New Proof of the Ostapenko–Tarasov Theorem

Abstract

We give a new proof of the Ostapenko–Tarasov unicellularity theorem for the classical Volterra integration operator on the space $C^{(n)}[0,1]$.

1. Introduction

Let $X$ be a Banach space, and let $B(X)$ be the Banach algebra of bounded linear operators $A\colon X\rightarrow X$. A closed subspace $E\subset X$ is called an invariant subspace of $A$ if $AE\subset E$, i.e., $Ax\in E$ for all $x\in E$. An operator $A\in B(X)$ is said to be unicellular if its lattice $\operatorname{Lat}(A)$ of invariant subspaces is linearly ordered, i.e., for any two $A$-invariant subspaces $E$ and $M$ one has either $E\subset M$ or $M\subset E$. A subspace $E\subset X$ is called a hyperinvariant subspace of $A$ if $BE\subset E$ for each operator $B$ such that $BA=AB$. The lattice of hyperinvariant subspaces of $A$ is denoted by $\operatorname{Hyplat}(A)$.

It is well known [1], [2] that the classical indefinite integration operator $V$ defined on the Lebesgue space $L^{p}[0,1]$ by

$$Vf(x)= \int_{0}^{x} f(t)\,dt$$

is unicellular for $p\in[1,\infty)$ and the lattice of invariant subspaces of $V$ is anti-isomorphic to the interval $[0,1]$. The same is true (see [1], [3]) for the Riemann and Liouville fractional integration operators

$$J^{\alpha}f(x)=\frac{1}{\Gamma(\alpha)} \int_{0}^{x} (x-t) ^{\alpha-1}f(t)\,dt,\quad \operatorname{Re} \alpha>0,$$

which are complex powers of $J$. Namely,

$$\operatorname{Lat}(J^{\alpha}) =\operatorname{Hyplat}(J^{\alpha}) =\{E_{a}=\varkappa _{[0,a]}L^{p}[0,1] \colon 0\leq a\leq1\}$$

(see [1], [4]). The results about the unicellularity of the operator $V$ on $L^{p}[0,1]$ (see Donoghue [5]) was extended to the Sobolev spaces $W_{2}^{(k)}[0,1]$ (see Tsekanovskii [6]), $W_{p}^{(k)}[0,1]$ and $C^{(n)}[0,1]$ (see Ostapenko–Tarasov [7] and also [3], [8]). For the results on some double integration operators

$$Wf( x,y)= \int_{0}^{x} \int_{0}^{y} f(t,\tau)\,d\tau\,dt,$$

we refer the reader to [9]. The Volterra integral operator

$$Vf(x)= \int_{0}^{x} f(t)\,dt $$

(1.1)

is well defined for functions $f$ in the space $C^{(n)}[0,1]$ of $n$ times continuously differentiable functions on the unit interval $[0,1]$ of the real axis $\mathbb{R}=(-\infty,\infty)$. A closed subspace $E\subset C^{(n)}[0,1]$ is said to be $V$-invariant if $VE\subset E$, i.e., $Vf\in E$ for every $f\in E$. The present paper is motivated by the papers of Ostapenko–Tarasov [7] and Tapdigoglu [8], where the unicellularity of the integration operator $V$ on the space $C^{(n)}[0,1]$ is proved. Here we propose a new proof of the Ostapenko–Tarasov theorem by using some of the ideas in the papers [8], [10], and [11]. Namely, we prove that

$$\operatorname{Lat}(V)=\{E_{\lambda},E^{(k)}\colon 0<\lambda<1,\quad k=1,\dots,n+1\},$$

where

$$E_{\lambda}=\{f\in C^{(n)}[ 0,1] \colon f(x) \equiv0\text{ on }[ 0,\lambda]\} $$

(1.2)

and

$$E^{(k)}=\{f\in C^{(n)}[0,1] \colon f(0)=f'(0)=\cdots=f^{(k-1)}(0)=0\}. $$

(1.3)

Clearly,

$$\{0\} \subset E_{\lambda}\subset E_{\mu}\subset E^{(n+1)}\subset E^{(n)}\subset\dotsm\subset E^{(1)}\subset C^{(n)}[0,1] \quad (\lambda>\mu), $$

(1.4)

and hence $V$ is a unicellular operator on $C^{( n)}[0,1]$.

