1 Introduction

1.1 Coboundaries

Let T be a bounded linear operator acting on a complex Banach space \(\mathcal X\). An element x of \(\mathcal X\) is called a coboundary for T if there is \(y\in \mathcal X\) such that \(x = y - Ty\). Coboundaries are related to the behavior of the ergodic sums

$$\begin{aligned} S_n(T) x := x + Tx + \dots + T^{n-1}x, \quad n \ge 1.\end{aligned}$$

A variant of the mean ergodic theorem for power bounded operators on reflexive Banach spaces has been proved by von Neumann for Hilbert spaces and by Lorch in the general case; see for instance [14]. Recall that T is said to be power bounded if \(\sup _{n\ge 1} \Vert T^n\Vert < \infty \). We have

$$\begin{aligned} \mathcal X= \left\rbrace x \in \mathcal X: \lim _{n\rightarrow \infty } \frac{1}{n} S_n(T) x \text { exists} \right\lbrace = \{ y \in \mathcal X: Ty = y \} \oplus \overline{(I-T)\mathcal X}.\end{aligned}$$

In particular, as a consequence of this ergodic decomposition, we have

$$\begin{aligned} x\in \overline{(I-T)\mathcal X} \quad \Leftrightarrow \quad \lim _{n\rightarrow \infty } \frac{1}{n} S_n(T) x = 0.\end{aligned}$$

One can say more about the rate of convergence of \((1/n)S_n(T) x\) to zero when x is a coboundary. Indeed, when there exists a solution y of the equation \(y - Ty = x\), the ergodic sums satisfy \(S_n(T) x = y - T^ny\). It follows that \((S_n(T)x)_{n\in \mathbb {N}}\) is bounded. Therefore

$$\begin{aligned} \left\Vert\frac{1}{n} S_n(T) x\right\Vert = O \left(\frac{1}{n}\right). \end{aligned}$$
(1.1)

This rate of convergence to zero, namely O(1/n), characterizes coboundaries of power bounded operators on reflexive spaces. Indeed, the converse result (whenever T is power bounded and \(\mathcal X\) is reflexive, an element x satisfying (1.1) is a coboundary for T) has been proved by Browder [1] and rediscovered by Butzer and Westphal [2].

We also note (see for instance [3, 4, 11] and the references therein) that if \((I-T) \mathcal X\) is not closed, then for every sequence \((a_n)_{n\ge 1}\) of positive real numbers converging to zero, there exists \(x \in \overline{(I-T)\mathcal X} \backslash (I-T)\mathcal X\) such that

$$\begin{aligned} \left\Vert\frac{1}{n} S_n(T) x\right\Vert \ge a_n, \quad \forall n \ge 1.\end{aligned}$$

In particular, there is no general rate of convergence in the mean ergodic theorem outside coboundaries.

1.2 Rochberg’s Theorem

Browder’s theorem has been extended to the case that T is a dual operator on a dual Banach space by Lin [16]; see also Lin and Sine [17]. We refer the reader to the introduction of [6], and the references cited therein, for the history of Browder’s theorem and for other extensions and generalizations. We mention here only two references, namely [19] and [13], dealing with the Hilbert space situation. Any of these Hilbert or Banach space abstract characterizations is not strong enough to obtain as consequences classical results of Fortet and Kac [9, 12] who dealt with the case \(\mathcal X= L^2(0,1)\) and \(Sf(x) = f(2x)\). This operator S is the Koopman operator associated with the doubling map on the torus; see the last section of this manuscript for more information about coboundaries of S. This situation has been remedied by Rochberg [20], who showed that a condition of \(o(\sqrt{n})\) growth of ergodic sums at x is sufficient to ensure that x is a coboundary for a unilateral shift on Hilbert space. Notice that the Koopman operator S acts as a unilateral shift on the subspace of \(L^2 (0,1)\) of functions whose zeroth Fourier coefficient vanishes.

We need the following classical definition in order to state Rochberg’s abstract coboundary theorem.

Definition 1.1

Let T be an isometry acting on Hilbert space \(\mathcal H\). A closed subspace \(\mathcal K\) of \(\mathcal H\) is called wandering for T whenever

$$\begin{aligned} T^p\mathcal K\perp T^q \mathcal K\quad \text {for} \quad p,q \in \mathbb {N}, p\ne q.\end{aligned}$$

The isometry T is called a (unilateral) shift if \(\mathcal H\) possess a closed subspace \(\mathcal K\), wandering for T and such that

$$\begin{aligned} \bigoplus _{n=0}^\infty T^n \mathcal K= \mathcal H.\end{aligned}$$

Theorem 1.2

([20]) Let S be a shift and let f be an element of \(\mathcal H\). Using the notation of the preceding definition, we denote by \(f_j\) the projection of f onto the closed subspace \(S^j \mathcal K\). Suppose that there exists \(\beta > 0\) such that

$$\begin{aligned} \Vert f_j\Vert = O (2^{-\beta j}).\end{aligned}$$

Then there exists g in \(\mathcal H\) such that \((I-S)g = f\) if and only if

$$\begin{aligned} \lim _{n\rightarrow \infty } \frac{1}{n} \left\Vert\sum _{k= 0}^n S^k f \right\Vert^2 = 0.\end{aligned}$$

Remark 1.3

The condition

$$\begin{aligned} \Vert f_j\Vert = O (2^{-\beta j})\end{aligned}$$

is of course dependent of the decomposition of \(\mathcal H\) associated with the unilateral shift S. It implies \(\Vert S^{* j} f \Vert = O (2^{-\beta j})\).

