Sequences with bounded iterated sums

The answer is no: not every such double sequence is of the form you describe. I will prove the following things:

1. Different (complex) measures give rise to different (complex) double sequences;
2. The domain of definition is larger than the set of measures specified by OP;
3. There exist double sequences meeting the requirements which are not of the form (x) for any complex measure $\mu$ on $\mathbb{T}$.

The main results of the second and third part are listed in blockquote. The textbooks I occasionally refer to are listed at the very end of this post.

Part 0: Prerequisites and notation

• Let $\mathbb{N} = \{0,1,2,\ldots\}$ denote the natural numbers with zero.
• Let us denote the space of real or complex double sequences satisfying (1.) and (2.) by $\mathcal S_{\mathbb{R}}$ and $\mathcal S_{\mathbb{C}}$, respectively.
• For $k\in\mathbb{Z}$, let $\lVert \:\cdot\: \rVert_k : \mathcal S_{\mathbb{C}} \to \mathbb{R}_{\geq 0}$ denote the seminorm given by $$ \lVert \{a_{n,m}\}_{n,m\in\mathbb{Z}} \rVert_k \: := \: \sup_{m\in\mathbb{Z}} |a_{k,m}|. $$ It follows from OP's uniqueness argument that every individual $\lVert \:\cdot\: \rVert_k$ already defines a norm on $\mathcal S_{\mathbb{C}}$, but choosing just one would not give us the right topology on $\mathcal S_{\mathbb{C}}$. More on that later.
• Let $\mathbb{T}$ denote the complex unit circle.
• Let $C(\mathbb{T})$ denote the $C^*$-algebra of all continuous functions $\mathbb{T} \to \mathbb{C}$.
• Let $M(\mathbb{T})$ denote the complex vector space of all (complex) Borel measures on $\mathbb{T}$, which becomes a Banach space with the total variation norm $\lVert \mu\rVert := |\mu|(\mathbb{T})$. Since $\mathbb{T}$ is compact and metrisable, it follows from [Cohn, proposition 7.2.3] that every $\mu \in M(\mathbb{T})$ is automatically regular. Consequently, it follows from the Riesz representation theorem that $M(\mathbb{T})$ is isometrically isomorphic with the Banach space dual of $C(\mathbb{T})$, where the dual pairing is given by $\langle f,\mu\rangle = \int f \: d\mu$.
• Let $\mathcal D(A) \subseteq M(\mathbb{T})$ denote the subspace consisting of those measures for which the formula (x) is well defined, and let $A : \mathcal D(A) \to \mathcal S_{\mathbb{C}}$ denote the linear map given by (x).

Part 1: Injectivity of $A$

Before we determine the domain of definition of $A$, we show that $A$ is necessarily injective. For this we prove the following slightly stronger statement:

Proposition. The map $\phi^t : M(\mathbb{T}) \to \ell^\infty(\mathbb{Z})$ given by $\phi^t(\mu)_m = \int \lambda^m \: d\mu$ is injective.

First proof. Let $\mu,\nu \in M(\mathbb{T})$ be given with $\phi^t(\mu) = \phi^t(\nu)$. Now let $f_\mu,f_\nu \in C(\mathbb{T})^*$ denote the linear functionals associated with $\mu$ and $\nu$. By assumption, $f_\mu$ and $f_\nu$ agree on the functions $\lambda^m \in C(\mathbb{T})$ for all $m\in\mathbb{Z}$. By linearity, $f_\mu$ and $f_\nu$ agree on the linear subspace $V \subseteq C(\mathbb{T})$ spanned by all $\lambda^m$. But this subspace is self-adjoint (we have $\overline\lambda = \lambda^{-1}$ on the unit circle), contains the unit, and separates points, so it follows from the Stone–Weierstrass theorem that $V$ is dense in $C(\mathbb{T})$. Now $f_\mu$ and $f_\nu$ coincide on a dense subspace, so they must be equal. Consequently, we have $\mu = \nu$, showing that $\phi^t$ is injective. $\quad\Box$

Second proof. Observe that $\phi^t$ is the transpose of the Gelfand representation $\phi : (\ell^1(\mathbb{Z}),*) \to C(\mathbb{T})$; hence the notation $\phi^t$. It is well known that $\phi$ is injective (this is the same as saying that $(\ell^1(\mathbb{Z}),*)$ is semisimple), and the image of $\phi$ is the space of those continuous functions $\mathbb{T} \to \mathbb{C}$ which have an absolutely convergent Fourier series. As such, the image of $\phi$ is dense in $C(\mathbb{T})$, but $\phi$ is not surjective. This gives an alternative proof of the above proposition: we have $\ker(\phi^t) = \text{ran}(\phi)^\perp = 0$; cf. [Conway, proposition VI.1.8 and the bipolar theorem, V.1.8]. Another immediate consequence is that the image of $\phi^t$ is weak-* dense in $\ell^\infty(\mathbb{Z})$, but not weak-* closed or even norm closed. (For the final claim, see [Conway, proposition VI.1.9 and theorem VI.1.10].) $\quad\Box$

It follows that $A$ is also injective.

