Commutators on power series spaces

Abstract

We prove that every continuous linear operator acting on a stable, nuclear power series space is a commutator, so in particular our theorem holds for the space of all entire functions, holomorphic functions on the unit disc or smooth functions. We also show that on the product of Fréchet spaces $\prod _{i=1}^\infty X$ all operators are commutators.

1 Introduction

A commutator of a pair of elements $A$ and $B$ in the algebra $\mathcal {L}(X)$ of linear, continuous operators on a locally convex space $X$ is given by

$$\begin{aligned}{}[A,B]:=AB-BA. \end{aligned}$$

The problem of representing operators as commutators comes from quantum mechanics, where the so-called commutator relation plays an important role. Commutators are also connected with derivations in algebra as they are the main examples of derivations (the so-called inner derivations).

As shown by Wintner [11], on a Banach space not every operator is a commutator. More precisely, all elements of the form $\lambda {1}\mathrm{l}+M$, where $\lambda \ne 0$, $M$ lies in a proper, closed ideal $\mathcal {M}$ of a Banach algebra $\mathcal {A}$ cannot be commutators.

We will show that in many classical Fréchet spaces (like spaces of holomorphic or smooth functions) every operator is a commutator (see Corollaries 2, 3). More precisely this is true for every nuclear stable power series space, both of finite and infinite type (see Theorem 1). We also prove that our method cannot be applied in the non-nuclear case although the question on which Fréchet spaces all operators are commutators remains open.

In the Banach space case a lot is known. Due to Wintner the question is whether or not in the algebra $\mathcal {L}(X)$ with the biggest, proper, closed ideal $\mathcal {M}$ all operators not of the form $\lambda I + M$ , $M\in \mathcal {M}$, $\lambda \ne 0$ are commutators. For many classical Banach spaces the answer is affirmative. First, in 1965 Brown and Pearcy [3] proved the above hypothesis for seperable Hilbert spaces. Then, in 1972–73 Apostol ([1, 2]) showed that the same holds for $\ell _p$, $1<p<\infty$ and $c_0$. Apostol also got some partial results for $\ell _1$ and $\ell _\infty$, but it was Dosev who generalized Apostol’s method and in 2009 gave a complete description of commutators on $\ell _1$ [4]. Two years later Dosev and Johnson [5] proved that on $\ell _\infty$ the commutators are all operators not of the form $\lambda I + S$, $\lambda \ne 0$, where $S$ is a strictly singular operator. This problem was also researched for non-sequence spaces. In 2011 Dosev, Johnson and Schechtman [6] showed that on the space $L_p[0,1]$, $1\le p< \infty$, the above hypothesis holds, with $\mathcal {M}=\{T\in \mathcal {L}(L_p) :\forall {A,B\in \mathcal {L}(L_p)} \; I\ne ATB \}$ as the largest closed ideal.

In this paper we study commutators on some Fréchet spaces. Using Dosev’s technique we show that on Fréchet spaces Wintner’s theorem does not hold in general and there exist spaces on which all operators are commutators.

2 Preliminaries

In this section we will introduce some notation and recall basic facts needed in the paper. We will also describe the method used by Dosev and adjusted to our needs.

Let a matrix $A=(a_{k,j})_{k,j\in \mathbb {N}}$ of non-negative numbers satisfy

• for each $j\in \mathbb {N}$ there exists a $k\in \mathbb {N}$ with $a_{k,j}>0$,

• $a_{k,j}\le a_{k+1,j}$ for all $j,k\in \mathbb {N}$.

We define

$$\begin{aligned} \lambda ^1(A)&=\left\{ x=(x_j)\in \mathbb {K}^{\mathbb {N}}: ||x ||_{k}=\sum _{j=0}^\infty \left| x_j \right| a_{k,j}<\infty \quad \forall k\in \mathbb {N}\right\} ,\\ \lambda ^\infty (A)&=\left\{ x\in \mathbb {K}^{\mathbb {N}}: ||x ||_{k}:=\underset{j\in \mathbb {N}}{{{\mathrm{sup}}}}\left| x_j \right| a_{k,j}< \infty \quad \forall k\in \mathbb {N}\right\} . \end{aligned}$$

We call the matrix $A$ a Köthe matrix and the spaces $\lambda ^1(A)$ and $\lambda ^\infty (A)$ Köthe sequence spaces associated with the matrix $A$. If $A$ is a Köthe matrix then spaces $\lambda ^1(A), \lambda ^\infty (A)$ are Fréchet spaces with the topology given by the family of seminorms $(||\cdot ||_k)$ [8, 27.1].

In our paper we consider one of the most important classes of Köthe spaces, the class of power series spaces.

For any non-negative monotonically increasing sequence $\alpha =(\alpha _j)$ tending to infinity and for $r = 0$ or $r =\infty$ we define

$$\begin{aligned} \Lambda _r(\alpha )=\left\{ x=(x_j)\in \mathbb {K}^\mathbb {N}: ||x ||_{t}=\sum _{j=0}^\infty \left| x_j \right| e^{t\alpha _j}<\infty \quad \text {for all t<r}\right\} . \end{aligned}$$

The space $\Lambda _r(\alpha )$ is called a power series space, for $r = 0$ of finite type and for $r =\infty$ of infinite type. For any sequence $t_k\nearrow r$ and the matrix $A=(\exp (t_k\alpha _j))$ we have that $\Lambda _r(\alpha )=\lambda ^1(A)$.

We say that the space $\Lambda _r(\alpha )$ is stable, if

$$\begin{aligned} \sup _{n\in \mathbb {N}} \frac{\alpha _{2n}}{\alpha _{n}}<\infty . \end{aligned}$$

This is equivalent to $\Lambda _r \times \Lambda _r \cong \Lambda _r$ (see [7]).

