of Functional Equations, Approximation and Convexity, Cluj-Napoca, May 21 – May 25, 2003, pp. 3–13.
On the pointwise convergence of a family of functionals on C ( I )
Mira-Cristiana Anisiu (Cluj-Napoca)
Valeriu Anisiu (Cluj-Napoca)
Abstract. Given a continuous functionu: [0,∞)→R, a family of function- alsϕα:C(I)→R,α >0, is defined byϕα(f) =α1
α
R
0
u(t)f(t/α) dt.It is proved that the necessary and sufficient conditions for the familyϕα, α >0 to satisfy
α→∞lim ϕα(f) =
α→∞lim
1 α
α
R
0
u(t)dt
·
1
R
0
f are:
I.∃ lim
α→∞
1 α
α
R
0
u(t)dt; II. sup
α>0 1 α
α
R
0
|u(t)|dt <∞.
Iff ∈ C1(I),condition I alone implies the existence of lim
α→∞ϕα(f).
A sequence of functionals (ϕn)n∈Nis attached to a numerical sequence (an)n∈N
which is Ces`aro-convergent toa,namely ϕn(f) = 1
n
n
X
k=1
akf(k/n), f Riemann integrable.
Additional conditions are imposed on the sequence (an)n∈N in order to prove that
n→∞lim ϕn(f) =a· Z 1
0
f.
MSC 2000: 47B38, 26E60
1 Introduction
For the intervalI = [0,1] and a Banach spaceF 6={0},let us denote by B(I, F) the Banach space of bounded functionsf :I → F endowed with the sup norm. The subspace ofB(I, F) of regular functions (which admit side limits at eacht∈I) will be denoted byR(I, F) ; the Banach space of continuous functions C(I, F) is a subspace of R(I, F).
Given a sequence of real numbers (an)n∈N, a sequence of operators ϕn:R(I, F)→F, n∈N,
(1.1) ϕn(f) = 1
n
n
X
k=1
akf k
n
can be generated. It was proved in [1] that:
A. The operatorsϕn are linear and continuous.
B. If the numeric sequence satisfies the conditions:
B1. (an)n∈N is Ces`aro-convergent toa( lim
n→∞
a1+...+an n =a);
B2. the sequence
|a1|+...+|an| n
n∈N is bounded, then the sequence (ϕn(f))n∈N is convergent and
(1.2) lim
n→∞ϕn(f) =a· Z 1
0
f.
C. If lim
n→∞ϕn(f) exists for everyf ∈ C(I, F) ⊆ R(I, F),the condi- tionsB1 and B2 from above are also necessary.
D. Iff ∈ C1(I, F) (i.e. f is continuous with a continuous derivative), the result in B holds even if conditionB2 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).
Letu : [0,∞)→ Rbe a continuous function. We define a family of
functionals associated tou, namelyϕα :C(I)→R, α >0, given by (1.3) ϕα(f) = 1
α
α
Z
0
u(t)f t
α
dt, f ∈ C(I).
Proposition 1.1 For each α >0, the functional ϕα is linear and con- tinuous, and its norm is given by
(1.4) kϕαk= 1
α Zα
0
|u(t)|dt.
Proof. This result is classical; see for a simple proof [3]. It also holds ifC(I) is replaced with the Banach spaceR(I) of regular functions.
2 Main results for families of functionals
As mentioned in B and C in the introduction, in the discrete case the conditionsB1 and 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) and u ∈ C([0,∞)). If the function u satisfies the conditions:
I. ∃ lim
α→∞
1 α
Rα 0
u(t)dt;
II.sup
α>0 1 α
Rα
0 |u(t)|dt <∞, there exists the limit lim
α→∞ϕα(f) and
(2.1) lim
α→∞ϕα(f) =
lim
α→∞
1 α
Zα
0
u(t)dt
· Z1
0
f.
The conditions I and II are also necessary in order to have (2.1) for each 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 functionf. To this end it suffices to prove (2.1) for f ∈ E with E ⊆ R(I) and spE = R(I) (condition II allows then to obtain (2.1) for any f ∈ R(I)). As a setE we choose the set of characteristic functions χ[a,b], [a, b]⊆I.
Forf =χ[a,b] we have ϕα(f) =
1
R
0
u(αt)f(t)dt=
b
R
a
u(αt)dt= α1
αb
R
αa
u(t)dt
= bαb1 Rαb 0
u(t)dt−aαa1
αaR
0
u(t)dt.
