots+|a_(n)|)/(n)=a^(**)\lim _{n \rightarrow \infty} \frac{\left|a_{1}\right|+\ldots+\left|a_{n}\right|}{n}=a^{*}. We can write
(iii) If the sequence (a_(n))_(n inN)\left(a_{n}\right)_{n \in \mathbb{N}} satisfies 1 and 2 , and is bounded from above ( a_(n) <= alphaa_{n} \leq \alpha ), the sequence (c_(n))_(n inN),c_(n)=alpha-a_(n)\left(c_{n}\right)_{n \in \mathbb{N}}, c_{n}=\alpha-a_{n} is Cesàro-convergent to alpha-a\alpha-a and has c_(n) >= 0c_{n} \geq 0. Applying again the result proved in (i) we obtain the conclusion.
(iv) Let us consider the sequence (a_(n))_(n inN)\left(a_{n}\right)_{n \in \mathbb{N}} with lim_(n rarr oo)(a_(1)+dots+a_(n))/(n)=a\lim _{n \rightarrow \infty} \frac{a_{1}+\ldots+a_{n}}{n}=a and lim_(n rarr oo)(|a_(1)|+dots+|a_(n)|)/(n)=a^(**)\lim _{n \rightarrow \infty} \frac{\left|a_{1}\right|+\ldots+\left|a_{n}\right|}{n}=a^{*}. We can write
a_(n)=(a_(n)+|a_(n)|)/(2)-(|a_(n)|-a_(n))/(2).a_{n}=\frac{a_{n}+\left|a_{n}\right|}{2}-\frac{\left|a_{n}\right|-a_{n}}{2} .
The sequence ((a_(n)+|a_(n)|)/(2))_(n inN)\left(\frac{a_{n}+\left|a_{n}\right|}{2}\right)_{n \in \mathbb{N}} is Cesàro-convergent to (a+a^(**))/(2)\frac{a+a^{*}}{2} and has nonnegative terms, hence it follows from (i) that
lim_(n rarr oo)(1)/(n)sum_(k=1)^(n)((a_(n)+|a_(n)|)/(2))f((k)/(n))=(a+a^(**))/(2)*int_(0)^(1)f\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n}\left(\frac{a_{n}+\left|a_{n}\right|}{2}\right) f\left(\frac{k}{n}\right)=\frac{a+a^{*}}{2} \cdot \int_{0}^{1} f
Similarly, for ((|a_(n)|-a_(n))/(2))_(n inN)\left(\frac{\left|a_{n}\right|-a_{n}}{2}\right)_{n \in \mathbb{N}} we have
lim_(n rarr oo)(1)/(n)sum_(k=1)^(n)((|a_(n)|-a_(n))/(2))f((k)/(n))=(a^(**)-a)/(2)*int_(0)^(1)f.\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n}\left(\frac{\left|a_{n}\right|-a_{n}}{2}\right) f\left(\frac{k}{n}\right)=\frac{a^{*}-a}{2} \cdot \int_{0}^{1} f .
It follows
lim_(n rarr oo)(1)/(n)sum_(k=1)^(n)a_(n)f((k)/(n))=(a+a^(**))/(2)*int_(0)^(1)f-(a^(**)-a)/(2)*int_(0)^(1)f=a*int_(0)^(1)f.\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n} a_{n} f\left(\frac{k}{n}\right)=\frac{a+a^{*}}{2} \cdot \int_{0}^{1} f-\frac{a^{*}-a}{2} \cdot \int_{0}^{1} f=a \cdot \int_{0}^{1} f .
Remark 3.1 The conditions in 3 of Theorem 3.1 are not consequences of 1 and 2, as the following examples show.
Example 3.1 AA Cesàro-convergent sequence (a_(n))_(n inN)\left(a_{n}\right)_{n \in \mathbb{N}} which is not bounded from above or from below, for which ((|a_(1)|+dots+|a_(n)|)/(n))_(n inN)\left(\frac{\left|a_{1}\right|+\ldots+\left|a_{n}\right|}{n}\right)_{n \in \mathbb{N}} is bounded, is given by
a_(n)={[k","" for "n=2^(k)","k" even "],[-k","" for "n=2^(k)","k" odd "],[0","" otherwise "]:}a_{n}=\left\{\begin{array}{l}
k, \text { for } n=2^{k}, k \text { even } \\
-k, \text { for } n=2^{k}, k \text { odd } \\
0, \text { otherwise }
\end{array}\right.
Example 3.2 AA Cesàro-convergent sequence (a_(n))_(n inN)\left(a_{n}\right)_{n \in \mathbb{N}}, for which ((|a_(1)|+dots+|a_(n)|)/(n))_(n inN)\left(\frac{\left|a_{1}\right|+\ldots+\left|a_{n}\right|}{n}\right)_{n \in \mathbb{N}} is bounded without being convergent, can be obtained from a bounded sequence d_(n) >= 0d_{n} \geq 0 for which (d_(1)+dots+d_(n))/(n)\frac{d_{1}+\ldots+d_{n}}{n} does not converge, as
a_(n)={[d_(n//2)","" for "n" even "],[-d_((n+1)//2)","" for "n" odd "]:}a_{n}=\left\{\begin{array}{l}
d_{n / 2}, \text { for } n \text { even } \\
-d_{(n+1) / 2}, \text { for } n \text { odd }
\end{array}\right.