1.3 Statement of the Main Results

In the next theorem the unilateral shift S is replaced by an arbitrary isometry T and the growth of the norm of the projection \(f_j\) by the convergence of the series \(\sum _{j=0}^\infty j\Vert T^{* j} f \Vert \). The statement of the result does not depend on the Wold decomposition, at least not in an explicit way. For the convenience of the reader, the Wold decomposition theorem is recalled below. Theorem 1.4 implies Rochberg’s theorem and it allows to recover Kac’s results about the coboundaries of the Koopman operator of the doubling map.

Theorem 1.4

Let T be an isometry acting on a Hilbert space \(\mathcal H\) and let \(x \in \mathcal H\). Suppose that

$$\begin{aligned} \sum _{k=0}^\infty k \Vert T^{*k} x \Vert < \infty . \end{aligned}$$
(1.2)

Then there exists \(y \in \mathcal H\) such that \(x = (I-T)y\) if and only if

$$\begin{aligned} \lim _{n\rightarrow \infty } \frac{1}{n} \left\Vert\sum _{k= 0}^n T^k x \right\Vert^2 = 0.\end{aligned}$$

Note however that the condition (1.2) implies that x is necessarily an element of the shift part of the isometry T.

Considering coboundaries of adjoints of isometries, we notice that the identity \(I-T = (T^*-I)T\) shows that every coboundary of the isometry T is also a coboundary for its adjoint \(T^*\). It follows from [7, Proposition 4.3] that when the isometry T is not invertible (i.e., not a unitary operator), there are coboundaries for \(T^*\) which are not coboundaries for T.

The following result, more general than Theorem 1.4, is about coboundaries of contractions (operators of norm no greater than one).

Theorem 1.5

Let T be a linear operator acting on a Hilbert space \(\mathcal H\) with \(\Vert T\Vert \le 1\). Let \(x \in \mathcal H\) and denote \(S_n(T) x := x + Tx + \dots + T^{n-1}x\). Suppose that (1.2) holds, as well as

$$\begin{aligned} \Vert S_n(T)x\Vert = o(\sqrt{n}), \quad n\rightarrow \infty \end{aligned}$$
(1.3)

and

$$\begin{aligned} \sum _{k=1}^{n} \left( \Vert S_k(T)x\Vert ^2 - \Vert TS_k(T)x\Vert ^2 \right) = o(n), \quad n\rightarrow \infty . \end{aligned}$$
(1.4)

Then there exists \(y \in \mathcal H\) such that \(x = (I-T)y\). In addition, y can be chosen such that \(\Vert Ty\Vert = \Vert y\Vert \).

We obtain the following consequence.

Corollary 1.6

Let T be a linear operator acting on a Hilbert space \(\mathcal H\) with \(\Vert T\Vert \le 1\). Let \(x \in \mathcal H\) and denote \(S_n(T) x := x + Tx + \dots + T^{n-1}x\). Suppose that (1.2) and (1.3) hold, as well as

$$\begin{aligned} \sum _{k=1}^{\infty } \frac{\left( \Vert S_k(T)x\Vert ^2 - \Vert TS_k(T)x\Vert ^2 \right) }{k} < \infty . \end{aligned}$$
(1.5)

Then there exists \(y \in \mathcal H\) such that \(x = (I-T)y\) and \(\Vert Ty\Vert = \Vert y\Vert \).

Some remarks are in order. Theorem 1.5 and its consequence Corollary 1.6 show that the coboundary equation can be solved within the maximal isometric subspace

$$\begin{aligned}M = \{x\in \mathcal H: \Vert T^nx\Vert = \Vert x\Vert \text { for every }n\ge 0\}.\end{aligned}$$

We refer to [18] and [15] for the canonical decomposition of a contraction into the maximal isometric subspace and its orthogonal.

Conditions (1.4) and (1.5) are easily verified when T is an isometry. The conditions (1.2) and (1.3) are always satisfied when \(\Vert T\Vert < 1\); however (1.5) is not, unless \(x=0\). In fact, \(\Vert Ty\Vert = \Vert y\Vert \) and \(\Vert T\Vert < 1\) imply that \(y=0\) and thus \(x=0\). Of course, as \((I-T)\) is invertible when \(\Vert T\Vert < 1\) by Carl Neumann’s lemma, the coboundary equation \(x = (I-T)y\) is always solvable in this case.

