On the pointwise convergence of a family of functionals on C(I)

Mira Anisiu
(original), paper

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

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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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[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.

2003-Anisiu-Anisiu-OnthePointwise

On the pointwise convergence of a family of functionals on C ( I ) C ( I ) C(I)\mathcal{C}(I)C(I)

Mira-Cristiana Anisiu Valeriu Anisiu(Cluj-Napoca) (Cluj-Napoca)

Abstract. Given a continuous function u : [ 0 , ∞ ) → R u : [ 0 , ∞ ) → R u:[0,oo)rarrRu:[0, \infty) \rightarrow \mathbb{R}u:[0,∞)→R, a family of functionals φ α : C ( I ) → R , α > 0 φ α : C ( I ) → R , α > 0 varphi_(alpha):C(I)rarrR,alpha > 0\varphi_{\alpha}: \mathcal{C}(I) \rightarrow \mathbb{R}, \alpha>0φα:C(I)→R,α>0, is defined by φ α ( f ) = 1 α ∫ 0 α u ( t ) f ( t / α ) d t φ α ( f ) = 1 α ∫ 0 α   u ( t ) f ( t / α ) d t 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} tφα(f)=1α∫0αu(t)f(t/α)dt. It is proved that the necessary and sufficient conditions for the family φ α , α > 0 φ α , α > 0 varphi_(alpha),alpha > 0\varphi_{\alpha}, \alpha>0φα,α>0 to satisfy lim α → ∞ φ α ( f ) = ( lim α → ∞ 1 α ∫ 0 α u ( t ) d t ) ⋅ ∫ 0 1 f lim α → ∞   φ α ( f ) = lim α → ∞   1 α ∫ 0 α   u ( t ) d t ⋅ ∫ 0 1   f 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} flimα→∞φα(f)=(limα→∞1α∫0αu(t)dt)⋅∫01f are:
I. ∃ lim α → ∞ 1 α ∫ 0 α u ( t ) d t ; ∃ lim α → ∞   1 α ∫ 0 α   u ( t ) d t ; 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∃limα→∞1α∫0αu(t)dt; II. sup α > 0 1 α ∫ 0 α | u ( t ) | d t < ∞ sup α > 0   1 α ∫ 0 α   | u ( t ) | d t < ∞ 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<\inftysupα>01α∫0α|u(t)|dt<∞.
If f ∈ C 1 ( I ) f ∈ C 1 ( I ) f inC^(1)(I)f \in \mathcal{C}^{1}(I)f∈C1(I), condition I alone implies the existence of lim α → ∞ φ α ( f ) lim α → ∞   φ α ( f ) lim_(alpha rarr oo)varphi_(alpha)(f)\lim _{\alpha \rightarrow \infty} \varphi_{\alpha}(f)limα→∞φα(f).
A sequence of functionals ( φ n ) n ∈ N φ n n ∈ N (varphi_(n))_(n inN)\left(\varphi_{n}\right)_{n \in \mathbb{N}}(φn)n∈N is attached to a numerical sequence ( a n ) n ∈ N a n n ∈ N (a_(n))_(n inN)\left(a_{n}\right)_{n \in \mathbb{N}}(an)n∈N which is Cesàro-convergent to a a aaa, namely
φ n ( f ) = 1 n ∑ k = 1 n a k f ( k / n ) , f Riemann integrable. φ n ( f ) = 1 n ∑ k = 1 n   a k f ( k / n ) , f  Riemann integrable.  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. }φn(f)=1n∑k=1nakf(k/n),f Riemann integrable. 
Additional conditions are imposed on the sequence ( a n ) n ∈ N a n n ∈ N (a_(n))_(n inN)\left(a_{n}\right)_{n \in \mathbb{N}}(an)n∈N in order to prove that
lim n → ∞ φ n ( f ) = a ⋅ ∫ 0 1 f lim n → ∞   φ n ( f ) = a ⋅ ∫ 0 1   f lim_(n rarr oo)varphi_(n)(f)=a*int_(0)^(1)f\lim _{n \rightarrow \infty} \varphi_{n}(f)=a \cdot \int_{0}^{1} flimn→∞φn(f)=a⋅∫01f
MSC 2000: 47B38, 26E60

