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On periodic oneparameter groups of linear operators in a Banach space and applications
Journal of Inequalities and Applications volume 2013, Article number: 172 (2013)
Abstract
Let D be the infinitesimal generator of a strongly continuous periodic oneparameter group of linear operators in a Banach space. Main results: An analog of the resolvent operator (= quasiresolvent operator of D) is defined for points of the spectrum of D and its evident form is given. The theorem on integral for the operator D, theorems on the existence of periodic solutions of a linear differential equation of the n th order with constant coefficients and systems of linear differential equations with constant coefficients in Banach spaces are obtained. In the case of the existence of periodic solutions, evident forms of all periodic solutions of a linear differential equation of the n th order with constant coefficients and systems of linear differential equations with constant coefficients in Banach spaces are given in terms of resolvent and quasiresolvent operators of D.
MSC:42A, 43, 47D.
1 Introduction
Oneparameter groups of linear operators and periodic oneparameter groups of linear operators in topological vector spaces were investigated by Stone [1], Dunford [2], Gelfand [3] and others (see [4–20]).
Let T be the onedimensional torus \{{e}^{it}:\pi \le t<\pi \}. Further we consider T as the additive group Q/2\pi Z\simeq \{t:\pi \le t<\pi \} with its Euclidean topology, where Q is the field of real numbers. Let \alpha (t) (t\in T) be a strongly continuous oneparameter group of bounded linear operators in a Banach space H, and let D be an infinitesimal generator of the group \alpha (t). The evident form of the resolvent operator R(\mu ,D) of D is known (see [7], Lemma 2.25)
where μ is an element of the resolvent set of D. For the element μ of the resolvent set of D and arbitrary element a\in H, the element R(\mu ,D)a is the evident form of the unique solution of the equation Dx\mu x=a.
In the present paper, we obtain conditions of the existence of a solution of the equation Dx\mu x=a for points of the spectrum of D. We define an analog of the resolvent operator (= quasiresolvent operator) for points of the spectrum of D and, in the case of the existence of a solution, we give the evident form of all solutions by using a quasiresolvent operator of D. We apply resolvent and quasiresolvent operators to a solution of a linear differential equation P(D)x=a of the n th order with constant coefficients and to a system of linear differential equations of the first order with constant coefficients in a Banach space H.
Contents of the present paper is the following. In Section 2 we give generalizations of Fejer’s theorem and RiemannLebesque’s lemma for a strongly continuous linear representation of T in a Banach space. These results are used in the next sections.
In Section 3 we give a definition of the infinitesimal generator D of a strongly continuous linear representation of T in a Banach space H and the domain H(D) of the definition of D. For D and any \lambda \in C, we introduce the operator {R}_{\lambda}:H\to H by formula (1) below for the point λ of the resolvent set of D and by (2) for the point of the spectrum of D. We show that the linear operator {R}_{\lambda} is bounded and has properties (3) and (4) below.
In Section 4 we prove that H(D)={R}_{\lambda}(H) for all \lambda \in C, the spectrum \sigma (D) of the operator D is a point spectrum and \sigma (D)=\{im\in Spec(H)\}, where Spec(H) is the spectrum of the linear representation α. It is proved that {R}_{\lambda} is equal to the resolvent operator of D for all points λ of the resolvent set of D. We obtain the theorem on an integral for D.
In Section 5 we give conditions of the existence of a periodic solution of a linear differential equation of the n th order with constant coefficients. In the case of existence, the evident form of all periodic solutions is given.
In Section 6 we give conditions of the existence of a periodic solution of a system of linear differential equations of the first order with constant coefficients. In the case of existence, the evident form of all periodic solutions is given.
For simplicity, we prove our main results for an isometric strongly continuous linear representation. But they are true for any strongly continuous linear representation.
2 Fejer’s theorem and RiemannLebesque’s lemma for a strongly continuous linear representation of T in a Banach space
Denote the group of all invertible bounded linear operators A:H\to H of a complex Banach space H by \mathit{GL}(H). The following Definitions 14 are known [21].
Definition 1 A homomorphism \alpha :T\to \mathit{GL}(H) is called a linear representation of T on a Banach space H.
Definition 2 Linear representations \alpha :T\to \mathit{GL}(H) and \beta :T\to \mathit{GL}(V) are called equivalent if there exists a bounded invertible linear operator B:H\to V such that B\alpha (t)=\beta (t)B for all t\in T.
Definition 3 A linear representation α of T on a Banach space H is called isometric if \parallel \alpha (t)x\parallel =\parallel x\parallel for all t\in T and x\in H.
Definition 4 A linear representation α of T on a Banach space H is called strongly continuous if {lim}_{t\to 0}\alpha (t)x=x for all x\in H.
It is known that every strongly continuous linear representation of T on a Banach space is equivalent to a strongly continuous isometric linear representation of T on a Banach space [21].
Let α be a strongly continuous linear representation of T on a Banach space H, and let Z be the ring of all integers and n\in Z. Put
Let x\in H and n\in Z. By the theorem in ([22], p.314), Riemann’s integral
exists and {F}_{n}(x)\in {H}_{n}.
Proposition 1 Let α be a strongly continuous isometric linear representation of T on a Banach space H. Then

1.
\alpha (t){F}_{n}(x)={F}_{n}(\alpha (t)x)={e}^{int}{F}_{n}(x)
for all n\in Z, x\in H and t\in T;

2.
{F}_{n}\cdot {F}_{m}=0
and {F}_{n}^{2}={F}_{n} for all m,n\in Z, m\ne n;