2. Lattice of $V$-Invariant Subspaces

In this section, we describe the lattice of invariant subspaces of the operator $V$ on the space $C^{(n)}[0,1]$ and prove its unicellularity. Our discussion is based on the Duhamel product of functions defined by

$$(f\circledast g)(x):=\frac{d}{dx} \int_{0}^{x} f(x-t) g(t)\,dt= \int_{0}^{x} f'(x-t) g(t)\,dt+f(0) g(x) $$

(2.1)

(see Wigley [12]).

Recall that the norm on $C^{(n)}[ 0,1]$ is defined by

$$\Vert f\Vert _{n}=\max_{0\leq i\leq n}\Vert f^{(i)}\Vert _{\infty}, $$

(2.2)

where

$$\Vert f^{(i)}\Vert _{\infty}:=\Vert f^{(i)}\Vert _{C^{(n)}[0,1]}=\max_{0\leq x\leq1}\vert f^{(i)}(x) \vert.$$

Theorem 1.

Let $V$ be the Volterra integration operator defined by (1.1) on the space $C^{(n)}[0,1]$. Then

$$\operatorname{Lat}(V)=\{E_{\lambda},E^{(k)}\colon \alpha<\lambda<1,\quad k=1,\dots,n+1\},$$

where $E_{\lambda}$ and $E^{(k)}$ are the nontrivial $V$-invariant subspaces defined by formulas (1.2) and (1.3), and $V$ is unicellular in $C^{(n)}[0,1]$.

Proof.

The proof is based on some of the arguments in the papers [3], [10], and [11]. Using (2.1), one can readily see that

$$(f\circledast g) ^{(k)}( x)= \int_{0}^{x} f^{(k)}(x-t) g'(t)\,dt+ \sum_{m=0}^{k-1} f^{(m)}(0) g^{(k-m)}(x) +g(0) f^{(k)}(x) $$

(2.3)

for all $f,g\in C^{(n)}[0,1]$ and $1\leq k\leq n$. Thus, it is an easy consequence of (2.3) that $f\circledast g$ belongs to $C^{(n)}[0,1]$ as well, and $(C^{(n)}[0,1],\circledast)$ becomes an algebra. One can use the results of operational calculus to show that $(C^{n}[0,1],\circledast)$ is a commutative and associative algebra with unit element $f(z)=\mathbf{1}$. Moreover, one can readily prove that $(C^{( n)}[0,1],\circledast)$ is a Banach algebra. Indeed, for all $f,g\in C^{(n)}[0,1]$, from (2.1)–(2.3) we have

$$\vert (f\circledast g) (x) \vert \leq\max\vert f'\vert \max\vert g\vert +\max\vert f\vert \max\vert g\vert$$

and

$$\begin{aligned} \, \vert (f\circledast g) ^{(k)}(x) \vert & \leq\max\vert f^{(k)}\vert \max\vert g'\vert + \sum_{m=0}^{k-1} \max\vert f^{(m)}\vert \max\vert g^{(k-m)}\vert +\max\vert f^{(k)}\vert \max\vert g\vert \\& \leq\Vert f\Vert _{n}\Vert g\Vert _{n}+k\Vert f\Vert _{n}\Vert g\Vert _{n}+\Vert f\Vert _{n}\Vert g\Vert _{n}\\& =(k+2) \Vert f\Vert _{n}\Vert g\Vert _{n}\leq(n+2) \Vert f\Vert _{n}\Vert g\Vert _{n} \end{aligned}$$

for each $k\in\{1,2,\dots,n\}$. Thus,

$$\Vert f\circledast g\Vert _{n}\leq(n+2) \Vert f\Vert _{n}\Vert g\Vert _{n}, $$

(2.4)

and by setting $M=n+2$ in (2.4), we obtain

$$\Vert f\circledast g\Vert _{n}\leq M\Vert f\Vert _{n}\Vert g\Vert _{n}. $$

(2.5)

