Selçuk J. Appl. Math. Selçuk Journal of Special Issue. pp. 27-38, 2010 Applied Mathematics
Local Estimation of Fractional Integrals Generated by the General-ized Shift Operators in the Weighted Spaces of Summable Functions Hüseyin Yıldırım
Department of Mathematics, Faculty of Science and Arts, Afyon Kocatepe University, Afyon, Türkiye
e-mail: hyildir@ aku.edu.tr
Presented in 2National Workshop of Konya Ere˘gli Kemal Akman College, 13-14 May 2010.
Abstract. In this paper, we consider the generalized shift operator, generated by the Bessel differential operator, defining the fractional integrals generated by the generalized shift operator, their local estimations are obtained.
Key words: Fractional Integral, Generalized Shift Operator, Maximal Func-tion, BMO Spaces.
2000 Mathematics Subject Classification: 42A38, 42B10, 42B20, 42B25. 1. Introduction
The purpose of this paper is to obtain local estimation of fractional integrals generated by the generalized shift operators for the classical fractional integrals introduced in N.K. Karapetyants and A.I. Ginsburg [3]. On the other hand, the local estimates of classical Riesz potentials were studied by N.K. Karapetyants and A.I. Ginsburg [1]. However, we studied the properties of generalized Riesz potentials generated by the generalized shift operator in [6] and [7]. Moreover, it is well known that the fractional integral
+ = +∗ provides the following conditions[5].
a. From (0 1) to (0 1) 1 ≤ ≤ where = (1 − )−1 for 1 b. From (0 1) to (0 1) for = 1
c. From (0 1) to −
1
(0 1) for 1
On the solution of integral equations of the first type + = is arising the necessity more exactknowledge about image of operator
+. For this reason, it is introduced a local characteristic [2]:
(1) ( ) = µ 1 || R I |()| ¶1 ⊂ (0 1) 1 ≤ ≤ ∞
and its asymptotic character is considered as || → 0 This estimations give necessary conditions for representation of function by the fractional integral. The local characteristics (1) are average of the absolute value of function and th order. These characteristics are also related to the maximal function of the Hardy-Littlewood.
On the other hand these local characteristics for fractional integrals and Riesz potentials were studied in [1],[2],[3] [5] and [8].
We will introduce and study similar characteristics on the weighted space (0 1) with weight () = which is generalization of their results [3].
2. Notations and Representations of the Main Results
= (0 1) is defined with respect to the Lebesque measure 2 as following = (0 1) = ⎧ ⎪ ⎨ ⎪ ⎩ : kk = ⎛ ⎝ 1 Z 0 |()|2 ⎞ ⎠ 1 ∞ ⎫ ⎪ ⎬ ⎪ ⎭ 1 + 1 0= 1
where 1 ≤ ∞ and 0 is a fixed parameter. On the other hand kk∞(01)= sup
∈(01) |()|
If ∈ () for any ⊆ (0 1) then is sait to be p-locall summable in (0 1) The collection of p-locall summable functions in (0 1) is denoted
(0 1) Denote by the generalized shift operator (B-shift operator)defined by
(2) () = Γ( + 1 2) Γ(1 2)Γ() Z 0 ³p2+ 2− 2 cos ´sin2−1 where ∈ R+ [4].
We note that is closely connected with the Bessel differential equation (see [4] for details) as 2 2 + 2 = 2 2 + 2 ( 0) = () ( 0) = 0
The convolution operator determined by the is as follow.
(3) ( ∗ )() =
∞ Z
0
This convolution known as a B-convolution. We note the following properties of the B-convolution and [4] and [6].
a. ∗ = ∗ b. 1 = 1
c. If () () ∈ (R+) , () is a bounded function all 0 and ∞ Z 0 () 2 ∞ then ∞ Z 0 ()()2 = ∞ Z 0 ()()2
d. Let () = 1 in (c) . Then there is a following equality with (b) ∞ Z 0 ()2 = ∞ Z 0 ()2 e. | ()| ≤ sup ≥0 |()|
We will show the operators of Riemann-Liouville fractional integrals by + which is generated by the generalized shift operator:
(+)() = R 0 ()||−2−12 0 1 0 2 + 1 where ||−2−1= Γ( + 1 2) Γ(1 2)Γ() Z 0 ³p 2+ 2− 2 cos ´−2−1sin2−1.