1.4 Outline of the Paper

A proof of Theorem 1.4 is given in the next section. The more general Theorem 1.5 and its consequence Corollary 1.6 are proved in Sect. 3. Some applications to the functional equation \(g(x) - g(2x) = f(x)\) are presented in the next section. The last section collects the acknowledgments, and (imposed) conflict of interest and data availability statements.

2 Proof of Theorem 1.4

We first recall Wold’s decomposition Theorem (see [18, Chapter 1]).

Theorem 2.1

(Wold decomposition) Let T be an isometry on a Hilbert \(\mathcal H\). Then \(\mathcal H\) decomposes as an orthogonal sum \(\mathcal H= \mathcal H_0 \oplus \mathcal H_1\) such that \(\mathcal H_0\) and \(\mathcal H_1\) are reducing for T, the restriction of T to \(\mathcal H_0\) is a unitary operator and the restriction of T to \(\mathcal H_1\) is a unilateral shift (one of the subspaces can eventually reduce to \(\{0\}\)). This decomposition is unique; in particular, we have

$$\begin{aligned} \mathcal H_0 = \bigcap _{n=0}^\infty T^n \mathcal H\quad \text { and }\quad \mathcal H_1 = \bigoplus _{n=0}^\infty T^n \mathcal K\,\text {, where } \quad \mathcal K= \mathcal H\ominus T\mathcal H.\end{aligned}$$

Proof of Theorem 1.4

If \(x = (I-T) y\), then \(\sum _{k= 0}^n T^k x = x - T^{n+1}x\). Therefore \(\sum _{k= 0}^n T^k x\) is bounded since the isometry T is clearly power-bounded. In particular,

$$\begin{aligned} \lim _{n\rightarrow \infty } \frac{1}{n} \left\Vert\sum _{k= 0}^n T^k x \right\Vert^2 = 0.\end{aligned}$$

Suppose now that

$$\begin{aligned} \lim _{n\rightarrow \infty } \frac{1}{n} \left\Vert\sum _{k= 0}^n T^k x \right\Vert^2 = 0.\end{aligned}$$

We want to show the existence of a solution y of the equation \((I-T)y = x\).

Let \(\mathcal H= \mathcal H_0 \oplus \mathcal H_1\) be the Wold’s decomposition associated with T. We notice that \(x \in \mathcal H_1\). Indeed, if \(x = x_0 + x_1\) according to Wold’s decomposition of \(\mathcal H\), then

$$\begin{aligned} \lim _{k\rightarrow \infty } \Vert T^{*k} x_1 \Vert = 0 \quad \text {and}\quad \Vert T^{*n} x_0 \Vert = \Vert x_0\Vert , \quad \forall n \in \mathbb {N}.\end{aligned}$$

Therefore

$$\begin{aligned}\lim _{n\rightarrow \infty } \Vert T^{*n} x \Vert = \Vert x_0\Vert .\end{aligned}$$

On the other hand, it follows from (1.2) that

$$\begin{aligned}\lim _{n\rightarrow \infty } \Vert T^{*n} x \Vert = 0.\end{aligned}$$

We obtain that \(x \in \mathcal H_1\). In particular, if \(\mathcal H_1\) is reduced to \(\{0\}\), then \(x = 0 = (I-T) 0\). Therefore, without loss of any generality, we can assume that T is a shift.

For each \(n \in \mathbb {N}\), we denote by \(P_n\) the projection onto the subspace \(T^n \mathcal K\). For \(u \in \mathcal H\), we set \(u_n:=P_n(u)\), \(u^n: = \sum _{j=0}^n u_j\) and \(R_n := u-u^n\).

Suppose that y is solution of the equation \((I-T)y = x\). We first obtain, by projecting to \(T^k \mathcal K\) for each \(k \in \mathbb {N}\), the following system of equations :

$$\begin{aligned}{\left\{ \begin{array}{ll} x_0 = y_0 \\ x_1 = y_1- T y_0 \\ \vdots \\ x_k = y_k - T y_{k-1} \\ \vdots \end{array}\right. }\end{aligned}$$

We then obtain

$$\begin{aligned}{\left\{ \begin{array}{ll} y_0 = x_0 \\ y_1 = x_1 + T y_0 = x_1 + T x_0\\ \vdots \\ y_k = x_k + T y_{k-1} = x_k + T x_{k-1} + \dots + T^{k-1}x_1 + T^k x_0 \\ \vdots \end{array}\right. }\end{aligned}$$

Consider now, for each \(r \in \mathbb {N}\), the element

$$\begin{aligned} y_r = \sum _{k=0}^r T^k x_{r-k} \in T^r \mathcal K.\end{aligned}$$

We will prove that \(\sum _{r=0}^\infty \Vert y_r\Vert ^2\) is convergent, thus showing that \(y = \sum _{r=0}^\infty y_r\) is well defined in \(\mathcal H\). In that case, for every \(r\in \mathbb {N}\), we have

$$\begin{aligned} P_r \big ( (I-T) y \big )&= y_r - T y_{r-1} \\&= \sum _{j=0}^r T^j x_{r-j} - \sum _{j=0}^{r-1} T^{j+1} x_{r-1-j} \\&= x_r. \end{aligned}$$

This shows that \((I-T)y = x\).