3.
\parallel {F}_{n}(x)\parallel \le \parallel x\parallel
for all n\in Z and x\in H.
Proof It is easy, so it is omitted. □
A series in the form {\sum}_{k=\mathrm{\infty}}^{\mathrm{\infty}}{x}_{k}, {x}_{k}\in {H}_{k}, is called the Fourier series of an element x\in H, if {x}_{k}={F}_{k}(x) for all k\in Z. It is written in the form
For x\in H and for an integer number n\ge 0, let us put
is a subspace of H.
In the present paper, we assume that Spec(H) is infinite. The case of the finite Spec(H) is investigated easy and it is omitted.
Theorem 1 Let α be a strongly continuous isometric linear representation of T on a Banach space H. Then {lim}_{n\to \mathrm{\infty}}{\psi}_{n}(x)=x for every x\in H and \overline{{H}_{f}}=H.
Proof
In a standard manner, we obtain the equality
Since {lim}_{t\to 0}\alpha (t)x=x, for every \epsilon >0, there exists \delta >0 such that 0<\delta <\frac{\pi}{2} and \parallel \alpha (t)xx\parallel <\frac{\epsilon}{2} for all t\in (\delta ,\delta ). We have
for all t\in [\delta ,\pi ]. Since α is a strongly continuous isometric linear representation, {K}_{n}(t)={K}_{n}(t) for all t\in [\pi ,\pi ], \frac{1}{2\pi}{\int}_{\pi}^{\pi}{K}_{n}(t)\phantom{\rule{0.2em}{0ex}}dt=1 and sin\frac{\delta}{2}\le sin\frac{t}{2} for all t\in [\delta ,\pi ), it follows that
Hence {lim}_{n\to \mathrm{\infty}}{\psi}_{n}(x)=x. This, in view of {\psi}_{n}(x)\in {H}_{f} for all n, implies that \overline{{H}_{f}}=H. □
Remark 1 Theorem 1 is known for the homogeneous Banach spaces on T ([6], p.87; [23], pp.1415). For a strongly continuous linear representation of T in a locally convex space, it is obtained in [10].
Corollary 1 Let α be a strongly continuous isometric linear representation of T on a Banach space H and x,y\in H. If {F}_{n}(x)={F}_{n}(y) for each n\in Z, then x=y.
Proof Since {F}_{n}(x)={F}_{n}(y) for each n\in Z, it follows that {\psi}_{n}(x)={\psi}_{n}(y) for each n\in Z. Hence Theorem 1 gives x=y. □
Theorem 2 Let α be a strongly continuous isometric linear representation of T on a Banach space H. Then {lim}_{n\to \mathrm{\infty}}{F}_{n}(x)=0 for each x\in H.
Proof Let \epsilon >0 be given. Since {lim}_{t\to 0}\alpha (t)x=x, there exists a natural number N(\epsilon ) such that \parallel \alpha (\frac{\pi}{n})xx\parallel <\epsilon for all natural numbers n\ge N(\epsilon ). Since
we have
So, \parallel {F}_{n}(x)\parallel \le \parallel \alpha (\frac{\pi}{n})(x)x\parallel <\epsilon for all n\ge N(\epsilon ). □
Remark 2 This theorem is a generalization of RiemannLebesque’s lemma ([23], p.13).
3 The operator {R}_{\lambda}
Let α be a strongly continuous isometric linear representation of T on a Banach space H.
Definition 5 ([7], p.45)
A point x\in H is called a differentiable point of α if there exists
Denote the set of all differentiable points of α by H(D). The set Spec(H) is called the spectrum of D. The set C\setminus Spec(H) is called the resolvent set of D.
Proposition 2 Let α be a strongly continuous isometric linear representation of T on a Banach space H. Then

(i)
H(D)
is a linear subspace of H, {H}_{f}\subset H(D) and \overline{H(D)}=H;

(ii)
H(D)
is \alpha (T)invariant and \alpha (t)Dx=D\alpha (t)x for all t\in T, x\in H(D);

(iii)
D{F}_{n}(x)={F}_{n}(Dx)=in{F}_{n}(x)
for all n\in Z and x\in H(D).
Proof (i) It is obvious that H(D) is a linear subspace of H. Let x\in {H}_{f}. Then x can be expressed in the form {\sum}_{\ell =m}^{m}{F}_{\ell}(x) for some m. Since {F}_{\ell}(x)\in {H}_{\ell}, we get
Hence x\in H(D) and Dx={\sum}_{\ell =m}^{m}i\ell {F}_{\ell}(x). Therefore {H}_{f}\subset H(D). By Theorem 1 \overline{{H}_{f}}=\overline{H(D)}=H.

(ii)
Let x\in H(D). Since \alpha (t) is strongly continuous, we have
\alpha (t)Dx=\alpha (t)\underset{s\to 0}{lim}\frac{\alpha (s)xx}{s}=\underset{s\to 0}{lim}\frac{\alpha (s)\alpha (t)x\alpha (t)x}{s}=D\alpha (t)x.
Hence \alpha (t)x\in H(D) and \alpha (t)Dx=D\alpha (t)x.