By passing to an equivalent norm on $C^{(n)}[0,1]$, from (2.5) we obtain the desired multiplicative inequality

$$\Vert f\circledast g\Vert _{n}\leq\Vert f\Vert _{n}\Vert g\Vert _{n},$$

which shows that $(C^{(n)}[ 0,1],\circledast)$ is a Banach algebra.

The set of $V$-cyclic vectors is denoted by $\operatorname{Cyc}(V)$; in other words,

$$\operatorname{Cyc}(V)=\{f\in C^{(n)}[0,1] \colon \operatorname{span}\{V^{m}f\colon m=0,1,\dots\}=C^{(n)}[ 0,1]\}.$$

Lemma 1.

One has $f\in\operatorname{Cyc}(V)$ if and only if $f(0) \neq0$.

Proof.

Indeed, it follows from (2.1) that $x\circledast f(x) =Vf(x)$ for all $f\in C^{(n)}[0,1]$. More generally,

$$V^{m}f(x)=\frac{x^{m}}{m!}\circledast f,\qquad m\geq0. $$

(2.6)

Then we have

$$\begin{aligned} \, \operatorname{span}\{V^{m}f\colon m\geq0\} &=\operatorname{span}\biggl\{f\circledast\frac{x^{m}}{m!} \colon m\geq0\biggr\} \\& =\operatorname{span}\biggl\{D_{f}\biggl( \frac{x^{m}}{m!}\biggr) \colon m\geq0\biggr\} \\& =\operatorname{clos}D_{f}\operatorname{span}\{x^{m}\colon m\geq0\} \\& =\operatorname{clos}D_{f}[\operatorname{span}\{z\mathbf{1} \colon z\in\mathbb{C}\} \oplus\operatorname{span}\{x^{m}\colon m\geq1\} ] \\& =\operatorname{clos}D_{f}C^{(n)}[ 0,1], \end{aligned}$$

where $D_{f}$ and $\oplus$ stand for the Duhamel operator and the direct sum of subspaces, respectively. Thus,

$$\operatorname{span}\{V^{m}f\colon m\geq0\}=\overline{D_{f}C^{(n)}[0,1]}, $$

(2.7)

which implies that $f\in\operatorname{Cyc}(V)$ if and only if the range of the Duhamel operator $D_{f}$ is dense, i.e.,

$$\overline{D_{f}C^{(n)}[0,1]}=C^{(n)}[0,1]. $$

(2.8)

Now if $f\in\operatorname{Cyc}(V)$, then it follows from (2.8) that there exists a sequence

$$\{g_{m}\} _{m\geq1}\subset C^{(n)}[0,1]$$

such that

$$\lim_{m\rightarrow \infty} f\circledast g_{m}=\mathbf{1}\qquad \text{in}\quad C^{(n)}[0,1].$$

Therefore, $(f\circledast g_{m}) (0) \rightarrow1$ as $m\rightarrow\infty$, or, equivalently, $f(0) g_{m}(0) \rightarrow1$ as $m\rightarrow\infty$, which shows that $f(0) \neq0$. Now it remains to prove that if $f\in C^{(n)}[0,1]$ and $f(0)\neq0$, then $f\in\operatorname{Cyc}(V)$. To this end, in view of (2.8), we must show that the range of the operator $D_{f}$ is dense under the condition $f(0)\neq0$. However, in the following Statement we prove even more than we actually need.

Statement.

If $f(0)\neq0$, then $D_{f}$ is invertible on $C^{(n)}[0,1]$.

