D O I: 1 0 .1 5 0 1 / C o m m u a 1 _ 0 0 0 0 0 0 0 7 6 0 IS S N 1 3 0 3 –5 9 9 1
SOME KOROVKIN TYPE RESULTS VIA POWER SERIES METHOD IN MODULAR SPACES
T. YURDAKADIM
Abstract. In this paper, we obtain a Korovkin type approximation result for a sequence of positive linear operators de…ned on modular spaces with the use of power series method . We also provide an example which satis…es our theorem.
1. Introduction
The classical Korovkin theorem states the uniform convergence of a sequence of positive linear operators in C[a; b], the space of all continuous real valued func-tions de…ned on [a; b] by providing the convergence only on three test funcfunc-tions f1; x; x2g. There are also trigonometric versions of this theorem with the test
func-tions f1; cos x; sin xg and abstract Korovkin type results have also been studied [13, 17]. These type of results let us to say the convergence with minimum cal-culations and also have important applications in the polynomial approximation theory, in various areas of functional analysis, in numerical solutions of di¤erential and integral equations [1, 2] . Recently some versions of Korovkin type theo-rems have been given in modular spaces that include as particular cases Lp, Orlicz and Musielak-Orlicz spaces [8, 19] with the use of more general convergences such as convergences generated by summability methods, statistical, …lter convergence [9, 10, 11, 14, 15, 16, 20].
In this paper, we give a Korovkin type theorem in modular spaces by power series method which includes both Abel and Borel methods. We also give an example which satis…es our theorems.
2. Notation and Definitions
Let us begin with recalling some basic de…nitions and notations used in the paper.
Received by the editors: February 22, 2016, Accepted: May 20, 2016.
2010 Mathematics Subject Classi…cation. Primary 40G10; Secondary 40C15, 41A36. Key words and phrases. Power series method, modular spaces, abstract Korovkin theory.
c 2 0 1 6 A n ka ra U n ive rsity C o m m u n ic a tio n s d e la Fa c u lté d e s S c ie n c e s d e l’U n ive rs ité d ’A n ka ra . S é rie s A 1 . M a th e m a t ic s a n d S t a tis t ic s .
Let (pj) be real sequence with p0 > 0 and p1; p2; p3; ::: 0, and such that
the corresponding power series p(t) := P1
j=0
pjtj has radius of convergence R with
0 < R 1. If, for all t 2 (0; R), lim t!R 1 p(t) 1 X j=0 xjpjtj= L
then we say that x = (xj) is convergent in the sense of power series method [18, 21].
Power series method includes many well known summability methods such as Abel and Borel. Both methods have in common that their de…nitions are based on power series and that they are not matrix methods (See [12, 22] for details ). In order to see that power series method is more e¤ective than ordinary convergence, let x = (1; 0; 1; 0; :::), R = 1, p (t) = etand for j 0, p
j=
1
j!. Then it is easy to see that lim t!1 1 et 1 X j=0 xjtj j! = limt!1 1 et 1 X j=0 t2j (2j)! = limt!1 1 etf et+ e t 2 g = 1 2:
So the sequence x = (xj) is convergent to 12 in the sense of power series method
but it is not convergent in the ordinary sense. Note that the power series method is regular if and only if
lim
t!R
pjtj
p(t) = 0; f or each j 2 N
hold [12]. Throughout the paper we assume that power series method is regular. Let G = [a; b] be a bounded interval of the real line R provided with the Lebesgue measure. We denote by X(G) the space of all real-valued measurable functions on G with equality almost everywhere, by C(G) the space of all continuous real valued functions on G, and by C1(G) the space of all in…nitely di¤erentiable functions
on G. A functional % : X(G) ! [0; 1] is a modular on X(G) provided that the following conditions hold:
(i) %[f ] = 0 if and only if f = 0 a.e on G, (ii) %[ f ] = %[f ] for every f 2 X(G),
(iii) %[ f + g] %[f ] + %[g] for every f; g 2 X(G) and for any ; 0 with + = 1.
A modular % is said to be Q-quasi convex if there exists a constant Q 1 such that the inequality
%[ f + g] Q %[Qf ] + Q %[Qg]
holds for every f; g 2 X(G), ; 0 with + = 1. In particular if Q = 1, then % is called convex.