(iii)
Let n\in Z and x\in H(D). Using the continuity of {F}_{n} and Proposition 1, we get
D{F}_{n}(x)={F}_{n}(Dx)=\underset{t\to 0}{lim}\frac{{e}^{int}{F}_{n}(x){F}_{n}(x)}{t}=in{F}_{n}(x).
□
Remark 3 It is easily seen that H=H(D) if and only if Spec(H) is finite.
Definition 6 The operator D is called an infinitesimal generator of a linear representation α (see [7], p.45).
Proposition 3 Let x\in H(D). Then the function {G}_{x}(t):=\alpha (t)x is differentiable on T and {G}_{x}^{\prime}(t)=\alpha (t){G}_{x}^{\prime}(0)=\alpha (t)Dx.
Proof Since α is strongly continuous, we have
□
Let α be a strongly continuous isometric linear representation of T on a Banach space H. For any x\in H and \lambda \in C, there exists the following vectorvalued Riemann’s integral:
We consider linear operators {R}_{\lambda} on H defined by
for all \lambda \in C such that \lambda \ne im, m\in Z and
for all m\in Z. In Theorem 4(iv) below, we prove that {R}_{\lambda} is equal to the resolvent operator of D for every point λ of the resolvent set of D. The operator {R}_{im}(x) is called the quasiresolvent operator of D for the point im of the spectrum of D. The operator {R}_{\lambda} was introduced in [9, 13].
Theorem 3 Let α be a strongly continuous isometric linear representation of T on a Banach space H. Then

(i)
the operator {R}_{\lambda}, defined by (1) and (2), is bounded for all \lambda \in C;

(ii)
{R}_{\lambda}({F}_{n}(x))={F}_{n}({R}_{\lambda}(x))=\frac{1}{in\lambda}{F}_{n}(x)(3)
for all \lambda \ne in, \lambda \in C, n\in Z;

(iii)
{R}_{in}({F}_{n}(x))={F}_{n}({R}_{in}(x))={F}_{n}(x)(4)
for all n\in Z and x\in H;

(iv)
{R}_{\lambda}{R}_{\mu}(x)={R}_{\mu}{R}_{\lambda}(x)
for all x\in H.
Proof (i) Let L:={\int}_{0}^{2\pi}\parallel {e}^{\lambda t}\parallel ({\int}_{0}^{t}\parallel {e}^{\lambda s}\parallel \phantom{\rule{0.2em}{0ex}}ds)\phantom{\rule{0.2em}{0ex}}dt. Then it is obvious that
Therefore {R}_{\lambda} is bounded in H for all \lambda \in C.

(ii)
Let x\in {H}_{f}, \lambda \in C and \lambda \ne im for all m\in Z. Then x={\sum}_{\ell =k}^{k}{F}_{\ell}(x) for some k. Since {F}_{\ell}(x)\in {H}_{\ell}, using Proposition 1, we get
\begin{array}{rcl}{R}_{\lambda}(x)& =& \frac{\lambda}{1{e}^{2\pi \lambda}}{\int}_{0}^{2\pi}{e}^{\lambda t}\left({\int}_{0}^{t}{e}^{\lambda s}(\sum _{\ell =k}^{k}\alpha (s){F}_{\ell}(x))\phantom{\rule{0.2em}{0ex}}ds\right)\phantom{\rule{0.2em}{0ex}}dt\\ +\frac{1}{2\pi (1{e}^{2\pi \lambda})}{\int}_{0}^{2\pi}(\sum _{\ell =k}^{k}\alpha (t){F}_{\ell}(x))\phantom{\rule{0.2em}{0ex}}dt\\ =& \frac{\lambda}{1{e}^{2\pi \lambda}}{\int}_{0}^{2\pi}{e}^{\lambda t}(\sum _{\ell =k}^{k}{\int}_{0}^{t}{e}^{(i\ell \lambda )s}{F}_{\ell}(x)\phantom{\rule{0.2em}{0ex}}ds)\phantom{\rule{0.2em}{0ex}}dt\\ +\frac{1}{2\pi (1{e}^{2\pi \lambda})}{\int}_{0}^{2\pi}\sum _{\ell =k}^{k}{e}^{i\ell t}{F}_{\ell}(x)\phantom{\rule{0.2em}{0ex}}dt\\ =& \frac{\lambda}{1{e}^{2\pi \lambda}}\sum _{\ell =k}^{k}{\int}_{0}^{2\pi}\frac{{e}^{i\ell t}}{i\ell \lambda}{F}_{\ell}(x)\phantom{\rule{0.2em}{0ex}}dt\frac{\lambda}{1{e}^{2\pi \lambda}}\sum _{\ell =k}^{k}{\int}_{0}^{2\pi}\frac{{e}^{\lambda t}}{i\ell \lambda}{F}_{\ell}(x)\phantom{\rule{0.2em}{0ex}}dt\\ +\frac{1}{2\pi (1{e}^{2\pi \lambda})}\sum _{\ell =k}^{k}{\int}_{0}^{2\pi}{e}^{i\ell t}{F}_{\ell}(x)\phantom{\rule{0.2em}{0ex}}dt\\ =& \frac{1}{{e}^{2\pi \lambda}1}{F}_{0}(x)+\sum _{\ell =k}^{k}\frac{1}{i\ell \lambda}{F}_{\ell}(x)\frac{1}{{e}^{2\pi \lambda}1}{F}_{0}(x)=\sum _{\ell =k}^{k}\frac{1}{i\ell \lambda}{F}_{\ell}(x).\end{array}
Hence, using {F}_{m}\circ {F}_{l}=0 for l\ne m (Proposition 1), we get {F}_{n}({R}_{\lambda}(x))=\frac{1}{in\lambda}{F}_{n}(x)={R}_{\lambda}({F}_{n}(x)).
Let x be an arbitrary element of H. By Theorem 1, {lim}_{p\to \mathrm{\infty}}{\psi}_{p}(x)=x. Since {\psi}_{p}(x)\in {H}_{f}, we have {F}_{n}({R}_{\lambda}({\psi}_{p}(x)))=\frac{1}{in\lambda}{F}_{n}({\psi}_{p}(x)) for all n\in Z and p\in N. On the other hand, from {lim}_{p\to \mathrm{\infty}}{\psi}_{p}(x)=x, using the continuity of operators {F}_{n} and {R}_{\lambda}, we get {F}_{n}({R}_{\lambda}(x))=\frac{1}{in\lambda}{F}_{n}(x)={R}_{\lambda}({F}_{n}(x)).
Now we prove equality (3) for \lambda =im, m\in Z, n\ne m. Let x\in {H}_{f} and {F}_{m}(x)=0. Then x={\sum}_{\ell \ne m,\ell =k}^{k}{F}_{\ell}(x) for some k\in N and \alpha (t)x={\sum}_{\ell \ne m,\ell =k}^{k}{e}^{i\ell t}{F}_{\ell}(x). Hence
It implies that
Let {F}_{m}(x)\ne 0. Then x= {F}_{m}(x)+{\sum}_{\ell \ne m,\ell =k}^{k}{F}_{\ell}(x) for some k\in N. Since {F}_{m}(x{F}_{m}(x))=0, using equalities (5) and (2), we have
Hence
This equality implies that
for all n,m\in Z, n\ne m, and {R}_{im}({F}_{m}(x))={F}_{m}({R}_{im}(x))={F}_{m}(x) for all m\in Z, x\in {H}_{f}.
Let x be an arbitrary element of H. By Theorem 1, {\psi}_{p}(x)\in {H}_{f} and equality (6), we obtain
for all n,m\in Z, n\ne m, x\in H.