To prove that \(\sum _{r=0}^\infty \Vert y_r\Vert ^2\) is finite, we need two more results.

Lemma 2.2

Let \(u\in \mathcal H\) be such that \(\sum _{j\ge 0} \Vert T^{*j}u \Vert < +\infty \). Then

$$\begin{aligned}\lim _{n\rightarrow \infty } \frac{1}{n} \left\| \sum _{k=0}^n T^k u \right\| ^2 = \Vert u\Vert ^2 + 2 Re \sum _{k=1}^\infty \langle u; T^k u \rangle .\end{aligned}$$

Proof

We first notice that the sum \(\sum _{k=1}^\infty \langle u; T^k u \rangle \) is absolutely convergent since \(( \Vert T^{*j}u \Vert )_{j\ge 0}\) is summable. For each \(n \in \mathbb {N}^*\), we have

$$\begin{aligned} \frac{1}{n} \left\| \sum _{k=0}^n T^k u \right\| ^2&= \frac{1}{n} \left( \sum _{i=0}^n \Vert T^i u\Vert ^2 + 2 Re \left( \sum _{0 \le i<j \le n } \langle T^i u; T^j u \rangle \right) \right) \\&= \frac{1}{n} \left( \sum _{i=0}^n \Vert u\Vert ^2 + 2 Re \left( \sum _{0 \le i <j \le n } \langle u; T^{j-i} u \rangle \right) \right) \\&= \frac{n+1}{n} \Vert u\Vert ^2 + \frac{2}{n} Re \left( \sum _{r=1}^n (n-r+1)\langle u; T^r u \rangle \right) \\&= \frac{n+1}{n} \Vert u\Vert ^2 + 2 Re \left( \sum _{r=1}^n \langle u; T^r u \rangle - \frac{1}{n} \sum _{r=1}^n (r-1)\langle u; T^r u \rangle \right) . \end{aligned}$$

On the other hand, we have

$$\begin{aligned} \left| \frac{1}{n} \sum _{r=1}^n (r-1)\langle u; T^r u \rangle \right| \le \frac{1}{n} \Vert u\Vert \sum _{r=1}^n (r-1) \Vert T^{*r} u \Vert .\end{aligned}$$

Using again the summability of the sequence \((\Vert T^{*j}u \Vert )_{j\ge 0}\) and the Kronecker’s lemma (see for instance [21, Lemma IV.3.2]), we get

$$\begin{aligned} \frac{1}{n} \sum _{r=1}^n (r-1)\langle u; T^r u \rangle \underset{ n\rightarrow \infty }{\longrightarrow } 0.\end{aligned}$$

As the series \(\sum _{k\ge 1} \langle u; T^k u \rangle \) is convergent, we obtain, as n tends to infinity,

$$\begin{aligned}\lim _{n\rightarrow \infty } \frac{1}{n} \left\| \sum _{k=0}^n T^k u \right\| ^2 = \Vert u\Vert ^2 + 2 Re \sum _{k=1}^\infty \langle u; T^k u \rangle .\end{aligned}$$

\(\square \)

Lemma 2.3

Let \(u \in \mathcal H\). For every \(r \in \mathbb {N}\) we have

$$\begin{aligned} \left\| \sum _{j=0}^r T^j u_{r-j} \right\| ^2 = \lim _{n\rightarrow \infty } \frac{1}{n} \left\| \sum _{j=0}^n T^j u^r\right\| ^2.\end{aligned}$$

Proof

Let \(n \ge r\). For \(k \in \mathbb {N}\) we have

$$\begin{aligned} P_k \left( \sum _{j=0}^n T^j u^r \right) = {\left\{ \begin{array}{ll} \sum _{j=0}^k T^j u_{k-j} \quad &{} \text {if}\quad 0 \le k< r, \\ \sum _{j=0}^r T^j u_{k-j} \quad &{} \text {if}\quad r \le k \le n, \\ \sum _{j=k-n}^r T^j u_{k-j} \quad &{} \text {if}\quad n < k \le n+r, \\ 0 &{} \text {if} \quad k > n+r. \end{array}\right. }\end{aligned}$$

