Let $ \mathcal{H}(\mathbb{D}) $ be the space of analytic functions on the unit disc $ \mathbb{D} $ of the complex plane $ \mathbb{C} $. For $ 0<p<\infty $, the Hardy space $ H^{p}(\mathbb{D}) $ consists of all functions $ f\in\mathcal{H}(\mathbb{D}) $ such that

$ \|f\|_{H^{p}}:=\bigg(\sup\limits_{0<r<1}\frac{1}{2\pi}\displaystyle{\int}_{0}^{2\pi}|f(re^{i\theta})|^{p}\, d\theta\bigg)^{\frac{1}{p}}<\infty. $

In this paper, we are more interested in Derivative Hardy spaces $ S^{p}(\mathbb{D}) $ which are defined by

$ S^{p}(\mathbb{D})=\{f\in H^{p}(\mathbb{D}):\, f'(z)\in H^{p}(\mathbb{D})\}, $

with norm

$ \|f\|_{S^{p}}=|f(0)|+\|f'\|_{H^{p}}. $

The main objective of this paper is to study the Duhamel product $ \circledast $ on the $ S^{p}(\mathbb{D}) $ space. The Duhamel product is defined as follows (see [1]):

$ (f\circledast g)(z):=\frac{d}{dz}\displaystyle{\int}_{0}^{z}f(z-t)g(t)\, dt=\displaystyle{\int}^{z}_{0}f'(z-t)g(t)\, dt+f(0)g(z). $

The problem about the Duhamel product is not new. Wigley[1] introduced this product and showed that the complex-valued function space is a Banach algebra with respect to the Duhamel product $ \circledast $. In 1975, Wigley[2] showed that the Hardy space $ H^{p}(\mathbb{D}) $ is a Banach algebra with respect to the Duhamel product $ \circledast $. Afterward, Merryfield and Watson extended this result to $ H^{p} $ of the polydisc[3]. Then, the study of Duhamel products $ \circledast $ attracted many scholars. Karaev and Tuna[4], showed that $ C^{(n)}(\overline{\mathbb{D}}) $ is a commutative unital Banach algebra with respect to the Duhamel product. Gürdal[5] extended some of Duhamel product properties to the analytic sequence space $ l_{A}^{p}(\mathbb{D}) $. Recently, Guediri and his collaborators[6] showed that the Bergman space is a commutative unital Banach algebra with respect to the Duhamel product. For more information about the Duhamel product, see [4, 6-10].

This paper gives a Banach algebra structure with respect to the Duhamel product $ \circledast $ in $ S^{p}(\mathbb{D}) $ spaces. Then, we consider the invertibility criterion of the Duhamel product $ \circledast $. For the invertibility criterion, Volterra integration operator $ V $ is essential in the proof process. Moreover, the invertibility criterion has two interesting results, the maximal ideal space consists of one homomorphism and the spectrum of $ f $ consists of one point $ f(0) $. In section 3, we will consider the extended eigenvalue of Volterra integration operator $ V $. In fact, the study of the extended eigenvalue of Volterra integration operator $ V $ has a lot of rich results. Biswas and his collaborators[11] considered the case of $ L^{2}(0, 1) $. Karaev[7] described the extended eigenvalues of $ V $ in $ n $-th differentiable functions space. For more information, we recommend [4, 6, 8-10]. Then, we also give some characterizations of the bounded linear operator on $ S^{p}(\mathbb{D}) $.

For convenience, we use the notation $ X\lesssim Y $ or $ Y\lesssim X $ to mean $ X<CY $ or $ Y<CX $, where $ C $ is positive constant. Similarly, we use the notation $ X\approx Y $ if both $ X\lesssim $ and $ Y\lesssim $ hold.

Elementary calculation shows that the Duhamel product-$ \circledast $ can also be given by

$ \begin{aligned} (f\circledast g)(z):&= \frac{d}{dz}\int^{z}_{0}f(z-t)g(t)\, dt\\ &=\int^{z}_{0}f'(z-t)g(t)\, dt+f(0)g(z)\\ &=\int^{z}_{0}f(z-t)g'(t)\, dt+f(z)g(0). \end{aligned} $

Apparently, it is a commutative product. Then, take the derivative of the above equation (see [1])