Part 2: What belongs to $\mathcal D(A)$?

The definition of $\mathcal D(A)$ is at present a little vague. This is in part due to the following technical difficulty: most textbooks on measure theory define integrals over signed or complex measures only for bounded functions, and consequently only define $L^p$ spaces over positive measures. However, there doesn't seem to be a serious problem; we may simply define $\mathscr{L}^1(\mu)$ for a complex measure $\mu$ by $$ \mathscr{L}^1(\mu) \: := \: \left\{f : \mathbb{T} \to \mathbb{C} \:\: \text{Borel measurable} \:\: \left| \:\:\: f\cdot \frac{d\nu}{d|\nu|} \in \mathscr{L}^1(|\mu|)\right.\right\}, $$ and the integral of a function $f \in \mathscr{L}^1(\mu)$ is given by $$ \int f\: d\mu \: := \: \int f\cdot\frac{d\nu}{d|\nu|}\: d|\nu|. $$ If $f$ is bounded, then this coincides with the definition of integration as given in [Cohn, section 4.1]; this is proven at the end of [Cohn, section 4.2]. Furthermore, it is shown that $\left|\frac{d\nu}{d|\nu|}\right|$ is $|\nu|$-almost everywhere equal to $1$, so we find $$ f \in \mathscr L^1(\mu) \quad \Longleftrightarrow \quad f \in \mathscr L^1(|\mu|). $$ Now it is clear that the domain of definition of the operator $A$ is $$ \mathcal D(A) \: = \: \left\{\mu \in M(\mathbb{T}) \:\: \left| \:\: \frac{1}{(1 - z)^n} \in \mathscr L^1(\mu) \:\: \text{for all $n\in\mathbb{N}$}\right.\right\}. $$

Main result. It is not so hard to see that $\mathcal D(A)$ is larger than the set conjectured by OP:

• There are complex measures $\mu\in M(\mathbb{T})$ for which $A(\mu)$ nonetheless defines a real double sequence. A concrete example is the measure given by $$ \mu(\{i\}) = \tfrac{1}{2i},\qquad \mu(\{-i\}) = -\tfrac{1}{2i},\qquad \mu(\mathbb{T} \setminus \{i,-i\}) = 0. $$ The double sequence it defines is given by $a_{n,m} = \text{Im}(i^{-m}(1 - i)^{-n})$. It seems that a complex measure $\mu \in M(\mathbb{T})$ gives rise to a real-valued double sequence if and only if $\mu(\overline X) = \overline{\mu(X)}$ holds for every Borel measurable set $X \subseteq \mathbb{T}$, whereas the real measures are given by the property $\overline{\mu(X)} = \mu(X)$.
• There are more serious counterexamples than the above. Let $\lambda \in M(\mathbb{T})$ denote the Lebesgue measure, and let us say that a positive function $f \in C(\mathbb{T})$ is of rapid decrease near $1$ if $f(z) > 0$ holds for all $z\in\mathbb{T} \setminus \{1\}$, and we have $$ \lim_{z \to 1} \frac{f(z)}{(1 - z)^n} \: = \: 0\tag*{(for all $n\in\mathbb{N}$).} $$ A concrete example of such a function is $f(z) = e^{-\frac{1}{|1 - z|}}$. If $f$ is any such a function, then the measure $\nu(X) := \int_X f \: d\lambda$ belongs to $\mathcal D(A)$, for we have $\frac{f(z)}{(1 - z)^n} \in C(\mathbb{T})$ for all $n\in\mathbb{N}$. However, the resulting measure $\nu$ always has $1$ in its support¹, so it is not of the form described by OP.

¹: See [Cohn, section 7.4] for the definition of the support of a regular Borel measure on a locally compact Hausdorff space.

Put in another way: in order for OP's equation (x) to be well defined, it is not necessary that $(1 - z)^{-1}$ is $\mu$-a.e. bounded. It suffices that $(1 - z)^{-n}$ is $\mu$-integrable for all $n \in\mathbb{N}$.

Part 3: $A$ is not surjective

It turns out that, even with the domain of definition stretched as far as possible, the map $A$ is not surjective, so there exist double sequences which are not of the form (x) for any measure $\mu$ on $\mathbb{T}$. We sketch the argument involved. (A full proof was given in a previous version of this answer, but edited out for being way too long.)