(iii)
The proof of equality (4) is similar to the proof of equality (3).

(iv)
Using (3) and (4), we obtain {F}_{n}{R}_{\lambda}{R}_{\mu}(x)={F}_{n}{R}_{\mu}{R}_{\lambda}(x) for x\in H, n\in Z, \lambda \in C, \mu \in C. Hence, by Corollary 1, we have {R}_{\lambda}{R}_{\mu}(x)={R}_{\mu}{R}_{\lambda}(x) for all x\in H, \lambda \in C, \mu \in C. □
Corollary 2 Let a\in H, im\in Spec(H) and \lambda \in C. Then {F}_{m}{R}_{\lambda}(a)=0 if and only if {F}_{m}(a)=0.
Proof It follows easily from Theorem 3. □
Proposition 4 Let α be a strongly continuous isometric linear representation of T on a Banach space H. Then

(i)
{R}_{\lambda}{R}_{0}=\lambda ({R}_{\lambda}\circ {R}_{0})\frac{1}{\lambda}{F}_{0}
for all \lambda \in C, \lambda \ne im, m\in Z;

(ii)
{R}_{im}{R}_{0}=im({R}_{im}\circ {R}_{0})\frac{1}{im}{F}_{0}\frac{1}{im}{F}_{m}
for all m\in Z, m\ne 0.
Proof (i) Let x\in H, \lambda \in C and \lambda \ne im, m\in Z. For n\ne 0, using Theorem 4 and {F}_{n}{F}_{0}=0, we obtain
Similarly,
Hence {F}_{n}({R}_{\lambda}(x){R}_{0}(x))={F}_{n}(\lambda {R}_{\lambda}\circ {R}_{0}(x)\frac{1}{\lambda}{F}_{0}(x)) for every n\in Z. By Corollary 1 we have {R}_{\lambda}(x){R}_{0}(x)=\lambda {R}_{\lambda}\circ {R}_{0}(x)\frac{1}{\lambda}{F}_{0}(x) for all x\in H, \lambda \in C and \lambda \ne im, m\in Z.
A proof of (ii) is similar. □
4 The theorem on resolvent and quasiresolvent operators
Theorem 4 Let α be a strongly continuous isometric linear representation of T on a Banach space H. Then

(i)
H(D)={R}_{\lambda}(H)
for all \lambda \in C;

(ii)
{R}_{im}(Dim)x=x
and (Dim){R}_{im}(y)=y for all im\in Spec(H) and x\in H(D), y\in H such that {F}_{m}(x)={F}_{m}(y)=0;

(iii)
{R}_{\lambda}(D\lambda )x=x
and (D\lambda ){R}_{\lambda}(y)=y for all \lambda \in C\setminus Spec(H) and x\in H(D), y\in H;

(iv)
{R}_{\lambda}(x)={(1{e}^{2\pi \lambda})}^{1}{\int}_{0}^{2\pi}{e}^{\lambda s}\alpha (s)x\phantom{\rule{0.2em}{0ex}}ds
for all \lambda \in C\setminus Spec(H);

(v)
the spectrum \sigma (D) of D is a point spectrum and \sigma (D)=Spec(H).
Proof (i) We need the following two lemmas.
Lemma 1 D{R}_{0}(x)=x{F}_{0}(x) for all x\in {H}_{f}.
Proof An element x\in {H}_{f} has the form x={\sum}_{\ell =k}^{k}{F}_{\ell}(x) for some k. Using Theorem 3 and Proposition 2, we obtain {F}_{n}(D{R}_{0}(x))={F}_{n}(x) for all n\ne 0 and {F}_{0}(D{R}_{0}(x))=0. Hence D{R}_{0}(x)=x{F}_{0}(x) for any x\in {H}_{f}. Lemma is proved. □
Lemma 2 Let x\in H such that {F}_{0}(x)=0. Then
Proof Let us define the functions {f}_{n},f:[0,2\pi )\to H by {f}_{n}(t):=\alpha (t){R}_{0}({\psi}_{n}(x)) and f(t):=\alpha (t){R}_{0}(x). Since α is a strongly continuous isometric linear representation, we have
and
Equality (8) implies that
Using {F}_{0}(x)=0, Proposition 3, Theorem 3, {R}_{0}({\psi}_{n}(x))\in {H}_{f}, {F}_{0}({\psi}_{n}(x))={F}_{0}(x) and Lemma 1, we get
Hence
where {C}_{n}\in H. Putting t=0, we obtain {C}_{n}={f}_{n}(0)=\alpha (0){R}_{0}({\psi}_{n}(x))={R}_{0}({\psi}_{n}(x)) and
Using equalities (8), (9), (11) and inequality (10), we obtain
for all t\in T. Since {lim}_{n\to \mathrm{\infty}}{\psi}_{n}(x)=x, it follows that
and Lemma 2 is proved. □
We continue the proof of the theorem.