Lemma 1

([8, 29.6]) The space $\Lambda _0(\alpha )$ is nuclear if and only if $\lim _{n\rightarrow \infty }\alpha _n^{-1}\ln n=0$. The space $\Lambda _\infty (\alpha )$ is nuclear if and only if ${{\mathrm{sup}}}_{n\in \mathbb {N}}\alpha _n^{-1}\ln n<\infty$.

Let $X$ be a Fréchet space with a Schauder basis $(e_j)_{j=0}^\infty$. For Köthe sequence spaces we will always use the unit vector basis. We define the shift operators in the following way:

$$\begin{aligned} R\biggl (\sum _{j=0}^\infty x_je_j\biggr )&=\sum _{j=0}^\infty x_j e_{j+1},\\ L\biggl (\sum _{j=0}^\infty x_je_j\biggr )&=\sum _{j=1}^\infty x_j e_{j-1}. \end{aligned}$$

For a given space $X$ it is not always true that these operators are well defined and continuous. Our first lemma shows on which Köthe spaces the operators $R$ and $L$ are continuous.

Lemma 2

Let $A=(a_{k,j})_{k,j\in \mathbb {N}}$ be a Köthe matrix.

• (a) Operator $L$ is continuous on $\lambda ^1(A)$ if and only if

  $$\begin{aligned} \forall k\in \mathbb {N}\; \exists p\in \mathbb {N}\; \exists c\in \mathbb {R} \; \forall j\in \mathbb {N}\;\; a_{k,j-1}\le c a_{p,j}. \end{aligned}$$

  (1)

• (b) Operator $R$ is continuous on $\lambda ^1(A)$ if and only if

  $$\begin{aligned} \forall k\in \mathbb {N}\; \exists p\in \mathbb {N}\; \exists c\in \mathbb {R} \; \forall j\in \mathbb {N}\;\; a_{k,j+1}\le c a_{p,j}. \end{aligned}$$

  (2)

Proof

$(a)\Leftarrow$: Let the matrix $A=(a_{k,j})_{k,j\in \mathbb {N}}$ be such that (1) holds. Then for all $k\in \mathbb {N}$, $x=\sum _{j=0}^\infty x_je_j\in \lambda ^1(A)$ we have

$$\begin{aligned} ||Lx ||_{k}=||\sum _{j=1}^\infty x_je_{j-1} ||_{k}=\sum _{j=1}^\infty \left| x_j \right| a_{k,j-1}\le \sum _{j=1}^\infty \left| x_j \right| c a_{p,j}\le c ||x ||_{p}, \end{aligned}$$

where $p$, $c$ satisfy (1). $\Rightarrow$: The operator $L$ is continuous, hence by taking $x=e_j$ we get (1).

The proof of $(b)$ is similar. $\square$

Corollary 1

If $\sup _j \frac{\alpha _{j+1}}{\alpha _j}<\infty$ then the operators $R$ and $L$ are continuous on $\Lambda _r(\alpha )$ for $r=0$ or $r=\infty$.

For our main theorem we will need the following lemma, which was stated and proved for Banach spaces by Dosev [4] and which can be easily transfered to the Fréchet case.

Lemma 3

([4, Lemma 3]) Let the operators $R$ and $L$ be well-defined and continuous on a Fréchet space $X$ with a Schauder basis $(e_j)_{j=0}^\infty$ and let $T\in \mathcal {L}(X)$. If the series $\sum _{n=0}^\infty R^nTL^n$ is pointwise convergent, then $T$ is a commutator.

Proof

Let $S$ be the pointwise limit of $\sum _{n=0}^\infty R^nTL^n$. By the Banach-Steinhaus theorem we get $S\in \mathcal {L}(X)$ and for all $x\in X$ we have

$$\begin{aligned} (L(RS)-(RS)L)x&=LR\sum _{n=0}^\infty R^nTL^nx-R\biggl (\sum _{n=0}^\infty R^nTL^n\biggr )Lx\\&=\sum _{n=0}^\infty R^nTL^nx-\sum _{n=1}^\infty R^nTL^nx=R^0TL^0x=Tx. \end{aligned}$$

Hence $T=L(RS)-(RS)L$ is a commutator. $\square$

It will be convenient for us to look at operators acting on a Fréchet space $X$ with the Schauder basis $(e_j)_{j\in \mathbb {N}}$ as infinite matrices. For any operator $T\in \mathcal {L}(X)$ and for all $j$ there exists the unique representation $Te_j=\sum _{i=0}^\infty t_{i,j}e_i$ for some sequence $(t_{i,j})_{i\in \mathbb {N}}$. Hence there is an infinite matrix $(t_{i,j})_{i,j\in \mathbb {N}}$ such that

$$\begin{aligned} Tx=\sum _{i=0}^\infty \biggl (\sum _{j=0}^\infty t_{i,j}x_{j}\biggr )e_i \quad \text {for all} \; x=\sum _{j=0}^\infty x_je_j. \end{aligned}$$

(3)

We will say that the operator $T\in \mathcal {L}(X)$ is defined by the matrix $(t_{i,j})$ whenever (3) holds.

From now on, an element $x=\sum _{j=0}^\infty x_je_j$ in a Fréchet space $X$ with the Schauder basis $(e_j)_{j=0}^\infty$ will be denoted by $(x_j)=(x_0,x_1,x_2,\ldots )$.

