1 Introduction

Recall that \(C^{\infty }:=C^{\infty }\left[ 0,1\right] \) denotes the Frechet space of all infinitely differentiable functions in \(\left[ 0,1\right] \). The topology in \(C^{\infty }\) is given by the family of seminorms \(\left\{ P_{n}\right\} _{n\ge 0},\) \(P_{n}(f)=\underset{0\le k\le n}{\max } \underset{x\in \left[ 0,1\right] }{\max }\left| f^{\left( k\right) } (x)\right| ,\) \(f\in C^{\infty }\). The so-called Duhamel product (see Wigley [20]) is the derivative of the classical Mikusinski convolution [10, 12]:

$$\begin{aligned} (f\circledast g)(x)&=\frac{d}{\mathrm{d}x}(f*g)(x)=\frac{\mathrm{d}}{\mathrm{d}x} {\textstyle \int \limits _{0}^{x}} f(x-t)g(t)\mathrm{d}t\\&= {\textstyle \int \limits _{0}^{x}} f^{\prime }(x-t)g(t)\mathrm{d}t+f(0)g(x)\\&=(f^{\prime }*g)(x)+f(0)g(x),\ f,g\in C^{\infty }. \end{aligned}$$

It is clear that with this multiplication, \(C^{\infty }\) becomes an algebra. So, let \((C^{\infty },\circledast )\) denote the algebra of infinitely differentiable functions in \(\left[ 0,1\right] \) with Duhamel product \(\circledast \) as multiplication. \((C^{\infty },\circledast )\) is commutative and associative, and the function \(f(x)\equiv 1\) is the identity for \((C^{\infty },\circledast )\) (see [6]).

Notice that the Duhamel product is widely applied in various questions of analysis, namely in the theory of differential equations and in the solution of boundary value problems of mathematical physics (see Wigley [20]). The Duhamel product is closely related to the Volterra integration operator. It is being studied quite intensively; see the paper of Karaev [8] and also the papers [3, 5,6,7, 9, 15,16,17, 21].

The integration operator is defined on \(C^{\infty }\) by the formula:

$$\begin{aligned} (Jf)(x)={\textstyle \int \limits _{0}^{x}}f(t)\mathrm{d}t. \end{aligned}$$

In this article, we describe in terms of Duhamel product commutant of the operator J. Also, we investigate Duhamel equations in \(C^{\infty }\) and study uniqueness of their solutions. Note that previously, Tkachenko [19] investigated properties of the commutant of the operator of generalized integration operator in a space of analytic functionals.

2 The Commutant of Operator J

Recall that for any continuous linear operator T on \(C^{\infty }\) (i.e., for any \(T\in {\mathcal {L}}(C^{\infty })\)), its commutant is

$$\begin{aligned} \left\{ T^{\prime }\right\} :=\left\{ A\in {\mathcal {L}}(C^{\infty }):AJ=JA\right\} . \end{aligned}$$

It is necessary to note that the study of commutant of the concrete operator T on \(C^{\infty }\) or, especially on Banach spaces, is one of the important , but generally, not-easy problems of operator theory. For this , it is enough to remember important results obtained in the works [1,2,3, 11, 13, 14]. In this section, we describe in terms of the Duhamel operators the commutant of the integration operator J on \(C^{\infty }\).

Theorem 1

Let \(T\in {\mathcal {L}}(C^{\infty })\). Then \(TJ=JT\), i.e. \(T\in \left\{ J\right\} ^{\prime }\), if and only if there exists a function \(f\in C^{\infty }\) such that \(T=D_{f}\), where \(D_{f}\) is the Duhamel operator defined by \(D_{f}g:=f\circledast g,g\in C^{\infty }\).