A modular % is said to be Q-quasi semiconvex if there exists a constant Q 1 such that the inequality
%[af ] Qa%[Qf ]
holds for every f 2 X(G); f 0 and a 2 (0; 1]. It is clear that every Q-quasi semiconvex modular is Q-quasi convex. A modular % is said to be monotone if %[f ] %[g] for all f; g 2 X(G) with jfj jgj:
We now consider some subspaces of X(G) by means of a modular % as follows L%(G) := ff 2 X(G) : lim
!0+%[ f ] = 0g
and
E%(G) := ff 2 L%(G) : %[ f ] < 1 for all > 0g
is called the modular space generated by % and is called the space of the …nite elements of L%(G) respectively. Observe that if % is Q-quasi semiconvex then the
space
ff 2 X(G) : %[ f] < 1 for some > 0g
coincides with L%(G). The notions about modulars have been introduced and widely
discussed in [4, 5, 6, 7, 8] .
Now we de…ne the convergences in the sense of power series method in modular spaces. Let ffjg be a function sequence whose terms belong to L%(G). Then, ffjg
is modularly convergent to a function f 2 L%(G) in the sense of power series method
if and only if lim t!R 1 p(t) 1 X j=0 pjtj%[ 0(fj f )] = 0 for some 0> 0:
Also, ffjg is strongly convergent to a function f 2 L%(G) in the sense of power
series method if and only if lim t!R 1 p(t) 1 X j=0 pjtj%[ (fj f )] = 0 for every > 0:
Recall that ffjg is modularly convergent to a function f 2 L%(G) if and only if
lim
j!1%[ 0(fj f )] = 0 for some 0> 0;
also ffjg is strongly convergent to a function f 2 L%(G) if and only if
lim
j!1%[ (fj f )] = 0 for every > 0:
If there exists a constant M > 0 such that %[2u] M %[u]
holds for all u 0 then it is said to be that % satis…es the 2-condition. A modular
% is said to be
absolutely …nite if % is …nite and for every " > 0; > 0 there exists > 0 such that %[ B] < " for any measurable subset B G with jBj < , strongly …nite if G2 E%(G),
absolutely continuous if there is a positive constant a with the property: for all f 2 X(G) with %[f] < 1, the following condition holds: for every " > 0 there is a > 0 such that %[af B] < " whenever B is any measurable subset of G with jBj < .
Recall that if a modular % is monotone and …nite, then we have C(G) L%(G) [4]. In a similar manner, if % is monotone and strongly …nite, then C(G) E%(G).
3. Modular Korovkin Theorem by Power Series Method
Let % be monotone and …nite modular on X(G). Assume that D is a set satisfying C1(G) D L%(G). We can construct such a subset D since % is monotone and
…nite. Assume further that T := fTjg is a sequence of positive linear operators
from D into X(G) for which there exists a subset XT D containing C1(G) such
that the inequality
lim sup t!R 1 p(t) 1 X j=0 pjtj%[ (Tjh)] P %( h) (3.1)
holds for every h 2 XT, > 0 and for an absolute positive constant P . Throughout
the paper we use the test functions de…ned by ei(x) = xi; i = 0; 1; 2; :::.
Theorem 1. Let % be a strongly …nite, monotone, absolutely continuous and Q-quasi semiconvex modular on X(G). Let Tj, j 2 N, be a sequence of positive linear
operators from D into X(G) satisfying (3.1). If lim t!R 1 p(t) 1 X j=0 pjtj%[ (Tjei ei)] = 0;
for every > 0 and i = 0; 1; 2, then for every f 2 L%(G) such that f g 2 X
T for every g 2 C1(G) lim t!R 1 p(t) 1 X j=0 pjtj%[ (Tjf f )] = 0; for some > 0.