Now, we will describe the sum $\sum _{n=0}^\infty R^nTL^nx$ for a given operator $T\in \mathcal {L}(X)$ and an element $x=(x_j)\in X$. Let the operator $T$ be defined by a matrix $(t_{i,j})_{i,j\in \mathbb {N}}$. For every $x=(x_j)_{j\in \mathbb {N}}\in X$ we have

$$\begin{aligned} R^nTL^nx=\left( \underbrace{0,\ldots ,0,}_{n} \sum _{j=0}^\infty t_{0,j} x_{j+n}, \sum _{j=0}^\infty t_{1,j}x_{j+n}, \sum _{j=0}^\infty t_{2,j}x_{j+n} \ldots \right) . \end{aligned}$$

Hence, for every $M,N\in \mathbb {N}$, $M>N$

$$\begin{aligned} \sum _{n=N}^{M}R^nTL^nx=&\Biggl (\underbrace{0,\ldots ,0}_{\text {N}},\sum _{j=0}^\infty t_{0,j}x_{j+N}, \sum _{j=0}^\infty t_{0,j}x_{j+N+1}+\sum _{j=0}^\infty t_{1,j}x_{i+N}, \ldots , \nonumber \\& \underbrace{\sum _{i=0}^{k-N}\sum _{j=0}^\infty t_{i,j}x_{j+k-i}}_{\text {k-th coordinate,} k\le M},\ldots , \underbrace{\sum _{i=k-M}^{k-N}\sum _{j=0}^\infty t_{i,j}x_{j+k-i}}_{\text {k-th coordinate,} k>M},\ldots \Biggr ). \end{aligned}$$

(4)

3 Main results

Now, we will prove our main theorem, which indicates the class of Fréchet spaces on which every operator is a commutator.

Theorem 1

Every continuous linear operator on a nuclear and stable power series space is a commutator.

Proof

The idea of the proof is to show that for any continuous operator $T$, the series $\sum _{n=0}^\infty R^nTL^n$ is pointwise convergent and so, due to Corollary 1 and Lemma 3 the operator $T$ is a commutator. First, we will prove the theorem for finite type power series spaces, and then for infinite type.

Let $X=\Lambda _0(\alpha )$ be a nuclear and stable power series space of finite type. Let us recall that stability of power series space means that for every $j\in \mathbb {N}$ we have $\alpha _{2j}\le C\alpha _j$ for some constant $C$. It will be more convenient for us to use an equivalent condition

$$\begin{aligned} \exists \; c \; \forall {j,k} \quad \alpha _{j+k}\le c(\alpha _j+\alpha _k). \end{aligned}$$

(5)

Take an operator $T\in \mathcal {L}(\Lambda _0(\alpha ))$. Since $\Lambda _0(\alpha )$ is a nuclear space we have that $\Lambda _0(\alpha )=\lambda ^1(\exp (-\frac{1}{k}\alpha _j))=\lambda ^\infty (\exp (-\frac{1}{k}\alpha _j))$ [8, 28.16] and so we can consider $T$ as an operator acting from $\Lambda _0(\alpha )$ to $\lambda ^\infty (\exp (-\frac{1}{k}\alpha _j))$. Let $T$ be given by the matrix $(t_{i,j})_{i,j\in \mathbb {N}}$. Then, for all $p\in \mathbb {N}$ there exist constants $M_p$, $m_p$ such that for all $m>m_p$

$$\begin{aligned} \sup _{i,j}\left| t_{i,j} \right| e^{\frac{1}{m}\alpha _i-\frac{1}{p}\alpha _j}\le M_p. \end{aligned}$$

(6)

Indeed, take $p\in \mathbb {N}$. Since the operator $T$ is continuous, there exists $M_p$, $m_p$ such that $||Tx ||_{p}\le M_p||x ||_{m_p}$ for all $x$. Consider the sequence $(x^j)_{j\in \mathbb {N}}\subset \Lambda _0(\alpha )$ defined as follows

$$\begin{aligned} x^j=(0,\ldots ,0,\underbrace{e^{\frac{1}{m_p}\alpha _j}}_{\text {j-th coordinate}},0,\ldots ). \end{aligned}$$

Then for all $j \in \mathbb {N}$ we have $||x^j ||_{m_p}=1$ and

$$\begin{aligned} \underset{i\in \mathbb {N}}{{{\mathrm{sup}}}}\left| t_{i,j} \right| e^{\frac{1}{m_p}\alpha _j}e^{-\frac{1}{p}\alpha _i}=||Tx^j ||_{p}\le M_p. \end{aligned}$$

Hence for all $m>m_p$

$$\begin{aligned} \underset{i,j\in \mathbb {N}}{{{\mathrm{sup}}}}\left| t_{i,j} \right| e^{\frac{1}{m}\alpha _j-\frac{1}{p}\alpha _i}\le \underset{i,j\in \mathbb {N}}{{{\mathrm{sup}}}}\left| t_{i,j} \right| e^{\frac{1}{m_p}\alpha _j-\frac{1}{p}\alpha _i}\le M_p. \end{aligned}$$

We will show that the series $\sum _{n=0}^\infty R^nTL^n$ is pointwise convergent which, by Lemma 3, completes the proof.

Take some $x\in \Lambda _0(\alpha )$. Using the formula (4) we will estimate the $p$-th norm of $\sum _{n=N}^MR^nTL^nx$ in $\lambda ^\infty ((\exp (-\frac{1}{k}\alpha _j)))$.

$$\begin{aligned} ||\sum _{n=N}^{M}R^nTL^nx ||_{p}&=\max \Biggl (\underset{N< k\le M}{{{\mathrm{sup}}}}\left| \sum _{i=0}^{k-N}\sum _{j=0}^\infty t_{i,j} x_{j+k-i} \right| e^{-\frac{1}{p}\alpha _k}, \\&\qquad \, \underset{k>M}{{{\mathrm{sup}}}}\left| \sum _{i=k-M}^{k-N}\sum _{j=0}^\infty t_{i,j} x_{j+k-i} \right| e^{-\frac{1}{p}\alpha _k}\Biggr )\\&\le \underset{N<k}{{{\mathrm{sup}}}}\sum _{i=0}^{k-N}\sum _{j=0}^\infty \left| t_{i,j} \right| \left| x_{j+k-i} \right| e^{-\frac{1}{p}\alpha _k}. \end{aligned}$$