Such a sequence (d_(n))_(n inN)\left(d_{n}\right)_{n \in \mathbb{N}} is, for example,
d_(n)={[1","" for "n in[2^(k),2^(k+1))","k" even "],[0","" for "n in[2^(k),2^(k+1))","k" odd "]:}d_{n}=\left\{\begin{array}{l}
1, \text { for } n \in\left[2^{k}, 2^{k+1}\right), k \text { even } \\
0, \text { for } n \in\left[2^{k}, 2^{k+1}\right), k \text { odd }
\end{array}\right.
Indeed, we have
lim_(k rarr oo,k" even ")(d_(1)+dots+d_(2^(k+1)-1))/(2^(k+1)-1)=lim_(k rarr oo)(2^(0)+2^(2)+dots+2^(k))/(2^(k+1)-1)=(2)/(3),\lim _{k \rightarrow \infty, k \text { even }} \frac{d_{1}+\ldots+d_{2^{k+1}-1}}{2^{k+1}-1}=\lim _{k \rightarrow \infty} \frac{2^{0}+2^{2}+\ldots+2^{k}}{2^{k+1}-1}=\frac{2}{3},
and
lim_(k rarr oo,k" odd ")(d_(1)+dots+d_(2^(k+1)-1))/(2^(k+1)-1)=lim_(k rarr oo)(2^(0)+2^(2)+dots+2^(k-1))/(2^(k+1)-1)=(1)/(3).\lim _{k \rightarrow \infty, k \text { odd }} \frac{d_{1}+\ldots+d_{2^{k+1}-1}}{2^{k+1}-1}=\lim _{k \rightarrow \infty} \frac{2^{0}+2^{2}+\ldots+2^{k-1}}{2^{k+1}-1}=\frac{1}{3} .
Remark 3.2 We mention, in connection with Theorem 3.1, the following result from [2] (see also [5]):
For k > 0k>0, if f,g:[0,oo)rarrRf, g:[0, \infty) \rightarrow \mathbb{R} satisfy:
-f inC^(1)([0,oo))-f \in \mathcal{C}^{1}([0, \infty)) is decreasing with f(oo)=0f(\infty)=0 and there exists lim_(alpha rarr oo)int_(0)^(alpha)f(x+k alpha)dx=a\lim _{\alpha \rightarrow \infty} \int_{0}^{\alpha} f(x+k \alpha) \mathrm{d} x=a;
-g inC([0,oo))-g \in \mathcal{C}([0, \infty)) is bounded and EElim_(alpha rarr oo)(1)/(alpha)int_(0)^(alpha)g=b\exists \lim _{\alpha \rightarrow \infty} \frac{1}{\alpha} \int_{0}^{\alpha} g=b,
then lim_(alpha rarr oo)int_(0)^(alpha)f(x+k alpha)g(x)dx=ab\lim _{\alpha \rightarrow \infty} \int_{0}^{\alpha} f(x+k \alpha) g(x) \mathrm{d} x=a b.
-f inC^(1)([0,oo))-f \in \mathcal{C}^{1}([0, \infty)) is decreasing with f(oo)=0f(\infty)=0 and there exists lim_(alpha rarr oo)int_(0)^(alpha)f(x+k alpha)dx=a\lim _{\alpha \rightarrow \infty} \int_{0}^{\alpha} f(x+k \alpha) \mathrm{d} x=a;
-g inC([0,oo))-g \in \mathcal{C}([0, \infty)) is bounded and EElim_(alpha rarr oo)(1)/(alpha)int_(0)^(alpha)g=b\exists \lim _{\alpha \rightarrow \infty} \frac{1}{\alpha} \int_{0}^{\alpha} g=b,
then lim_(alpha rarr oo)int_(0)^(alpha)f(x+k alpha)g(x)dx=ab\lim _{\alpha \rightarrow \infty} \int_{0}^{\alpha} f(x+k \alpha) g(x) \mathrm{d} x=a b.
References
[1] M.-C. Anisiu, V. Anisiu. Sequences of linear operators related to Cesàro-convergent sequences. Revue d'Analyse Num. Th. Approx. 31 (2) (2002), 139-145.
[2] R. Gologan. On the convergence of some partial sums and a contest problem. Gazeta Matematică seria A, Anul XIX nr. 2 (2001), 70-73.
[3] I. J Maddox, The norm of a linear functional. Amer. Math. Monthly 96 (1989), 434-436.
[4] G. Szász, L. Gehér, I. Kovács, L. Pintér (editors). Contests in Higher Mathematics. Akadémiai Kiadó, Budapest 1968.
[5] E. C. Titchmarch. The Theory of Functions. Oxford University Press 1939.
[2] R. Gologan. On the convergence of some partial sums and a contest problem. Gazeta Matematică seria A, Anul XIX nr. 2 (2001), 70-73.
[3] I. J Maddox, The norm of a linear functional. Amer. Math. Monthly 96 (1989), 434-436.
[4] G. Szász, L. Gehér, I. Kovács, L. Pintér (editors). Contests in Higher Mathematics. Akadémiai Kiadó, Budapest 1968.
[5] E. C. Titchmarch. The Theory of Functions. Oxford University Press 1939.