(i)
From equality (7) we obtain that the function f(t) is differentiable and {f}^{\prime}(t)=\alpha (t)x. Using Proposition 3, we have
\alpha (t)x={f}^{\prime}(t)=\alpha (t){f}^{\prime}(0)=\alpha (t)\underset{s\to 0}{lim}\frac{\alpha (s){R}_{0}(x){R}_{0}(x)}{s}=\alpha (t)D({R}_{0}(x)).
Hence {R}_{0}(x)\in H(D) for all x\in H such that {F}_{0}(x)=0. Now let x\in H be an arbitrary element. Since {F}_{0}(x{F}_{0}(x))=0, we have {R}_{0}(x{F}_{0}(x))\in H(D). On the other hand, by Theorem 3 and Proposition 2, {R}_{0}{F}_{0}(x)={F}_{0}(x)\in {H}_{0}\subset H(D). Since x=(x{F}_{0}(x))+{F}_{0}(x) and H(D) is a linear subspace of H, we get {R}_{0}(x)\in H(D). Hence {R}_{0}(H)\subset H(D).
Conversely, let x\in H(D). By Proposition 2 and Theorem 3, Dx\sim {\sum}_{n=\mathrm{\infty}}^{\mathrm{\infty}}in{F}_{n}(x) and {R}_{0}(Dx)\sim {\sum}_{n\ne 0,n=\mathrm{\infty}}^{\mathrm{\infty}}{F}_{n}(x). Hence {R}_{0}(Dx)+{F}_{0}(x)\sim {\sum}_{n=\mathrm{\infty}}^{\mathrm{\infty}}{F}_{n}(x). Using Corollary 1 and equality (4), we get {R}_{0}(Dx)+{F}_{0}(x)=x, {R}_{0}({F}_{0}(x))={F}_{0}(x) and {R}_{0}(Dx+{F}_{0}(x))=x. Let us put a:=Dx+{F}_{0}(x). Then {R}_{0}(a)=x. This means that x\in {R}_{0}(H), that is, H(D)\subset {R}_{0}(H); and consequently H(D)={R}_{0}(H).
Now we prove that {R}_{0}(H)={R}_{im}(H) for all m\in Z. By claim (ii) of Proposition 4 and claim (iv) of Theorem 3, {R}_{im}(x)={R}_{0}(x)+im{R}_{0}{R}_{im}(x)\frac{1}{im}{F}_{0}(x)\frac{1}{im}{F}_{m}(x). Since {F}_{0}(x)\in {H}_{0}, {F}_{m}(x)\in {H}_{m} and {H}_{f}\subset H(D)={R}_{0}(H), we get {R}_{im}(H)\subset {R}_{0}(H). By Theorem 3, {R}_{im}({H}_{f})={H}_{f}\subset {R}_{im}(H). Claim (ii) of Proposition 4 implies that {R}_{0}(x)={R}_{im}(x)im{R}_{im}{R}_{0}(x)+\frac{1}{im}{F}_{0}(x)+\frac{1}{im}{F}_{m}(x). Hence {R}_{0}(H)\subset {R}_{im}(H) and H(D)={R}_{im}(H). Similarly, using claims (i) and (ii) of Proposition 4, we obtain H(D)={R}_{\lambda}(H) for all \lambda \in C, \lambda \ne im, m\in Z. Thus H(D)={R}_{\lambda}(H) for all \lambda \in C.

(ii)
Let x\in H(D), {F}_{m}(x)=0. By Proposition 2, Dx\sim {\sum}_{k=\mathrm{\infty}}^{\mathrm{\infty}}ik{F}_{k}(x). Hence Dximx\sim {\sum}_{k=\mathrm{\infty}}^{\mathrm{\infty}}(ikim){F}_{k}(x). Using {F}_{m}(x)=0 and Theorem 3, we get {R}_{im}(Dim)x\sim {\sum}_{k\ne m,k=\mathrm{\infty}}^{\mathrm{\infty}}{F}_{k}(x). By Corollary 1, {R}_{im}(Dim)x=x. Thus {R}_{im}(Dim)x=x for all x\in H(D) such that {F}_{m}(x)=0.
Let x\in H, {F}_{m}(x)=0. Since x\sim {\sum}_{k\ne m,k=\mathrm{\infty}}^{\mathrm{\infty}}{F}_{k}(x), we have {R}_{im}(x)\sim {\sum}_{k\ne m,k=\mathrm{\infty}}^{\mathrm{\infty}}\frac{1}{ikim}{F}_{k}(x). According to the statement (i) of this theorem, {R}_{im}(x)\in H(D). Using Proposition 2, we get (Dim){R}_{im}(x)\sim {\sum}_{k\ne m,k=\mathrm{\infty}}^{\mathrm{\infty}}{F}_{k}(x), and hence (Dim){R}_{im}(x)=x. Thus (Dim){R}_{im}(x)=x for all x\in H such that {F}_{m}(x)=0.

(iii)
The proof of claim (iii) is similar to the one of (ii).

(iv)
Equalities (iii) mean that {R}_{\lambda} is the resolvent operator for all \lambda \in C\setminus Spec(H). Hence equality (iv) follows from the form of the resolvent operator in ([7], Lemma 2.25).