$ \begin{aligned} (\mathcal{D}_{f}g)'(z)&=(f\circledast g)'(z)=\int^{z}_{0}f'(z-t)g'(t)\, dt+f(0)g'(z)+f'(z)g(0)\\ &=\int^{\frac{z}{2}}_{0}f'(z-t)g'(t)\, dt+\int^{\frac{z}{2}}_{0}f'(t)g'(z-t)\, dt+f(0)g'(z)+f'(z)g(0)\\ &=\int^{\frac{z}{2}}_{0}\big(f'(z-t)g'(t)+f'(t)g'(z-t)\big)\, dt+f(0)g'(z)+f'(z)g(0). \end{aligned} $

Lemma 2.1 Let $ 1\leq p<\infty $, for any $ f\in S^{p}(\mathbb{D}) $, then (see [12])

$ |f'(z)|\lesssim\frac{\|f\|_{S^{p}}}{(1-|z|^{2})^{\frac{1}{p}}}. $

Since $ t\in[0, \frac{z}{2}] $, by the above Lemma, we have $ |f'(t)|\lesssim\|f\|_{S^{p}} $.

$ \begin{equation} \begin{aligned} \big|(\mathcal{D}_{f}g)'(z)\big|\lesssim &\|g\|_{S^{p}}\int^{\frac{z}{2}}_{0}|f'(z-t)|\, |dt|+|f'(z)||g(0)|\\ &+\|f\|_{S^{p}}\int^{\frac{z}{2}}_{0}|g'(z-t)|\, |dt|+|f(0)||g'(z)|. \end{aligned} \end{equation} $

To simplify the notation, we denote $ \int^{\frac{z}{2}}_{0}|f'(z-t)|\, |dt| $ and $ \int^{\frac{z}{2}}_{0}|g'(z-t)|\, |dt| $ by $ F(z) $ and $ G(z) $. Next we consider

$ M^{p}_{p}(F, r)=\frac{1}{2\pi}\displaystyle{\int}_{0}^{2\pi}|F(re^{i\theta})|^{p}\, d\theta. $

Lemma 2.2 Let $ p>1 $, then $ M_{p}(F, r)\lesssim(1-|z|)^{1-\frac{1}{p}}\|f\|_{S^{p}} $.

Proof Let $ z=re^{i\theta} $, $ 0\leq r<1 $, and $ t=\rho e^{i\theta} $, then we have

$ \begin{equation} \begin{aligned}\nonumber F(z)&=\int^{\frac{z}{2}}_{0}|f'(z-t)|\, |dt|=\int_{\frac{r}{2}}^{r}|f'(\rho e^{i\theta})|\, d\rho\\ &\leq\|f\|_{S^{p}}\int_{\frac{r}{2}}^{r}\frac{d\rho}{(1-\rho^{2})^{\frac{1}{p}}}\leq\|f\|_{S^{p}}\int_{\frac{r}{2}}^{r}\frac{d\rho}{(1-\rho)^{\frac{1}{p}}}\\ &\lesssim(1-|z|)^{1-\frac{1}{p}}\|f\|_{S^{p}}.\\ \end{aligned} \end{equation} $

Thus, we get

$ \begin{equation} \begin{split}\nonumber M_{p}^{p}(F, r)&=\frac{1}{2\pi}\int_{0}^{2\pi}\bigg(\int_{0}^{\frac{r}{2}e^{i\theta}}|f'(re^{i\theta}-t)|^{p}|d\, t|\bigg)^{p}\, d\theta\\ &\lesssim(1-|z|)^{p-1}\|f\|_{S^{p}}^{p}. \end{split} \end{equation} $

The proof of Lemma is complete.

Theorem 2.3 Suppose that $ 1<p<\infty $, the $ S^{p}(\mathbb{D}) $ space is a unital and commutative Banach algebra with respect to the Duhamel product $ \circledast $, which we will denote $ (S^{p}(\mathbb{D}), \circledast) $.

Proof Firstly, $ S^{p}(\mathbb{D}) $ is a Banach space. By (2.1), Lemma 2.1 and Minkowski inequality, we have

$ \begin{equation} \begin{aligned}\nonumber \|\mathcal{D}_{f}g\|_{S^{p}}&=|f(0)||g(0)|+\|(f\circledast g)'\|_{H^{p}}\\ &=|f(0)||g(0)|+\sup\limits_{0<r<1}M_{p}((f\circledast g)', r)\\ &\leq3\|f\|_{S^{p}}\|g\|_{S^{p}}+\|f\|_{S^{p}}\sup\limits_{0<r<1}M_{p}(F, r)+\|g\|_{S^{p}}\sup\limits_{0<r<1}M_{p}(G, r)\\ &\lesssim\|f\|_{S^{p}}\|g\|_{S^{p}}. \end{aligned} \end{equation} $

The proof is complete.