Proof. Let g 2 C(G) and …rst we show that lim t!R 1 p(t) 1 X j=0 pjtj%[ (Tjg g)] = 0; f or every > 0: (3.2)
Since g is uniformly continuous on G then there exists a constant M > 0 such that jg(x)j M for every x 2 G: Given " > 0; we can choose > 0 such that jy xj <
implies jg(y) g(x)j < " where x; y 2 G: One can see that for all x; y 2 G jg(y) g(x)j < " +2M2 (y x)
2:
Since fTjg is a sequence of positive linear operators, we get
jTj(g; x) g(x)j = jTj(g(:) g(x); x) + g(x)(Tj(e0(:); x) e0(x))j Tj(jg(:) g(x)j; x) + jg(x)jjTj(e0(:); x) e0(x)j Tj(" + 2M 2 (: x) 2 ; x) + M jTj(e0(:); x) e0(x)j "Tj(e0(:); x) + 2M 2 Tj((: x) 2 ; x) + M jTj(e0(:); x) e0(x)j " + (" + M )jTj(e0(:); x) e0(x)j +2M2 [Tj(e2(:); x) 2e1(x)Tj(e1(:); x) + e2(x)Tj(e0(:); x)] " + (" + M )jTj(e0(:); x) e0(x)j + 2M 2 jTj(e2(:); x) e2(x)j +4M je21(x)jjTj(e1(:); x) e1(x)j + 2M e2(x) 2 jTj(e0(:); x) e0(x)j " + (" + M +2M r 2 2 )jTj(e0(:); x) e0(x)j +4M r2 jTj(e1(:); x) e1(x)j + 2M 2 jTj(e2(:); x) e2(x)j
where r := maxfjaj; jbjg: So the last inequality gives for any > 0 that
jTj(g; x) g(x)j " + KjTj(e0(:); x) e0(x)j + KjTj(e1(:); x) e1(x)j
+ KjTj(e2(:); x) e2(x)j
where K := maxf" + M +2M r22;4M r2 ;2M2 g: By applying the modular % in the both
sides of the above inequality, since % is monotone, we have
%[ (Tj(g; :) g(:))] %[ " + KjTje0 e0j + KjTje1 e1j + KjTje2 e2j]:
So we may write that
%[ (Tj(g; :) g(:))] %[4 "]+%[4 K(Tje0 e0)]+%[4 K(Tje1 e1)]+%[4 K(Tje2 e2)]:
Since % is Q-quasi semiconvex and strongly …nite, we have
%[ (Tj(g; :) g(:))] Q"%[4 Q] + %[4 K(Tje0 e0)] + %[4 K(Tje1 e1)]
without loss of generality where 0 < " 1. Hence 1 p(t) 1 X j=0 pjtj%[ (Tj(g; :) g(:))] Q"%[4 Q] + 1 p(t) 1 X j=0 pjtj%[4 K(Tje0 e0)] + 1 p(t) 1 X j=0 pjtj%[4 K(Tje1 e1)] + 1 p(t) 1 X j=0 pjtj%[4 K(Tje2 e2)]
and taking limit superior as t ! R in the both sides, by using hypothesis, we get lim t!R 1 p(t) 1 X j=0 pjtj%[ (Tjg g)] = 0
which proves our claim. Now let f 2 L%(G) satisfying f g 2 X
T for every
g 2 C1(G): Since jGj < 1 and % is strongly …nite and absolutely continuous, it is
known that % is also absolutely …nite on X(G) (see [3]). Using the properties of % and it is also known from [8] that the space C1(G) is modularly dense in L%(G),
i.e., there exists a sequence fgkg C1(G) such that
lim
k %[3 0(gk f )] = 0 f or some 0> 0:
This means that, for every " > 0, there is a positive number k0= k0(") so that
%[3 0(gk f )] < " f or every k k0:
On the other hand, by linearity and positivity of the operators Tj we may write
that
0jTjf f j 0jTj(f gk0)j + 0jTjgk0 gk0j + 0j(gk0 f )j:
Applying the modular % in the both sides of the above inequality, since % is monotone %[ 0(Tjf f )] %[3 0(Tj(f gk0))] + %[3 0(Tjgk0 gk0)] + %[3 0((gk0 f ))]:
Then it follows from the above inequalities that
%[ 0(Tjf f )] %[3 0(Tj(f gk0))] + %[3 0(Tjgk0 gk0)] + ":
Hence, using the facts that gk02 C1(G) and f gk0 2 XT; we have
1 p(t) 1 X j=0 pjtj%[ 0(Tjf f )] 1 p(t) 1 X j=0 pjtj%[3 0(Tj(f gk0))] + 1 p(t) 1 X j=0 pjtj%[3 0(Tjgk0 gk0)] + ":
Taking limit superior as t ! R in both sides, we obtain that lim sup t!R 1 p(t) 1 X j=0 pjtj%[ 0(Tjf f )] " + P %[3 0(f gk0)] + lim sup t!R 1 p(t) 1 X j=0 pjtj%[3 0(Tjgk0 gk0)] (3.3) which gives lim sup t!R 1 p(t) 1 X j=0 pjtj%[ 0(Tjf f )] "+"P +lim sup t!R 1 p(t) 1 X j=0 pjtj%[3 0(Tjgk0 gk0)]: By (3.2), we get lim sup t!R 1 p(t) 1 X j=0 pjtj%[3 0(Tjgk0 gk0)] = 0
and this implies
lim sup t!R 1 p(t) 1 X j=0 pjtj%[ 0(Tjf f )] " + "P:
Since " is arbitrary positive real number, we have lim sup t!R 1 p(t) 1 X j=0 pjtj%[ 0(Tjf f )] = 0 and also p(t)1 P1 j=0 pjtj%[ 0(Tjf f )] is nonnegative then lim t!R 1 p(t) 1 X j=0 pjtj%[ 0(Tjf f )] = 0:
This completes the proof.