(v)
follows from (ii) and (iii). The proof of the theorem is completed. □
Remark 4 Equality (iv) means that for points λ of the resolvent set of D, {R}_{\lambda} is the other form of the resolvent operator of D.
Theorem 5 Let α be a strongly continuous isometric linear representation of T on a Banach space H, a\in H and m\in Z. Then the equation Dximx=a has a solution if and only if {F}_{m}(a)=0. In the case {F}_{m}(a)=0, a general solution of the equation Dximx=a has the form x={R}_{im}(a)+c, where c is an arbitrary element of {H}_{m}.
Proof Suppose that the equation Dximx=a has a solution x. By Proposition 2 we have Dx\sim {\sum}_{k=\mathrm{\infty}}^{\mathrm{\infty}}ik{F}_{k}(x) and Dximx\sim {\sum}_{k=\mathrm{\infty}}^{\mathrm{\infty}}(ikim){F}_{k}(x). Hence {F}_{m}(a)={F}_{m}(Dximx)=0.
For the converse, we assume that {F}_{m}(a)=0. By Theorem 4 we have (Dim){R}_{im}(a)=a. Therefore x={R}_{im}(a) is a solution of the equation Dximx=a. Let y\in H be a solution of the equation Dyimy=0. Then Dyimy\sim {\sum}_{k=\mathrm{\infty}}^{\mathrm{\infty}}(ikim){F}_{k}(y)={\sum}_{k\ne m,k=\mathrm{\infty}}^{\mathrm{\infty}}(ikim){F}_{k}(y)=0. Hence {F}_{k}(y)=0 for all k\ne m, that is, y={F}_{m}(y)\in {H}_{m}. On the other hand, Dyimy=0 for all y\in {H}_{m}. Thus a general solution of the equation Dximx=a has the form x={R}_{im}(a)+c, where c is an arbitrary element of {H}_{m}. □
Remark 5 This theorem is the theorem on integral for periodic oneparameter groups of operators.
The following theorem is known (see [7], Theorem 2.26)
Theorem 6 Let α be a strongly continuous isometric linear representation of T on a Banach space H and a\in H(D). Then the Fourier series of the element a is convergent to a in H.
Proof
In a standard manner, we have the equality
Using \frac{1}{2\pi}{\int}_{\pi}^{\pi}{D}_{n}(t)\phantom{\rule{0.2em}{0ex}}dt=1, we obtain {s}_{n}(a)a=\frac{1}{2\pi}{\int}_{\pi}^{\pi}{D}_{n}(t)(\alpha (t)aa)\phantom{\rule{0.2em}{0ex}}dt. Let us consider the function
Since {lim}_{t\to 0}g(t)=0, defining g(0)=0, we obtain that g(t) is a continuous function on [\pi ,\pi ]. Using the equality
we get
where \psi (t):=\frac{\alpha (t)aa}{t} for t\ne 0 and \psi (0):=D(a). Functions \psi (t), \alpha (t)aa and g(t)(\alpha (t)aa) are continuous vectorvalued functions on T. So, using Theorem 2 and equality (12), we obtain {lim}_{n\to \mathrm{\infty}}{s}_{n}=a. □
Remark 6 Our proof of this theorem differs from that in ([7], Theorem 2.26).
Corollary 3 Let α be a strongly continuous isometric linear representation of T on a Banach space H, x be an arbitrary element of H and x\sim {\sum}_{k=\mathrm{\infty}}^{\mathrm{\infty}}{x}_{k}. Then the series {\sum}_{k=\mathrm{\infty}}^{\mathrm{\infty}}\frac{1}{ik\lambda}{x}_{k} is convergent to {R}_{\lambda}(x) in Hfor all \lambda \in C\setminus Spec(H) and the series {x}_{m}+{\sum}_{k\ne m,k=\mathrm{\infty}}^{\mathrm{\infty}}\frac{1}{ikm}{x}_{k} is convergent to {R}_{im}(x) in H for all im\in Spec(H).
Proof Let \lambda \in C\setminus Spec(H). According to Theorem 4, we have {R}_{\lambda}(x)\in H(D). By Theorem 6 and {R}_{\lambda}(x)\sim {\sum}_{k=\mathrm{\infty}}^{\mathrm{\infty}}\frac{1}{ik\lambda}{x}_{k}, the series {\sum}_{k=\mathrm{\infty}}^{\mathrm{\infty}}\frac{1}{ik\lambda}{x}_{k} is convergent to {R}_{\lambda}(x) in H. The proof of the second statement is similar. □
Corollary 4 Let α be a strongly continuous isometric linear representation of T in a Banach space H and x\in H.

(i)
Let \lambda \in C\setminus Spec(H). Then x\in H(D) if and only if there exists y\in H such that x={\sum}_{k=\mathrm{\infty}}^{\mathrm{\infty}}\frac{1}{ik\lambda}{F}_{k}(y);

(ii)
Let \lambda \in Spec(H). Then x\in H(D) if and only if there exists y\in H such that x={y}_{m}+{\sum}_{k\ne m,k=\mathrm{\infty}}^{\mathrm{\infty}}\frac{1}{ikim}{F}_{k}(y).
The proof follows from Theorem 4, Corollary 2 and Proposition 2. □
5 Periodic solutions of the linear differential equation P(D)x=aof the n th order with constant coefficients
Let \alpha (t) be a strongly continuous linear representation of T in a Banach space H and D be the infinitesimal generator of α. For a\in H, we consider a solution of the linear differential equation
in H, where
, i=0,\dots ,n1, I is the unit operator in H.
Theorem 7 Let P(D) be a linear differential operator (14), a\in H, let {\lambda}_{1},\dots ,{\lambda}_{k} be the roots of the polynomial P(z), and let {m}_{i} be the multiplicity of {\lambda}_{i}, {m}_{1}+\cdots +{m}_{k}=n. Then

(i)
In the case {\lambda}_{1},\dots ,{\lambda}_{n}\notin Spec(H), for any a\in H, there exists the unique solution of equation (13) in H, and it is x={R}_{{\lambda}_{1}}^{{m}_{1}}\cdots {R}_{{\lambda}_{n}}^{{m}_{k}}(a).