Using the decomposition of \(\mathcal H\) as \(\mathcal H= \bigoplus _{n=0}^\infty T^n \mathcal K\), we obtain

$$\begin{aligned}\frac{1}{n} \left\| \sum _{j=0}^n T^j u^r\right\| ^2= & {} \frac{1}{n} \sum _{k = 0}^{r-1} \left\Vert \sum _{j=0}^k T^j u_{k-j} \right\Vert^2 + \frac{1}{n} \sum _{k = r}^{n} \left\| \sum _{j=0}^r T^j u_{k-j} \right\| ^2 \\ {}{} & {} + \frac{1}{n} \sum _{k = n+1}^{n+r} \left\| \sum _{j=k-n}^r T^j u_{k-j}\right\| ^2.\end{aligned}$$

We have

$$\begin{aligned} \frac{1}{n} \left( \sum _{k = 0}^{r-1} \left\Vert \sum _{j=0}^k T^j u_{k-j} \right\Vert^2 \right) \underset{ n\rightarrow \infty }{\longrightarrow } 0\end{aligned}$$

and

$$\begin{aligned}\frac{1}{n} \sum _{k = n+1}^{n+r} \left\| \sum _{j=k-n}^r T^j u_{k-j}\right\| ^2 = \frac{1}{n} \left( \sum _{k = 1}^{r} \left\| \sum _{j=k}^r T^j u_{k-j}\right\| ^2 \right) \underset{ n\rightarrow \infty }{\longrightarrow } 0,\end{aligned}$$

as well as

$$\begin{aligned}\frac{1}{n} \sum _{k = r}^{n} \left\| \sum _{j=0}^r T^j u_{k-j} \right\| ^2 = \frac{n-r+1}{n} \left\| \sum _{j=0}^r T^j u_{r-j} \right\| ^2 \underset{ n\rightarrow \infty }{\longrightarrow } \left\| \sum _{j=0}^r T^j u_{r-j} \right\| ^2.\end{aligned}$$

We thus obtain

$$\begin{aligned} \lim _{n\rightarrow \infty } \frac{1}{n} \left\| \sum _{j=0}^n T^j u^r\right\| ^2 = \left\| \sum _{j=0}^r T^j u_{r-j} \right\| ^2 .\end{aligned}$$

\(\square \)

We finally show that \(\sum _{r \ge 0} \Vert y_r\Vert ^2 < \infty \). Using Lemma 2.3, we have for each \(r \in \mathbb {N}\),

$$\begin{aligned} \Vert y_r\Vert ^2 = \left\| \sum _{i=0}^r T^i x_{r-i} \right\| ^2 = \lim _{n\rightarrow \infty } \frac{1}{n} \left\| \sum _{i=0}^n T^i x^r\right\| ^2.\end{aligned}$$

Using the parallelogram identity for the vectors \(x^r + R_r = x\), we get

$$\begin{aligned} \frac{2}{n} \left\| \sum _{i=0}^n T^i x^r\right\| ^2 = \frac{1}{n} \left\| \sum _{i=0}^n T^i x\right\| ^2 + \frac{1}{n} \left\| \sum _{i=0}^n T^i (x^r - R_r)\right\| ^2 \\ - \frac{2}{n} \left\| \sum _{i=0}^n T^i R_r\right\| ^2. \end{aligned}$$

Make now n tends to infinity. Using Lemma 2.2 for \(R_r\) and \(x^r - R_r\), and the hypothesis \(\frac{1}{n} \left\| \sum _{k=0}^n T^k x \right\| ^2 \underset{ n\rightarrow \infty }{\longrightarrow } 0\), we obtain

$$\begin{aligned} 2 \Vert y_r\Vert ^2&= \lim _{n\rightarrow \infty } \frac{2}{n} \left\| \sum _{i=0}^n T^i x^r\right\| ^2 \\&= \lim _{n\rightarrow \infty } \frac{1}{n} \left\| \sum _{i=0}^n T^i (x^r - R_r)\right\| ^2 - 2 \lim _{n \rightarrow \infty } \frac{1}{n} \left\| \sum _{i=0}^n T^i R_r\right\| ^2\\&= \Vert x^r - R_r \Vert ^2 - 2 \Vert R_r\Vert ^2 \\&\quad + 2 Re \sum _{k=1}^\infty \Big ( \langle x^r - R_r; T^k(x^r - R_r) \rangle - 2 \langle R_r; T^k R_r \rangle \Big ) \\&= \Vert x^r\Vert ^2 - \Vert R_r\Vert ^2 + 2 Re \sum _{k=1}^\infty \Big ( \langle x^r; T^k x^r \rangle - \langle x^r; T^k R_r \rangle \\&\quad - \langle R_r; T^k x^r \rangle - \langle R_r; T^k R_r \rangle \Big ) \\&= \Vert x^r\Vert ^2 - \Vert R_r\Vert ^2 + 2 Re \sum _{k=1}^\infty \langle x^r; T^k x^r \rangle - 2 Re \sum _{k=1}^\infty \langle R_r; T^k x \rangle . \end{aligned}$$