If we take $m=\max (m_{2p},2p)$ and $s=2mc$, with $c\in \mathbb {N}$ satisfying (5), we get

$$\begin{aligned} ||\sum _{n=N}^{M}R^nTL^nx ||_{p}&\le \underset{N<k}{\sup }\sum _{i=0}^{k-N}\sum _{j=0}^\infty \left| t_{i,j} \right| e^{\frac{1}{m}\alpha _j- \frac{1}{2p}\alpha _i} \left| x_{j+k-i} \right| e^{-\frac{1}{s}\alpha _{k+j-i}} \nonumber \\&\quad \;\cdot e^{-\frac{1}{m}\alpha _j+\frac{1}{2p}\alpha _i+\frac{1}{s}\alpha _{k+j-i} -\frac{1}{p}\alpha _k} \nonumber \\&\le M_{2p}||x ||_{s}\underset{N<k}{\sup }\sum _{i=0}^{k-N}\sum _{j=0}^\infty e^{-\frac{1}{m}\alpha _j+\frac{1}{2p}\alpha _i+\frac{1}{s}\alpha _{k+j-i} -\frac{1}{p}\alpha _k} \nonumber \\&\le M_{2p}||x ||_{s}\underset{N<k}{\sup }\sum _{i=0}^{k-N} e^{-\frac{1}{2p}\alpha _k}\sum _{j=0}^\infty e^{-\frac{1}{m}\alpha _j+\frac{1}{s}\alpha _{k+j-i}}. \end{aligned}$$

(7)

The last inequality follows from $\alpha _i\le \alpha _k$ and $\frac{1}{2p}\alpha _i-\frac{1}{p}\alpha _k\le -\frac{1}{2p}\alpha _k$. Applying (5) we obtain

$$\begin{aligned} -\frac{1}{m}\alpha _j+\frac{1}{s}\alpha _{k+j-i}&\le -\frac{1}{m}\alpha _j+ \frac{c}{s}\alpha _j+\frac{c}{s}\alpha _{k-i} = -\frac{1}{m}\alpha _j+ \frac{1}{2m}\alpha _j+\frac{1}{2m}\alpha _{k-i} \nonumber \\&= -\frac{1}{2m}\alpha _j+\frac{1}{2m}\alpha _{k-i}. \end{aligned}$$

(8)

We denote by ${1}\mathrm{l}$ the vector with all coordinates equal to $1$. It is easy to see that if $\Lambda _0(\alpha )$ is nuclear then, by Lemma 1, ${1}\mathrm{l}\in \Lambda _0(\alpha )$. We apply (8) and (7) and as a result we get

$$\begin{aligned} ||\sum _{n=N}^{M}R^nTL^nx ||_{p}&\le M_{2p}||x ||_{s}\underset{N<k}{{{\mathrm{sup}}}}\sum _{i=0}^{k-N} e^{-\frac{1}{2p}\alpha _k + \frac{1}{2m}\alpha _k}\sum _{j=0}^\infty e^{-\frac{1}{2m}\alpha _j} \\&= M_{2p}||x ||_{s}||{1}\mathrm{l} ||_{2m}\underset{N<k}{\sup }\sum _{i=0}^{k-N} e^{-\frac{1}{2}(\frac{1}{p}-\frac{1}{m})\alpha _k} \\&\le M_{2p}||x ||_{s}||{1}\mathrm{l} ||_{2m}\underset{N<k}{\sup }\sum _{i=0}^{k-N} e^{-\frac{1}{4p}\alpha _k}\\&\le M_{2p}||x ||_{s}||{1}\mathrm{l} ||_{2m} \underset{N<k}{\sup }\, k e^{-\frac{1}{4p}\alpha _k}.\\ \end{aligned}$$

The sequence $(\alpha _j)$ satisfies

$$\begin{aligned} \lim _{j\rightarrow \infty } \alpha _j^{-1}\ln j=0, \end{aligned}$$

hence for $k$ big enough we have $\ln k \le \frac{1}{8p}\alpha _k$ and

$$\begin{aligned} \left\| {\sum _{n=N}^{M}R^nTL^nx}\right\| _{p} \rightarrow 0, \quad as \; N\rightarrow \infty , \end{aligned}$$

which finishes the proof.

Now we will show a similar proof for nuclear and stable power series space of infinite type.

Analogously to the previous case, we take an operator $T\in \mathcal {L}(\Lambda _\infty (\alpha ))$ and consider it as an operator acting from $\Lambda _\infty (\alpha )$ to $\lambda ^\infty (\exp (k\alpha _j))$. Let $T$ be given by the matrix $(t_{i,j})_{i,j\in \mathbb {N}}$. Then for all $p\in \mathbb {N}$ there exist constants $M_p$, $m_p$ such that for all $m>m_p$

$$\begin{aligned} \sup _{i,j}\left| t_{i,j} \right| e^{-m\alpha _i+p\alpha _j}\le M_p. \end{aligned}$$

(9)

Indeed, fix $p\in \mathbb {N}$. Because $T$ is continuous there exist $M_p$, $m_p$ such that $||Tx ||_{p}\le M_p||x ||_{m_p}$ for all $x$. Consider the sequence $(x^j)_{j\in \mathbb {N}}\subset \lambda _\infty (\alpha )$ defined by

$$\begin{aligned} x^j=(0,\ldots ,0,\underbrace{e^{-{m_p}\alpha _j}}_{\text {j-th coordinate}},0,\ldots ). \end{aligned}$$

Then for all $j\in \mathbb {N}$ we have $||x^j ||_{m_p}=1$ and

$$\begin{aligned} \underset{i\in \mathbb {N}}{{{\mathrm{sup}}}}\left| t_{i,j} \right| e^{-{m_p}\alpha _j}e^{{p}\alpha _i}=||Tx^j ||_{p}\le M_p. \end{aligned}$$

Hence

$$\begin{aligned} \underset{i,j\in \mathbb {N}}{{{\mathrm{sup}}}}\left| t_{i,j} \right| e^{-{m}\alpha _j+{p}\alpha _i}\le \underset{i,j\in \mathbb {N}}{{{\mathrm{sup}}}}\left| t_{i,j} \right| e^{-{m_p}\alpha _j+{p}\alpha _i}\le M_p, \end{aligned}$$

which shows (9).