It follows
αlim→∞ϕα(f) = (b−a) lim
α→∞
1 α
Rα 0
u(t)dt
=
αlim→∞
1 α
Rα 0
u(t)dt
· R1 0
f.
Let us suppose now that (2.1) takes place for each f ∈ C(I). For f(x) = 1, ∀x ∈ I we obtain condition I. To prove II, we apply the Banach-Steinhaus principle for sequences αn → ∞, because it cannot be used directly for generalized sequences. (In fact, if X is an infinite dimensional Banach space, then its dual X∗ is sequentially closed and dense in X#,weak∗
, see [6, p. 138].) It follows that sup
α>0kϕαk < ∞, and using the expression ofkϕαk given in (1.4) we obtain II.
Remark 2.1 If the function u is periodic withu(t+T) =u(t) for each t >0, then
αlim→∞
1 α
Zα
0
u(t)dt= 1 T
ZT
0
u(t)dt
and condition II automatically holds because of the boundedness ofu. In
this special case we obtain
(2.2) lim
α→∞ϕα(f) = 1 T
ZT
0
u(t)dt· Z1
0
f,
which is a result due to L. Fej´er, see [4, p. 114].
Remark 2.2 Condition II does not follow from condition I. Indeed, let u∈ C([0,∞))be given by
u(x) =
2√
n+ 1 (x−2n), x∈[2n,2n+ 1/2)
−2√
n+ 1 (x−2n−1), x∈[2n+ 1/2,2n+ 3/2) 2√
n+ 1 (x−2n−2), x∈[2n+ 3/2,2n+ 2) (n∈N). For this function we have
sup
α>0
1 α
α
Z
0
|u(t)|dt≥ sup
n∈N∗
1 2n
2n
Z
0
|u(t)|dt = sup
n∈N∗
1 2n
n
X
k=1
√k=∞.
For2n≤α <2n+ 2 it follows 1α Rα 0
u(t)dt= α1 Rα 2n
u(t)dt≤ √2αn+1 ≤ √4nn+1 and lim
α→∞
1 α
α
R
0
u(t)dt= 0.
If we consider only functions in C1(I),we obtain the corresponding property as in D mentioned in the introduction for sequences of opera- tors.
Theorem 2.2 Let there be given a function f ∈ C1(I) and u ∈ C([0,∞)). If the function u satisfies condition I from Theorem 2.1
∃ lim
α→∞
1 α
Rα 0
u(t)dt=a
, then there exists lim
α→∞ϕα(f) and its value is given by (2.1).
Proof. Let U be an antiderivative of u with U(0) = 0 (i.e. U(x) = Rx
0
u(t)dt). We have forf ∈ C1(I)
ϕα(f) = α1
α
R
0
U0(t)f αt dt
= α1U(α)f(1)−α12 Rα 0
U(t)f0 αt dt
= α1U(α)f(1)−α1
1
R
0
U(αt)f0(t) dt.
But α1U(αt) = α1 Rαt 0
u(s)ds and we get lim
α→∞
1
αU(αt) = ta. Using the theorem of dominated convergence we get
αlim→∞ϕα(f) =af(1)−a Z1
0
tf0(t) dt=a· Z1
0
f.
3 A new result for sequences of operators
In connection with the result mentioned inB in the introduction for the sequence of operators ϕn : R(I, F) → F, n ∈ N, given by (1.1), an open question was formulated in [1]: Is the conclusion in B true if F =Rforf Riemann integrable (instead of regular)?
We shall prove that this result holds if the sequence (an)n∈N is bounded from above or below, or if
|a1|+···+|an| n
n∈N is convergent.
Theorem 3.1 Let there be given a Riemann integrable function f :I → R and a sequence(an)n∈N of real numbers satisfying the conditions:
1. lim
n→∞
a1+...+an
n =a;
2. the sequence
|a1|+...+|an| n
n∈N is bounded;
3. the sequence (an)n∈N is bounded from above or from below, or |a1|+...+|an|
n
n∈N is convergent.
Then the sequence(ϕn(f))n∈N given by (1.1) is convergent toa· R1 0
f.
Proof. (i) Let us consider a sequence (an)n∈N with an ≥ 0, which satisfies condition 1 (hence also condition 2). Given ε > 0, for the Riemann integrable functionf there exist the continuous functionsu, v: I →Rsuch that
(3.1) u≤f ≤v and
Z
(v−u)< ε.
Then the functionalsϕn given by (1.1) will satisfy
(3.2) ϕn(u)≤ϕn(f)≤ϕn(v).