Using now Lemma 2.2 applied to \(x^r\) and Lemma 2.3, we get

$$\begin{aligned} \Vert x^r\Vert ^2 + 2 Re \sum _{k=1}^\infty \langle x^r; T^k x^r\rangle&= \lim _{n\rightarrow \infty } \frac{1}{n} \left\| \sum _{i=0}^\infty T^i x^r\right\| ^2 \\&= \Vert y_r\Vert ^2. \end{aligned}$$

We can infer that

$$\begin{aligned} 2 \Vert y_r\Vert ^2 = \Vert y_r\Vert ^2 - \Vert R_r\Vert ^2 - 2 Re \sum _{k=1}^\infty \langle R_r; T^k x\rangle ,\end{aligned}$$

so

$$\begin{aligned} \Vert y_r\Vert ^2 = - \Vert R_r\Vert ^2 - 2 Re \sum _{k=1}^\infty \langle R_r; T^k x\rangle .\end{aligned}$$

For each fixed r we have \(R_r = T^{r+1} T^{*(r+1)}x\). Thus \( \Vert R_r\Vert = \Vert T^{*(r+1)}x\Vert \). As

$$\begin{aligned}\sum _{j=1}^{+\infty } j \Vert T^{*j}x\Vert < +\infty ,\end{aligned}$$

we obtain that \((\Vert R_r\Vert ^2)_r\) is summable. It suffices to show that

$$\begin{aligned}\sum _{r=0}^\infty \left| \sum _{k=1}^\infty \langle R_r; T^k x \rangle \right| < \infty .\end{aligned}$$

We have

$$\begin{aligned} \left| \sum _{k=1}^\infty \langle R_r; T^k x \rangle \right|&= \left| \sum _{k=1}^r \langle R_r; T^k x \rangle + \sum _{k = r+1}^\infty \langle R_r; T^k x \rangle \right| \\&= \left| \sum _{k=1}^r \langle R_r; T^k x \rangle + \sum _{k=r+1}^\infty \langle R_k; T^k x \rangle \right| \\&\le \sum _{k= 1}^r \Vert R_r\Vert \Vert T^k x\Vert + \sum _{k=r+1}^\infty \Vert R_k\Vert \Vert T^k x\Vert \\&\le \Vert x\Vert \left( r \Vert R_r\Vert + \sum _{k=r+1}^\infty \Vert R_k\Vert \right) \\&\le \Vert x\Vert \left( r \Vert T^{*(r+1)}x\Vert + \sum _{k=r+1}^\infty \Vert T^{*(k+1)}x\Vert \right) . \end{aligned}$$

Using again the summability of \((r \Vert T^{*r}x\Vert )_r\), we get

$$\begin{aligned} \sum _{r=0}^\infty \sum _{k=r+1}^\infty \Vert T^{*k}x\Vert = \sum _{k=0}^\infty k \Vert T^{*k}x\Vert < \infty .\end{aligned}$$

Therefore \(\sum _{r=0}^\infty \Vert y_r\Vert ^2 <\infty .\) \(\square \)

3 The Case of Contractions

We now prove Theorem 1.5 and its consequence Corollary 1.6.

Proof of Theorem 1.5

Let D denote the defect operator \(D = (I-T^*T)^{1/2}\), which is well defined since T is a contraction. As

$$\begin{aligned} \Vert Tx\Vert ^2 + \Vert Dx\Vert ^2 = {\langle T^*Tx,x\rangle } + {\langle (I-T^*T)x,x\rangle } = \Vert x\Vert ^2,\end{aligned}$$

the operator \(R : \ell ^2(\mathcal H) \mapsto \ell ^2(\mathcal H)\) given by

$$\begin{aligned} R(x_0, x_1, x_2, \ldots ) = (Tx_0, Dx_0, x_1, x_2, \ldots )\end{aligned}$$

and with matrix representation

$$\begin{aligned} R = \begin{bmatrix} T &{} &{} \\ D &{} &{} \\ &{} I &{} \\ &{} &{} I \\ &{} &{} &{} \ddots \end{bmatrix}, \end{aligned}$$
(3.1)

is an isometry. We can thus apply Theorem 1.4 to R.