Corollary 2.4 $ (S^{p}(\mathbb{D}), \circledast) $ is an involutive algebra (cf. [13]), the involution is a map

$ \begin{equation} \begin{aligned}\nonumber \ast\, :\quad\big(S^{p}(\mathbb{D})\big), \circledast)&\rightarrow \big(S^{p}(\mathbb{D}), \circledast\big)\\ f&\mapsto f^{\ast} \end{aligned} \end{equation} $

where $ f^{\ast}=\overline{f(\overline{z})} $.

Proof Let $ f\in S^{p}(\mathbb{D}) $, then we have

$ \begin{equation} \begin{aligned}\nonumber \|f\|_{S^{p}}&=|f(0)|+\|f'(z)\|_{H^{p}}\\ &=|f(0)|+\bigg(\sup\limits_{0<r<1}\int_{0}^{2\pi}|f'(re^{i\theta})|^{p}\, d\theta\bigg)^{\frac{1}{p}}\\ &=|f(0)|+\bigg(\sup\limits_{0<r<1}\int_{0}^{2\pi}|f'(re^{-i\theta})|^{p}\, d\theta\bigg)^{\frac{1}{p}}\\ &=|f(0)|+\bigg(\sup\limits_{0<r<1}\int_{0}^{2\pi}|\overline{f'(re^{-i\theta})}|^{p}\, d\theta\bigg)^{\frac{1}{p}}=\|f^{\ast}\|_{S^{p}}. \end{aligned} \end{equation} $

We can get $ f^{\ast}\in S^{p}(\mathbb{D}) $. Clearly, the involution on $ (S^{p}(\mathbb{D}), \circledast) $ is satisfying $ (f^{\ast})^{\ast}=f $, and $ (\alpha\circledast f+\beta\circledast g)^{\ast}=\overline{\alpha}\circledast f^{\ast}+\overline{\beta}\circledast g^{\ast} $ for all $ f\, , g\in S^{p}(\mathbb{D}) $, $ \alpha\, , \beta\in\mathbb{C} $. Next we prove that $ (f\circledast g)^{\ast}=f^{\ast}\circledast g^{\ast} $,

$ \begin{equation} \begin{aligned}\nonumber (f^{\ast}\circledast g^{\ast})(z)&=\int_{0}^{z}\overline{f'(\overline{z-t})}\, \overline{g(\overline{t})}\, dt+\overline{f(0)}\, \overline{g(\overline{z})}\\ &=\overline{\int_{0}^{z}f'(\overline{z-t})g(\overline{t})\, d\overline{t}}+\overline{f(0)}\, \overline{g(\overline{z})}\\ &=\overline{\int_{0}^{\overline{z}}f'(\overline{z}-t)g(t)\, dt}+\overline{f(0)}\, \overline{g(\overline{z})}=(f\circledast g)^{\ast}(z), \\ \end{aligned} \end{equation} $

which implies that $ (S^{p}(\mathbb{D}), \circledast) $ is an involutive algebra, or a $ \ast $-algebra.

In [6], Guediri and his collaborators, see the Duhamel product as a special operator, which is called the Duhamel convolution operator with analytic symbol $ f $. For convenience, we call this operators as Duhamel convolution operator and denote it by

$ \mathcal{D}_{f}g:=(f\circledast g)(z)=\displaystyle{\int}_{0}^{z}f'(z-t)g(t)\, dt+f(0)g(z). $

Clearly, if $ f(z)=z $ then $ \mathcal{D}_{f}g=Vg=\int_{0}^{z}g(t)dt $ is the Volterra integration operator. Moreover, $ \mathcal{D}_{f}V=V\mathcal{D}_{f} $. Next we consider the operator norm of $ \mathcal{D}_{f} $.