1 Introduction

For the interval I = [ 0 , 1 ] I = [ 0 , 1 ] I=[0,1]I=[0,1]I=[0,1] and a Banach space F ≠ { 0 } F ≠ { 0 } F!={0}F \neq\{0\}F≠{0}, let us denote by B ( I , F ) B ( I , F ) B(I,F)\mathcal{B}(I, F)B(I,F) the Banach space of bounded functions f : I → F f : I → F f:I rarr Ff: I \rightarrow Ff:I→F endowed with the sup norm. The subspace of B ( I , F ) B ( I , F ) B(I,F)\mathcal{B}(I, F)B(I,F) of regular functions (which admit side limits at each t ∈ I t ∈ I t in It \in It∈I ) will be denoted by R ( I , F ) R ( I , F ) R(I,F)\mathcal{R}(I, F)R(I,F); the Banach space of continuous functions C ( I , F ) C ( I , F ) C(I,F)\mathcal{C}(I, F)C(I,F) is a subspace of R ( I , F ) R ( I , F ) R(I,F)\mathcal{R}(I, F)R(I,F).
Given a sequence of real numbers ( a n ) n ∈ N a n n ∈ N (a_(n))_(n inN)\left(a_{n}\right)_{n \in \mathbb{N}}(an)n∈N, a sequence of operators φ n : R ( I , F ) → F , n ∈ N φ n : R ( I , F ) → F , n ∈ N varphi_(n):R(I,F)rarr F,n inN\varphi_{n}: \mathcal{R}(I, F) \rightarrow F, n \in \mathbb{N}φn:R(I,F)→F,n∈N,
(1.1) φ n ( f ) = 1 n ∑ k = 1 n a k f ( k n ) (1.1) φ n ( f ) = 1 n ∑ k = 1 n   a k f k 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*}(1.1)φn(f)=1n∑k=1nakf(kn)
can be generated. It was proved in [1] that:
A A AAA. The operators φ n φ n varphi_(n)\varphi_{n}φn are linear and continuous.
B B BBB. If the numeric sequence satisfies the conditions:
B 1 ⋅ ( a n ) n ∈ N B 1 ⋅ a n n ∈ N B_(1)*(a_(n))_(n inN)B_{1} \cdot\left(a_{n}\right)_{n \in \mathbb{N}}B1⋅(an)n∈N is Cesàro-convergent to a ( lim n → ∞ a 1 + … + a n n = a ) a lim n → ∞   a 1 + … + a n n = a 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)a(limn→∞a1+…+ann=a);
B 2 B 2 B_(2)B_{2}B2. the sequence ( | a 1 | + … + | a n | n ) n ∈ N a 1 + … + a n n n ∈ N ((|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}}(|a1|+…+|an|n)n∈N is bounded, then the sequence ( φ n ( f ) ) n ∈ N φ n ( f ) n ∈ N (varphi_(n)(f))_(n inN)\left(\varphi_{n}(f)\right)_{n \in \mathbb{N}}(φn(f))n∈N is convergent and
(1.2) lim n → ∞ φ n ( f ) = a ⋅ ∫ 0 1 f (1.2) lim n → ∞   φ n ( f ) = a ⋅ ∫ 0 1   f {:(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*}(1.2)limn→∞φn(f)=a⋅∫01f
C C CCC. If lim n → ∞ φ n ( f ) lim n → ∞   φ n ( f ) lim_(n rarr oo)varphi_(n)(f)\lim _{n \rightarrow \infty} \varphi_{n}(f)limn→∞φn(f) exists for every f ∈ C ( I , F ) ⊆ R ( I , F ) f ∈ C ( I , F ) ⊆ R ( I , F ) f inC(I,F)subeR(I,F)f \in \mathcal{C}(I, F) \subseteq \mathcal{R}(I, F)f∈C(I,F)⊆R(I,F), the conditions B 1 B 1 B_(1)B_{1}B1 and B 2 B 2 B_(2)B_{2}B2 from above are also necessary.
D D DDD. If f ∈ C 1 ( I , F ) f ∈ C 1 ( I , F ) f inC^(1)(I,F)f \in \mathcal{C}^{1}(I, F)f∈C1(I,F) (i.e. f f fff is continuous with a continuous derivative), the result in B B BBB holds even if condition B 2 B 2 B_(2)B_{2}B2 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 = R F = R F=RF=\mathbb{R}F=R ).
Let u : [ 0 , ∞ ) → R u : [ 0 , ∞ ) → R u:[0,oo)rarrRu:[0, \infty) \rightarrow \mathbb{R}u:[0,∞)→R be a continuous function. We define a family of
functionals associated to u u uuu, namely φ α : C ( I ) → R , α > 0 φ α : C ( I ) → R , α > 0 varphi_(alpha):C(I)rarrR,alpha > 0\varphi_{\alpha}: \mathcal{C}(I) \rightarrow \mathbb{R}, \alpha>0φα:C(I)→R,α>0, given by
(1.3) φ α ( f ) = 1 α ∫ 0 α u ( t ) f ( t α ) d t , f ∈ C ( I ) . (1.3) φ α ( f ) = 1 α ∫ 0 α   u ( t ) f t α d t , f ∈ C ( I ) . {:(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*}(1.3)φα(f)=1α∫0αu(t)f(tα)dt,f∈C(I).
Proposition 1.1 For each α > 0 α > 0 alpha > 0\alpha>0α>0, the functional φ α φ α varphi_(alpha)\varphi_{\alpha}φα is linear and continuous, and its norm is given by
(1.4) ‖ φ α ‖ = 1 α ∫ 0 α | u ( t ) | d t (1.4) φ α = 1 α ∫ 0 α   | u ( t ) | d t {:(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*}(1.4)‖φα‖=1α∫0α|u(t)|dt
Proof. This result is classical; see for a simple proof [3]. It also holds if C ( I ) C ( I ) C(I)\mathcal{C}(I)C(I) is replaced with the Banach space R ( I ) R ( I ) R(I)\mathcal{R}(I)R(I) of regular functions.