Recall from the second proof in part 1 that the natural map $\phi^t : M(\mathbb{T}) \to \ell^\infty(\mathbb{Z})$ fails to be an open mapping. It turns out that a similar statement is true for the map $A : \mathcal D(A) \to \mathcal S_{\mathbb{C}}$. To that end, note that the following (infinite) diagram commutes: $$\begin{array}{ccccccccc} \cdots & \stackrel{D}{\longleftarrow} & M(\mathbb{T}) & \stackrel{D}{\longleftarrow} & M(\mathbb{T}) & \stackrel{D}{\longleftarrow} & M(\mathbb{T}) & \stackrel{D}{\longleftarrow} & \cdots \\[1ex] & & \bigg\downarrow{\phi^t} & & \bigg\downarrow{\phi^t} & & \bigg\downarrow{\phi^t} & & \\ \cdots & \stackrel{\Delta}{\longleftarrow} & \ell^\infty(\mathbb{Z}) & \stackrel{\Delta}{\longleftarrow} & \ell^\infty(\mathbb{Z}) & \stackrel{\Delta}{\longleftarrow} & \ell^\infty(\mathbb{Z}) & \stackrel{\Delta}{\longleftarrow} & \cdots \end{array}\tag*{$(*)$}$$ Here $\Delta : \ell^\infty(\mathbb{Z}) \to \ell^\infty(\mathbb{Z})$ denotes the discrete backwards difference, and likewise $D : M(\mathbb{T}) \to M(\mathbb{T})$ denotes the corresponding "downward" map that takes a measure $\mu$ and multiplies is by the function $(1 - z)$, or to put it more formally, that turns $\mu$ into the measure $\mu'$ given by $$ \mu'(X) \: := \: \int_X (1 - z)\cdot \frac{d\mu}{d|\mu|} \: d|\mu|. $$ (Note: in $(*)$, one should not take the maps in the other direction, for they are not continuous!)

The spaces $\mathcal D(A)$ and $\mathcal S_{\mathbb{C}}$ can be considered as the inverse (or projective) limits of the $\mathbb{Z}$-indexed systems defined by the rows of $(*)$, and the corresponding inverse limit topologies turn them into Fréchet spaces. Furthermore, one can show that not only $\phi^t : M(\mathbb{T}) \to \ell^\infty(\mathbb{Z})$ but also $A : \mathcal D(A) \to \mathcal S_{\mathbb{C}}$ fails to be an open mapping. Therefore in particular it cannot be surjective (by the open mapping theorem).

References

• [Cohn]: Donald L. Cohn, Measure Theory, second edition (2013), Birkäuser Advanced Texts, Birkhäuser, Basel.
• [Conway]: John B. Conway, A Course in Functional Analysis, second edition, third printing (2007), Springer Graduate Texts in Mathematics 96, Springer-Verlag, New York.

I think that there are solutions which are not of the form (x). I only use the following lemma (see Sequence with bounded sums for an explicit construction).

Lemma : Let us consider the operator $T : \mathbf{a} \in \mathbb{R}^{\mathbb{N}} \mapsto \big(\sum \limits_{k=0}^n a_k\big)_{n \ge 0}$. There exists $\mathbf{a} \in \mathbb{R}^{\mathbb{N}}$ such that for all $p \in \mathbb{N}$, $T^p \big(\mathbf{a}\big)$ is bounded, and $\sup_m |\big(T^p \mathbf{a}\big)_m| \ge p^p$.

If you have such a sequence, which we call $(b_n)$, you can define a double sequence $(a_{n,m})$ including $(b_n)$. It is easy to see that it is enough to define only $(a_{0,m})_{m \in \mathbb{Z}}$, and $(a_{n,0})_{n \ge 1}$.

Let $(a_{n,m})_{(n,m) \in \mathbb{Z}^2}$ be the only sequence satisfying $(1)$ and such that :

• for $m > 0$, $a_{0,m} = b_m$

• for $n \ge 0$, $a_{n,0} = b_0$

• for $m < 0$, $a_{0,m} = 0$.

As you pointed out in the comments, as $(a_{0,m})_{m \in \mathbb{Z}}$ is bounded, for all $n < 0$, $(a_{n,m})_{m \in \mathbb{Z}}$ is bounded (proof by induction).

Then, it is easy to prove by induction that for $p \ge 0$, $(a_{p,m})_{m \in \mathbb{N}} = T^p \big( (b_m)_{m \in \mathbb{N}} \big)$, so $(a_{p,m})_{m \in \mathbb{N}}$ is bounded, by definition of $(b_m)$.

Finally, by induction on $m$, it is easy to prove that for all $m < 0$, $\forall n \ge 0,\ a_{n,m} = 0$ (the lower-right quarter contains only zeroes). Thus we conclude that for all $n \ge 0$, $(a_{n,m})_{m \in \mathbb{Z}}$ is bounded. Hence

$ $

$$(a_{n,m})_{(n,m) \in \mathbb{Z}^2} \mbox{ satisfies } (1) \mbox{ and } (2).$$

To conclude, note that as mathworker21 pointed out, if $(a_{n,m})$ were of the form (x), we'd have $\sup_n \big( \sup_m |a_{n,m}| \big)^{1/n} < \infty$. However, by definition of $(b_n)$, $\big( \sup_m |a_{n,m}| \big)^{1/n} \ge n$.

Hence $(a_{n,m})$ is not of the form (x).

$ $

Remark : I can change a bit the lemma to conclude that the space of double sequences of the form (x) has a complement of uncountable dimension in the space of double sequences satisfying $(1)$ and $(2)$.