1 Introduction and Preliminaries

Let X a Banach space and \(\mathcal {B}(X)\) the Banach algebra of all bounded linear operators on X. for \(T\in \mathcal {B}(X),\) by \(T^*\), N(T), R(T), \( R^{\infty }(T)=\bigcap _{n\ge 0}R(T^n)\), \(N^{\infty }(T)=\bigcup _{n\ge 0}N(T^n)\), \(\rho (T)\) and \(\sigma (T)\), we denote, respectively, the adjoint, the null space, the range, the hyper-range, the hyper-kernel, the resolvent set and the spectrum of T.

Let \(T \in B(X)\), the ascent a(T) and the descent d(T) of T are defined in [16] by \(a(T) = \inf \{n\in {\mathbb {N}} : N(T^n) = N(T^{n+1})\}\) and \(d(T) = \inf \{n \in {\mathbb {N}} : R(T^n) = R(T^n+1)\},\) respectively.

The operator T is said to be Drazin invertible if \(d(T) < \infty \) and \(a(T) < \infty .\) It is well known that T is Drazin invertible if and only if \(T = T_1 \oplus T_2\) where \(T_1\) is invertible and \(T_2\) is nilpotent, see [9, Corollary 2.2]. This is equivalent to the fact that there exists an integer n such that the space \(R(T^n)\) is closed and the restriction of T of \(R(T^n)\) viewed as a map from \(R(T^n)\) into \(R(T^n)\) is invertible, see [2, Theorem 2.5]. The Drazin spectrum of T is defined by

$$\begin{aligned} \sigma _{D}(T ) = \{\lambda \in {\mathbb {C}} : \lambda -T \hbox { is not Drazin invertible}\}. \end{aligned}$$

Similarly, from [7], T is left (respect right) generalized Drazin invertible if and only if \(T = T_1 \oplus T_2\) such that \(T_1\) is left (respect right) invertible and \(T_2\) is quasi-nilpotent.

Recall that T is said to be semi-regular or Kato operator, if R(T) is closed and \(N(T) \subset R^{\infty }(T),\) see for example [1]. In addition, T is said to be pseudo-Fredholm operator if there exist two closed T-invariant subspaces M and N such that \(X=M\oplus N\) and \(T=T_{\mid M } \oplus T_{\mid N}\) with \(T_{\mid M}\) is semi-regular and \(T_{\mid N}\) is nilpotent. This is equivalent to the fact that there exists an integer n such that \(R(T^n)\) is closed and the restriction of T of \(R(T^n)\) viewed as a map from \(R(T^n)\) into \(R(T^n)\) is semi-regular. The quasi-Fredholm spectrum is defined by

$$\begin{aligned} \sigma _{QF}(T ) = \{\lambda \in {\mathbb {C}} : \lambda -T \hbox { is not quasi-Fredholm}\}, \end{aligned}$$

see [7, 10, 11] for more information.

Similarly, from [3] T is said to be a B-Fredholm operator, if there exists an integer n such that the space \(R(T^{n})\) is closed and the restriction of T of \( R(T^{n})\) viewed as a map from \(R(T^{n})\) into \( R(T^{n})\) is Fredholm.

Let \(T\in B(X)\) and \(x\in X,\) the local resolvent of T at x noted \(\rho _{T}(x)\) is defined as the union of all open subset U of \({\mathbb {C}}\) for which there is an analytic function \(f: U\rightarrow X\) such that the equation \((T-\mu I)f(\mu )=x\) holds for all \( \mu \in U\). The local spectrum \(\sigma _T(x)\) of T at x is defined by \(\sigma _T(x)={\mathbb {C}}{\setminus } \rho _T(x).\) Evidently \(\rho _T(x)\) is an open subset of \({\mathbb {C}}\) and \(\sigma _T(x)\) is closed. If \(f(z)={\sum \nolimits _{i=0}^{\infty }}x_i(z-\mu )^i\) ( in a neighborhood of \(\mu \)), then \(\mu \in \rho _{T}(x)\) if and only if there exists a sequence \((x_i)_{i\ge 0}\subseteq X\), \(x_0=x\), \((T-\mu )x_{i+1}=x_i\), and \({\sup _i}||x_i||^{\frac{1}{i}}<\infty \), see [8].

