In [27], Srivastava and Owa provided the definitions for fractional operators (derivative and integral) in the complex *z*-plane ℂ as follows.

**Definition 2.1** The fractional derivative of order *α* is defined for a function f(z) by

{D}_{z}^{\alpha}f(z):=\frac{1}{\Gamma (1-\alpha )}\frac{d}{dz}{\int}_{0}^{z}\frac{f(\zeta )}{{(z-\zeta )}^{\alpha}}\phantom{\rule{0.2em}{0ex}}d\zeta ,\phantom{\rule{1em}{0ex}}0\le \alpha <1,

where the function f(z) is analytic in a simply-connected region of the complex *z*-plane ℂ containing the origin, and the multiplicity of {(z-\zeta )}^{-\alpha} is removed by requiring log(z-\zeta ) to be real when (z-\zeta )>0.

**Definition 2.2** The fractional integral of order *α* is defined, for a function f(z), by

{I}_{z}^{\alpha}f(z):=\frac{1}{\Gamma (\alpha )}{\int}_{0}^{z}f(\zeta ){(z-\zeta )}^{\alpha -1}\phantom{\rule{0.2em}{0ex}}d\zeta ,\phantom{\rule{1em}{0ex}}\alpha >0,

where the function f(z) is analytic in a simply-connected region of the complex *z*-plane (ℂ) containing the origin, and the multiplicity of {(z-\zeta )}^{\alpha -1} is removed by requiring log(z-\zeta ) to be real when (z-\zeta )>0.

**Remark 2.1** From Definitions 2.1 and 2.2, we have

{D}_{z}^{\alpha}{z}^{\beta}=\frac{\Gamma (\beta +1)}{\Gamma (\beta -\alpha +1)}{z}^{\beta -\alpha},\phantom{\rule{1em}{0ex}}\beta >-1,0\le \alpha <1

and

{I}_{z}^{\alpha}{z}^{\beta}=\frac{\Gamma (\beta +1)}{\Gamma (\beta +\alpha +1)}{z}^{\beta +\alpha},\phantom{\rule{1em}{0ex}}\beta >-1,\alpha >0.

In this note, we are concerned about the following fractional Cauchy problem (in the sense of the Srivastava-Owa operator):

\begin{array}{c}{D}_{z}^{\alpha}u(z)+C(z)Au(z)=\varphi (z),\phantom{\rule{1em}{0ex}}\alpha \in (0,1),\hfill \\ {I}_{z}^{1-\alpha}u(z){|}_{z=0}=0,\hfill \end{array}

(1)

where A:D(A)=:\tilde{H}\to H is a closed densely defined linear operator on a complex Hilbert space *H*, C(\cdot ) is a bounded operator defined everywhere in {L}^{2}(U;H), \varphi \in {L}^{2}(U;H) and u\in {L}^{2}(U;\tilde{H}), \tilde{H}\subset H, U:=\{z\in \mathbb{C}:|z|<1\}. Denote by \partial U:=\{z\in \mathbb{C}:|z|=1\} and \overline{U}:=U\cup \partial U. For complex Hilbert spaces {H}^{\ast} and *H* with the inner product {(\cdot )}_{H} and {(\cdot )}_{{H}^{\ast}} respectively, let B({H}^{\ast},H) be the space of all bounded linear operators from {H}^{\ast} to *H*; if H={H}^{\ast}, we write B(H). Recall that the operator *P* is called *accretive* if \mathrm{\Re}{(Pu,u)}_{H}\ge 0, \mathrm{\forall}u\in {H}^{\ast}, and *m*-*accretive* if Rang(\lambda I+P)=H, \lambda >0. Denoted by \rho (A):=\{\lambda \in \mathbb{C}:\lambda \text{is a regular value of}A\}, the *resolvent* set of the operator *A*. Note that the resolvent set \rho (A)\subseteq \mathbb{C} of a bounded linear operator *A* is an open set. Moreover, the space {L}^{2}(U;H) is a Hilbert space with the inner product

{(f,g)}_{{L}^{2}(U;H)}={\int}_{0}^{1}{(f(z),g(z))}_{H}\phantom{\rule{0.2em}{0ex}}dz,\phantom{\rule{1em}{0ex}}z\in U.