2 Main results for families of functionals

As mentioned in B B BBB and C C CCC in the introduction, in the discrete case the conditions B 1 B 1 B_(1)B_{1}B1 and B 2 B 2 B_(2)B_{2}B2 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 ∈ C ( I ) f ∈ C ( I ) f inC(I)f \in \mathcal{C}(I)f∈C(I) and u ∈ C ( [ 0 , ∞ ) ) u ∈ C ( [ 0 , ∞ ) ) u inC([0,oo))u \in \mathcal{C}([0, \infty))u∈C([0,∞)). If the function u u uuu satisfies the conditions:
I. ∃ lim α → ∞ 1 α ∫ 0 α u ( t ) d t ∃ lim α → ∞   1 α ∫ 0 α   u ( t ) d t 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∃limα→∞1α∫0αu(t)dt;
II. sup α > 0 1 α ∫ 0 α | u ( t ) | d t < ∞ sup α > 0   1 α ∫ 0 α   | u ( t ) | d t < ∞ 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<\inftysupα>01α∫0α|u(t)|dt<∞,
there exists the limit lim α → ∞ φ α ( f ) lim α → ∞   φ α ( f ) lim_(alpha rarr oo)varphi_(alpha)(f)\lim _{\alpha \rightarrow \infty} \varphi_{\alpha}(f)limα→∞φα(f) and
(2.1) lim α → ∞ φ α ( f ) = ( lim α → ∞ 1 α ∫ 0 α u ( t ) d t ) ⋅ ∫ 0 1 f (2.1) lim α → ∞   φ α ( f ) = lim α → ∞   1 α ∫ 0 α   u ( t ) d t ⋅ ∫ 0 1   f {:(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*}(2.1)limα→∞φα(f)=(limα→∞1α∫0αu(t)dt)⋅∫01f
The conditions I and II are also necessary in order to have (2.1) for each f ∈ C ( I ) f ∈ C ( I ) f inC(I)f \in \mathcal{C}(I)f∈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 f f fff. To this end it suffices to prove (2.1) for f ∈ E f ∈ E f in Ef \in Ef∈E with E ⊆ R ( I ) E ⊆ R ( I ) E subeR(I)E \subseteq \mathcal{R}(I)E⊆R(I) and sp ― E = R ( I ) sp ¯ E = R ( I ) bar(sp)E=R(I)\overline{\mathrm{sp}} E=\mathcal{R}(I)sp―E=R(I) (condition II allows then to obtain (2.1) for any f ∈ R ( I ) f ∈ R ( I ) f inR(I)f \in \mathcal{R}(I)f∈R(I) ). As a set E E EEE we choose the set of characteristic functions χ [ a , b ] , [ a , b ] ⊆ I χ [ a , b ] , [ a , b ] ⊆ I chi_([a,b]),[a,b]sube I\chi_{[a, b]},[a, b] \subseteq Iχ[a,b],[a,b]⊆I.
For f = χ [ a , b ] f = χ [ a , b ] f=chi_([a,b])f=\chi_{[a, b]}f=χ[a,b] we have
φ α ( f ) = ∫ 0 1 u ( α t ) f ( t ) d t = ∫ a b u ( α t ) d t = 1 α ∫ α a α b u ( t ) d t = b 1 α b ∫ 0 α b u ( t ) d t − a 1 α a ∫ 0 α a u ( t ) d t φ α ( f ) = ∫ 0 1   u ( α t ) f ( t ) d t = ∫ a b   u ( α t ) d t = 1 α ∫ α a α b   u ( t ) d t = b 1 α b ∫ 0 α b   u ( t ) d t − a 1 α a ∫ 0 α a   u ( t ) d t {:[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}φα(f)=∫01u(αt)f(t)dt=∫abu(αt)dt=1α∫αaαbu(t)dt=b1αb∫0αbu(t)dt−a1αa∫0αau(t)dt
It follows
lim α → ∞ φ α ( f ) = ( b − a ) lim α → ∞ 1 α ∫ 0 α u ( t ) d t = ( lim α → ∞ 1 α ∫ 0 α u ( t ) d t ) ⋅ ∫ 0 1 f lim α → ∞   φ α ( f ) = ( b − a ) lim α → ∞   1 α ∫ 0 α   u ( t ) d t = lim α → ∞   1 α ∫ 0 α   u ( t ) d t ⋅ ∫ 0 1   f {:[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}limα→∞φα(f)=(b−a)limα→∞1α∫0αu(t)dt=(limα→∞1α∫0αu(t)dt)⋅∫01f