Let \(\mathcal {T} = (T (t))_{t\ge 0}\) be a strongly continuous semigroup (\(C_0\)-semigroup in short) with infinitesimal generator A on X. We will denote the type (growth bound) of \(\mathcal {T}\) by \(\omega _{0}\):

$$\begin{aligned} \omega _{0} =\inf \big \{\omega \in \mathbb {R}~~: \hbox {there exists M such that}\,\, \Vert T(t)\Vert \le Me^{\omega t },t\ge 0 \big \}, \end{aligned}$$

see [4, 5, 13] for more information. Also, in [4, 5, 13] the authors showed that

$$\begin{aligned} e^{t\nu (A)}\subseteq \nu (T(t))\subseteq e^{t\nu (A)}\bigcup \{0\} \end{aligned}$$

where \(\nu \left( .\right) \in \{\sigma _p(\cdot ),\sigma _{ap}() ,\sigma _r(\cdot )\}\) is the point spectrum, approximative spectrum or residual spectrum.

The semigroup T(t) is called differentiable for \(t> t_0\) if for every \(x \in X, ~t \rightarrow T(t)x\) is differentiable for \(t > t_0\). T(t) is called differentiable if it is differentiable for \(t > 0.\) If \(B(\lambda ,t)x = \int _{0}^{t}e^{\lambda (t-s)}T(s)xds,\) then \(B(\lambda ,t)x\) is differentiable in t with \(B^{\prime }(\lambda ,t)x = T(t)x + \lambda B(\lambda ,t)x\) and \(B_{\lambda }^{\prime }(t)\) is a bounded linear operator in X,  see [5, 13].

Spectral inclusions for various reduced spectra of a \(C_0\)-semigroup were studied by authors in [4, 5] for point spectrum, approximative spectrum and residual spectrum. Also, the spectral equality for a \(C_0\)-semigroup was studied by authors in [15] for semi-regular, essentially semi-regular and semi-Fredholm spectrum, respectively. In this work, we will continue in this direction, we will prove that the spectral inclusion for Drazin and quasi-Fredholm spectra. Also, we will show that if \((T(t))_{t\ge 0}\) is a \(C_0\)-semigroup with infinitesimal generator A such that the equality \(\lim _{t\rightarrow \infty }\frac{1}{t^n}\Vert T(t)\Vert =0\) holds for some \(n\in {\mathbb {N}}\), then the infinitesimal generator A is quasi-Fredholm if and only if it is Drazin invertible if and only if it is B-Fredholm.

2 Main Results

We start by given the following two lemmas which are proved in [5]. They will be used to prove our main result.

Lemma 2.1

[5] Let (A, D(A)) be the infinitesimal generator of a strongly continuous semigroup \((T(t))_{t\ge 0}\) and \(B(\lambda ,t)=\int _{0}^{t}e^{\lambda (t-s)}T(s)xds\) is a bounded operator from X to D(A). Then, for every \(\lambda \in {\mathbb {C}}\), \(t > 0\) and \(n\in {\mathbb {N}}\), the following statements hold:

  1. (1)

    \((a):(e^{\lambda t}- T(t))^{n}(x)=(\lambda -A)^{n}B(\lambda ,t)^{n}(x) ,\quad \lambda \in {\mathbb {C}},x\in X;\)

    \((b):(e^{\lambda t}- T(t))^{n}(x)=B(\lambda ,t)^{n}(\lambda -A)^{n}(x) , \lambda \in {\mathbb {C}},x\in D(A).\)

  2. (2)

    \( R(e^{\lambda t} -T(t))^{n}\subseteq R(\lambda -A)^{n} .\)

  3. (3)

    \( N(\lambda -A)^{n}\subseteq N(e^{\lambda t} -T(t))^{n}.\)

Let Y be a Banach space that is continuously embedded in X (in symbols:\(Y\hookrightarrow X\)). We need the following lemma.

Lemma 2.2

[5] Let (A, D(A)) be the infinitesimal generator of a strongly continuous semigroup \((T(t))_{t\ge 0}\) on X and assume that the restricted semigroup \((T(t)_{|})_{t\ge 0}\) is strongly continuous on some \((T(t))_{t\ge 0}\)-invariant Banach space \(Y\hookrightarrow X\). Then the infinitesimal generator of \((T(t)|)_{t\ge 0}\) is the part \((A_{|},D(A_{|}))\) of A in Y.

The following two lemmas which are proved in [15] will be used in the sequel.

Lemma 2.3

[15, Lemma 3.1] Let A the infinitesimal generator of \(C_0\)-semigroup \(T(t)_{t\ge 0}\) and \(B(\lambda ,t)=\int _{0}^{t}e^{\lambda (t-s)}T(s)ds \) is a linear bounded operator on X. Then there exist C and D tow operator such that \((\lambda -A),B(\lambda ,t),C,D\) are mutually commuting operators, for all \(x\in D(A)\) and \(C(\lambda -A)+DB(\lambda ,t)=I,\) for \(t>0\).