(ii)
In the case {\lambda}_{1},\dots ,{\lambda}_{r}\in Spec(H) (r>0) and {\lambda}_{r+1},\dots ,{\lambda}_{k}\notin Spec(H), a solution of equation (13) exists if and only if
{F}_{i{\lambda}_{1}}(a)=\cdots ={F}_{i{\lambda}_{r}}(a)=0.(15)
For a\in H, satisfying condition (15), a general solution of equation (13) has the form
where {b}_{i} is an arbitrary element of {H}_{i{\lambda}_{i}}, i=1,\dots ,r.
Proof (i) P(D) may be written in the form P(D)=(D{\lambda}_{1}I)\cdots (D{\lambda}_{n}I).
Then equation (13) has the form
By Theorem 4(iii),
for all x\in H(D) and \lambda \notin Spec(H). Using equality (18) to equation (17), we obtain x={R}_{{\lambda}_{1}}\cdots {R}_{{\lambda}_{n}}a.

(ii)
Let {\lambda}_{1},\dots ,{\lambda}_{r}\in Spec(H) (r>0) and {\lambda}_{r+1},\dots ,{\lambda}_{n}\notin Spec(H). Using {R}_{{\lambda}_{r+1}},\dots ,{R}_{{\lambda}_{n}} to (17), we obtain
(D{\lambda}_{1}I)\cdots (D{\lambda}_{r}I)x={R}_{{\lambda}_{r+1}}\cdots {R}_{{\lambda}_{n}}a.(19)
This equation we can be written in the form
where {\lambda}_{i}\ne {\lambda}_{j} for i\ne j.
Lemma 3 Let a\in H, \lambda \in Spec(H) and m>1. Then the equation
has a solution if and only if {F}_{i\lambda}(a)=0. In the case {F}_{i\lambda}(a)=0, a general solution of this equation is x={R}_{\lambda}^{m}(a)+b, where b is an arbitrary element of {H}_{i\lambda}.
Proof Put y={(D\lambda I)}^{m1}x. Then equation (21) has the form (D\lambda I)y=a. By Theorem 5 this equation has a solution if and only if {F}_{i\lambda}(a)=0. In the case {F}_{i\lambda}(a)=0, a general solution is y={R}_{\lambda}(a)+{b}_{1}, where {b}_{1} is an arbitrary element of {H}_{i\lambda}. Let {F}_{i\lambda}(a)=0. Then equation (21) reduces to the equation
Put z={(D\lambda I)}^{m2}x. Then this equation has the form
By Theorem 5, this equation is solvable if and only if {F}_{i\lambda}({R}_{\lambda}(a)+{b}_{1})={F}_{i\lambda}({R}_{\lambda}(a))+{F}_{i\lambda}({b}_{1})={F}_{i\lambda}(a)+{F}_{i\lambda}({b}_{1})={F}_{i\lambda}({b}_{1})=0. Therefore a general solution of equation (23) has the form z={R}_{\lambda}^{2}(a)+{b}_{2}, where {b}_{2} is an arbitrary element of {H}_{i\lambda}. By induction, we obtain that a general solution of equation (21) has the form x={R}_{\lambda}^{m}(a)+b, where b is an arbitrary element of {H}_{i\lambda}. The lemma is proved. □
By this lemma, a solution of equation (20) reduces to a solution of the equation
where {b}_{1} is an arbitrary element of {H}_{i{\lambda}_{1}}. Using Lemma 3, by induction we obtain that a general solution of equation (20) has the form (16). □
6 Periodic solutions of the system of linear differential equations with constant coefficients
Let ({a}_{1},\dots ,{a}_{n}) be a vectorline, where {a}_{i}\in H, i=1,\dots ,n. Denote by ⊤ the operator of transposition of a matrix. The {({a}_{1},\dots ,{a}_{n})}^{\mathrm{\top}} denotes a vectorcolumn of {a}_{i}\in H, i=1,\dots ,n. Denote by {H}^{\mathrm{\top}n} the set of all vectors {({a}_{1},\dots ,{a}_{n})}^{\mathrm{\top}}, where {a}_{i}\in H, i=1,\dots ,n. For a={({a}_{1},\dots ,{a}_{n})}^{\mathrm{\top}}\in {H}^{\mathrm{\top}n} put Da={(D{a}_{1},\dots ,D{a}_{n})}^{\mathrm{\top}}.
We consider the following system of linear differential equations:
where a\in {H}^{\mathrm{\top}n}, x={({x}_{1},\dots ,{x}_{n})}^{\mathrm{\top}} and A is a complex n\times n matrix.
Definition 7 Systems Dx=Ax+a and Dy=By+b, where A, B are complex n\times nmatrices, a,b\in {H}^{\mathrm{\top}n} are called equivalent, if there exists a complex n\times nmatrix K such that detK\ne 0, A={K}^{1}BK and a={K}^{1}b.
In this case y=Kx.
Proposition 5 Every system (25) is equivalent to the system of the form Dy=By+b, where the matrix B has the Jordan form.
Proof For A there exists a complex n\times nmatrix K such that A={K}^{1}BK, where B is the Jordan normal form of A. From Dx=Ax+a, we obtain Dy=By+b, where y=Kx and b=Ka. □
By this proposition, a solution of system (25) reduces to a solution of the system of the form (25), where B has the Jordan normal form. Let B have the form
where {B}_{j} is a Jordan block of the {n}_{j}th order, {n}_{1}+\cdots +{n}_{m}=n. Then a solution of system (25), where B has the form (26), reduces to a solution of the following system of equations:
where {B}_{j} is a Jordan block of the {n}_{j}th order, {b}_{j}\in {H}^{\mathrm{\top}{n}_{j}}, {u}_{j}\in {H}^{\mathrm{\top}{n}_{j}} and x={({u}_{1},{u}_{2},\dots ,{u}_{m})}^{\mathrm{\top}}. Therefore a solution of system (27) reduces to a solution of the equation of the form
where B is a Jordan block of the q th order with eigenvalue λ: {g}_{i}^{i}=\lambda, i=1,\dots ,q; {g}_{i+1}^{i}=1, i=1,\dots ,q1; {g}_{j}^{i}=0, ji<0 and ji>1.
Theorem 8 Let Dx=Gx+b be a system of the form (28), where G is a Jordan block of the qth order with eigenvalue λ and b={({b}_{1},\dots ,{b}_{q})}^{\mathrm{\top}}\in {H}^{\mathrm{\top}q}. Then