Theorem 2.5 Suppose that $ 1<p<\infty $, then $ \mathcal{D}_{f} $ from $ S^{p}(\mathbb{D}) $ to $ S^{p}(\mathbb{D}) $ is bounded if and only if $ f\in S^{p}(\mathbb{D}) $. Moreover, the norm of operator satisfies $ \|\mathcal{D}_{f}\|\approx\|f\|_{S^{p}}. $

Proof Firstly, by Theorem 2.3 and Minkowski inequality we have

$ \|\mathcal{D}_{f}g\|_{S^{p}}\lesssim\|g\|_{S^{p}}. $

Conversely, if $ \mathcal{D}_{f} $ is bounded. Take test function $ g=1 $, then

$ \begin{equation} \nonumber \|\mathcal{D}_{f}\|\geq\|\mathcal{D}_{f}1\|_{S^{p}}=\|f\|_{S^{p}}. \end{equation} $

So we get $ \|\mathcal{D}_{f}\|\approx\|f\|_{S^{p}} $. The proof of Theorem 2.5 is complete.

For the invertibility criterion of the Duhamel product, the Volterra integration operator acts as a bond. In the next Theorem we need the following convolutions(see [14])

$ \begin{equation} (V^{n}g)(z)=\displaystyle{\int}_{0}^{z}\frac{(z-t)^{n-1}}{(n-1)!}g(t)\, dt=\bigg(\frac{\omega^{n-1}}{(n-1)!}*g\bigg)(z), \end{equation} $

(2.2)

where the notation * means

$ (f*g)(z):=\displaystyle{\int}_{0}^{z}f(z-t)g(t)\, dt. $

Before we start proving the theorem, we need the following lemma.

Lemma 2.6 (see [15]) The Taylor series of every function $ f\in S^{p}(\mathbb{D}) $ with $ f(0)=0 $ converges in norm if and only if $ 1<p<\infty $.

Proof Let $ 1<p<\infty $, if $ f\in S^{p}(\mathbb{D}) $, then $ f'(z)\in H^{p}(\mathbb{D}) $. By Corollary 3 of [15], the Taylor series of every function in $ H^{p}(\mathbb{D}) $ converges in norm $ \|\cdot\|_{H^{p}} $. We get the Taylor series of $ f'(z) $ converges in norm $ \|\cdot\|_{H^{p}} $, i.e.

$ \lim\limits_{n\rightarrow \infty}\|f(z)-f_{n}(z)\|_{S^{p}}=\lim\limits_{n\rightarrow \infty}\|f'(z)-f'_{n}(z)\|_{H^{p}}=0. $

Theorem 2.7 Let $ 1<p<\infty $. Then $ f\in S_{p}(\mathbb{D}) $ is $ \circledast $-invertible if and only if $ f(0)\neq0 $.

Proof If $ f(z)\in S^{p}(\mathbb{D}) $ is $ \circledast $-invertible, then there exists $ g\in S^{p}(\mathbb{D}) $ such that

$ (f\circledast g)(z)=1, \quad \forall z\in\mathbb{D}. $

Clearly, if $ z=0 $ we have $ f(0)g(0)=1 $, i.e. $ f(0)\neq0 $.

On the other hand, if $ f(0)\neq0 $, let $ F(z)=f(z)-f(0) $. Then, we consider the operator $ \mathcal{D}_{F}\, :\, S^{p}(\mathbb{D})\rightarrow S^{p}(\mathbb{D}) $, defined by as follow

$ \mathcal{D}_{F}g=(F\circledast g)(z)=\displaystyle{\int}_{0}^{z}F'(z-t)g(t)\, dt=\displaystyle{\int}_{0}^{z}f'(z-t)g(t)\, dt. $

Next, we show that $ \mathcal{D}_{F} $ is compact. For the partial Taylor series

$ F_{n}=\sum\limits_{n=0}^{N}\hat{F}(n)z^{n}=\sum\limits_{n=1}^{N}\hat{f}(n)z^{n}. $

Then,

$ \begin{equation} \begin{aligned}\nonumber \mathcal{D}_{F_{N}}g(z)&=\int_{0}^{z}F'_{N}(z-t)g(t)\, dt=\int_{0}^{z}F'_{N}(t)g(z-t)\, dt\\ &=\sum\limits_{n=1}^{N}(n!)\hat{f}(n)\int_{0}^{z}\frac{(z-t)^{n-1}}{(n-1)!}g(z-t)\, dt. \end{aligned} \end{equation} $