Lemma 2.4

[15, Lemma 3.2] Let \((\lambda -A),B(\lambda ,t),~~C,~~D\) be mutually commuting operators in D(A) such that \(C(\lambda -A)+DB(\lambda ,t)=I\), \(t>0.\) Then we have:

  1. (1)

    For every positive integer n there are \(C_n,D_n \in D(A)\) such that \((\lambda -A)^{n},B^{n}(\lambda ,t),C_{n},D_{n}\) are mutually commuting and

    $$\begin{aligned} (\lambda -A)^{n}C_{n} + B^{n}(\lambda ,t)D_n = I. \end{aligned}$$
  2. (2)

    For every positive integer n,  \(R(e^{\lambda t}-T(t))^{n} = R(\lambda -A)^{n} \bigcap R(B^{n}(\lambda ,t))\) and \(N((e^{\lambda t}-T(t))^{n}) = N((\lambda -A)^{n}) + N(B^{n}(\lambda ,t)).\) Further \(R^{\infty }(e^{\lambda t}-T(t)) = R^{\infty }(\lambda -A) \bigcap R^{\infty }(B(\lambda ,t))\) and \(N^{\infty }(e^{\lambda t}-T(t)) = N^{\infty }(\lambda -A) + N^{\infty }(B(\lambda ,t)).\)

  3. (3)

    \(N^{\infty }(\lambda -A) \subset R^{\infty }(B(\lambda ,t)) \)and \(N^{\infty }(B(\lambda ,t))\subset R^{\infty }(\lambda -A).\)

Now, we give some spectral results for differentiable \(C_0\)-semigroup.

Lemma 2.5

Let (A, D(A)) be the infinitesimal generator of a strongly continuous semigroup \((T(t))_{t\ge 0}\), \(B(\lambda ,t)=\int _{0}^{t}e^{\lambda (t-s)}T(s)xds\) is a bounded operator from X to D(A). If T(t) is differentiable for \(t > t_0,\) then for every \(\lambda \in {\mathbb {C}}\), \(t > t_0\) and \(n\in {\mathbb {N}}\), the following statements hold:

  1. (1)

    \((a):(\lambda e^{\lambda t}- AT(t))^{n}(x)=(\lambda -A)^{n}B(\lambda ,t)^{'n}(x) , \quad \lambda \in {\mathbb {C}},x\in X;\)

    \((b):(\lambda e^{\lambda t}- T(t))^{n}(x)=B(\lambda ,t)^{'n}(\lambda -A)^{n}(x) \lambda \in {\mathbb {C}},x\in D(A).\)

  2. (2)

    \( R(\lambda e^{\lambda t} -AT(t))^{n}\subseteq R(\lambda -A)^{n}. \)

  3. (3)

    \( N(\lambda -A)^{n}\subseteq N(\lambda e^{\lambda t} -AT(t))^{n}.\)

Proof

Assuming now that \(t > t_0\) and differentiating (a) and (b) in (1) of Lemma 2.1 with respect to t and \(n=1\), we obtain

$$\begin{aligned} \lambda e^{\lambda t}x - AT(t)x= & {} (\lambda I - A)B^{\prime }(\lambda ,t)x ~~~~\text{ for } \text{ every }~~ x \in X;\\ \lambda e^{\lambda t}x - AT(t)x= & {} B^{\prime }(\lambda ,t)x(\lambda I - A) ~~~~\text{ for } \text{ every }~~ x \in D(A) . \end{aligned}$$

This gives (a) and (b) for \(n=1.\) By induction, we obtain (a) and (b) for all \(n\in {\mathbb {N}}.\) The rest of Lemma follows from (1). \(\square \)

Proposition 2.1

Let \(T( t)_{t>0}\) be a \(C_0\)-semigroup and let A be its infinitesimal generator. If T(t) is differentiable for \(t > t_0\) and \(\lambda \in \sigma _{A}(x)\) for \(x\in X,\) then

$$\begin{aligned} \lambda e^{\lambda t} \in \sigma _{AT(t)}(x). \end{aligned}$$

Proof

Let \(t>t_{0}\) be fixed and suppose that \(\lambda e^{\lambda t}\notin \sigma _{AT(t_0)}(x)\), then there exist a sequence \((x_i)_{i\in {\mathbb {N}}}\) of X such that \(x_0=x\), \((\lambda e^{\lambda t}-AT(t))x_i=x_{i-1}\) and \({\sup _i}\Vert x_i\Vert ^{\frac{1}{i}}<\infty .\)