Throughout the paper, we consider \mathrm{\Re}{(Au,u)}_{H}>0, u\in \tilde{H} and {\parallel x\parallel}_{\tilde{H}}={\parallel Ax\parallel}_{H}.

**Definition 2.3** Equation (1) has maximal regularity in {L}^{2}(U;H) if for every \varphi \in {L}^{2}(U;H), \mathrm{\exists}u(z) such that

u\in {L}^{2}(U;\tilde{H}),\phantom{\rule{1em}{0ex}}{I}_{z}^{1-\alpha}u\in {W}_{0}^{1,2}(U,H),

where {W}_{0}^{1,2}(U,H) is the Sobolev space defined by

{W}_{0}^{1,2}(U,H)=\{f:\mathrm{\exists}\psi \in {L}^{2}(U;H);f(z)={\int}_{0}^{z}\psi (\zeta )\phantom{\rule{0.2em}{0ex}}d\zeta ,\zeta \in U\}

or

{W}_{0}^{1,2}(U,H)=\{f:\mathrm{\exists}\psi \in {L}^{2}(U;H);f(z)\in {L}^{1}(U,\psi )\}.

By employing the concept of sums of accretive operators, we shall prove the maximal regularity of problem (1).

We proceed to extend the fractional integral operator {I}_{z}^{\alpha} to the space {L}^{2}(U;H). Define the fractional integral operator {\mathcal{I}}_{z}^{\alpha}:{L}^{2}(U;H)\to {L}^{2}(U;H) by

{\mathcal{I}}_{z}^{\alpha}u(z):={\int}_{0}^{z}\frac{{(z-\zeta )}^{\alpha -1}}{\Gamma (\alpha )}u(\zeta )\phantom{\rule{0.2em}{0ex}}d\zeta ,\phantom{\rule{1em}{0ex}}\alpha >0,

where u\in {L}^{2}(U;H). We have the following property.

**Lemma 2.1** {\mathcal{I}}_{z}^{\alpha}\in B({L}^{2}(U;H)).

*Proof* By making use of the Young inequality, it follows that

{\parallel {\mathcal{I}}_{z}^{\alpha}\parallel}_{{L}^{2}(U;H)}\le {\parallel {\mu}_{\alpha}\parallel}_{{L}^{1}(U)}{\parallel u\parallel}_{{L}^{2}(U;H)}\le C{\parallel u\parallel}_{{L}^{2}(U;H)}.

Similarly, we extend the fractional integral operator {D}_{z}^{\alpha} to the space {L}^{2}(U;H) by the operator

{\mathcal{D}}_{z}^{\alpha}:{L}^{2}(U;H)\to {L}^{2}(U;H)

such that

{\mathcal{D}}_{z}^{\alpha}u(z):=\frac{1}{\Gamma (1-\alpha )}\frac{d}{dz}{\int}_{0}^{z}\frac{u(\zeta )}{{(z-\zeta )}^{\alpha}}\phantom{\rule{0.2em}{0ex}}d\zeta ,\phantom{\rule{1em}{0ex}}0\le \alpha <1.

Furthermore, we define the space {\mathcal{W}}_{0}^{\alpha ,2}(U,H) as follows:

{\mathcal{W}}_{0}^{\alpha ,2}(U,H):=\{u\in {L}^{2}(U;H):{I}_{z}^{1-\alpha}u\in {W}_{0}^{1,2}(U,H)\}.

□

**Lemma 2.2** *Let* f\in {L}^{2}(U;H), *then*

{\mathcal{I}}_{z}^{\alpha +\beta}f={\mathcal{I}}_{z}^{\alpha}{\mathcal{I}}_{z}^{\beta}f,\phantom{\rule{1em}{0ex}}\alpha >0,\beta >0.