The iterates of R are given by

$$\begin{aligned} R^k(x_0, x_1, x_2, \ldots ) = (T^kx_0, DT^{k-1}x_0, DT^{k-2}x_0,\ldots ,DTx_0, Dx_0, x_1, x_2, \ldots )\end{aligned}$$

while their adjoints are given by

$$\begin{aligned}{} & {} R^{*k}(x_0, x_1, x_2, \cdots )\\ {}{} & {} \quad = (T^{*k}x_0 + T^{*(k-1)}Dx_1 + \cdots + T^*Dx_{k-1}, Dx_k, x_{k+1}, x_{k+2}, \cdots ).\end{aligned}$$

Denote \(\tilde{x} = (x,0,0, \cdots ) \in \ell ^2(\mathcal H)\) and \(\tilde{y} = (y,y_1,y_2, \cdots ) \in \ell ^2(\mathcal H)\). The equation

$$\begin{aligned}\tilde{x} = (I-R)\tilde{y}\end{aligned}$$

reduces to the system of equations \(x = (I-T)y\), \(y_1 = Dy\), \(y_2 = y_1\), \(y_3 = y_2\), etc. As \(\tilde{y} \in \ell ^2(\mathcal H)\), we obtain \(y_1 = y_2 = \cdots = 0\). Therefore the equation \(\tilde{x} = (I-R)\tilde{y}\) in \(\ell ^2(\mathcal H)\) is equivalent to

$$\begin{aligned} x = (I-T)y \quad \text {and} \quad Dy = 0 .\end{aligned}$$

Every positive (i.e. positive semi-definite) operator has the same kernel as its positive square-root; thus \((I-T^*T)y = 0\). Therefore \(\Vert Ty\Vert = \Vert y\Vert \).

An easy computation shows that the summability condition \(\sum _{k=0}^\infty k \Vert R^{*k} \tilde{x} \Vert < \infty \) is equivalent to \(\sum _{k=0}^\infty k \Vert T^{*k}x \Vert < \infty \).

Notice now that

$$\begin{aligned} R^k \tilde{x} = R^k(x,0,0, \ldots ) = (T^kx, DT^{k-1}x, \ldots , Dx, 0,0, \ldots ).\end{aligned}$$

Therefore

$$\begin{aligned} \sum _{k=0}^n R^k \tilde{x} = \left( \sum _{k=0}^n T^kx, D\left( \sum _{k=0}^{n-1} T^kx\right) , D\left( \sum _{k=0}^{n-2} T^kx\right) , \ldots , Dx, 0,0, \ldots \right) .\end{aligned}$$

Hence, using the notation \(S_n(T) x = x + Tx + \dots + T^{n-1}x\), the \(o(\sqrt{n})\) condition

$$\begin{aligned}\left| \left| \sum _{k=0}^{n}R^k\tilde{x}\right| \right| = o(\sqrt{n})\end{aligned}$$

is equivalent to

$$\begin{aligned} \left| \left| \sum _{k=0}^{n}T^k x\right| \right| = o(\sqrt{n}) \quad \text {and} \quad \sum _{k=0}^{n} \Vert D(S_k(T)x)\Vert ^2 = o(n).\end{aligned}$$

The proof is now complete using the identity \(\Vert Du\Vert ^2 = \Vert u\Vert ^2 - \Vert Tu\Vert ^2\). \(\square \)

Corollary 1.6 follows from Theorem 1.5 and Kronecker’s lemma, already used in the proof of Theorem 1.4.

4 Coboundaries of the Doubling Map

Let \({{\,\textrm{val}\,}}_2(n)\) be the 2-valuation of n, that is

$$\begin{aligned}{{\,\textrm{val}\,}}_2(n) = k \quad \text {if} \quad n = m 2^k\quad \text {with}\quad m \notin 2 \mathbb {Z}.\end{aligned}$$

For \(n\in \mathbb {Z}\), we denote by \( \hat{f}(n) = \int _0^1 f(t)e^{-int} \, dt\) the n-th Fourier coefficient of \(f \in L^2 (0,1)\).

Corollary 4.1

Suppose f is a periodic function of period 1 such that \(f \in L^2 (0,1)\),

$$\begin{aligned} \int _0^1 f(t) \, dt = 0 \end{aligned}$$
(4.1)

and there exists \(\varepsilon >0\) such that

$$\begin{aligned} \sum _{n=-\infty }^\infty {{\,\textrm{val}\,}}_2 (n)^{4+\varepsilon } \left| \hat{f}(n)\right| ^2 < \infty . \end{aligned}$$
(4.2)

Then there is a function g in \(L^2(0,1)\) of period one such that

$$\begin{aligned} f(t) = g(t) - g(2t) \quad a.e. \end{aligned}$$

if and only if

$$\begin{aligned} \lim _{n\rightarrow \infty } \frac{1}{n} \int _0^1 \left| \sum _{i=0}^n f(2^i t)\right|^2 dt = 0\end{aligned}$$

Proof

We use Theorem 1.4 applied to the isometry \(T : L^2(0,1) \longrightarrow L^2(0,1)\) defined by

$$\begin{aligned} Tf (t) = f(2t), \quad t \in (0,1)\ \text {mod} \ 1.\end{aligned}$$