We put \(y_i=B^{'i}(\lambda ,t)x_i\), as \(B^{\prime }(\lambda ,t)\) is a bounded linear operator in X, we have \(y_0=x_0=x,~~y_0\in D(A)\).

$$\begin{aligned} (\lambda -A)y_i= & {} (\lambda -A)B^{\prime }(\lambda ,t)x_i B^{'(i-1)}(\lambda ,t)x_i \\= & {} (\lambda e^{\lambda t}-AT(t))B^{'(i-1)}(\lambda ,t)x_i \\= & {} B^{'(i-1)}(\lambda ,t)(\lambda e^{\lambda t}-AT(t))x_i \\= & {} B^{'(i-1)}(\lambda ,t)x_{i-1}\\= & {} y_{i-1} \end{aligned}$$

Therefore \((\lambda -A)y_i=y_{i-1}.\) On the other hand

$$\begin{aligned} \Vert y_i\Vert =\Vert B^{'i}(\lambda ,t)x_i\Vert<\Vert B^{'i}(\lambda ,t)\Vert \Vert x_i\Vert <M^i\Vert x_i\Vert , \end{aligned}$$

then

$$\begin{aligned} \displaystyle {\sup _i}\Vert y_i\Vert ^{\frac{1}{i}}< \displaystyle {\sup _i} M\Vert x_i\Vert ^{\frac{1}{i}}<\infty . \end{aligned}$$

So that \(\lambda \notin \sigma _A(x)\)\(\square \)

Denote by \(\sigma _{su}(A)\) the subjectivity spectrum of A. It is known that \({\bigcup _{x\in X}}\sigma _{A}(x)= \sigma _{su}(A)\) for a closed operator A. Hence the following corollary holds.

Corollary 2.1

Let \(T( t)_{t>0}\) be a \(C_0\)-semigroup and let A be its infinitesimal generator. If T(t) is differentiable for \(t > t_0\) and \(\lambda \in \sigma _{su}(A),\) then \(\lambda e^{\lambda t} \in \sigma _{su}(AT(t)).\)

Proposition 2.2

Let \(T( t)_{t>0}\) be a \(C_0\)-semigroup and let A be its infinitesimal generator. If T(t) is differentiable for \(t > t_0\) and \(\lambda \in \sigma _{ap}(A)\), then \(\lambda e^{\lambda t} \in \sigma _{ap}(AT(t)).\)

Proof

For \(t > t_0\), since \(\lambda e^{\lambda t}-AT(t)=(\lambda -A)B^{\prime }(\lambda ,t)\) and \(B^{\prime }(\lambda ,t)\) is bounded linear operator, then \(\lambda e^{\lambda t}-AT(t)\) is a bounded linear operator. It is easy to check that for \(x\in D(A)\), \(B^{\prime }(\lambda ,t)Ax=AB^{\prime }(\lambda ,t)x.\) If \(\lambda \in \sigma _{ap}(A)\), then there exists sequence \((x_n)_{n\in {\mathbb {N}}}\in D(A)\) satisfying \(||x_n||=1\) and \(||(\lambda -A)x_n||\rightarrow 0.\) From (1) of Lemma 2.5, we obtain the result. \(\square \)

By an outline of the proof of [6, Theorem 2.1], we obtain the following result.

Proposition 2.3

Let \(T( t)_{t>0}\) be a \(C_0\)-semigroup and let A be its infinitesimal generator. If T(t) is differentiable for \(t > t_0\) and \(\lambda \in \nu (A)\), then \(\lambda e^{\lambda t} \in \nu (AT(t)).\) where \(\nu (\cdot ) \in \{\sigma _{\gamma }(\cdot ),\sigma _{\pi }(\cdot ),\sigma _{\gamma e}(\cdot )\}\) and \(\sigma _{\gamma }(\cdot ),\sigma _{\pi }(\cdot ),\sigma _{\gamma e}(\cdot )\) denote the regular spectrum, essential regular spectrum and left essential spectrum.

In the next theorem, we will prove that the spectral inclusion of \(C_0\)-semigroups remains true for the Drazin invertible and quasi-Fredholm spectra.