(2)

*Proof* For a function *f*, using the Dirichlet technique yields

\begin{array}{rl}{\mathcal{I}}_{z}^{\alpha}{\mathcal{I}}_{z}^{\beta}f(z)& =\frac{1}{\Gamma (\alpha )}{\int}_{0}^{z}{(z-\zeta )}^{\alpha -1}{\mathcal{I}}_{\zeta}^{\beta}f(\zeta )\phantom{\rule{0.2em}{0ex}}d\zeta \\ =\frac{1}{\Gamma (\alpha )\Gamma (\beta )}{\int}_{0}^{z}{(z-\zeta )}^{\alpha -1}({\int}_{\xi}^{\zeta}{(\zeta -\xi )}^{\beta -1}f(\xi ))\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\zeta \\ =\frac{1}{\Gamma (\alpha )\Gamma (\beta )}{\int}_{0}^{z}f(\xi )\left({\int}_{\xi}^{\zeta}{(z-\zeta )}^{\alpha -1}{(\zeta -\xi )}^{\beta -1}\right)\phantom{\rule{0.2em}{0ex}}d\zeta \phantom{\rule{0.2em}{0ex}}d\xi .\end{array}

(3)

Let \omega :=\frac{\zeta -\xi}{z-\xi}, we impose

\begin{array}{rl}{\int}_{\xi}^{\zeta}{(z-\zeta )}^{\alpha -1}{(\zeta -\xi )}^{\beta -1}\phantom{\rule{0.2em}{0ex}}d\zeta & ={(z-\xi )}^{\alpha +\beta -1}{\int}_{0}^{1}{(1-\omega )}^{\alpha -1}{\omega}^{\beta -1}\phantom{\rule{0.2em}{0ex}}d\omega \\ ={(z-\xi )}^{\alpha +\beta -1}\frac{\Gamma (\alpha )\Gamma (\beta )}{\Gamma (\alpha +\beta )}.\end{array}

(4)

Thus we have

\begin{array}{rl}{\mathcal{I}}_{z}^{\alpha}{\mathcal{I}}_{z}^{\beta}f(z)& =\frac{1}{\Gamma (\alpha )\Gamma (\beta )}{\int}_{0}^{z}{(z-\zeta )}^{\alpha +\beta -1}\frac{\Gamma (\alpha )\Gamma (\beta )}{\Gamma (\alpha +\beta )}f(\xi )\phantom{\rule{0.2em}{0ex}}d\xi \\ =\frac{1}{\Gamma (\alpha )\Gamma (\beta )}{\int}_{0}^{z}{(z-\xi )}^{\alpha +\beta -1}\frac{\Gamma (\alpha )\Gamma (\beta )}{\Gamma (\alpha +\beta )}f(\xi )\phantom{\rule{0.2em}{0ex}}d\xi \\ =\frac{1}{\Gamma (\alpha +\beta )}{\int}_{0}^{z}{(z-\xi )}^{\alpha +\beta -1}f(\xi )\phantom{\rule{0.2em}{0ex}}d\xi \\ ={\mathcal{I}}_{z}^{\alpha +\beta}f(z).\end{array}

(5)

□

**Lemma 2.3** {\mathcal{D}}_{z}^{\alpha}{\mathcal{I}}_{z}^{\alpha}u(z)=u(z).

*Proof* By Lemma 2.2, we have

{\mathcal{D}}_{z}^{\alpha}{\mathcal{I}}_{z}^{\alpha}u(z)={\mathcal{D}}_{z}^{1}{\mathcal{I}}_{z}^{1-\alpha}{\mathcal{I}}_{z}^{\alpha}u(z)={\mathcal{D}}_{z}^{1}{\mathcal{I}}_{z}^{1}u(z)=u(z).

From the last assertion, we conclude that {\mathcal{D}}_{z}^{1}={\mathcal{I}}_{z}^{-1}. □

**Lemma 2.4** *Let* \alpha \in [k-1,k), *then* {\mathcal{I}}_{z}^{\alpha}{\mathcal{D}}_{z}^{\alpha}u(z)=u(z)-{\sum}_{j=1}^{k}{[{\mathcal{D}}_{z}^{\alpha -j}u(z)]}_{z=0}\frac{{z}^{\alpha -j}}{\Gamma (\alpha -j+1)}.