Indeed, for $F=f-f(0)$ we have $D_{f}=f(0) I_{n}+D_{F}$, where $I_{n}$ is the identity operator on $C^{(n)}[ 0,1]$. To prove that $D_{f}$ is invertible, it suffices to show that $D_{F}$ is a quasinilpotent operator on $C^{(n)}[0,1]$, i.e., $\sigma(D_{F}) =\{0\}$, or, equivalently, the spectral radius $r(D_{F})$ of the operator $D_{F}$ is zero. In fact, by the well-known Gelfand formula (e.g., see Dunford–Schwartz [13]),

$$r(D_{F})=\lim_{m\rightarrow\infty} \Vert D^{m}_{F}\Vert ^{\frac{1}{m}},$$

and so we will estimate $\Vert D^{m}_{F}\Vert$. For every $g\in C^{(n)}[0,1]$, we have

$$\begin{aligned} \, D_{F}g(x) &=\frac{d}{dx} \int_{0}^{x} F(x-t) g(t)\,dt= \int_{0}^{x} F'(x-t) g(t)\,dt\\& \overset{\operatorname{def}}{=}(F'\ast g) (x) \overset{\operatorname{def}}{=}(K_{F'}g) (x). \end{aligned}$$

Hence we obtain

$$\begin{aligned} \, \vert (K_{F'}g) (x) \vert &=\biggl\vert \int_{0}^{x} F'(x-t) g(t)\,dt\biggr\vert \leq \int_{0}^{x} \vert F'(x-t) \vert \vert g(t) \vert\,dt\leq \int_{0}^{x} \Vert F'\Vert _{\infty}\Vert g\Vert _{\infty}\,dt \\& \leq\Vert F\Vert _{n}\Vert g\Vert _{n}(x), \end{aligned}$$

$$\begin{aligned} \, \vert (K_{F'}^{2}g) (x) \vert &=\biggl(\int_{0}^{x} F'(x-t) (K_{g}) (t) \,dt\biggr)=\biggl\vert \int_{0}^{x} F'(x-t) \biggl( \int_{0}^{t} F'(t-\tau) g(\tau)\,d\tau\biggr) \,dt\biggr\vert \\& \leq \int_{0}^{x} \vert F'(x-t) \vert \biggl(\biggl\vert \int_{0}^{t} F'(t-\tau) g(\tau)\,d\tau\biggr\vert \biggr)\,dt\\& \leq\Vert F\Vert _{n}^{2}\Vert g\Vert _{n} \frac{x^{2}}{2!}. \end{aligned}$$

Thus, by induction,

$$\vert (K_{F'}^{m}g) (x) \vert \leq\Vert F\Vert _{n}^{m}\Vert g\Vert _{n}\frac{x^{m}}{m!}$$

for each integer $m\geq0$. Further, we can prove that

$$\vert (K_{F'}^{m}g) '(x) \vert \leq\Vert F\Vert _{n}^{m}\Vert g\Vert _{n}\frac{(x+1) ^{m}}{m!}$$

for all $x\in[0,1]$ and $m\geq0$. The proof is similar to that in the paper [11] and hence is omitted. In a similar way, one can also prove by induction (we omit the proof) that

$$\vert (K_{F'}^{m}g) ^{(j)}(x) \vert \leq\Vert F\Vert _{n}^{m}\Vert g\Vert _{n}\frac{(x+j) ^{m}}{m!},\qquad j=0,1,\dots,n. $$

(2.9)

It follows from (2.9) that

$$\Vert K_{F'}^{m}g\Vert _{n}\leq\Vert F\Vert _{n}^{m}\Vert g\Vert _{n}\frac{(n+1) ^{m}}{m!},$$

and hence

$$\Vert K_{F'}^{m}\Vert ^{\frac{1}{m}}\leq\Vert F\Vert _{n}\frac{n+1}{(m!) ^{\frac{1}{m}}}\rightarrow 0\qquad \text{as} \quad m\rightarrow\infty.$$

Consequently, $r(K_{\alpha,F'})=0$, i.e., $K_{\alpha,F'}$ is quasinilpotent, and hence $D_{\alpha,f}$ is an invertible operator on $C^{(n)}[ 0,1]$, which proves the Statement. The proof of the lemma is complete.