Theorem 2.1

Let \((T(t))_{t\ge 0}\) a \(C_0-\)semigroup, with infinitesimal generator A. Then

$$\begin{aligned} e^{t\sigma _{D}(A)}\subseteq \sigma _{D}(T(t)). \end{aligned}$$

Proof

Let \(t_{0}>0\) be fixed and suppose that \((e^{\lambda t_0} -T(t_0))\) is Drazin invertible for some \(\lambda \in {\mathbb {C}}{\setminus } \{0\}\). Then \(M:= R(e^{\lambda t_0} -T(t_0))^n\) is closed and the restricted semigroup \((e^{\lambda t_0} -T(t_0)_{\mid M})\) is invertible. We show that \((\lambda -A)\) is Drazin invertible. To this end, in the first we show that \(R(\lambda -A)^{n}\) is closed. Let \(x\in \overline{R((\lambda -A)^n))}\), that is, there exist \(u_{k}\in D(A^n),\)\( k=1,2,...,\) such that \((\lambda -A)^{n}u_{k}\rightarrow x,\) hence

$$\begin{aligned} (e^{\lambda t_0} -T(t_0))^{n}u_{k} := B(\lambda ,t_0)^{n}(\lambda -A)^{n}u_{k} \end{aligned}$$

by Lemma 2.1. Also,

$$\begin{aligned} B(\lambda ,t_0)^{n}(\lambda -A)^{n}u_{k}~~\rightarrow B(\lambda ,t_0)^{n}x. \end{aligned}$$

Hence

$$\begin{aligned} (e^{\lambda t_0} -T(t_0))^{n}u_{k}~~\rightarrow B(\lambda ,t_0)^{n}x. \end{aligned}$$

Since \(M:= R(e^{\lambda t_0} -T(t_0))^n\) is closed, then

$$\begin{aligned} B(\lambda ,t_{0})^{n}x \in R(e^{\lambda t_0} -T(t_0))^n. \end{aligned}$$

Hence there exists \(u\in D(A^n)\) such that

$$\begin{aligned} B(\lambda ,t_0)^{n}x= (e^{\lambda t_0} -T(t_0))^n u. \end{aligned}$$

In the other hand, From Lemma 2.1, we have that

$$\begin{aligned} (e^{\lambda t_0} -T(t_0))^n u = B(\lambda ,t_0)^{n}(\lambda -A)u, \end{aligned}$$

hence

$$\begin{aligned} B(\lambda ,t_0)^{n}x= B(\lambda ,t_0)^{n}(\lambda -A)u. \end{aligned}$$

This implies that,

$$\begin{aligned} x- (\lambda -A)^{n}u \in N(B(\lambda ,t_{0}))\subseteq R(\lambda -A)^n. \end{aligned}$$

So \(x\in R(\lambda -A)^n\), and hence \(R(\lambda -A)^n\) is closed.

Now, let us to show that \((\lambda -A_{ \mid R(\lambda -A)^n})\) is invertible. For this, as \((e^{\lambda t_0} -T(t_0)_{\mid M})\) is invertible, then \((e^{\lambda t_0} -T(t_0)_{\mid M})\) is bounded below and \(R(e^{\lambda t_0} -T(t_0)_{\mid M})=R(e^{\lambda t_0} -T(t_0))^{n+1}\) is onto. We show that \((\lambda -A_{\mid R(\lambda -A)^{n} \bigcap D(A)})\) is bounded below. Since \((e^{\lambda t_0} -T(t_0)_{\mid M})\) is bounded below, then \((e^{\lambda t_0} -T(t_0)_{\mid M})\) is injective and \(R(e^{\lambda t_0} -T(t_0)_{\mid M})\) is closed. We show that \((\lambda -A_{\mid R(\lambda -A)^{n} \bigcap D(A)})\) is injective and \((\lambda -A_{\mid R(\lambda -A)^{n} \bigcap D(A)})\) is closed. For all \(x\in D(A)\) we have

$$\begin{aligned} \{0\}= & {} N(e^{\lambda t_0} -T(t_0)_{\mid M \bigcap D(A)})= N(\lambda -A_{\mid R(\lambda -A)^{n} \bigcap D(A)})\\&+\,N(B(\lambda ,t_0))\bigcap R(B^{n}(\lambda ,t_0)), \end{aligned}$$

then \(N(\lambda -A_{\mid R(\lambda -A)^{n} \bigcap D(A)})=\{0\},\) therefore \((\lambda -A_{\mid R(\lambda -A)^{n} \bigcap D(A)})\) is injective. As \(R(e^{\lambda t_0} -T(t_0))^{n+1}\) is closed, then \((\lambda -A_{\mid R(\lambda -A)^{n} \bigcap D(A)})=R(\lambda -A)^{n+1}\) is also closed. On the other hand, as \(R(e^{\lambda t_0} -T(t_0)_{\mid M})\) is onto, we can easily verify that \(R(\lambda -A_{\mid R(\lambda -A)^{n} \bigcap D(A)})\) is onto. In fact one can verify that

$$\begin{aligned} R(\lambda -A_{\mid R(\lambda -A)^{n} \bigcap D(A)})=R(\lambda -A)^{n+1}=R(\lambda -A)^{n}. \end{aligned}$$