Here the operator + is a generalized fractional integral generated by the generalized shift operator. Furthermore, 0 2 + 1 1 ≤ ≤ ∞ 0 = ( − 1)−1 = (1 + 2) 1 + 2 − for 2 + 1 (4) (0)( ) = µ 2+1 2+1 R 0 |()| 2 ¶1 (5) ()( ) = Ã 2+1 () +R −|()| 2 !1
where 0 ≤ 1, ( ) = ( + )2+1− ( − )2+1 Moreover, in the last case we assume that + ∈ (0 1) 2−
Lemma 1: Let ∈ (R+) Then, there is the following inequality | ()|≤ |()| for 1
+ 1
0 = 1 1 ∞
Proof : From Hölder’s inequality, | ()| = ¯ ¯ ¯ ¯ R 0 (p2+ 2− 2 cos ) sin2−1 ¯ ¯ ¯ ¯ ≤ µ R 0 ¯ ¯ ¯(p2+2−2 cos )¯¯¯sin2−1 ¶ µ R 0 sin2−1 ¶ 0 = (|()| ) where = Γ( +12) Γ(12)Γ()
Lemma 2: Let ∈ (R+) and 1 ∞ Then we k k≤ kk
Proof : From Lemma 1 there is the following inequality k k = ∞ R 0 | ()| 2 ≤ ∞R 0 (|()| ) 2
If we consider the properties (b) and (d) of the operator , then we have the following inequality k k ≤ µR∞ 0 |()| 2 ¶1 = kk Letter will denote all the constants we use in the theorem.
Theorem 1. Let = + 0 2 + 1, ∈ (0 1) 1 ≤ ≤ ∞ () = = 0 ()() ∈ (0 1) 0 1 + 2 Then, (6) (0)( ) ≤ −2 + 1 µR 0|| 2 ¶1
where if 1 2 + 1
1 ≤ ≤ if 2 + 1
≤ ∞ 1 ≤ ≤ ∞ if = 1 1 ≤ if = 2 + 1 1 ≤ ∞ According to these conditions, (7) ()( ) ≤ () kk 0 1 where, () = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ −2 + 1 for 1 2 + 1 1 ≤ ≤ −(2+1) for = 1 1 ≤ (1 + |ln |)01 for = 2 + 1 1 ≤ ∞ 1 for 2 + 1 ≤ ∞ 1 ≤ ≤ ∞
Proof : We will prove the theorem by several steps for 0 and 0 When = 0, we will consider the characteristic (0)( ) and ()( ) while 0 1
I. The case = 0. In this case, we have
(8) (0)( ) = µ 2+1 2+1 R 0 || 2 ¶1 = µ 2+1 2+1 R 0 ¯ ¯ ¯ ¯ R 0 ()||−2−12 ¯ ¯ ¯ ¯ 2 ¶1 ≤ µ 2+1 2+1 R 0 ¯ ¯ ¯ ¯ R 0 ∙µ ¶ −1 ¸ Ψ()||−2−12 ¯ ¯ ¯ ¯ 2 ¶1 + µ 2+1 2+1 R 0 ¯ ¯ ¯ ¯ R 0 Ψ()||−2−12 ¯ ¯ ¯ ¯ 2 ¶1 = (0) µ + ∙µ ¶ − 1 ¸ Ψ ¶ + (0)¡+Ψ ¢ where Ψ() = () ∈ (0 1)