Let us suppose now that (2.1) takes place for each f ∈ C ( I ) f ∈ C ( I ) f inC(I)f \in \mathcal{C}(I)f∈C(I). For f ( x ) = 1 , ∀ x ∈ I f ( x ) = 1 , ∀ x ∈ I f(x)=1,AA x in If(x)=1, \forall x \in If(x)=1,∀x∈I we obtain condition I. To prove II, we apply the Banach-Steinhaus principle for sequences α n → ∞ α n → ∞ alpha_(n)rarr oo\alpha_{n} \rightarrow \inftyαn→∞, because it cannot be used directly for generalized sequences. (In fact, if X X XXX is an infinite dimensional Banach space, then its dual X ∗ X ∗ X^(**)X^{*}X∗ is sequentially closed and dense in ( X # X # (X^(#):}\left(X^{\#}\right.(X#, weak ∗ ) ∗ {:^(**))\left.^{*}\right)∗), see [6, p. 138].) It follows that sup ‖ φ α ‖ < ∞ sup φ α < ∞ s u p||varphi_(alpha)|| < oo\sup \left\|\varphi_{\alpha}\right\|<\inftysup‖φα‖<∞, and using the expression of ‖ φ α ‖ φ α ||varphi_(alpha)||\left\|\varphi_{\alpha}\right\|‖φα‖ given in (1.4) we obtain II.
Remark 2.1 If the function u u uuu is periodic with u ( t + T ) = u ( t ) u ( t + T ) = u ( t ) u(t+T)=u(t)u(t+T)=u(t)u(t+T)=u(t) for each t > 0 t > 0 t > 0t>0t>0, then
lim α → ∞ 1 α ∫ 0 α u ( t ) d t = 1 T ∫ 0 T u ( t ) d t lim α → ∞   1 α ∫ 0 α   u ( t ) d t = 1 T ∫ 0 T   u ( t ) d t 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} tlimα→∞1α∫0αu(t)dt=1T∫0Tu(t)dt
and condition II automatically holds because of the boundedness of u u uuu. In
this special case we obtain
(2.2) lim α → ∞ φ α ( f ) = 1 T ∫ 0 T u ( t ) d t ⋅ ∫ 0 1 f (2.2) lim α → ∞   φ α ( f ) = 1 T ∫ 0 T   u ( t ) d t ⋅ ∫ 0 1   f {:(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*}(2.2)limα→∞φα(f)=1T∫0Tu(t)dt⋅∫01f
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 ∈ C ( [ 0 , ∞ ) ) u ∈ C ( [ 0 , ∞ ) ) u inC([0,oo))u \in \mathcal{C}([0, \infty))u∈C([0,∞)) be given by
u ( x ) = { 2 n + 1 ( x − 2 n ) , x ∈ [ 2 n , 2 n + 1 / 2 ) − 2 n + 1 ( x − 2 n − 1 ) , x ∈ [ 2 n + 1 / 2 , 2 n + 3 / 2 ) 2 n + 1 ( x − 2 n − 2 ) , x ∈ [ 2 n + 3 / 2 , 2 n + 2 ) u ( x ) = 2 n + 1 ( x − 2 n ) , x ∈ [ 2 n , 2 n + 1 / 2 ) − 2 n + 1 ( x − 2 n − 1 ) , x ∈ [ 2 n + 1 / 2 , 2 n + 3 / 2 ) 2 n + 1 ( x − 2 n − 2 ) , x ∈ [ 2 n + 3 / 2 , 2 n + 2 ) 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.u(x)={2n+1(x−2n),x∈[2n,2n+1/2)−2n+1(x−2n−1),x∈[2n+1/2,2n+3/2)2n+1(x−2n−2),x∈[2n+3/2,2n+2)
( n ∈ N ) ( n ∈ N ) (n inN)(n \in \mathbb{N})(n∈N). For this function we have
sup α > 0 1 α ∫ 0 α | u ( t ) | d t ≥ sup n ∈ N ∗ 1 2 n ∫ 0 2 n | u ( t ) | d t = sup n ∈ N ∗ 1 2 n ∑ k = 1 n k = ∞ sup α > 0   1 α ∫ 0 α   | u ( t ) | d t ≥ sup n ∈ N ∗   1 2 n ∫ 0 2 n   | u ( t ) | d t = sup n ∈ N ∗   1 2 n ∑ k = 1 n   k = ∞ 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}=\inftysupα>01α∫0α|u(t)|dt≥supn∈N∗12n∫02n|u(t)|dt=supn∈N∗12n∑k=1nk=∞