We have \(R(\lambda -A)^{n+1}\subset R(\lambda -A)^{n}\) and if \(y\in R(\lambda -A)^{n},\) then there exist \(x\in D(A^{n})\) such that \(y= R(\lambda -A)^{n}x\) and by (1) of Lemma 2.3, we have

$$\begin{aligned} (\lambda -A)^{n}x= & {} (\lambda -A)^{n}C_{n}(\lambda -A)^{n}x+D_{n}B^{n}(\lambda ,t)(\lambda -A)^{n}x\\= & {} (\lambda -A)^{n+1}C_{n}(\lambda -A)^{n-1}x+D^{n}(e^{\lambda t_0}-T(t_0))^{n}x \end{aligned}$$

and as \(R(e^{\lambda t_0}-T(t_0)\mid _{(e^{\lambda t_0}-T(t_0))^{n}})=R(e^{\lambda t_0}-T(t_0))^{n+1}\) is onto, then there exist \(x^{\prime }\in X\) such that \((e^{\lambda t_0}-T(t_0))^{n}x=(e^{\lambda t_0}-T(t_0))^{n+1}x^{\prime }=(\lambda -A)^{n+1}B^{n+1}(\lambda ,t)x^{\prime },\) therefore \(y\in R(\lambda -A)^{n}\) and then \(R(\lambda -A_{\mid R(\lambda -A)^{n} \bigcap D(A)})\) is onto.

Finally, \((\lambda -A_{\mid R(\lambda -A)^{n} \bigcap D(A)})\) is Drazin invertible. \(\square \)

Corollary 2.2

For the infinitesimal generator A of a strongly continuous semigroup \((T(t))_{t\ge 0}\), one has the spectral inclusion

$$\begin{aligned} e^{t\sigma _{\nu }(A)}\subseteq \sigma _{\nu }(T(t)) \end{aligned}$$

where \(\sigma _{\nu }(\cdot )\) is the left Drazin and right Drazin spectra.

The following example shows that the inclusion in Theorem 2.1 is strict.

Example 1

Let X be the Banach space of continuous functions on [0, 1] which are equal to zero at \(x = 1\) with the supremum norm. Define

$$\begin{aligned}(T(t)f)(x):=\left\{ \begin{array}{lll} f(x+t) &{} \quad \hbox {if} \,\,x+t\le 1; \\ 0 &{} \quad \hbox {if}\,\, x+t>1. \end{array} \right. \end{aligned}$$

T(t) is obviously a \(C_0\)-semigroup on X. Its infinitesimal generator A is given on

$$\begin{aligned} D(A) = \{f:f\in C^{1}([0, 1])\cup X,f^{\prime } \in X\} \end{aligned}$$

by

$$\begin{aligned} Af=f^{\prime } \text{ for } f\in D(A). \end{aligned}$$

One checks easily that for every \(\lambda \in {\mathbb {C}} \) and \(g \in X\) the equation \(\lambda f - f^{\prime } = g\) has a unique solution \(f \in X\) given by

$$\begin{aligned} f(t) = \int _{t}^{1}\hbox {e}^{\lambda (t-S)}g(s) \hbox {d}s. \end{aligned}$$

Therefore \(\sigma (A) = \emptyset \) see [13], hence \(\sigma _{D}(A)=\emptyset .\)

On the other hand, T(t) is a bounded linear operator for every \(t \ge 0.\) Suppose that \(f\in X\) is not periodic, then we have \(0\in \sigma _{D}(T(t)).\) Indeed, if \(0\notin \sigma _{D}(T(t))\), then there exist a finite integer \(n\in {\mathbb {N}}\) such that \((T^{n}(t)f)(x)=(T^{n+1}(t)f)(x)\), this implies that \(f(x+nt)=f(x+nt+t)\); therefore, f is periodic; this is a contradiction. Hence the inclusion

$$\begin{aligned} e^{t\sigma _{D}(A)}\subseteq \sigma _{D}(T(t)). \end{aligned}$$

is strict.