From the result in [1] mentioned at B in the introduction we have
nlim→∞ϕn(v) = a· R1 0
v, hence there exists n1 ∈N so that for any n≥n1, ϕn(v)< a·
R1 0
v+ε. Condition (3.1) implies that R1 0
v <
R1 0
f+ε,hence
(3.3) ϕn(v)< a· Z1
0
f+ε(a+ 1).
Similarly, there exists n2 ∈Nso that for any n≥n2,
(3.4) a·
Z1
0
f−ε(a+ 1)< ϕn(u).
From (3.2), (3.3) and (3.4) we obtain
a· Z1
0
f−ε(a+ 1)< ϕn(f)< a· Z1
0
f +ε(a+ 1),
hence
ϕn(f)−a· R1 0
f
≤ε(a+ 1) for n≥max{n1, n2} and the conclu- sion holds.
(ii) Let us consider (an)n∈N which satisfies the conditions 1 and 2 and is bounded from below, i. e. an ≥ −α. The sequence (bn)n∈N, bn=an+α is Ces`aro-convergent toa+α and has bn≥0; applying the result proved in (i) it follows that
nlim→∞
1 n
n
X
k=1
(an+α)f k
n
= (a+α)·
1
Z
0
f,
hence lim
n→∞ϕn(f) =a· R1 0
f.
(iii) If the sequence (an)n∈N satisfies 1 and 2, and is bounded from above (an≤α),the sequence (cn)n∈N, cn=α−an is Ces`aro-convergent to α−a and has cn ≥ 0. Applying again the result proved in (i) we obtain the conclusion.
(iv) Let us consider the sequence (an)n∈N with lim
n→∞
a1+...+an
n = a
and lim
n→∞
|a1|+...+|an|
n =a∗.We can write an= an+|an|
2 −|an| −an
2 .
The sequence
an+|an| 2
n∈N is Ces`aro-convergent to a+a2∗ and has non- negative terms, hence it follows from (i) that
nlim→∞
1 n
n
X
k=1
an+|an| 2
f
k n
= a+a∗
2 ·
Z1
0
f.
Similarly, for
|an|−an 2
n∈N we have
nlim→∞
1 n
n
X
k=1
|an| −an
2
f k
n
= a∗−a
2 ·
1
Z
0
f.
It follows
nlim→∞
1 n
n
X
k=1
anf k
n
= a+a∗
2 ·
Z1
0
f−a∗−a 2 ·
Z1
0
f =a· Z1
0
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 A Ces`aro-convergent sequence (an)n∈N which is not bounded from above or from below, for which
|a1|+...+|an| n
n∈N is bounded, is given by
an=
k, forn= 2k, keven
−k, forn= 2k, k odd 0, otherwise.
Example 3.2 A Ces`aro-convergent sequence (an)n∈N, for which |a1|+...+|an|
n
n∈N is bounded without being convergent, can be obtained from a bounded sequence dn ≥0 for which d1+...+dn n does not converge, as
an=
dn/2, forneven
−d(n+1)/2, forn odd.
Such a sequence (dn)n∈N is, for example,
dn=
1, forn∈
2k,2k+1
, keven 0, forn∈
2k,2k+1
, k odd.
Indeed, we have
k→∞lim, k even
d1+...+d2k+1−1
2k+1−1
= lim
k→∞
20+ 22+...+ 2k
2k+1−1
= 2 3, and
k→∞lim, k odd
d1+...+d2k+1−1
2k+1−1
= lim
k→∞
20+ 22+...+ 2k−1
2k+1−1
= 1 3. Remark 3.2 We mention, in connection with Theorem 3.1, the follow- ing result from [2] (see also [5]):
Fork >0, if f, g: [0,∞)→R satisfy:
− f ∈ C1([0,∞)) is decreasing with f(∞) = 0 and there exists
αlim→∞
Rα 0
f(x+kα)dx=a;
−g∈ C([0,∞))is bounded and ∃ lim
α→∞
1 α
Rα 0
g=b, then lim
α→∞
Rα 0
f(x+kα)g(x)dx=ab.
References
[1] M.-C. Anisiu, V. Anisiu. Sequences of linear operators related to Ces`aro-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 Matematic˘a 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. Sz´asz, L. Geh´er, I. Kov´acs, L. Pint´er (editors).Contests in Higher Mathematics. Akad´emiai Kiad´o, 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 Analysis 37, Republicii st., 3400 Cluj-Napoca
Romania
Babe¸s-Bolyai University, Faculty of Mathematics 1, Kog˘alniceanu st., 3400 Cluj-Napoca
Romania