For 2 n ≤ α < 2 n + 2 2 n ≤ α < 2 n + 2 2n <= alpha < 2n+22 n \leq \alpha<2 n+22n≤α<2n+2 it follows 1 α ∫ 0 α u ( t ) d t = 1 α ∫ 2 n α u ( t ) d t ≤ n + 1 2 α ≤ n + 1 4 n 1 α ∫ 0 α   u ( t ) d t = 1 α ∫ 2 n α   u ( t ) d t ≤ n + 1 2 α ≤ n + 1 4 n (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}1α∫0αu(t)dt=1α∫2nαu(t)dt≤n+12α≤n+14n and lim α → ∞ 1 α ∫ 0 α u ( t ) d t = 0 lim α → ∞   1 α ∫ 0 α   u ( t ) d t = 0 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=0limα→∞1α∫0αu(t)dt=0.
If we consider only functions in C 1 ( I ) C 1 ( I ) C^(1)(I)\mathcal{C}^{1}(I)C1(I), we obtain the corresponding property as in D D DDD mentioned in the introduction for sequences of operators.
Theorem 2.2 Let there be given a function f ∈ C 1 ( I ) f ∈ C 1 ( I ) f inC^(1)(I)f \in \mathcal{C}^{1}(I)f∈C1(I) and u ∈ C ( [ 0 , ∞ ) ) u ∈ C ( [ 0 , ∞ ) ) u inC([0,oo))u \in \mathcal{C}([0, \infty))u∈C([0,∞)). If the function u u uuu satisfies condition I from Theorem 2.1 ( ∃ lim α → ∞ 1 α ∫ 0 α u ( t ) d t = a ) ∃ lim α → ∞   1 α ∫ 0 α   u ( t ) d t = a (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)(∃limα→∞1α∫0αu(t)dt=a), then there exists lim α → ∞ φ α ( f ) lim α → ∞   φ α ( f ) lim_(alpha rarr oo)varphi_(alpha)(f)\lim _{\alpha \rightarrow \infty} \varphi_{\alpha}(f)limα→∞φα(f) and its value is given by (2.1).
Proof. Let U U UUU be an antiderivative of u u uuu with U ( 0 ) = 0 U ( 0 ) = 0 U(0)=0U(0)=0U(0)=0 (i.e. U ( x ) = ∫ ∫ x u ( t ) d t ) U ( x ) = ∫ ∫ x   u ( t ) d t U(x)={:int_(int)^(x)u(t)dt)U(x)= \left.\int_{\int}^{x} u(t) \mathrm{d} t\right)U(x)=∫∫xu(t)dt). We have for f ∈ C 1 ( I ) f ∈ C 1 ( I ) f inC^(1)(I)f \in \mathcal{C}^{1}(I)f∈C1(I)
0
φ α ( f ) = 1 α ∫ 0 α U ′ ( t ) f ( t α ) d t = 1 α U ( α ) f ( 1 ) − 1 α 2 ∫ 0 α U ( t ) f ′ ( t α ) d t = 1 α U ( α ) f ( 1 ) − 1 α ∫ 0 1 U ( α t ) f ′ ( t ) d t φ α ( f ) = 1 α ∫ 0 α   U ′ ( t ) f t α d t = 1 α U ( α ) f ( 1 ) − 1 α 2 ∫ 0 α   U ( t ) f ′ t α d t = 1 α U ( α ) f ( 1 ) − 1 α ∫ 0 1   U ( α t ) f ′ ( t ) d t {:[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}φα(f)=1α∫0αU′(t)f(tα)dt=1αU(α)f(1)−1α2∫0αU(t)f′(tα)dt=1αU(α)f(1)−1α∫01U(αt)f′(t)dt
But 1 α U ( α t ) = 1 α ∫ 0 α t u ( s ) d s 1 α U ( α t ) = 1 α ∫ 0 α t   u ( s ) d s (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} s1αU(αt)=1α∫0αtu(s)ds and we get lim α → ∞ 1 α U ( α t ) = t a lim α → ∞   1 α U ( α t ) = t a lim_(alpha rarr oo)(1)/(alpha)U(alpha t)=ta\lim _{\alpha \rightarrow \infty} \frac{1}{\alpha} U(\alpha t)=t alimα→∞1αU(αt)=ta. Using the theorem of dominated convergence we get
lim α → ∞ φ α ( f ) = a f ( 1 ) − a ∫ 0 1 t f ′ ( t ) d t = a ⋅ ∫ 0 1 f . lim α → ∞   φ α ( f ) = a f ( 1 ) − a ∫ 0 1   t f ′ ( t ) d t = a ⋅ ∫ 0 1   f . 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 .limα→∞φα(f)=af(1)−a∫01tf′(t)dt=a⋅∫01f.