Theorem 2.2

For the infinitesimal generator A of a strongly continuous semigroup \((T(t))_{t\ge 0}, \) we have the following inclusion:

$$\begin{aligned} e^{t\sigma _{QF}(A)}\subseteq \sigma _{QF}(T(t)). \end{aligned}$$

Proof

Let \(t_{0}>0\) be fixed and suppose that \((e^{\lambda t_0} -T(t_0))\) is quasi-Fredholm, for some \(\lambda \in {\mathbb {C}}{\setminus } \{0\}\). Then \(M:= R(e^{\lambda t_0} -T(t_0))^n\) is closed and the restricted semigroup \((e^{\lambda t_0} -T(t_0)_{\mid M})\) is semi-regular.

We show that \((\lambda -A)\) is quasi-Fredholm, to this end we show that \(R(\lambda -A)^{n}\) is closed and \((\lambda -A_{|R(\lambda -A)^{n}\bigcap D(A)})\) is semi-regular. That is, (\(R(\lambda -A_{|R(\lambda -A)^{n}\bigcap D(A)})\) is closed and \(N(\lambda -A_{|R(\lambda -A)^{n}\bigcap D(A)})\subseteq R^{\infty }(\lambda -A_{|R(\lambda -A)^{n}\bigcap D(A)})\)).

As \(M:= R(e^{\lambda t_0} -T(t_0))^n\) is closed, then by the same argument as in the proof of Theorem 2.1, we conclude that \(R(\lambda -A)^{n}\) is closed. Since \((e^{\lambda t_0} -T(t_0)_{\mid M})\) is semi-regular, then \(R(e^{\lambda t_0} -T(t_0)_{\mid M})=R(e^{\lambda t_0} -T(t_0))^{n+1}\) is closed, this implies that \(R(\lambda -A_{|R(\lambda -A)^{n}\bigcap D(A)})=R(\lambda -A)^{n+1}\) is closed. On the other hand, by Lemma 2.1 we have that

$$\begin{aligned} N(\lambda -A_{|R(\lambda -A)^{n}\bigcap D(A)})\subseteq N(e^{\lambda t_0} -T(t_0)_{\mid M\bigcap D(A)}\subseteq N(e^{\lambda t_0} -T(t_0)_{\mid M}). \end{aligned}$$

Since \((e^{\lambda t_0} -T(t_0)_{\mid M})\) is semi-regular, then \( N(e^{\lambda t_0} -T(t_0)_{\mid M}) \subseteq R^{\infty }(e^{\lambda t} -T(t_0)_{\mid M})=R^{\infty }(\lambda -A_{|R(\lambda -A)\bigcap D(A)})\bigcap R^{\infty }B(\lambda ,t_0)\subset R^{\infty }(\lambda -A_{|R(\lambda -A)^{n}\bigcap D(A)}),\) hence \((\lambda -A_{|R(\lambda -A)^{n}\bigcap D(A)})\) is semi-regular. Finally, we conclude that \((\lambda -A)\) is quasi-Fredholm. \(\square \)

Remark 1

For an operator \(T\in B(X)\) we have \(\sigma _{D}(T)\subseteq \sigma _{QF}(T),\) this implies by Example 1 the inclusion spectral in Theorem 2.2 is strict.

Let A be the infinitesimal generator of a \(C_0\)-semigroup \((T(t))_{t\ge 0}.\)

In the next Theorem, we will use the concept of the \(weak^*\) -integral. for more information of This integral, see [17, Appendix 1]

In the following, we will give condition on \((T(t))_{t\ge 0}\) under which facts A is Drazin invertible, A is \(B-\)Fredholm and A is \(Q-\)Fredholm are equivalent.

Theorem 2.3

Let A be the infinitesimal generator of a \(C_0\)-semigroup \((T(t))_{t\ge 0}\).

If \(\lim _{t\rightarrow \infty }\frac{1}{t^n}\Vert T(t)\Vert =0\), for some \(n\in {\mathbb {N}}\), the following assertions are equivalent:

  1. (1)

    A is quasi-Fredholm;

  2. (2)

    A is Drazin invertible;

  3. (3)

    A is B-Fredholm.