*Proof* Since

\begin{array}{rl}{\mathcal{I}}_{z}^{\alpha}{\mathcal{D}}_{z}^{\alpha}u(z)& =\frac{1}{\Gamma (\alpha )}{\int}_{0}^{z}{(z-\zeta )}^{\alpha -1}{\mathcal{D}}_{z}^{\alpha}u(\zeta )\phantom{\rule{0.2em}{0ex}}d\zeta \\ =\frac{d}{dz}[\frac{1}{\Gamma (\alpha +1)}{\int}_{0}^{z}{(z-\zeta )}^{\alpha}{\mathcal{D}}_{z}^{\alpha}u(\zeta )\phantom{\rule{0.2em}{0ex}}d\zeta ],\end{array}

then, by using integration by parts, we get

\begin{array}{rl}\frac{1}{\Gamma (\alpha +1)}{\int}_{0}^{z}{(z-\zeta )}^{\alpha}{\mathcal{D}}_{z}^{\alpha}u(\zeta )\phantom{\rule{0.2em}{0ex}}d\zeta =& \frac{1}{\Gamma (\alpha +1)}{\int}_{0}^{z}{(z-\zeta )}^{\alpha}\frac{{d}^{k}}{d{\zeta}^{k}}{\mathcal{D}}_{z}^{-(k-\alpha )}u(\zeta )\phantom{\rule{0.2em}{0ex}}d\zeta \\ =& \frac{1}{\Gamma (\alpha -k+1)}{\int}_{0}^{z}{(z-\zeta )}^{\alpha -k}{\mathcal{D}}_{z}^{-(k-\alpha )}u(\zeta )\phantom{\rule{0.2em}{0ex}}d\zeta \\ -\sum _{j=1}^{k}\frac{{d}^{k-j}}{d{\zeta}^{k-j}}{[{\mathcal{D}}_{z}^{-(k-\alpha )}u(z)]}_{z=0}\frac{{z}^{\alpha -j+1}}{\Gamma (\alpha -j+2)}\\ =& \frac{1}{\Gamma (\alpha -k+1)}{\int}_{0}^{z}{(z-\zeta )}^{\alpha -k}{\mathcal{D}}_{z}^{-(k-\alpha )}u(\zeta )\phantom{\rule{0.2em}{0ex}}d\zeta \\ -\sum _{j=1}^{k}{[{\mathcal{D}}_{z}^{\alpha -j}u(z)]}_{z=0}\frac{{z}^{\alpha -j+1}}{\Gamma (\alpha -j+2)}\\ =& {\mathcal{I}}_{z}^{\alpha -k+1}({\mathcal{D}}_{z}^{-(k-\alpha )}u(z))-\sum _{j=1}^{k}{[{\mathcal{D}}_{z}^{\alpha -j}u(z)]}_{z=0}\frac{{z}^{\alpha -j+1}}{\Gamma (\alpha -j+2)}\\ =& {\mathcal{I}}_{z}^{1}u(z)-\sum _{j=1}^{k}{[{\mathcal{D}}_{z}^{\alpha -j}u(z)]}_{z=0}\frac{{z}^{\alpha -j+1}}{\Gamma (\alpha -j+2)}.\end{array}

Combining the last two assertions, we end the proof. □

**Remark 2.2** For a special case \alpha \in (0,1), we have the relation

{\mathcal{I}}_{z}^{\alpha}{\mathcal{D}}_{z}^{\alpha}u(z)=u(z)-{[{\mathcal{D}}_{z}^{\alpha -1}u(z)]}_{z=0}\frac{{z}^{\alpha -1}}{\Gamma (\alpha )}.