Proof

Indeed, let \(TJ=JT\), then \(TJ^{n}=J^{n}T\) for all \(n\ge 0\), that is,

$$\begin{aligned} TJ^{n}g=J^{n}Tg \end{aligned}$$
(1)

for all \(g\in C^{\infty }\) and \(n\ge 0\). It is easy to see from (1) that

$$\begin{aligned} J^{n}g=\frac{x^{n}}{n!}\circledast g\ \ (n=1,2,3,\ldots ). \end{aligned}$$
(2)

Then, we have, in particular, that \(TJ^{n}\mathbf {1}=J^{n}T\mathbf {1}\), hence

$$\begin{aligned} T\left( \frac{x^{n}}{n!}\circledast \mathbf {1}\right) =\frac{x^{n}}{n!} \circledast T\mathbf {1} \end{aligned}$$

for all \(n\ge 0\). This implies that

$$\begin{aligned} T\left( \frac{x^{n}}{n!}\right) =\frac{x^{n}}{n!}\circledast T\mathbf {1}, \end{aligned}$$

or equivalently,

$$\begin{aligned} T(x^{n})=x^{n}\circledast T\mathbf {1}=D_{T\mathbf {1}}(x^{n})=D_{f}(x^{n}), \end{aligned}$$

where \(f:=T\mathbf {1}\in C^{\infty }\) and \(D_{f}\) is the Duhamel operator on \(C^{\infty }.\) It is clear from [6] that \(D_{f}\) is the continuous linear operator on \(C^{\infty }\). From the last equality, we have that

$$\begin{aligned} Tp(x)=D_{f}p(x) \end{aligned}$$

for any polynomial p, and thus by Weierstrass approximation theorem, we deduce that

$$\begin{aligned} Tg=D_{f}g \end{aligned}$$

for all \(g\in C^{\infty }\), that is, \(T=D_{f}\), as desired. This proves the theorem. \(\square \)

For the related results in Banach spaces, we refer the readers to the papers [9, 15, 16].

Corollary 1

We have \(\left\{ J\right\} ^{\prime \prime }=\left\{ J\right\} ^{\prime }\), where \(\left\{ J\right\} ^{\prime \prime }\) denotes the bicommutant of the operator J, i.e.,

$$\begin{aligned} \left\{ J\right\} ^{\prime \prime }=\left\{ X\in {\mathcal {L}}(C^{\infty }): XT=TX\text { for all }T\in \left\{ J\right\} ^{\prime }\right\} . \end{aligned}$$

Proof

It suffices to prove that \(T_{1}T_{2}=T_{2}T_{1}\) for every \(T_{1},T_{2}\) in \(\left\{ J\right\} ^{\prime }\). In fact, by Theorem 2.1, there exist functions \(f,g\in C^{\infty }\) such that

$$\begin{aligned} (T_{1}h)(x)&=(D_{f}h)(x)=f(0)h(x) +{\textstyle \int \limits _{0}^{x}} f^{\prime }(x-t)h(t)dt\nonumber \\&=(f(0)I+D_{f-f(0)})h(x) \end{aligned}$$
(3)

and

$$\begin{aligned} (T_{2}h)(x)&=(D_{g}h)(x)=g(0)h(x) +{\textstyle \int \limits _{0}^{x}} g^{\prime }(x-t)h(t)\mathrm{d}t\nonumber \\&=(g(0)I+D_{g-g_{0}})h(x) \end{aligned}$$
(4)

for all \(h\in C^{\infty }\), where I is an identity operator on \(C^{\infty } \). Since \((C^{\infty },\circledast )\) is the commutative (topological) algebra, we get from (3) and (4) that \(T_{1}T_{2} =T_{2}T_{1}\), as desired.

\(\square \)

3 Isomorphisms in the Commutant of J and Uniqueness of Solutions of Duhamel Equations

In this section, we characterize isomorphisms in the commutant \(\left\{ J\right\} ^{\prime }\) of the integration operator J on \(C^{\infty }\).

Theorem 2

Let \(T\in {\mathcal {L}}(C^{\infty })\) be an operator. Then, it will be an isomorphism of the space \(C^{\infty }\) into itself (i.e., T is invertible) and commutes with J if and only if it can be written in the form

$$\begin{aligned} (Tf)(x)=\varphi (0)f(x)+{\displaystyle \int \limits _{0}^{x}} \varphi ^{\prime }(x-t)f(t)\mathrm{d}t \end{aligned}$$
(5)

and \(\varphi (0)=(T\mathbf {1})(0)\ne 0.\)

Proof

If \(T\in {\mathcal {L}}(C^{\infty })\) is an isomorphism of the space \(C^{\infty }\) into itself and commutes with J, i.e., \(TJ=JT\), then by Theorem 2.1 we have representation (5) for T. Since \(TC^{\infty }=C^{\infty }\), it follows that \(\varphi (0)\ne 0\), that is, \((T\mathbf {1})(0)\ne 0\).