So, we deduced that

$ \mathcal{D}_{F_{N}}g(z)=\sum\limits_{n=1}^{N}(n!)\hat{f}(n)V^{n}g(z). $

Thus, $ \mathcal{D}_{F_{N}} $ is compact on $ S^{p}(\mathbb{D}) $. Since

$ \begin{equation} \nonumber \|(\mathcal{D}_{F}-\mathcal{D}_{F_{N}})g\|_{S^{p}}=\|\mathcal{D}_{(F-F_{N})}g(z)\|_{S^{p}}\lesssim\|F-F_{N}\|_{S^{p}}\|g\|_{S^{p}}, \end{equation} $

we have

$ \|\mathcal{D}_{F}-\mathcal{D}_{F_{N}}\|\lesssim\|F-F_{N}\|_{S^{p}}. $

Let $ N\rightarrow \infty $, then by Lemma 2.6 and Theorem 1.18 of [18], we deduced that $ \mathcal{D}_{F} $ is compact. Now, suppose that $ g\in {\rm ker}\mathcal{D}_{f} $, i.e.

$ \mathcal{D}_{f}g(z)=\displaystyle{\int}_{0}^{z}f'(z-t)g(t)\, dt+f(0)g(z)=0. $

We get $ g(0)=0 $. Then derivative of the above equation, we have

$ \frac{d}{dz}\mathcal{D}_{f}g(z)=\displaystyle{\int}_{0}^{z}f''(z-t)g(t)dt+f(0)g'(z)+f'(0)g(z)=0. $

We get $ g'(0)=0 $. Similarly, by repeating the derivative process, we deduced that $ g^{(n)}(0)=0 $. So we get $ g=0 $, i.e. $ {\rm ker}\mathcal{D}_{f}=\{0\} $. Notice that

$ \mathcal{D}_{f}=f(0)I+\mathfrak{D}_{F}. $

Since $ \mathcal{D}_{F} $ is compact, $ \mathcal{D}_{f} $ is injective and $ f(0)\neq 0 $. Then, according to Fredholm alternative theorem (see [16, Theorem 3.22]). We infer that $ \mathcal{D}_{f} $ is invertible in $ S^{p}(\mathbb{D}) $. The proof is complete.

Corollary 2.8 The maximal ideal space $ \mathcal{M}(S^{p}, \circledast) $ consists of only one homomorphism.

Proof Apparently, the spectrum of any $ f\in S^{p}(\mathbb{D}) $ consists of only one point $ f(0) $. The functions which vanish at the origin form a maximal ideal, since no proper ideal of $ (S^{p}(\mathbb{D}), \circledast) $ contains any invertible $ f $, and hence $ \mathcal{M} $ is only one homomorphism, and Gelfand transform is trivial.

In this section, we consider the extended eigenvalue of Volterra integration operator $ V $. Let $ \lambda $ be a complex number, if there exists a nonzero operator $ X $ satisfying the equation $ XV=\lambda VX $, then we say $ \lambda $ is an extended eigenvalue of $ V $. In the next assertion, we consider the extended eigenvector of the Volterra integration operator $ V $. For all bounded linear operators in the $ S^{p}(\mathbb{D}) $ space we denote by $ \mathcal{B}(S^{p}(\mathbb{D})) $.

Theorem 3.1 Let $ \lambda\neq0\in\mathbb{C} $, $ 1\leq p<\infty $, and $ X\in\mathcal{B}(S^{p}(\mathbb{D})) $, then we have

(1) Suppose that $ |\lambda|\leq1 $, then, $ XV=\lambda VX $ if and only if $ X=\mathcal{D}_{X1}{\mathcal{C}_{\lambda}}. $

(2) Suppose that $ |\lambda|>1 $, then, $ XV=\lambda VX $ if and only if $ X\mathcal{C}_{\frac{1}{\lambda}}=\mathcal{D}_{X1}. $

Proof Firstly we consider (1), if $ XV=\lambda VX $, then $ XV^{n}=\lambda^{n}V^{n}X $. In particular, take the function $ 1 $, we have $ XV^{n}1=\lambda^{n}V^{n}X1 $. By (2.2), we get

$ X\frac{z^{n}}{n!}=\lambda^{n}\frac{z^{n-1}}{(n-1)!}*X1, $

then

$ 1*X\frac{z^{n}}{n!}=1*\lambda^{n}\frac{z^{n-1}}{(n-1)!}*X1=\frac{(\lambda n)^{n}}{n!}*X1. $