Lemma 2.

Let $f\in C^{(n)}[ 0,1]$.

• (i) If $f\in E^{(k)}$, $1\leq k\leq n$, then $f\in\operatorname{Cyc}(V\mid E^{(k)})$ if and only if $f\in E^{k}\backslash E^{k+1}$, i.e., if

  $$f(0)=f' (0)=\dotsm=f^{k-1}(0)=0, \qquad f^{(k)}(0) \neq0.$$

• (ii) If $f\in E^{(n+1)}$, then $f\in\operatorname{Cyc}(V\mid E^{(n+1)})$ if only if $f\in E^{(n+1)}\backslash E_{\lambda}$ for every $\lambda \in(0,1)$.

• (iii) If $f\in E_{\lambda}$, then $f\in\operatorname{Cyc}(V\mid E_{\lambda})$ if and only if $f\in X_{\lambda}\backslash X_{\mu}$ for all $\mu>\lambda$.

Proof.

Let us define the following convolution products on the subspaces $E^{(k)}$, $1\leq k\leq n+1$:

$$(f\overset{k}{\circledast}g) (x)=\frac{d}{dx} \int_{0}^{x} \frac{f(x-t)}{(x-t) ^{k}}g(t) \,dt,\qquad f,g\in E^{(k)}. $$

(2.10)

(i) Let $f\in E^{(k)}$ and $f\notin E^{(k+1)}$, where $1\leq k\leq n$. It is easily seen from (2.10) that

$$\frac{x^{k+m}}{m!}\overset{k}{\circledast}g =V^{m}g,\qquad g\in E^{(k)}, $$

(2.11)

for all $m\geq0$. The Maclaurin series expansion of $f$ gives

$$f(x)=\frac{f^{(k)}(0)}{k!}x^{k}+\frac{f^{( k+1)}(0)}{(k+1) !}x^{k+1} +\dotsm+\frac{f^{(n)}(0)}{n!}x^{n}+q(x),$$

whence it follows that

$$f(x)=\frac{f^{(k)}(0)}{k!}x^{k}+\widetilde{q}(x), $$

(2.12)

where $f^{(k)}(0) \neq0$ and

$$\widetilde{q}(x) =\frac{f^{(k+1)}(0)}{(k+1)!}x^{k+1} +\dotsm+\frac {f^{(n)}(0)}{n!}x^{n}+q(x) \in E^{(k)}.$$

We define the $k$-Duhamel operator $D_{k,f}$ on the subspace $E^{(k)}$ by the formula

$$D_{k,f}g=f\overset{k}{\circledast}g,\qquad g\in E^{(k)}.$$

It is obvious from (2.11) and (2.12) that

$$D_{k,f}=f^{(k)}(0) I_{E^{(k)}}+D_{k,\widetilde{q}}.$$

Since $f^{(k)}(0) \neq0$, it can be proved by the same argument as in the papers [3, Lemma 2] and [10, Lemma 1] that $D_{k,f}$ is invertible on $E^{(k)}$ (we omit the proof). On the other hand,

$$\operatorname{span}\{x^{k+m}\colon m\geq0\}=E^{(k)}.$$

In view of the representation (2.11), we have

$$\begin{aligned} \, E_{f}(V\mid E_{n}^{(k)}) & :=\operatorname{span} \{V^{m}f\colon m\geq0\}=\operatorname{span}\biggl\{\frac{x^{k+m}} {m!}\overset{k}{\circledast}f\colon m\geq0\biggr\} \\& =\operatorname{span}\biggl\{D_{k,f}\frac{x^{k+m}}{m!} \colon m\geq0\biggr\} \\& =\operatorname{clos}D_{k,f}\operatorname{span}\{x^{k+m}\colon m\geq0\} \\& =\operatorname{clos}D_{k,f}E^{(k)}=E^{(k)}; \end{aligned}$$

i.e., if $f\in E^{(k)}\backslash E^{(k+1)}$, then $f\in\operatorname{Cyc}(V\mid E^{(k)})$.