If 1 ≤ ≤ 2 + 1 ≤ and by applying the changes of variables instead of and instead of in the above stated expression
(Ψ ) = (0) µ + ∙µ ¶ − 1 ¸ Ψ ¶
then, we can also obtain ( (Ψ )) = (2 + 1) 1 R 0 ¯ ¯ ¯ ¯ R 0 Ψ() || −2−1∙µ ¶ − 1 ¸ 2 ¯ ¯ ¯ ¯ 2 = (2 + 1) 1 R 0 |( )| 2
Dividing the set of integration in the interior integral ( ) to interval (+1 ) where
= 2−
= 0 1 2 we represent ( ) in the form
( ) = R 1 Ψ() ∙µ ¶ − 1 ¸ ||−2−12 + P∞ =1 R +1 Ψ() ∙µ ¶ − 1 ¸ ||−2−12 = 1+ 2
Then we get the following inequality for (Ψ )
(Ψ ) ≤ µ (2 + 1) 1 R 0 | 1( )|2 ¶1 + µ (2 + 1) 1 R 0 | 2( )|2 ¶1 = 1(Ψ ) + 2(Ψ ) Note that in the integral 1
µ ¶ − 1 ≤ 1 Therefore we have |1| ≤ R 1 |Ψ()| ||−2−12 Where we are applying Hölder’s inequality for the factors |Ψ()|1− ³ ||−2−1´ 1 −2−1(− 2+1 0 ) and |Ψ()| ³ ||−2−1´ 1 −2−1(−1) with 1 = 2= − 3= 011 +12 +13 = 1 Moreover, 2 = 1−2+1 By Lemma 1 and Lemma 2 we have the following inequality,
|1| ≤ R 1 |Ψ()| ||−2−12 ≤ Ã R 1 |Ψ()|1 ³ ||−2−1´ 1 −2−1(− 1 ) 2 !1
× Ã R 1 |Ψ()|2(1−)2 !1 2Ã R 1 ³ ||−2−1´ 3 −2−1(− 2+1 0 ) 2 !1 3 ≤ Ã R 1 |Ψ()|||−12 !1 × Ã R 1 |Ψ()|2 !1 2Ã R 1 ||0−2+12 !1 3 ≤ −(2+1)(1− 1 )kΨk1− 1 Ã R 1 |Ψ()|||−12 !1 where R 1 ||0−2+12 ∞ Then, we obtain 1(Ψ ) = µ (2 + 1) 1 R 0| 1( )|2 ¶1 ≤ −(2+1)(1− 1 )kΨk1− Ã R 1 |Ψ()| µR1 0 ||−12 ¶ 2 !1 ≤ −2+1 kΨk
Now, let us consider the estimation of 2 By 0 2 + 1 µ ¶ − 1 ≤ 2+1− 1 and ||−2−1≤ 22+1− 2+1−, we have |2| = ∞ P =1 R +1 |Ψ()| ∙µ ¶ − 1 ¸ ||−2−12 ≤ P∞ =1 (2+1− 1) R +1 |Ψ()| ||−2−12 ≤ P∞ =1 (2+1− 1)−2−1 R +1 |Ψ()| 2 ≤ −2−1 P∞ =1 (2+1− 1) R +1 |Ψ()| 2 ≤ −2+1 −2−1kΨk ∞ P =1 (2+1− 1) Ã R +1 2 !1 0 ≤ −2+1 −2−1 kΨk ∞ P =1 (2+1− 1)2−(2+1)0 ≤ −2+1 − 2−1 kΨk
Therefore, we have 2(Ψ ) ≤ µR1 0 | 2(Ψ )|2 ¶1 ≤ −2+1 kΨk µ1 R 0 (−2+1 )2 ¶1 ≤ −2+1 kΨk The case 1 ≤ 2+1 follows from inequality
(0)( ) ≤ (0)( )
In the case 1 ≤ ≤ ∞ and 2+1 ≤ ∞ we have the following inequality.