Similarly, for any polynomial $ P(z) $

$ 1*XP(z)=P(\lambda z)*X1. $

Since the polynomial is dense, we get

$ 1*Xf(z)=f(\lambda z)*X1. $

We deduce that

$ VXf=\displaystyle{\int}_{0}^{z}X1(z-t)C_{\lambda}f(t)\, dt. $

Then take derivative of the above equation,

$ Xf(z)=\frac{d}{dz}\displaystyle{\int}_{0}^{z}X1(z-t)C_{\lambda}f(t)\, dt=\mathcal{D}_{X1}C_{\lambda}f(z). $

Conversely, suppose that $ X=\mathcal{D}_{X1}C_{\lambda} $, then

$ \begin{equation} \begin{aligned}\nonumber XVf(z)&=\mathcal{D}_{X1}C_{\lambda}Vf(z)=X1\circledast(Vf)(\lambda z)\\ &=X1\circledast\lambda z\circledast f(\lambda z)=\lambda z\circledast X1\circledast C_{\lambda}f(z)\\ &=\lambda\big(z\circledast\mathcal{D}_{X1}C_{\lambda}f(z)\big)=\lambda VXf(z). \end{aligned} \end{equation} $

Thus we have $ XV=\lambda VX $.

The proof of (2) is similar to that of (1), and a brief proof is given for completeness. Suppose that $ |\lambda|>1 $, then $ V^{n}X=\frac{1}{\lambda^{n}}XV^{n} $. Choose the function $ 1 $, we get $ V^{n}X1=\frac{1}{\lambda^{n}}XV^{n}1 $. By (2.2) we have

$ 1*V^{n}X1=1*\frac{1}{\lambda^{n}}XV^{n}1, $

i.e.

$ \frac{z^{n}}{n!}*X1=1*X\frac{\frac{z^{n}}{\lambda^{n}}}{n!}. $

Since ploynoimal is dense, we deduce that

$ 1*Xf(\frac{z}{\lambda})=f(z)*X1. $

Taking the derivative of the above equation, we get $ XC_{\frac{1}{\lambda}}=\mathcal{D}_{X1} $.

Conversely, suppose that $ XC_{\frac{1}{\lambda}}=\mathcal{D}_{X1} $, then

$ \lambda VXz=\lambda VXC_{\frac{1}{\lambda}}C_{\lambda}z=\lambda V\mathcal{D}_{X1}C_{\lambda}z. $

Since

$ V\mathcal{D}_{X1}f=z\circledast X1\circledast f=X1\circledast z\circledast f=\mathcal{D}_{X1}Vf, $

so, for any polynomials $ P(z) $ we get

$ \begin{equation} \begin{aligned}\nonumber \lambda VXP(z)&=\lambda V\mathcal{D}_{X1}C_{\lambda}P(z)=\lambda\mathcal{D}_{X1}VP(\lambda z)\\ &=\lambda XC_{\frac{1}{\lambda}}VP(\lambda z)=XC_{\frac{1}{\lambda}}(\lambda z\circledast P(\lambda z))\\ &=X(z\circledast P(z))=XVP(z). \end{aligned} \end{equation} $

By density, we get $ \lambda VXf=XVf $. The proof of the Theorem 2.5 is complete.

If $ \lambda=1 $, we have $ X=\mathcal{D}_{X1} $, i.e the commutant of the Volterra operator on the $ S^{p}(\mathbb{D}) $ is the Duhamel product.

Remark 1 The commutants of the Volterra integration operator on the $ S^{p}(\mathbb{D}) $ is characterized by $ \{V\}'=\{\mathcal{D}_{f}:\, f\in S^{p}(\mathbb{D})\} $.

Corollary 3.2 Let $ f\in S^{2}(\mathbb{D}) $, $ \rm the Duhamel convolution operator $ $ \mathfrak{D}_{f}=\overline{{\rm alg}\{V\}} $, where

$ {\rm alg}\{V\}:=\{p(V):p{\rm\enspace is\enspace a\enspace poloynomial}\}. $

Clearly, $ V\mathfrak{D}_{f}=\mathfrak{D}_{f}V $ and $ S^{2}(\mathbb{D}) $ is a Hilbert space. By Theorem 2.5 we have $ \mathfrak{D}_{f}\in\mathcal{B}(S^{2}(\mathbb{D})) $. Every bounded linear operator commuting with $ V $ is in the strongly closed algebra generated by $ V $ (see [14, 17]). So, $ \mathfrak{D}_{f}=\overline{{\rm alg}\{V\}} $.