Conversely, the equality $E_{f}(V)=E^{(k)}$ readily implies that $f^{(k)}(0)\neq0$, whence $f\notin E^{(k+1)}$. Thus, if $f\in E^{(k)}$ and $f\in\operatorname{Cyc}(V\mid E^{(k)})$, then $f^{(k)}(0)\neq0$, which proves (i).

The proof of (ii) is very similar to that of (i).

Indeed, the Maclaurin formula for the function $f\in E^{(n+1)}$ has the integral representation

$$\begin{aligned} \, f(x) &=\frac{1}{(n-1) !} \int_{0}^{x} f^{(n)}(t) (x-t) ^{n-1}\,dt\\& =\frac{x^{n}}{n!}+\biggl(\frac{1}{(n-1) !} \int_{0}^{x} f^{(n)}(t) (x-t) ^{n-1} \,dt-\frac{x^{n}}{n!}\biggr) \\& =\frac{x^{n}}{n!}+q_{1}(x). \end{aligned}$$

Clearly, $q_{1}\in E^{(n)}$ and $f\overset{n}{\circledast}g\in E^{(n+1)}$ for every $g\in E^{(n)}$. Since $f\notin E_{\lambda}$ for all $0<\lambda<1$, we have $\ker(D_{n,f})=\{0\}$ by the Titchmarsh convolution theorem. Further, the condition $q_{1}\in E^{(n)}$ implies that $D_{n,q_{1}}$ is a compact operator on the space $E^{(n+1)}$ (see [3, Lemma 2]). Since

$$D_{n,f}=D_{n,\frac{x^{n}}{n!}+q_{1}(x)} =D_{n,\frac{x^{n}}{n!}}+D_{n,q_{1}(x)} =\frac{1}{n!}I_{E^{(n+1)}}+D_{n,q_{1}(x)}$$

and $D_{n,f}$ is invertible on $E^{(n+1)}$, we have

$$D_{n,f}E^{(n+1)}=E^{(n+1)}.$$

Now

$$\begin{aligned} \, E_{f}(V\mid E^{(n)}) &=\operatorname{span}\{V^{m}f\colon m\geq0\} =\operatorname{span}\biggl\{\frac{x^{n+m}}{m!}\overset{n}{\circledast} f\colon m\geq0\biggr\} \\& =\operatorname{span}\biggl\{D_{n,f}\frac{x^{n+m}}{m!} \colon m\geq0\biggr\} \supset\operatorname{span}\biggl\{D_{n,f}\frac{x^{n+1+m}}{(m+1) !}\colon m\geq0\biggr\} \\& =\operatorname{clos}D_{n,f}\operatorname{span}\biggl\{\frac{x^{n+1+m}}{(m+1) !}\colon m\geq0\biggr\} =\operatorname{clos}D_{n,f}E^{(n+1)}=E^{(n+1)}. \end{aligned}$$

Hence $E_{f}(V\mid E^{(n)}) \supset E^{(n+1)}$. On the other hand, $VE^{(n+1)}\subset E^{(n+1)}$, and so $E_{f}(V\mid E^{(n)}) \subset E^{(n+1)}$. Thus, $E_{f}(V\mid E^{(n)})=E^{(n+1)}$, whence $f\in\operatorname{Cyc}(V\mid E^{(n+1)})$. Conversely, if $f\in\operatorname{Cyc}(V\mid E^{( n+1)})$, then $f\in E^{(n+1)}\backslash E_{\lambda}$ for all $\lambda\in(0,1)$, which proves (ii).

The proof of (iii) can be obtained from Lemma 1 by a standard argument based on a simple change of variables (e.g., see Ostapenko–Tarasov [7], Kalisch [14], and Gohberg–Krein [1]) and hence it is omitted. The proof of Lemma 2 is complete.