Now, we estimate the $p$-th norm of $\sum _{n=N}^MR^nTL^nx$ in $\lambda ^\infty (\exp (k\alpha _j))$.

$$\begin{aligned} ||\sum _{n=N}^{M}R^nTL^nx ||_{p}&=\max \Biggl ( \underset{N< k\le M}{{{\mathrm{sup}}}}\left| \sum _{i=0}^{k-N}\sum _{j=0}^\infty t_{i,j} x_{j+k-i} \right| e^{{p}\alpha _k}, \nonumber \\&\qquad \underset{k>M}{{{\mathrm{sup}}}}\left| \sum _{i=k-M}^{k-N}\sum _{j=0}^\infty t_{i,j} x_{j+k-i} \right| e^{{p}\alpha _k}\Biggr ) \nonumber \\&\le \underset{N<k}{{{\mathrm{sup}}}}\sum _{i=0}^{k-N}\sum _{j=0}^\infty \left| t_{i,j} \right| \left| x_{j+k-i} \right| e^{{p}\alpha _k}. \end{aligned}$$

(10)

Let $m,s,u\in \mathbb {N}$ be such that

• $u-cp>A$, where $A=\sup _{j\in \mathbb {N}}\alpha _j^{-1}\ln j$, and $c$ satisfies $\alpha _{j+k}\le c(\alpha _j+\alpha _k)$ for all $j$,$k$,

• $m>m_u$, where $m_u$ satisfies (9) for $p=u$,

• $s-2m>2A$,

• $s-2cp>0$.

We further estimate (10) by

$$\begin{aligned} ||\sum _{n=N}^{M}R^nTL^nx ||_{p}&\le \underset{N<k}{{{\mathrm{sup}}}}\sum _{i=0}^{k-N}\sum _{j=0}^\infty \left| t_{i,j} \right| e^{-m\alpha _j+u\alpha _i} \left| x_{j+k-i} \right| e^{s\alpha _{j+k-i}} \cdot \nonumber \\&\qquad \cdot e^{m\alpha _j -u\alpha _i-s\alpha _{j+k-i}+ {p}\alpha _k} \nonumber \\&\le M_{u}||x ||_{s} \underset{N<k}{{{\mathrm{sup}}}}\sum _{i=0}^{k-N}\sum _{j=0}^\infty \ e^{m\alpha _j -u\alpha _i-s\alpha _{j+k-i}+ {p}\alpha _k}. \end{aligned}$$

(11)

Since the sequence $(\alpha _n)$ is increasing it satisfies $\alpha _{a+b}> \frac{1}{2}(\alpha _a+\alpha _b)$ for all $a,b\in \mathbb {N}$. We apply this to the element $s\alpha _{j+k-i}$ in (11) and get the following

$$\begin{aligned} ||\sum _{n=N}^{M}R^nTL^nx ||_{p}&\le M_{u}||x ||_{s} \underset{N<k}{{{\mathrm{sup}}}}\sum _{i=0}^{k-N}\sum _{j=0}^\infty e^{m\alpha _j -u\alpha _i-\frac{s}{2}\alpha _{j}-\frac{s}{2}\alpha _{k-i}+{p}\alpha _k} \nonumber \\&\le M_{u}||x ||_{s} \sum _{j=0}^\infty e^{-(\frac{s}{2}-m)\alpha _j} \underset{N<k}{{{\mathrm{sup}}}}\sum _{i=0}^{k-N} e^{-u\alpha _i-\frac{s}{2}\alpha _{k-i}+{p}\alpha _k}. \end{aligned}$$

(12)

By the choice of $c$, we have $\alpha _k \le c(\alpha _{k-i}+\alpha _i)$, so

$$\begin{aligned} \underset{N<k}{{{\mathrm{sup}}}}\sum _{i=0}^{k-N} e^{-u\alpha _i-\frac{s}{2}\alpha _{k-i}+{p}\alpha _k}&\le \underset{N<k}{{{\mathrm{sup}}}}\sum _{i=0}^{k-N} e^{-u\alpha _i-\frac{s}{2}\alpha _{k-i}+cp\alpha _{k-i}+cp\alpha _i}\\&\le \underset{N<k}{{{\mathrm{sup}}}}\sum _{i=0}^{k-N} e^{-(u-cp)\alpha _i -(\frac{s}{2}-cp)\alpha _{k-i}} \\&\le e^{ -(\frac{s}{2}-cp)\alpha _N} \sum _{i=0}^{\infty } e^{-(u-cp)\alpha _i}. \end{aligned}$$

Applying the above to (12) we get that

$$\begin{aligned} \left\| {\sum _{n=N}^{M}R^nTL^nx}\right\| _{p}&\le M_{u}||x ||_{s} \biggl (\sum _{j=0}^\infty e^{-(\frac{s}{2}-m)\alpha _j}\biggr ) \biggl ( \sum _{i=0}^{\infty } e^{-(u-cp)\alpha _i}\biggr ) e^{ -(\frac{s}{2}-cp)\alpha _N} \end{aligned}$$

Notice that for all $k> \sup _{j\in \mathbb {N}}\alpha _j^{-1}\ln j$ the series $\sum _{j=0}^\infty e^{-k\alpha _j}$ converges. Hence, due to the choice of the constants $s,m,u$ we have

$$\begin{aligned} \left\| {\sum _{n=N}^{M}R^nTL^nx}\right\| _{p} \xrightarrow [N\rightarrow \infty ]{} 0. \end{aligned}$$

$\square$

As the stable and nuclear power series spaces form a wide class of spaces we get in particular

Corollary 2

For the following spaces all continuous linear operators acting on them are commutators:

• The space of all holomomorphic functions on the unit disc ${H}(\mathbb {D})$ or the polydisc ${H}(\mathbb {D}^n)$,

• The space of all entire functions ${H}(\mathbb {C})$, ${H}(\mathbb {C}^n)$,

• The space of rapidly decreasing sequences $s$,

• The space $\mathbb {C}_{2\pi }^\infty (\mathbb {R})$ of all $2\pi$-periodic $C^\infty$-fucntions on $\mathbb {R}$,

• The Schwartz space of all rapidly decreasing functions $S(\mathbb {R})$.