Note that the initial condition of problem (1) implies that u(z) of the form

u(z)=\sum _{n=1}^{\mathrm{\infty}}{a}_{n}{z}^{n},\phantom{\rule{1em}{0ex}}z\in U

(this class of analytic functions has wide applications in the geometric function theory and the univalent function theory when {a}_{1}=1 (see [28])); hence we obtain

{\mathcal{I}}_{z}^{\alpha}{\mathcal{D}}_{z}^{\alpha}u(z)=u(z),\phantom{\rule{1em}{0ex}}\alpha \in (0,1).

By virtue of the last discussion, we have the following result.

**Lemma 2.5** *Let* u\in {L}^{2}(U;H), *then* {\mathcal{D}}_{z}^{\alpha}u(z), \alpha \in (0,1) *is an accretive operator*.

*Proof* To prove that {\mathcal{D}}_{z}^{\alpha}u(z) is an accretive operator, it suffices to show that \mathrm{\Re}{({\mathcal{D}}_{z}^{\alpha}u,u)}_{{L}^{2}(U;H)}\ge 0, where *u* is in the domain of {\mathcal{D}}_{z}^{\alpha}. By the definition of {\mathcal{I}}_{z}^{\alpha}u(z), we receive that \mathrm{\Re}({(z-\zeta )}^{\alpha -1})>0; consequently, this implies that

\mathrm{\Re}{({\mathcal{I}}_{z}^{\alpha}v,v)}_{{L}^{2}(U;H)}\ge 0,

where v\in {L}^{2}(U;H). Hence, by Remark 2.2, we have

\mathrm{\Re}{({\mathcal{D}}_{z}^{\alpha}u,u)}_{{L}^{2}(U;H)}=\mathrm{\Re}{({\mathcal{D}}_{z}^{\alpha}u,{\mathcal{I}}_{z}^{\alpha}{\mathcal{D}}_{z}^{\alpha}u)}_{{L}^{2}(U;H)}:=\mathrm{\Re}{({u}^{(\alpha )},{\mathcal{I}}_{z}^{\alpha}{u}^{(\alpha )})}_{{L}^{2}(U;H)}\ge 0,

but *u* is in the domain of {\mathcal{D}}_{z}^{\alpha}, so, consequently, {\mathcal{D}}_{z}^{\alpha}u(z) is an accretive operator. □

**Lemma 2.6** *Let* u\in {L}^{2}(U;H), *then* {\mathcal{D}}_{z}^{\alpha}u(z) *is an* *m*-*accretive operator*.

*Proof* To prove that Rang(\lambda I+{\mathcal{D}}_{z}^{\alpha})=H, it suffices to show that the function {(\lambda I+{\mathcal{D}}_{z}^{\alpha})}^{-1}\varphi is well defined for all \lambda >0, u\in {L}^{2}(U;H) and bounded in {L}^{2}(U;H). A simple computation shows that

\mathrm{\Phi}(z):=\left[{(\lambda I+{\mathcal{D}}_{z}^{\alpha})}^{-1}\varphi \right](z)={\int}_{0}^{z}{(z-\zeta )}^{\alpha -1}{E}_{\alpha ,\alpha}(-\lambda {(z-\zeta )}^{\alpha})\varphi (\zeta )\phantom{\rule{0.2em}{0ex}}d\zeta ,

(6)

where \varphi \in {L}^{2}(U;H) and

{E}_{\alpha ,\beta}:=\sum _{m=0}^{\mathrm{\infty}}\frac{{z}^{m}}{\Gamma (m\alpha +\beta )}

is a Mittag-Leffler function. Therefore, by applying the Young inequality, we conclude that

{\parallel \mathrm{\Phi}(z)\parallel}_{{L}^{2}(U;H)}\le {L}_{\alpha ,\lambda}{\parallel \varphi \parallel}_{{L}^{2}(U;H)},\phantom{\rule{1em}{0ex}}|z|<1,

where {L}_{\alpha ,\lambda}>0. Thus \mathrm{\Phi}(z) is well defined for all \lambda >0, u\in {L}^{2}(U;H) and bounded in {L}^{2}(U;H). This implies that {\mathcal{D}}_{z}^{\alpha}u(z) is an *m*-accretive operator. □