Now let us return to the proof of Theorem 1. We will show that there exist no $V$-invariant subspaces other than those in the chain (1.4) and hence

$$\operatorname{Lat}(V)=\{E_{\lambda},E^{(k)}\colon 0<\lambda<1;\quad k=1,2,\dots,n+1\}.$$

Indeed, assume the contrary: there exists a nontrivial $V$-invariant subspace $E\subset C^{(n)}[0,1]$ different from the invariant subspaces in (1.4). It is clear that

$$E= \bigcup_{g\in E} E_{g}(V\mid E),$$

where, as before,

$$E_{g}(V\mid E) =\operatorname{span}\{V^{m}g\colon m=0,1,\dots\}.$$

Then it is clear by Lemma 1 that there exists a function $f\in E$ such that $f(0)\neq0$. Consequently, by Lemma 1, we conclude that $E=C^{(n)}[0,1]$, which contradicts our assumption that $E$ is a nontrivial subspace. Since the set of subspaces in (1.4) is linearly ordered, it follows that the operator $V$ is unicellular. The proof of the theorem is complete.

Recall that the commutant of an operator $A$ on a Banach space $X$ is the set

$$\{A\}'=\{B\in B(X) \colon BA=AB\},$$

where $B(X)$ is the Banach algebra of all bounded linear operators on $X$. Set

$$\operatorname{Hyplat}(B):=\{E\subset X\colon AE\subset E\text{ for each }A\in\{B\}'\}.$$

Corollary 1.

$\operatorname{Hyplat}(V)=\{E_{\lambda},E^{(k)}\colon 0<\lambda<1;\quad k=1,\dots,n+1\}.$

The proof is immediate by the well-known general theorem stating that the lattice of hyperinvariant subspaces of any unicellular operator $A$ coincides with the lattice of nontrivial $A$-invariant subspaces (e.g., see Radjavi–Rosenthal [2]).

The following corollary describes the commutant of the operator $V$.

Corollary 2.

$\{V\}'=\{D_{f}\colon f\in C^{(n)}[0,1]\}$.

Proof.

Indeed, it is clear from the formula

$$Vf(x)=x \circledast f\qquad \forall f\in C^{(n)}[0,1]$$

that $D_{f}V=VD_{f}$ for all $f\in C^{(n)}[0,1]$ and hence

$$\{D_{f}\colon f\in C^{(n)}[ 0,1]\} \subset\{V\}'. $$

(2.13)

Let $T\in\{V\}'$. Then $TV=VT$, and hence $TV^{m} =V^{m}T$ for all $m\geq0$. Applying formula (2.6) to the identity function $f=\mathbf{1}$, we obtain

$$TV^{m}\mathbf{1}=\frac{x^{m}}{m!}\circledast T\mathbf{1},\qquad m\geq0,$$

i.e.,

$$T\biggl(\frac{x}{m!}\circledast\mathbf{1}\biggr)=D_{T\mathbf{1}}\biggl(\frac{x^{m}}{m!}\biggr),\qquad m\geq0.$$

Since

$$\frac{x^{m}}{m!}\circledast\mathbf{1}=\frac{x^{m}}{m!}, \qquad m\geq0,$$

we have

$$T\biggl(\frac{x^{m}}{m!}\biggr)=D_{T\mathbf{1}}\biggl( \frac{x^{m}} {m!}\biggr),$$

or, equivalently,

$$T(x^{m})=D_{T\mathbf{1}}(x^{m}),\qquad m\geq0.$$

Since

$$\operatorname{span}\{x^{m}\colon m\geq0\} =\operatorname{span}\{z\mathbf{1}\colon z\in\mathbb{C}\} \oplus\operatorname{span}\{x^{m} \colon m\geq1\} =C^{(n)}[0,1],$$

it follows from the last equalities that $Tf( x)=D_{T\mathbf{1}}f(x)$ for all $f\in C^{(n)}[0,1]$ and hence $T=D_{T\mathbf{1}}$. Thus, $\{V\}' \subset\{D_{f}\colon f\in C^{(n)}[ 0,1]\}$, which, together with (2.13), proves the corollary.