Proof

It is known that ${H}(\mathbb {D})$ is isomorphic to $\Lambda _0(j)$ [8, 27.27] and $H(\mathbb {D}^n)\simeq \Lambda _0(\root n \of {j})$ [9, 8.3.2]. The space ${H}(\mathbb {C}^n)$ is isomorphic to $\Lambda _\infty (\root n \of {j})$ [9, 8.3.2] and the rest of the spaces are isomorphic as Fréchet spaces to $s=\Lambda _\infty (\ln j)$ [8, 29.5]. $\square$

Let $(X_i, (||\cdot ||^i_n)_{n\in \mathbb {N}})_{i\in N}$ be a family of Fréchet spaces. By $\prod _{i=1}^\infty X_i$ we denote its product

$$\begin{aligned} \prod _{i=1}^\infty X_i=\{(x_i): \; x_i\in X_i\}. \end{aligned}$$

The space $\prod _{i=0}^\infty X_i$ with the topology generated by the family of seminorms $(||| \cdot |||_{n})_{n\in \mathbb {N}}$, $||| x |||_{n}=\max _{i\le n} ||x_i ||_n^i$ is a Fréchet space.

The following theorem shows another group of spaces on which every operator is a commutator.

Theorem 2

Let $(X, (||\cdot ||_n)_{n\in \mathbb {N}})$ be a Fréchet space. All continuous linear operators acting on $\prod _{i=0}^\infty X_i$, where $X_i=X$ for all $i\in \mathbb {N}$, are commutators.

Proof

For $x=(x_i)\in \prod _{i=0}^\infty X_i$ we define the operators $R$ and $L$ as follows

$$\begin{aligned} R(x)&=R(x_0,x_1,x_1,\ldots )=(0,x_0,x_1,\ldots ), \\ L(x)&=L(x_0,x_1,x_1,\ldots )=(x_1,x_2,x_3,\ldots ), \end{aligned}$$

where $0$ denotes the zero vector in $X$. Notice that from the above definitions we have that $LR=I$ and these operators are continuous. Morover, for all $x=(x_i)\in \prod _{i=0}^\infty X_i$ and all $n, k\in \mathbb {N}$, $n>k$ the norm $||R^nx ||_k=0$.

Let $T\in \mathcal {L}(\prod _{i=0}^\infty X_i)$. Consider the series $\sum _{n=0}^\infty R^nTL^n$. The remark stated above implies that it is pointwise convergent. For all $x\in \prod _{i=0}^\infty X_i$, $k\in \mathbb {N}$ and $M\ge N>k$ we have that

$$\begin{aligned} \left\| {\sum _{n=N}^M R^nTL^nx}\right\| _{k}=0. \end{aligned}$$

Hence the series $\sum _{n=0}^\infty R^nTL^nx$ is convergent and by the Banach-Steinhaus theorem there exists the operator $S=\sum _{n=0}^\infty R^nTL^n\in \mathcal {L}(\prod _{i=0}^\infty X_i)$. Moreover, for all $x\in \prod _{i=0}^\infty X_i$

$$\begin{aligned} (LRS-RSL)x&=LR\sum _{n=0}^\infty R^nTL^nx-R\biggl (\sum _{n=0}^\infty R^nTL^n\biggr )Lx \\&=\sum _{n=0}^\infty R^nTL^nx -\sum _{n=1}^\infty R^nTL^nx =R^0TL^0x=Tx. \end{aligned}$$

Hence $T=L(RS)-(RS)L$ is a commutator. $\square$

Corollary 3

On the space of all continuous functions $C(\mathbb {R})$, the space of all sequences $\omega$ and the space $C^\infty (\Omega )$ of all smooth functions on an open subset $\Omega \in \mathbb {R}^n$ every continuous linear operator is a commutator.

Proof

This follows from $C(\mathbb {R})\simeq \prod _{i=0}^\infty C[0,1]$ and $C^\infty (\Omega )\simeq \prod _{i=0}^\infty s$ [10, p.383]. $\square$

4 Remarks

In this section we will show that the technique used in the proof of Theorem 1 can be applied only to nuclear power series spaces.

Proposition 1

Let $X=\Lambda _0(\alpha )$ or $X=\Lambda _\infty (\alpha )$. The series $\sum _{n=0}^\infty R^nIL^n$ is pointwise convergent if and only if the space $X$ is nuclear.

Remark 1

The above proposition implies that if $\sup _{j\in \mathbb {N}}{\frac{\alpha _{j+1}}{\alpha _j}}<\infty$ (i.e. $R$ and $L$ are continuous) then the identity map on $\Lambda _0(\alpha )$ or $\Lambda _\infty (\alpha )$ is a commutator in the nuclear case.

Proof

The proof will be done separately for $X=\Lambda _0(\alpha )$ and $X=\Lambda _\infty (\alpha )$.

We will start with the finite type spaces.