(i)
for the case \lambda \in C\setminus Spec(H), the system has the unique solution x={({x}_{1},\dots ,{x}_{q})}^{\mathrm{\top}}\in {H}^{\mathrm{\top}q}, where:

(ii)
for the case \lambda \in Spec(H), \lambda =ip, p\in Z, the system has a solution if and only if {F}_{p}{b}_{q}=0,

(iii)
for the case \lambda \in Spec(H), \lambda =ip, p\in Z and {F}_{p}{b}_{q}=0, a general solution x={({x}_{1},\dots ,{x}_{q})}^{\mathrm{\top}}\in {H}^{\mathrm{\top}q} of the system has the form
\{\begin{array}{c}{x}_{1}={\sum}_{k=1}^{q1}{R}_{ip}^{k}({b}_{k}{F}_{p}{b}_{k})+{R}_{ip}^{q}({b}_{q})+c;\hfill \\ {x}_{qr}={\sum}_{k=1}^{r}{R}_{ip}^{k}({b}_{qr1+k}{F}_{p}{b}_{qr1+k})+{R}_{ip}^{r+1}({b}_{q}{F}_{p}{b}_{qr1}),\hfill \\ \phantom{\rule{1em}{0ex}}r=1,\dots ,q2;\hfill \\ {x}_{q}={R}_{ip}{b}_{q}{F}_{p}{b}_{q1},\hfill \end{array}(29)
where c is an arbitrary element of {H}_{p}.
Proof System (28) has the form
Let \lambda \in C\setminus Spec(H). From (30), using the operator {R}_{\lambda}, we obtain
Let \lambda \in Spec(H). Then \lambda =ip for some p\in Z. By Theorem 5, the equation
is solvable if and only if {F}_{p}{b}_{q}=0.
Let {F}_{p}{b}_{q}=0. By Theorem 5, a general solution of (31) has the form {x}_{q}={R}_{ip}{b}_{q}+{c}_{q}, where {c}_{q} is an arbitrary element of {H}_{ip}. In this case, the equation D{x}_{q1}ip{x}_{q1}={x}_{q}+{b}_{q1} has the form
By Theorem 5, this equation has a solution if and only if
By {F}_{ip}({b}_{q})=0 and Theorem 3(ii), (iii), we have {F}_{ip}{R}_{ip}{b}_{q}={R}_{ip}{F}_{ip}{b}_{q}=0. Hence (33) is equivalent to {F}_{ip}({c}_{q}+{b}_{q1})=0. Then {F}_{ip}{c}_{q}={F}_{ip{b}_{q1}}. By {c}_{q}\in {H}_{ip}, we obtain {c}_{q}={F}_{ip}{c}_{q}={F}_{ip}{b}_{q1}. Thus equation (33) is solvable if and only if {c}_{q}={F}_{ip}{b}_{q1}. Then equation (32) has the form D{x}_{q1}ip{x}_{q1}=({b}_{q1}{F}_{ip}{b}_{q1})+{R}_{ip}{b}_{q}. By Theorem 5, a general solution of this equation is {x}_{q1}={R}_{ip}({b}_{q1}{F}_{ip}{b}_{q1})+{R}_{ip}^{2}{b}_{q}+{c}_{q1}, where {c}_{q1} is an arbitrary element of {H}_{ip}. Then the equation D{x}_{q2}ip{x}_{q2}={x}_{q1}+{b}_{q2} has the form D{x}_{q2}ip{x}_{q2}=({b}_{q2}+{c}_{q1})+{R}_{ip}({b}_{q1}{F}_{ip}{b}_{q1})+{R}_{ip}^{2}{b}_{q}. As above, this equation is solvable if and only if {c}_{q1}={F}_{ip}{b}_{q2}. Similarly, by induction, we obtain a solution of system (30) in the form (29). □
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This work has been supported by the commission of Scientific Research Projects of Karadeniz Technical University, Project number: 2010.111.3.1/Faculty of Science.
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Çavuş, A., Khadjiev, D. & Kunt, M. On periodic oneparameter groups of linear operators in a Banach space and applications. J Inequal Appl 2013, 172 (2013). https://doi.org/10.1186/1029242X2013172
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DOI: https://doi.org/10.1186/1029242X2013172
Keywords
 Fourier series
 infinitesimal generator
 resolvent operator
 periodic solution
 theorem on integral