**Journal of Applied Mathematics and Physics**

Vol.06 No.10(2018), Article ID:87729,9 pages

10.4236/jamp.2018.610169

Is ?A^{−1} an Infinitesimal Generator?

Ru Liu^{ }

College of Information Science and Engineering, Chengdu University, Chengdu, China

Copyright © 2018 by author and Scientific Research Publishing Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

Received: September 12, 2018; Accepted: October 7, 2018; Published: October 10, 2018

ABSTRACT

There are some researchers considering the problem whether is the generator of a bounded C_{0}-semigroup if A generates a bounded C_{0}-semigroup. Actually, it is a very basic and important problem. In this paper, we discuss whether is the generator of a bounded α-times resolvent family if generates a bounded α-times resolvent family.

**Keywords:**

α-Times Resolvent Family, Analytic α-Times Resolvent Family, Fractional Power of Generator

1. Introduction

In paper [1] , the author studies the problem whether is the generator of a bounded C_{0}-semigroup if A generates a bounded C_{0}-semigroup. We know that α-times resolvent operator family is generalization of C_{0}-semigroup and C_{0}-semigroup is 1-times resolvent operator family. So, in this paper, we will show that when the operator generates a bounded α-times resolvent operator family, under certain condition, is also the generator of a bounded α-times resolvent operator family. The representation of such bounded α-times resolvent operator family will be obtained, too. Furthermore, we will consider the problem whether owns this property.

Let us first recall the definitions of α-times resolvent operator family. Let A be a closed densely defined linear operator on a Banach space X and. is a Mittag-Leffler function.

Definition 1.1 [2] A family is called an α-times resolvent operator family for A if the following conditions are satisfied:

1) is strongly continuous for and;

2) and for and ;

3) For, satisfies

where,.

If where, we write as (or shortly). Then we give the definitions of analytic α-times resolvent operator family.

Definition 1.2 [2] An α-times resolvent family is called analytic if admits an analytic extension to a sector for some, where. An analytic solution operator is said to be of analyticity type if for each and, there is such that.

Then we give a Lemma which will be used later.

Lemma 1.1 [2] . Then if and only if and there is a strongly continuous operator-valued function satisfying, and such that

2. Main Theorem and Conclusion

Theorem 2.1. On a Hilbert space H, the following statements are equivalent:

(1),;

(2) A is a closed, densely defined operator, , and for,;

(3) A is a closed, densely defined operator, for, and is invertible for some.

Proof. (2) Þ (3) For, , then we have for,

(1)

hence we know is invertible from the proposition 1.5 of chapter 3 in book [3] . While, from equation (1), we can also have for, , then.

(3) Þ (2) Since, , then for,. is invertible for some imply that is invertible for any. Together with A is closed and densely defined, we have, hence and .

(1) Þ (2) From lemma 1.3 of [4] , we know that A is a closed, densely defined operator. And we can get the other conclusion from theorem 2.8 of [2] .

(2) Þ (1). Firstly, set. For every, is a bounded operator and can commute with one another. It follows from Theorem 2.5 of [2] that generates an α-times resolvent family which is also uniformly continuous and exponential bounded.

For,. There exists a, such that, that is. Then

Since and, then we have that. It means that for,. Consequently, , and.

From Lemma II, 3.4(ii) of [5] , we have that converges to A pointwise on. If we can get the following properties, we will have.

(a) (*) exists for;

(b) is an α-times resolvent family which is generated by A;

(c).

(a) For is bounded, we can only need to prove (*) on. For,

where ((2.52) and (2.53) of [2] ). Together with, we can get that. Thus for,

By Lemma II, 3.4(ii) of [5] , is a Cauchy sequence for each. Therefore converges uniformly on each interval.

(b) I. For, is the uniformly continuous functions and so is continuous itself. For each, is uniformly bounded on every interval and, then so is. By Lemma I, 5.2 of [5] , is strongly continuous and.

II. For, , and. Together with that A is an closed operator, we have that. That is.

We have, and converge to A and pointwise, respectively. So, we have.

III. We know that

And for, converges uniformly on the interval, then

For all the above, we can obtain that is an α-times resolvent family which is generated by A.

(c) For each, and converges to pointwise, so, too. That is.

To sum up the above (a), (b) and (c), we can conclude that.

Theorem 2.2., Û A is a closed, densely defined operator, , and for,.

Proof. In the proof of the previous theorem, we have only used the properties of Hilbert space in the acquisition of and we can get without the properties of Hilbert space.

In fact, on a Banach space, for each, can generate a C_{0}-semigroup. Moreover,

From the subordination principle, we have, where. So

We can obtain that this theorem is tenable from the proof of the previous theorem.

Theorem 2.3. On a Hilbert space H, if, and exists as a closed, densely defined operator, then.

Proof. From the above Theorem 2.1, we have, and for,. And exists as a closed, densely defined operator, so it is easy to show that is bounded invertible for some, from (8) of [1] . Further more, for, then, thus there exists an, such that. Then . By Theorem 2.1, we can obtain that.

Theorem 2.4. If, , is the α-times resolvent family generated by it and. And if exists as a closed, densely defined operator, then generates an α-times resolvent family, which is given by

where is the first order Bessel function [6] . Moreover, there exists an, such that.

Proof. Since, then. Together with the assumption that is a closed, densely defined operator, we have that. Because of the property of Bessel function and for large t, , then

Thus, the integral is well defined. Set . Obviously, is strongly continuous and. From the above discussion, we can get that. For,

Consequently, we can obtain a conclusion that generates an α-times resolvent family from Lemma 2.1 and ,.

Theorem 2.5. A satisfies the assumption of Theorem 2.4, for, generates a bounded analytic α-times resolvent family. If, then

where

and

is oriented counterclockwise, where

and.

Proof., so and. For,. It follows from the Remark 2.8(a) of [7] that generates a bounded analytic α-times resolvent family. If, we set

Since, then

From [8] , we have that there exists an, such that. Next we estimate, it follows from (2.4) of [8] that

The same estimate holds for the integral on.

The same estimate holds for the integral on.

To sum up, we can conclude that there exists an, such that. So. Then we should show that is strongly continuous at 0. It following from the dominated convergence Theorem and Fubini Theorem that

For, it follows from Fubini Theorem that

From all the above, we can obtain a conclusion that if, generates a bounded analytic α-times resolvent family,

3. Conclusion

In this paper, we considered when the operator generates a bounded α-times resolvent operator family, under certain condition, as well as is also the generator of a bounded α-times resolvent operator family. Through the study of the problem whether is the generator of a bounded α-times resolvent operator family if A generates a bounded α-times resolvent operator family, we can know the generator A more clearly. Furthermore, this work can improve the study of the inverse problem.

Acknowledgements

The author was supported by Scientific Research Starting Foundation of Chengdu University, No. 2081915055.

Conflicts of Interest

The author declares no conflicts of interest regarding the publication of this paper.

Cite this paper

Liu, R. (2018) Is ?A^{−1} an Infinitesimal Generator? Journal of Applied Mathematics and Physics, 6, 1979-1987. https://doi.org/10.4236/jamp.2018.610169

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