$\Rightarrow$: Assume that $X=\Lambda _0(\alpha )$ is not nuclear. We will show that the series $\sum _{n=0}^\infty R^nIL^n$ is divergent. We start with the observation that $R^nL^n=I-\sum _{j=0}^{n-1} P_j$, where $P_j$ is the projection on the $j$-th coordinate. Hence the operator $R^nL^n$ is well defined. For $x=(x_j)_{j\in \mathbb {N}}$ we have that

$$\begin{aligned} \sum _{n=0}^NR^nL^nx=\left( x_0,2x_1,\ldots ,Nx_{N-1},(N+1)x_N, (N+1)x_{N+1}, \ldots \right) . \end{aligned}$$

(13)

Now consider two cases.

Let $(\frac{1}{j+1})_{j\in \mathbb {N}}\in X$. Since $X$ is non-nuclear, by the Grothendieck-Pietsch theorem [8, 28.15] it is easy to show that ${1}\mathrm{l}\notin X$. Hence, there exists $K\in \mathbb {N}$ such that

$$\begin{aligned} \sum _{j=0}^\infty e^{-\frac{1}{K}\alpha _j}=\infty . \end{aligned}$$

Taking $x=(\frac{1}{j+1})_{j\in \mathbb {N}}$, we get by (13) that

$$\begin{aligned} \sum _{n=0}^NR^nL^nx=\left( \underbrace{1,\ldots ,1}_{\text {N+1}},\frac{N+1}{N+2},\frac{N+1}{N+3},\ldots \right) . \end{aligned}$$

Hence

$$\begin{aligned} \lim _{N\rightarrow \infty }||\sum _{n=0}^N R^nL^nx ||_{K}&= \lim _{N\rightarrow \infty }\left( \sum _{j=0}^N e^{-\frac{1}{K}\alpha _j}+\sum _{j=N+1}^\infty \frac{N+1}{j+1}e^{-\frac{1}{K}\alpha _j}\right) \\&\ge \lim _{N\rightarrow \infty }\left( \sum _{j=0}^N e^{-\frac{1}{K}\alpha _j}\right) =\infty . \end{aligned}$$

Now assume that $(\frac{1}{j+1})_{j\in \mathbb {N}}\notin X$. Then there exists a $K\in \mathbb {N}$, such that

$$\begin{aligned} \sum _{j=0}^\infty \frac{1}{j+1} e^{-\frac{1}{K}\alpha _j} =\infty . \end{aligned}$$

Let $x=\left( \frac{1}{(j+1)^2}\right) _{j\in \mathbb {N}}$. The vector $x$ belongs to $X$ because for all $m$ and $j$

$$\begin{aligned} e^{-\frac{1}{m}\alpha _j}\le 1 \end{aligned}$$

and

$$\begin{aligned} \sum _{j=J_m}^\infty \Bigl (\frac{1}{j+1}\Bigr )^2 e^{-\frac{1}{m}\alpha _j} \le \sum _{j=J_m}^\infty \Bigl (\frac{1}{j+1}\Bigr )^2 <\infty . \end{aligned}$$

We will show that the sequence $(\sum _{n=0}^N R^nL^nx)_{N\in \mathbb {N}}$ is unbounded. Indeed, we have

$$\begin{aligned} \sum _{n=0}^NR^nL^nx=\left( 1,\frac{1}{2},\frac{1}{3},\ldots ,\frac{1}{N+1},\frac{N+1}{(N+2)^2},\frac{N+1}{(N+3)^2},\ldots \right) \end{aligned}$$

and

$$\begin{aligned} \lim _{N\rightarrow \infty }||\sum _{n=0}^N R^nL^nx ||_{k}&= \lim _{N\rightarrow \infty }\left( \sum _{j=0}^N \frac{1}{j+1}e^{-\frac{1}{K}\alpha _j}+(N+1)\sum _{j=N+1}^\infty (\frac{1}{j+1})^2 e^{-\frac{1}{K}\alpha _j}\right) \\&\ge \lim _{N\rightarrow \infty }\left( \sum _{j=0}^N \frac{1}{j+1} e^{-\frac{1}{K}\alpha _j}\right) =\infty . \end{aligned}$$

$\Leftarrow$: Let $X$ be nuclear. Then

$$\begin{aligned} \lim _{j\rightarrow \infty } \alpha _j^{-1}\ln j=0. \end{aligned}$$

Hence

$$\begin{aligned} \forall k\in \mathbb {N}\; \exists J_k \; \forall j>J_k \quad \ln j\le \frac{1}{k}\alpha _j. \end{aligned}$$

Take an arbitrary $x\in X$. By (13) we have

$$\begin{aligned} \sum _{n=N}^MR^nL^nx=\left( \underbrace{0\ldots ,0}_{N}, x_{N+1},2x_{N+2},\ldots , (M-N)x_M, (M-N)x_{M+1}, \ldots \right) . \end{aligned}$$

Fix $p\in \mathbb {N}$. For all $J_{2p}<N<M$ we have the following

$$\begin{aligned} ||\sum _{n=N}^MR^nL^nx ||_{p}&= \sum _{j=N+1}^M(j-N)\left| x_j \right| e^{-\frac{1}{p}\alpha _j}+ \sum _{j=M+1}^\infty (M-N)\left| x_j \right| e^{-\frac{1}{p}\alpha _j} \nonumber \\&\le \sum _{j=N+1}^\infty j \left| x_j \right| e^{-\frac{1}{p}\alpha _j} = \sum _{j=N+1}^\infty \left| x_j \right| e^{-\frac{1}{p}\alpha _j+ \ln j} \nonumber \\&\le \sum _{j=N+1}^\infty \left| x_j \right| e^{-\frac{1}{p}\alpha _j+\frac{1}{2p} \alpha _j}= \sum _{j=N+1}^\infty \left| x_j \right| e^{-\frac{1}{2p}\alpha _j} \end{aligned}$$

(14)

and the right side tends to $0$, when $N\rightarrow \infty$.

Now we will prove the proposition for the infinite type spaces.