3 A new result for sequences of operators

In connection with the result mentioned in B B BBB in the introduction for the sequence of operators φ n : R ( I , F ) → F , n ∈ N φ n : R ( I , F ) → F , n ∈ N varphi_(n):R(I,F)rarr F,n inN\varphi_{n}: \mathcal{R}(I, F) \rightarrow F, n \in \mathbb{N}φn:R(I,F)→F,n∈N, given by (1.1), an open question was formulated in [1]: Is the conclusion in B B BBB true if F = R F = R F=RF=\mathbb{R}F=R for f f fff Riemann integrable (instead of regular)?
We shall prove that this result holds if the sequence ( a n ) n ∈ N a n n ∈ N (a_(n))_(n inN)\left(a_{n}\right)_{n \in \mathbb{N}}(an)n∈N is bounded from above or below, or if ( | a 1 | + ⋯ + | a n | n ) n ∈ N a 1 + ⋯ + a n n n ∈ N ((|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}}(|a1|+⋯+|an|n)n∈N is convergent.
Theorem 3.1 Let there be given a Riemann integrable function f : I → R f : I → R f:I rarrRf: I \rightarrow \mathbb{R}f:I→R and a sequence ( a n ) n ∈ N a n n ∈ N (a_(n))_(n inN)\left(a_{n}\right)_{n \in \mathbb{N}}(an)n∈N of real numbers satisfying the conditions:
  1. lim n → ∞ a 1 + … + a n n = a lim n → ∞   a 1 + … + a n n = a lim_(n rarr oo)(a_(1)+dots+a_(n))/(n)=a\lim _{n \rightarrow \infty} \frac{a_{1}+\ldots+a_{n}}{n}=alimn→∞a1+…+ann=a;
  2. the sequence ( | a 1 | + … + | a n | n ) n ∈ N a 1 + … + a n n n ∈ N ((|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}}(|a1|+…+|an|n)n∈N is bounded;
  3. the sequence ( a n ) n ∈ N a n n ∈ N (a_(n))_(n inN)\left(a_{n}\right)_{n \in \mathbb{N}}(an)n∈N is bounded from above or from below, or ( | a 1 | + … + | a n | n ) n ∈ N a 1 + … + a n n n ∈ N ((|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}}(|a1|+…+|an|n)n∈N is convergent.
Then the sequence ( φ n ( f ) ) n ∈ N φ n ( f ) n ∈ N (varphi_(n)(f))_(n inN)\left(\varphi_{n}(f)\right)_{n \in \mathbb{N}}(φn(f))n∈N given by (1.1) is convergent to a ⋅ ∫ 0 1 f a ⋅ ∫ 0 1   f a*int_(0)^(1)fa \cdot \int_{0}^{1} fa⋅∫01f.
Proof. (i) Let us consider a sequence ( a n ) n ∈ N a n n ∈ N (a_(n))_(n inN)\left(a_{n}\right)_{n \in \mathbb{N}}(an)n∈N with a n ≥ 0 a n ≥ 0 a_(n) >= 0a_{n} \geq 0an≥0, which satisfies condition 1 (hence also condition 2). Given ε > 0 ε > 0 epsi > 0\varepsilon>0ε>0, for the Riemann integrable function f f fff there exist the continuous functions u , v u , v u,vu, vu,v : I → R I → R I rarrRI \rightarrow \mathbb{R}I→R such that
(3.1) u ≤ f ≤ v and ∫ ( v − u ) < ε . (3.1) u ≤ f ≤ v  and  ∫ ( v − u ) < ε . {:(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*}(3.1)u≤f≤v and ∫(v−u)<ε.
Then the functionals φ n φ n varphi_(n)\varphi_{n}φn given by (1.1) will satisfy
(3.2) φ n ( u ) ≤ φ n ( f ) ≤ φ n ( v ) (3.2) φ n ( u ) ≤ φ n ( f ) ≤ φ n ( v ) {:(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*}(3.2)φn(u)≤φn(f)≤φn(v)
From the result in [1] mentioned at B B BBB in the introduction we have lim n → ∞ φ n ( v ) = a ⋅ ∫ 0 1 v lim n → ∞   φ n ( v ) = a ⋅ ∫ 0 1   v lim_(n rarr oo)varphi_(n)(v)=a*int_(0)^(1)v\lim _{n \rightarrow \infty} \varphi_{n}(v)=a \cdot \int_{0}^{1} vlimn→∞φn(v)=a⋅∫01v, hence there exists n 1 ∈ N n 1 ∈ N n_(1)inNn_{1} \in \mathbb{N}n1∈N so that for any n ≥ n 1 n ≥ n 1 n >= n_(1)n \geq n_{1}n≥n1, φ n ( v ) < a ⋅ ∫ 0 1 v + ε φ n ( v ) < a ⋅ ∫ 0 1   v + ε varphi_(n)(v) < a*int_(0)^(1)v+epsi\varphi_{n}(v)<a \cdot \int_{0}^{1} v+\varepsilonφn(v)<a⋅∫01v+ε. Condition (3.1) implies that ∫ 0 1 v < ∫ 0 1 f + ε ∫ 0 1   v < ∫ 0 1   f + ε int_(0)^(1)v < int_(0)^(1)f+epsi\int_{0}^{1} v<\int_{0}^{1} f+\varepsilon∫01v<∫01f+ε, hence
(3.3) φ n ( v ) < a ⋅ ∫ 0 1 f + ε ( a + 1 ) (3.3) φ n ( v ) < a ⋅ ∫ 0 1   f + ε ( a + 1 ) {:(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*}(3.3)φn(v)<a⋅∫01f+ε(a+1)
Similarly, there exists n 2 ∈ N n 2 ∈ N n_(2)inNn_{2} \in \mathbb{N}n2∈N so that for any n ≥ n 2 n ≥ n 2 n >= n_(2)n \geq n_{2}n≥n2,
(3.4) a ⋅ ∫ 0 1 f − ε ( a + 1 ) < φ n ( u ) (3.4) a ⋅ ∫ 0 1   f − ε ( a + 1 ) < φ n ( u ) {:(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*}(3.4)a⋅∫01f−ε(a+1)<φn(u)