(9) ¯ ¯ ¯ ¯+ µ∙µ ¶ − 1 ¸ Ψ ¶¯¯¯¯ ≤ ¯ ¯ ¯ ¯ ¯ R 0 |Ψ()| ∙µ ¶ − 1 ¸0 ||−2−12 ¯ ¯ ¯ ¯ ¯ ≤ µR 0 |Ψ()| 2 ¶1 µR 0 ∙µ ¶ − 1¸ ³||−2−1´ 0 2 ¶1 0 ≤ µR 0 |Ψ()| 2 ¶1 µR 0 ³ ||−2−1´ 0 2 ¶1 0 ≤ −2+1 µR 0|Ψ()| 2 ¶1
Thus, there is the following inequality for the (0)( )
(0)( ) ≤ −2+1 kΨk µR1 0 (−2+1 )2 ¶1 ≤ −2+1 kk
Since = , 1 2+1 and 1 = (2+1)(2+1)− =1−2+1 we can show that + is a bounded operator from (0 1) to (0 1) by applying
Hardy-Littlewood theorem. The boundedness properties of the classical fractional in-tegrals were studied by Karapetians [1], [2] and [3]. On the other hand, smiler boundedness was studied by Yildirim for the generalized Riesz potentials gen-erated by the shift operators in [6].
II. The case 0 Firstly consider the cases 1 ≤ ≤ ∞ 2 + 1
From (9), we have (by using 2 + 1) ()( ) ≤ kΨk ⎛ ⎜ ⎜ ⎜ ⎝ (2 + 1) ( + ) (−2 + 1 ) ( ) +R − (−2+1 )2 ⎞ ⎟ ⎟ ⎟ ⎠ 1 ≤ kΨk
where 2 ∈ (0 1) Here we write the following inequality for the other cases. (10) () ∙ + µ∙µ ¶ − 1 ¸ Ψ ¶¸ ≤ à 2+1 () +R − ¯ ¯ ¯ ¯ ¯ R − ∙µ ¶ − 1 ¸ Ψ()||−2−12 ¯ ¯ ¯ ¯ ¯ 2 !1 + à 2+1 () +R − ¯ ¯ ¯ ¯ R− 0 ∙µ ¶ − 1 ¸ Ψ()||−2−12 ¯ ¯ ¯ ¯ 2 !1 = 1+ 2
If we apply the change of variables = − + = − + in the integral 1 then we obtain 1 = à 2+1 () +R − ¯ ¯ ¯ ¯ ¯ R − ∙µ ¶ − 1 ¸ Ψ()||−2−12 ¯ ¯ ¯ ¯ ¯ 2 !1 = à 2+1 () 2 R 0 ¯ ¯ ¯ ¯ ¯ R 0 h³ +− +− ´ − 1i+−| + − |−2−1 × Ψ( + − )( + − )2¯¯( + − )2¢1 = (0)h+³h³++−−´− 1iΨ( + − ) ( )´i ≤ (0) ∙ + µ∙µ ¶ − 1 ¸ Ψ( + − ) ( ) ¶¸ ≤ −2+1 kΨk
Due to proved case above for the = 0.
2 ≤ ⎛ ⎝ 2+1 () +R − ¯ ¯ ¯ ¯ ¯ − 2 R 0 ∙µ ¶ − 1 ¸ ||−2−1Ψ()2 ¯ ¯ ¯ ¯ ¯ 2 ⎞ ⎠ 1 + ⎛ ⎝ 2+1 () +R − ¯ ¯ ¯ ¯ ¯ ¯ R− − 2 ∙µ ¶ − 1 ¸ ||−2−1Ψ()2 ¯ ¯ ¯ ¯ ¯ ¯ 2 ⎞ ⎠ 1 = 20 + 200 In the 20 we have 2 and therefore || −2−1 ≤ 21+2−−2−1 Then, 20 = ⎛ ⎝ 2+1 () +R − ¯ ¯ ¯ ¯ ¯ − 2 R 0 ∙µ ¶ − 1 ¸ ||−2−1Ψ()2 ¯ ¯ ¯ ¯ ¯ 2 ⎞ ⎠ 1 ≤ ⎛ ⎝ 2+1 () +R − 21+2−(−2−1+) ¯ ¯ ¯ ¯ ¯ − 2 R 0 −Ψ()2 ¯ ¯ ¯ ¯ ¯ 2 ⎞ ⎠ 1 ≤ ⎛ ⎝ 2+1 ()( − ) (−2−1)( + )+R − ¯ ¯ ¯ ¯ ¯ − 2 R 0 −Ψ()2 ¯ ¯ ¯ ¯ ¯ 2 ⎞ ⎠ 1 = ( − )−2−1( + ) − 2 R 0 −Ψ()2 ≤ ( − )−2−1( + )( − )2+10 −kΨk ≤ −2+1 ³ + − ´ kΨk ≤ −2+1 kΨk where we used 2 + − 3 and 0 = 2 + 1 For estimation 200 we use