If the modular % satis…es the 2-condition, then one can get the following result
from the above theorem.
Theorem 2. Let % and T = fTjg be as in the above theorem. If % satis…es the 2-condition, then the followings are equivalent:
lim t!R 1 p(t) 1 P j=0 pjtj%[ (Tjei ei)] = 0; f or every > 0 and i = 0; 1; 2: lim t!R 1 p(t) 1 P j=0
pjtj%[ (Tjf f )] = 0; f or every > 0 then every f 2
L%(G) such that f g 2 X
4. Concluding Remarks
Take G = [0; 1] and let ' : [0; 1) ! [0; 1) be a continuous function for which the following conditions hold:
' is convex,
'(0) = 0; '(u) > 0 f or u > 0 and lim
u!+1'(u) = 1:
Here, consider the functional ' on X(G) de…ned by '(f ) :=
1
Z
0
'(jf(x)j)dx; for f 2 X(G):
In this case, ' is a convex modular on X(G) (see [4]). Consider the Orlicz space generated by ' as follows
L'(G) := ff 2 L0(G) : '( f ) < 1 for some > 0g:
Then, consider the following classical Bernstein-Kantorovich operator U := fUjg
on the space L'(G) (see [4]) which is de…ned by
Uj(f ; x) := j X k=0 j r x k(1 x)j k(j + 1) k+1 j+1 Z k j+1 f (t)dt; x 2 G:
Observe that the operators Uj map the Orlicz space L'(G) into itself. Moreover,
it is also known that the property lim sup
j!1
%( (Tjh)) P %( h) is satis…ed with the
choice of XU := L'(G) and for every function f 2 L'(G)such that f g 2 XU for
every g 2 C1(G), fU
jf g is modularly convergent to f. Using the operators fUjf g
de…ne the sequence of positive linear operators V := Vj on L'(G) as follows:
Vj(f ; x) = (1 + sj)Uj(f ; x); f or f 2 L'(G); x 2 [0; 1] and j 2 N; (4.1)
where fsjg is a sequence of zeros and ones which is not convergent but convergent
to 0 in the sense of power series method. By Lemma 5.1 of [4], for every h 2 XV :=
L'(G), all > 0 and for an absolute positive constant P , we get
'( V jh) = '[ (1 + sj)Ujh] '(2 Ujh) + '(2 sjUjh) = '(2 Ujh) + sj '(2 Ujh) = (1 + sj) '(2 Ujh) (1 + sj)P '(2 h): Then, we get lim sup t!R 1 p(t) 1 X j=0 pjtj '( Vjh) P '(2 h):
Vj(e0; x) = 1 + sj Vj(e1; x) = (1 + sj)f jx j + 1+ 1 2(j + 1)g Vj(e2; x) = (1 + sj)f j(j 1)x2 (j + 1)2 + 2jx (j + 1)2 + 1 3(j + 1)2g
where ei(t) = ti. So for any > 0, we can see, that
jVj(e0; x) e0(x)j = j1 + sj 1j = sj; which implies '[ (V j(e0) e0)] = '( sj) = 1 Z 0 '( sj)dx = '( sj) = sj'( )
because of the de…nition of fsjg: Since fsjg is convergent to 0 in the sense of power
series method, for every > 0 lim sup t!R 1 p(t) 1 X j=0 pjtj '( Vj(e0) e0) = lim sup t!R 1 p(t) 1 X j=0 pjtjsj'( ) = 0: Also jVj(e1; x) e1(x)j = x( j j + 1+ jsj j + 1 1) + 1 2(j + 1)+ sj 2(j + 1) f2(j + 1)3 + sj( 2j + 1 2(j + 1))g; we may write that