Conversely, suppose that T has representation (5) with \(\varphi (0)=(T\mathbf {1})(0)\ne 0\) and prove then that \(T\in \left\{ J\right\} ^{\prime }\) and T is an isomorphism. In fact, the fact that \(T\in \left\{ J\right\} ^{\prime }\), i.e., \(TJ=JT\), is immediate from Theorem 2.1. On the other hand, it is easy to see from (5) that

$$\begin{aligned} T=D_{\varphi }=\varphi (0)I+D_{\varphi -\varphi (0),} \end{aligned}$$
(6)

where \(D_{\varphi -\varphi (0)}\) is the usual Mikusinski convolution operator on \(C^{\infty },\) i.e.,

$$\begin{aligned} D_{\varphi -\varphi (0)}f={\displaystyle \int \limits _{0}^{x}} \varphi ^{\prime }(x-t)f(t)\mathrm{d}t=\varphi ^{\prime }*f:=C_{\varphi ^{\prime }}f. \end{aligned}$$
(7)

It is well known that the convolution operator \(C_{\varphi ^{\prime }}\) is compact in \(C^{\infty }\) (for the proof, the method of the paper [18] can also be used).

Let us continue the proof of the theorem. Now we show that \(\ker (D_{\varphi })=\left\{ 0\right\} \). Indeed, let \(f\in \ker (D_{\varphi }).\) Then,

$$\begin{aligned} (\varphi (0)I+D_{\varphi -\varphi (0)})f=0, \end{aligned}$$

that is,

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x} {\displaystyle \int \limits _{0}^{x}} \varphi (x-t)f(t)\mathrm{d}t=0,\ x\in \left[ 0,1\right] . \end{aligned}$$

Then,

$$\begin{aligned} {\displaystyle \int \limits _{0}^{x}} \varphi (x-t)f(t)\mathrm{d}t=c \end{aligned}$$

for all \(x\in \left[ 0,1\right] \) and some constant c. In particular, for \(x=0\), we have \(c=0.\) Hence,

$$\begin{aligned} {\displaystyle \int \limits _{0}^{x}}\varphi (x-t)f(t)dt =0\text { for all }x\in \left[ 0,1\right] \end{aligned}$$
(8)

since by condition \(\varphi \in C^{\infty }\) and \(\varphi (0)\ne 0,\) we have that \(\varphi \) does not vanish identically in any right neighborhood of 0. Then, by the Titchmarsh convolution theorem [12], we conclude from (7) that \(f(x)=0\) for all \(x\in \left[ 0,1\right] \). This means that \(\ker (D_{\varphi })=\left\{ 0\right\} \). Since \(D_{\varphi -\varphi (0)}\) is compact, it follows from Williamson theorem [22] that \(D_{\varphi }\) is invertible in \(C^{\infty },\) that is, T is an isomorphism. The theorem is proven. \(\square \)

We have from Theorems 2.1 and 3.1 the following.

Corollary 2

For any function \(\varphi \in C^{\infty }\) with \(\varphi (0)\ne 0\), there exists a unique isomorphism T of the \(C^{\infty }\) such that T commutes with J and \(T\mathbf {1}=\varphi \).

Corollary 3

If \(\varphi \in C^{\infty }\) and \(\varphi (0)\ne 0\), then the Duhamel equation

$$\begin{aligned} \varphi (0)f(x)+{\displaystyle \int \limits _{0}^{x}} \varphi ^{\prime }(x-t)f(t)\mathrm{d}t=g(x) \end{aligned}$$
(9)

has a unique solution for any fixed function \(g\in C^{\infty }.\)

Proof

Equation (9) means the following:

$$\begin{aligned} D_{\varphi }f=g. \end{aligned}$$
(10)

Since \(\varphi (0)\ne 0,\) it follows from the proof of Theorem 3.1 that the Duhamel operator \(D_{\varphi }\) is invertible in \(C^{\infty }\); hence, we have from (10) that \(f=D_{\varphi }^{-1}g;\) that is, (9) has a unique solution. \(\square \)

The authors of the paper [4] study the generalized Duhamel product in space of all holomorphic functions on a convex domain in the complex plane, containing the origin.