≤ 2 + − ≤ 6 , 2 Then, we have 200 ≤ ⎛ ⎝ 2+1 () +R − ⎛ ⎝R− − 2 ||−2−1|Ψ()| 2 ⎞ ⎠ 2 ⎞ ⎠ 1 ≤ −2+1 kΨk
Finally, consider the case = 2 + 1 In the (10), the estimations for 1can be realized in the similar way as mentioned above. For estimation 2 consider
0
and 200 Moreover, it is easy to see the estimation
0
2≤ kΨk For 1+ 1 0 = 1 and − 2 − 1 = −2+10 we have following inequality.
200≤ ⎛ ⎝ 2+1 () +R − 2 ¯ ¯ ¯ ¯ ¯ ¯ −R − 2 ||−2−1|Ψ()| 2 ¯ ¯ ¯ ¯ ¯ ¯ ⎞ ⎠ 1 ≤ ⎛ ⎝ 2+1 () +R − 2 ¯ ¯ ¯ ¯ ¯ ¯ R− − 2 ||−2+10 |Ψ()| 2 ¯ ¯ ¯ ¯ ¯ ¯ ⎞ ⎠ 1 ≤ ⎛ ⎜ ⎝ ()2+1 +R − 2 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ⎛ ⎝R− − 2 ||−(2+1)2 ⎞ ⎠ 1 0 ⎛ ⎝R− − 2 |Ψ()|2 ⎞ ⎠ 1 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ⎞ ⎟ ⎠ 1 ≤ kΨk ⎛ ⎜ ⎝ ()2+1 +R − 2 ¯ ¯ ¯ ¯ ¯ ¯ R− − 2 ||−(2+1)2 ¯ ¯ ¯ ¯ ¯ ¯ 0 ⎞ ⎟ ⎠ 1 ≤ kΨk ¯ ¯ln(− 2 ) − ln( − ) ¯ ¯01 à 2+1 () +R − 2 !1 ≤ kΨk ¯ ¯ln¡− 2 ¢¯¯ 1 0
where¯¯ln¡−2 ¢− ln( − )¯¯ ≤¯¯ln¡2−¢¯¯ − ∈ (0 1) and 2 Then, we have
200 ≤ kΨk(1 + |ln |)01
This completes the proof for the case 0 III. The case 0 The case is easy, since
(0)( ) ≤ µ 2+1 () R 0 2 ¯ ¯ ¯ ¯ R 0 |Ψ()| ||−2−12 ¯ ¯ ¯ ¯ ¶1 ≤ (0)(|Ψ| ) By analogy ()( ) ≤ ()(|Ψ| ) and it is possible use "without weight" estimations.
IV.As final remarks, first of all, note that the constant independent of in the Theorem 1 and for = 1 + 2 it hold estimation
= ³
1+20
´
Moreover, the estimation of Theorem 1 is the best possible in the scale of loga-rithmic degree.
Theorem 2: In conditions of theorem 1 for 1 ≤ ∞ it is true that (11) (0)( ) = ³−2+1
´
→ 0 By analogy in the case 1 ≤ ≤ 2+1 we have,
(12) ()( ) = (()) → 0
Proof: The first relation of Theorem 2 follows from (6). For proving the second, the following inequality is used,
()( ) ≤ ()( ) + ()( + ( − ) ) ≤ ³ kk∞+ () k( − )k ´ where = + ∈ ∞[ ] and k − k → 0 as → ∞ References
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