'[ (V j(e1) e1)] ' fsj( 2j + 1 2(j + 1)) + 3 2(j + 1)g sj '( ( 2j + 1 (j + 1))) + '( 3 (j + 1))
by the de…nitions of fsjg and '. Since f(j+1)2j+1g is convergent, there exists a constant
M > 0 such that f(j+1)2j+1 M g, for every j 2 N: Then using the monotonicity of ', we have
'[ 2j + 1
(j + 1)]
'( M )
for any > 0, which implies
'[ (V j(e1) e1)] sj '( M ) + '( 3 j + 1) = sj'( M ) + '( 3 j + 1):
Since ' is continuous, we have lim j '( 3 j+1) = '(limj 3 j+1) = '(0) = 0: So we get
'(j+13 ) is convergent to 0 in the sense of power series method. Using this and by the de…nition of fsjg, we obtain
lim sup t!R 1 p(t) 1 X j=0 pjtj '( Vj(e1) e1) lim sup t!R 1 p(t) 1 X j=0 pjtj[sj'( M ) + '( 3 j + 1)] = '( M ) lim sup t!R 1 p(t) 1 X j=0 pjtjsj+ lim sup t!R 1 p(t) 1 X j=0 pjtj'( 3 j + 1) = 0 Finally, since jVj(e2; x) e2(x)j = x2j(j 1) (j + 1)2 + 2jx (j + 1)2 + 1 3(j + 1)2 + sj j(j 1)x2 (j + 1)2 + sj 2jx (j + 1)2 + sj 1 3(j + 1)2 x 2 f 15j + 4 3(j + 1)2 + sj( 3j2+ 3j + 1 3(j + 1)2 )g:
Since f3j3(j+1)2+3j+12 g is convergent, there exists a constant K > 0 such that j
3j2+3j+1 3(j+1)2 j
K, for every j 2 N. Then using the monotonicity of ' and the de…nition of fs jg, we have '[ (V j(e2) e2)] ' 2 ( 15j + 4 3(j + 1)2) + ' 2 s j( 3j2+ 3j + 1 3(j + 1)2 ) '( ( 30j + 8 3(j + 1)2)) + '(2 s jK);
where which yields
'[ (V
j(e2) e2)] '( (
30j + 8
3(j + 1)2)) + sj'(2 K):
Since ' is continuous, we have lim
j '( 30j+8 3(j+1)2) = '( lim j 30j+8 3(j+1)2) = '(0) = 0: So
we get '( 3(j+1)30j+82) is convergent to 0 in the sense of power series method. Using
this and by the de…nition of fsjg, we obtain
lim sup t!R 1 p(t) 1 X j=0 pjtj '( Vj(e2) e2) = 0; f or every > 0:
So we can say that our sequence V := fVjg satis…es all assumptions of Theorem 1.
Therefore we conclude that lim sup t!R 1 p(t) 1 X j=0 pjtj '( 0Vj(f ) f ) = 0; f or some 0> 0
holds for every f 2 L'(G) such that f g 2 XV for every g 2 C1(G). However since
fsjg is not convergent to zero, it is clear that fVj(f )g is not modularly convergent
to f . Note that
in the case of R = 1, p (t) = 1
1 t and for j 0, pj = 1 the power series method coincides with Abel method which is a sequence-to-function transformation,
in the case of R = 1, p (t) = et and for j 0, p
j =
1
j! the power series method coincides with Borel method.
We can therefore give all of the theorems of this paper for Abel and Borel con-vergences.
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Current address : Department of Mathematics, Hitit University, Çorum, Turkey E-mail address : tugbayurdakadim@hotmail.com
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