From (3.2), (3.3) and (3.4) we obtain
a ⋅ ∫ 0 1 f − ε ( a + 1 ) < φ n ( f ) < a ⋅ ∫ 0 1 f + ε ( a + 1 ) a ⋅ ∫ 0 1   f − ε ( a + 1 ) < φ n ( f ) < a ⋅ ∫ 0 1   f + ε ( a + 1 ) 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)a⋅∫01f−ε(a+1)<φn(f)<a⋅∫01f+ε(a+1)
hence | φ n ( f ) − a ⋅ ∫ 0 1 f | ≤ ε ( a + 1 ) φ n ( f ) − a ⋅ ∫ 0 1   f ≤ ε ( a + 1 ) |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)|φn(f)−a⋅∫01f|≤ε(a+1) for n ≥ max { n 1 , n 2 } n ≥ max n 1 , n 2 n >= max{n_(1),n_(2)}n \geq \max \left\{n_{1}, n_{2}\right\}n≥max{n1,n2} and the conclusion holds.
(ii) Let us consider ( a n ) n ∈ N a n n ∈ N (a_(n))_(n inN)\left(a_{n}\right)_{n \in \mathbb{N}}(an)n∈N which satisfies the conditions 1 and 2 and is bounded from below, i. e. a n ≥ − α a n ≥ − α a_(n) >= -alphaa_{n} \geq-\alphaan≥−α. The sequence ( b n ) n ∈ N b n n ∈ N (b_(n))_(n inN)\left(b_{n}\right)_{n \in \mathbb{N}}(bn)n∈N, b n = a n + α b n = a n + α b_(n)=a_(n)+alphab_{n}=a_{n}+\alphabn=an+α is Cesàro-convergent to a + α a + α a+alphaa+\alphaa+α and has b n ≥ 0 b n ≥ 0 b_(n) >= 0b_{n} \geq 0bn≥0; applying the result proved in (i) it follows that
lim n → ∞ 1 n ∑ k = 1 n ( a n + α ) f ( k n ) = ( a + α ) ⋅ ∫ 0 1 f lim n → ∞   1 n ∑ k = 1 n   a n + α f k n = ( a + α ) ⋅ ∫ 0 1   f 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} flimn→∞1n∑k=1n(an+α)f(kn)=(a+α)⋅∫01f
hence lim n → ∞ φ n ( f ) = a ⋅ ∫ 0 1 f lim n → ∞   φ n ( f ) = a ⋅ ∫ 0 1   f lim_(n rarr oo)varphi_(n)(f)=a*int_(0)^(1)f\lim _{n \rightarrow \infty} \varphi_{n}(f)=a \cdot \int_{0}^{1} flimn→∞φn(f)=a⋅∫01f.
(iii) If the sequence ( a n ) n ∈ N a n n ∈ N (a_(n))_(n inN)\left(a_{n}\right)_{n \in \mathbb{N}}(an)n∈N satisfies 1 and 2 , and is bounded from above ( a n ≤ α a n ≤ α a_(n) <= alphaa_{n} \leq \alphaan≤α ), the sequence ( c n ) n ∈ N , c n = α − a n c n n ∈ N , c n = α − a n (c_(n))_(n inN),c_(n)=alpha-a_(n)\left(c_{n}\right)_{n \in \mathbb{N}}, c_{n}=\alpha-a_{n}(cn)n∈N,cn=α−an is Cesàro-convergent to α − a α − a alpha-a\alpha-aα−a and has c n ≥ 0 c n ≥ 0 c_(n) >= 0c_{n} \geq 0cn≥0. Applying again the result proved in (i) we obtain the conclusion.
(iv) Let us consider the sequence ( a n ) n ∈ N a n n ∈ N (a_(n))_(n inN)\left(a_{n}\right)_{n \in \mathbb{N}}(an)n∈N with lim n → ∞ a 1 + … + a n n = a lim n → ∞   a 1 + … + a n n = a lim_(n rarr oo)(a_(1)+dots+a_(n))/(n)=a\lim _{n \rightarrow \infty} \frac{a_{1}+\ldots+a_{n}}{n}=alimn→∞a1+…+ann=a and lim n → ∞ | a 1 | + … + | a n | n = a ∗ lim n → ∞   a 1 + … + a n n = a ∗ 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^{*}limn→∞|a1|+…+|an|n=a∗. We can write
a n = a n + | a n | 2 − | a n | − a n 2 . a n = a n + a n 2 − a n − a n 2 . 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} .an=an+|an|2−|an|−an2.
The sequence ( a n + | a n | 2 ) n ∈ N a n + a n 2 n ∈ N ((a_(n)+|a_(n)|)/(2))_(n inN)\left(\frac{a_{n}+\left|a_{n}\right|}{2}\right)_{n \in \mathbb{N}}(an+|an|2)n∈N is Cesàro-convergent to a + a ∗ 2 a + a ∗ 2 (a+a^(**))/(2)\frac{a+a^{*}}{2}a+a∗2 and has nonnegative terms, hence it follows from (i) that
lim n → ∞ 1 n ∑ k = 1 n ( a n + | a n | 2 ) f ( k n ) = a + a ∗ 2 ⋅ ∫ 0 1 f lim n → ∞   1 n ∑ k = 1 n   a n + a n 2 f k n = a + a ∗ 2 ⋅ ∫ 0 1   f 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} flimn→∞1n∑k=1n(an+|an|2)f(kn)=a+a∗2⋅∫01f
Similarly, for ( | a n | − a n 2 ) n ∈ N a n − a n 2 n ∈ N ((|a_(n)|-a_(n))/(2))_(n inN)\left(\frac{\left|a_{n}\right|-a_{n}}{2}\right)_{n \in \mathbb{N}}(|an|−an2)n∈N we have
lim n → ∞ 1 n ∑ k = 1 n ( | a n | − a n 2 ) f ( k n ) = a ∗ − a 2 ⋅ ∫ 0 1 f . lim n → ∞   1 n ∑ k = 1 n   a n − a n 2 f k n = a ∗ − a 2 ⋅ ∫ 0 1   f . 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 .limn→∞1n∑k=1n(|an|−an2)f(kn)=a∗−a2⋅∫01f.
It follows
lim n → ∞ 1 n ∑ k = 1 n a n f ( k n ) = a + a ∗ 2 ⋅ ∫ 0 1 f − a ∗ − a 2 ⋅ ∫ 0 1 f = a ⋅ ∫ 0 1 f . lim n → ∞   1 n ∑ k = 1 n   a n f k n = a + a ∗ 2 ⋅ ∫ 0 1   f − a ∗ − a 2 ⋅ ∫ 0 1   f = a ⋅ ∫ 0 1   f . 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 .limn→∞1n∑k=1nanf(kn)=a+a∗2⋅∫01f−a∗−a2⋅∫01f=a⋅∫01f.