$\Rightarrow$: Let $X=\Lambda _\infty (\alpha )$ be non-nuclear. Then the sequence $\alpha =(\alpha _j)_{j\in \mathbb {N}}$ satisfies

$$\begin{aligned} \sup _{0<j} \alpha _j^{-1} \ln j =\infty . \end{aligned}$$

Hence there exists a subsequence $(j_n)$ and an increasing, unbounded sequence $(\beta _{j_n})$, such that

$$\begin{aligned} \beta _{j_n}=\alpha _{j_n}^{-1}\ln (j_n). \end{aligned}$$

Consider the element $x=(x_j)_{j\in \mathbb {N}}$ defined by

$$\begin{aligned} x_j={\left\{ \begin{array}{ll} \frac{1}{j+1} &{} \text {if } j=j_n, \\ 0 &{} \text {if } j\ne j_n \end{array}\right. } \end{aligned}$$

and let $y=(y_j)_{j\in \mathbb {N}}$ be defined by $y_j=x_j^2$. Since for all $k\in \mathbb {N}$ we have

$$\begin{aligned} ||y ||_{k}=\sum _{n\in \mathbb {N}}\left( \frac{1}{j_n+1}\right) ^2e^{k\alpha _{j_n}} =\sum _{n\in \mathbb {N}} \left( \frac{1}{j_n+1}\right) ^2 e^{k\beta _{j_n}^{-1} \ln j_n} \le \sum _{n\in \mathbb {N}}\left( \frac{1}{j_n}\right) ^{2-k\beta _{j_n}^{-1}} <\infty , \end{aligned}$$

the vector $y$ belongs to $X$.

Recall that for $w=(w_j)_{j\in \mathbb {N}}$ we have

$$\begin{aligned} \sum _{n=0}^NR^nL^nw=\left( w_0,2w_1,\ldots ,Nw_{N-1},(N+1)w_N, (N+1)w_{N+1}, \ldots \right) \end{aligned}$$

and consider two cases.

Let $x\in X$. Then

$$\begin{aligned} \left( \sum _{n=0}^NR^nL^nx\right) _j= {\left\{ \begin{array}{ll} 0 &{} \text {if } j\ne j_n, \\ 1 &{} \text {if } j= j_n, j_n\le N, \\ \frac{N+1}{j_n+1} &{} \text {if } j= j_n, j_n> N. \end{array}\right. } \end{aligned}$$

Hence, for all $k\in \mathbb {N}$

$$\begin{aligned} \lim _{N\rightarrow \infty }||\sum _{n=0}^N R^nL^nx ||_{k}&= \lim _{N\rightarrow \infty }\left( \sum _{j_n\le N} e^{{k}\alpha _{j_n}}+\sum _{j_n>N}^\infty \frac{N+1}{j_n+1}e^{{k}\alpha _{j_n}}\right) \\&\ge \lim _{N\rightarrow \infty }\left( \sum _{j_n\le N} e^{{k}\alpha _{j_n}}\right) =\infty . \end{aligned}$$

Now assume that $x\notin X$. Then there exists $K\in \mathbb {N}$, such that

$$\begin{aligned} \sum _{n=0}^\infty \frac{1}{j_n+1} e^{{K}\alpha _{j_n}} =\infty \end{aligned}$$

and the sequence $(\sum _{n=0}^N R^nL^ny)_{N\in \mathbb {N}}$ is unbounded. Indeed, we have that

$$\begin{aligned} \sum _{n=0}^NR^nL^ny= {\left\{ \begin{array}{ll} 0 &{} \text {if } j\ne j_n, \\ \frac{1}{j_n+1} &{} \text {if } j= j_n, j_n\le N, \\ \frac{N+1}{(j_n+1)^2} &{} \text {if } j= j_n, j_n> N \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned} \lim _{N\rightarrow \infty }||\sum _{n=0}^N R^nL^ny ||_{K}&= \lim _{N\rightarrow \infty }\Biggl (\sum _{j_n\le N} \frac{1}{j_n+1}e^{{K}\alpha _{j_n}}+(N+1)\sum _{j_n>N}^\infty \frac{1}{(j_n+1)^2} e^{{K}\alpha _{j_n}}\Biggr )\\&\ge \lim _{N\rightarrow \infty }\Biggl (\sum _{j_n\le N} \frac{1}{j+1} e^{{K}\alpha _{j_n}}\Biggr ) =\infty . \end{aligned}$$

$\Leftarrow$: Let $X$ be a nuclear space. Then, there exists a constant $c\in \mathbb {N}$, such that

$$\begin{aligned} \forall j\in \mathbb {N}\!\setminus \!\{0\} \quad \ln j\le c\alpha _j. \end{aligned}$$

Recall that for all $x\in X$ and $N<M$ we have

$$\begin{aligned} \sum _{n=N}^MR^nL^nx=\bigl (\underbrace{0\ldots ,0}_{N}, x_{N+1},2x_{N+2},\ldots , (M-N)x_M, (M-N)x_{M+1}, \ldots \bigr ). \end{aligned}$$

and

$$\begin{aligned} ||\sum _{n=N}^MR^nL^nx ||_{p}&= \sum _{j=N+1}^M(j-N)\left| x_j \right| e^{p\alpha _j}+ \sum _{j=M+1}^\infty (M-N)\left| x_j \right| e^{p\alpha _j} \nonumber \\&\le \sum _{j=N+1}^\infty j \left| x_j \right| e^{p\alpha _j} = \sum _{j=N+1}^\infty \left| x_j \right| e^{p\alpha _j+ \ln j} \nonumber \\&\le \sum _{j=N+1}^\infty \left| x_j \right| e^{p\alpha _j+c \alpha _j}= \sum _{j=N+1}^\infty \left| x_j \right| e^{(p+c)\alpha _j}. \end{aligned}$$

Hence the series $\sum _{n=0}^\infty R^nL^nx$ is convergent. $\square$