We first remark that condition (4.1) is justified by the fact that T acts as a shift operator on the subspace of \(L^2 (0,1)\) of functions whose zeroth Fourier coefficient vanishes. The condition

$$\begin{aligned} \lim _{n\rightarrow \infty } \frac{1}{n} \int _0^1 \left| \sum _{i=0}^n f(2^i t)\right|^2 dt = 0\end{aligned}$$

is exactly the condition

$$\begin{aligned} \lim _{n\rightarrow \infty } \frac{1}{n} \left\Vert\sum _{k= 0}^n T^k f \right\Vert^2 = 0,\end{aligned}$$

which appears in Theorem 1.4. We want to show that

$$\begin{aligned} \sum _{k=0}^\infty k \Vert T^{*k} f \Vert < \infty .\end{aligned}$$

Recall that T acts as a shift operator on the subspace of \(L^2 (0,1)\) of functions whose zeroth Fourier coefficient vanishes. Let \((a_n) = (\hat{f}(n))\) be the sequence of Fourier coefficients of f. We have \(a_0 = 0\). The iterates of the adjoint of T at f can be computed as

$$\begin{aligned} T^{*k}f (t) = \sum _{j=-\infty }^\infty a_{j 2^k} e^{2i\pi j t}, \quad k \in \mathbb {N}.\end{aligned}$$

For \(\varepsilon >0\), using the change \(n = j2^k\) in the order of summation, we get

$$\begin{aligned} \sum _{k=0}^\infty k \Vert T^{*k} f \Vert&= \sum _{k=0}^\infty k \left( \sum _{j=-\infty }^\infty | a_{j 2^k} |^2 \right)^{1/2} \\&= \sum _{k=0}^\infty k^{-(1+\varepsilon )/2} \left( k^{3+\varepsilon } \sum _{j=-\infty }^\infty | a_{j 2^k} |^2 \right)^{1/2} \\&\le \left( \sum _{k=0}^\infty k^{-(1+\varepsilon )} \right)^{1/2} \left( \sum _{k=0}^\infty k^{3+\varepsilon } \sum _{j=-\infty }^\infty |a_{j 2^k}|^2 \right)^{1/2} \\&= \left( \sum _{k=0}^\infty k^{-(1+\varepsilon )} \right)^{1/2} \left( \sum _{n =-\infty }^\infty |a_n|^2 \sum _{k=0}^{{{\,\textrm{val}\,}}_2(n)} k^{3+\varepsilon }\right)^{1/2} \\&\le 2 \left( \sum _{k=0}^\infty k^{-(1+\varepsilon )} \right)^{1/2} \left( \sum _{n =-\infty }^\infty {{\,\textrm{val}\,}}_2(n)^{4+\varepsilon } |a_n|^2 \right)^{1/2} . \end{aligned}$$

Thus, under our hypothesis about the Fourier coefficients, we have

$$\begin{aligned} \sum _{k=0}^\infty k \Vert T^{*k} f \Vert < \infty .\end{aligned}$$

\(\square \)

Corollary 4.2

[20] Let f be a periodic function of period 1 such that \(f \in L^2 (0,1)\),

$$\begin{aligned}\int _0^1 f(t) \, dt = 0\end{aligned}$$

and there exists \(\alpha > 0\) such that

$$\begin{aligned} \sum _{k =-\infty }^\infty |\hat{f}((2k+1))2^i|^2 = O (2^{-\alpha i}). \end{aligned}$$
(4.3)

Then there is a function g in \(L^2(0,1)\) of period one such that

$$\begin{aligned} f(t) = g(t) - g(2t) \quad a.e. \end{aligned}$$

if and only if

$$\begin{aligned} \lim _{n\rightarrow \infty } \frac{1}{n} \int _0^1 \left| \sum _{i=0}^n f(2^i t)\right|^2 dt = 0\end{aligned}$$

Proof

The result follows from Corollary 4.1 with \(\varepsilon = 1\), say. Indeed, using the condition (4.3), one can estimate

$$\begin{aligned} \sum _{n=-\infty }^\infty {{\,\textrm{val}\,}}_2 (n)^{5} \left| \hat{f}(n)\right| ^2&= \sum _{i=1}^{\infty }\sum _{k=-\infty }^\infty i^5|\hat{f}((2k+1))2^i|^2\\&\lesssim \sum _{i=1}^{\infty } \frac{i^5}{2^{\alpha i}} < \infty . \end{aligned}$$

\(\square \)

Remark 4.3

Condition (4.3) is condition (a) from Theorem 4 in [20]. It has been proved in [20] that each of other three conditions of Hölder type, called there (b), (c) and (d), implies the condition (4.3). Mark Kac has already considered in [12] the case when f is in the Hölder class \(C^{0,\alpha }\) for some \(\alpha > 1/2\). We refer to [5, 9, 10] for other contributions concerning the functional equation \(f(t) = g(t) - g(2t)\).

Remark 4.4

All the remarks at the end of the paper [20] apply also in our situation. In particular, the generalization to the functional equation \(f(t) = g(t) - g(nt)\) (for a fixed integer n) is immediate.