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 A A AAA Cesàro-convergent sequence ( a n ) n ∈ N a n n ∈ N (a_(n))_(n inN)\left(a_{n}\right)_{n \in \mathbb{N}}(an)n∈N which is not bounded from above or from below, for which ( | a 1 | + … + | a n | n ) n ∈ N a 1 + … + a n n n ∈ N ((|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}}(|a1|+…+|an|n)n∈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 = k ,  for  n = 2 k , k  even  − k ,  for  n = 2 k , k  odd  0 ,  otherwise  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.an={k, for n=2k,k even −k, for n=2k,k odd 0, otherwise 
Example 3.2 A A AAA Cesàro-convergent sequence ( a n ) n ∈ N a n n ∈ N (a_(n))_(n inN)\left(a_{n}\right)_{n \in \mathbb{N}}(an)n∈N, for which ( | a 1 | + … + | a n | n ) n ∈ N a 1 + … + a n n n ∈ N ((|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}}(|a1|+…+|an|n)n∈N is bounded without being convergent, can be obtained from a bounded sequence d n ≥ 0 d n ≥ 0 d_(n) >= 0d_{n} \geq 0dn≥0 for which d 1 + … + d n n d 1 + … + d n n (d_(1)+dots+d_(n))/(n)\frac{d_{1}+\ldots+d_{n}}{n}d1+…+dnn does not converge, as
a n = { d n / 2 , for n even − d ( n + 1 ) / 2 , for n odd a n = d n / 2 ,  for  n  even  − d ( n + 1 ) / 2 ,  for  n  odd  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.an={dn/2, for n even −d(n+1)/2, for n odd 
Such a sequence ( d n ) n ∈ N d n n ∈ N (d_(n))_(n inN)\left(d_{n}\right)_{n \in \mathbb{N}}(dn)n∈N is, for example,
d n = { 1 , for n ∈ [ 2 k , 2 k + 1 ) , k even 0 , for n ∈ [ 2 k , 2 k + 1 ) , k odd d n = 1 ,  for  n ∈ 2 k , 2 k + 1 , k  even  0 ,  for  n ∈ 2 k , 2 k + 1 , k  odd  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.dn={1, for n∈[2k,2k+1),k even 0, for n∈[2k,2k+1),k odd 
Indeed, we have
lim k → ∞ , k even d 1 + … + d 2 k + 1 − 1 2 k + 1 − 1 = lim k → ∞ 2 0 + 2 2 + … + 2 k 2 k + 1 − 1 = 2 3 , lim k → ∞ , k  even    d 1 + … + d 2 k + 1 − 1 2 k + 1 − 1 = lim k → ∞   2 0 + 2 2 + … + 2 k 2 k + 1 − 1 = 2 3 , 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},limk→∞,k even d1+…+d2k+1−12k+1−1=limk→∞20+22+…+2k2k+1−1=23,
and
lim k → ∞ , k odd d 1 + … + d 2 k + 1 − 1 2 k + 1 − 1 = lim k → ∞ 2 0 + 2 2 + … + 2 k − 1 2 k + 1 − 1 = 1 3 . lim k → ∞ , k  odd    d 1 + … + d 2 k + 1 − 1 2 k + 1 − 1 = lim k → ∞   2 0 + 2 2 + … + 2 k − 1 2 k + 1 − 1 = 1 3 . 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} .limk→∞,k odd d1+…+d2k+1−12k+1−1=limk→∞20+22+…+2k−12k+1−1=13.
Remark 3.2 We mention, in connection with Theorem 3.1, the following result from [2] (see also [5]):
For k > 0 k > 0 k > 0k>0k>0, if f , g : [ 0 , ∞ ) → R f , g : [ 0 , ∞ ) → R f,g:[0,oo)rarrRf, g:[0, \infty) \rightarrow \mathbb{R}f,g:[0,∞)→R satisfy:
− f ∈ C 1 ( [ 0 , ∞ ) ) − f ∈ C 1 ( [ 0 , ∞ ) ) -f inC^(1)([0,oo))-f \in \mathcal{C}^{1}([0, \infty))−f∈C1([0,∞)) is decreasing with f ( ∞ ) = 0 f ( ∞ ) = 0 f(oo)=0f(\infty)=0f(∞)=0 and there exists lim α → ∞ ∫ 0 α f ( x + k α ) d x = a lim α → ∞   ∫ 0 α   f ( x + k α ) d x = a 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=alimα→∞∫0αf(x+kα)dx=a;
− g ∈ C ( [ 0 , ∞ ) ) − g ∈ C ( [ 0 , ∞ ) ) -g inC([0,oo))-g \in \mathcal{C}([0, \infty))−g∈C([0,∞)) is bounded and ∃ lim α → ∞ 1 α ∫ 0 α g = b ∃ lim α → ∞   1 α ∫ 0 α   g = b EElim_(alpha rarr oo)(1)/(alpha)int_(0)^(alpha)g=b\exists \lim _{\alpha \rightarrow \infty} \frac{1}{\alpha} \int_{0}^{\alpha} g=b∃limα→∞1α∫0αg=b,
then lim α → ∞ ∫ 0 α f ( x + k α ) g ( x ) d x = a b lim α → ∞   ∫ 0 α   f ( x + k α ) g ( x ) d x = a b 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 blimα→∞∫0αf(x+kα)g(x)dx=ab.

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.

[6] A. Wilanski. Modern Methods in Topological Vector Spaces. McGraw-Hill 1978.

T. Popoviciu Institute of Numerical Analysis37, Republicii st., 3400 Cluj-NapocaRomaniaBabeş-Bolyai University, Faculty of Mathematics1, Kogălniceanu st., 3400 Cluj-NapocaRomania

2003

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