Proof

\((1)\Rightarrow (2):\)

Since A is quasi-Fredholm, then \(R(A^n)\) closed and \(A_{\mid (R(A^n)\bigcap D(A))}\) is semi-regular. Let \(y\in N(A_{\mid (R(A^n)\bigcap D(A))}),\) then there exists \(x \in (R(A^n)\bigcap D(A^{n}))\) such that \(y=A^{n}x\). We integer by parts in the following formula:

$$\begin{aligned} T(t)x-x =\int _{0}^{t}T(s)Ax\hbox {d}s , \end{aligned}$$

we obtain that

$$\begin{aligned} T(t)x=x+tA+\frac{t^2}{2!}A^{2}+\int _{0}^{t}\frac{(t-s)^{2}}{2!}T(s)A^{3}x\hbox {d}s. \end{aligned}$$

We repeat this operation for n times, we obtain that

$$\begin{aligned} T(t)x=\sum _{k=0}^{n-1}\frac{t^k}{k!}A^{k}x+\int _{0}^{t}\frac{(t-s)^{n-1}}{(n-1)!}T(s)A^{n}x\hbox {d}s. \end{aligned}$$

Hence,

$$\begin{aligned} T(t)x= & {} \sum _{k=0}^{n-1}\frac{t^k}{k!}A^{k}x+y\int _{0}^{t}\frac{(t-s)^{n-1}}{(n-1)!}ds\\= & {} \sum _{k=0}^{n-1}\frac{t^k}{k!}A^{k}x+\frac{t^n}{n!}y. \end{aligned}$$

As \({\lim _{t\rightarrow \infty }} \frac{1}{t^n}\Vert T(t)\Vert =0\), then \(y=0\), this implies that

$$\begin{aligned} N(A_{\mid (R(A^n)\bigcap D(A))})=\{0\}. \end{aligned}$$

On the other hand, let \((T(t)^{\prime })_{t\ge 0}\) with infinitesimal generator \(A^{\prime }\) the adjoint semigroup of \((T(t))_{t\ge 0}\). Since \(A_{\mid (R(A^n)\bigcap D(A))}\) is semi-regular, then \(A^{\prime }_{\mid (R(A^{'n})\bigcap D(A^{\prime }))}\) is also semi-regular, see [12, Proposition 1.6]. Using the following formula

$$\begin{aligned} T(t)^{\prime }x^{\prime }-x^{\prime } =\hbox {weak}^{*}\int _{0}^{t}T(s)^{\prime }A^{\prime }x^{\prime }\hbox {d}s,~~\forall x^{\prime }\in (R(A^{'n})\bigcap D(A^{\prime })),~~ \text{ for } \text{ all } t\ge 0 \end{aligned}$$

which is proved in [17, Proposition 1.2.2] and by the same argument as above, we get that \(N(A^{\prime }_{\mid (R(A^{'n})\bigcap D(A^{\prime }))})=\{0\}\). This is equivalent to the fact that

$$\begin{aligned} \overline{R(A_{\mid (R(A^n)\bigcap D(A))})}= (R(A^n)\bigcap D(A)). \end{aligned}$$

Hence \(R(A_{\mid (R(A^n)\bigcap D(A))})= (R(A^n)\bigcap D(A)),\) since \((R(A^n)\bigcap D(A))\) is closed. From this it follows that \(A_{\mid (R(A^n)\bigcap D(A))}\) is surjective and hence it is invertible. Finally, A is Drazin invertible.

\((2)\Rightarrow (1)\): is clear.

(1),  (2) and (3) are equivalent, since the class of Drazin invertible operator is a subclass of B-Fredholm operator and the class of B-Fredholm operator is a subclass of quasi-Fredholm operator. \(\square \)

The following example shows that the condition \(\lim _{t\rightarrow \infty }\frac{1}{t^n}\Vert T(t)\Vert =0\), for some \(n\in {\mathbb {N}}\) in Theorem 2.3 is needed for conclusion.

Example 2

Let H be a Hilbert space with an orthonormal basis \(\{e_n\}_{1}^{\infty }\) and T be unilateral weighted shift operator on H defined by \(Ae_{n}=w_{n}e_{n+1},n=1,2,\dots \) [14, p. 51]. Let \(T(t)=e^{tA}\) be the semigroup generated by A, \(t\ge 0\) and \(w_{n}=1\). Easy to see that \(||T(t)||=e^{t}\), thus, \({\lim _{t\rightarrow \infty }}\frac{1}{t^n}\Vert T(t)\Vert \ne 0\), for all \(n\in {\mathbb {N}}^{*}\). We know (see [1, Theorem 2.86 and Example 3.30]) that \(\sigma (A)=\{\lambda \in {\mathbb {C}}, |\lambda |\le 1\}\) and \(\sigma _{ap}(A)=\{\lambda \in {\mathbb {C}}, |\lambda |=1\},\) thus, A is semi-regular then A quasi-Fredholm. On the other hand A is not surjective so \(R(A^n)\supset R(A^{n+1})\) this implies that \(A^n:R(A^n) \rightarrow R(A^n)\) is not invertible; therefore, A is not Drazin invertible.