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Consider double sequences $a_{n,m}\in\mathbb R$ where $n,m\in\mathbb Z,$ satisfying

  1. $a_{n,m}=a_{n-1,m}+a_{n,m-1}$ for all $n,m\in\mathbb Z,$ and
  2. $\sup_\limits{m\in\mathbb Z}|a_{n,m}|<\infty$ for all $n\in\mathbb Z.$

An example solution is $a_{n,m}=(-1)^m2^{-n}.$ A more general solution is $$a_{n,m}=\int z^{-m}(1-z)^{-n} d\mu(z),\tag{x}$$ for a finite signed measure $\mu$ supported on an arc of the unit circle $\{\exp(2\pi i\theta)\mid \theta\in[\epsilon,2\pi-\epsilon]\}$ with $\epsilon>0.$ I am curious if there is a characterization, but to make a specific question:

Is there a solution of (1.) and (2.) not of the form (x)?


My thoughts:

  • Decreasing $n$ is taking the discrete backwards difference. Increasing $n$ is like choosing a discrete integral.
  • Given a row $(a_{n,m})_{m\in\mathbb Z}$ such that $\sup_{m}|a_{n,m}|$ is finite, the previous row is always ok: $\sup_{m}|a_{n-1,m}|$ is automatically finite because $a_{n-1,m}=a_{n,m}-a_{n,m-1}$
  • Any row $(a_{n,m})_{m\in\mathbb Z}$ determines $(a_{n,m})_{n,m\in\mathbb Z}$ uniquely if a solution exists. To prove this, by linearity it suffices to show there is a unique solution of conditions 1 and 2 satisfying $a_{0,m}=0$ for all $m.$ Condition 1 gives $a_{1,m}=a_{1,m-1}$ so $a_{1,m}=a_{1,0}$ for all $m.$ This then gives $a_{2,m}=a_{1,0}+a_{2,m-1},$ so $a_{2,m}=a_{1,0}m+a_{2,0}$ for all $m.$ Condition 2 with $n=2$ forces $a_{1,0}=0,$ so $a_{1,m}=0$ for all $m.$ By induction, we get $a_{n,m}=0$ for all $n\geq 0.$ And $a_{n-1,m}=a_{n,m}-a_{n,m-1}$ implies $a_{n,m}=0$ for all $n=-1$ and hence all negative $n$ by induction.
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  • $\begingroup$ Can you explain why "row $N+2$ is determined up to an affine function"? $\endgroup$ Dec 30, 2017 at 23:57
  • $\begingroup$ Also, it might be interesting to think about how the model, associating to each solution a measure $\mu$, behaves under limits. Pointwise convergence in $\mathbb{Z}^2$ corresponds to weak* convergence in measures. For example, if there are solutions with corresponding measures $\mu$ which have supports tending toward all of $[0,2\pi)$, and the solutions have a limit, then we get a contradiction. Similarly, if the corresponding measures to the solutions increase to infinity in norm $||\mu||$ (since any weakly convergent sequence must be bounded in norm). $\endgroup$ Dec 31, 2017 at 0:00
  • $\begingroup$ We can partially ensure that a sequence of solutions has a limit by insisting the 0th row is contained in $[0,1]$ (so that we can appeal to compactness), and then by what you said, the row 0 - solution would possibly generate a whole solution. But I'm not convinced of your claim of that yet... $\endgroup$ Dec 31, 2017 at 0:01
  • $\begingroup$ @mathworker21: Thanks for your interest. I've reworded that point to make it clearer. Note it's just uniqueness for given boundary conditions $a_{0,m},$ not existence. I don't think weak* convergence of measures on $[0,2\pi)$ is enough - the integrand isn't bounded. For example $\int_{|z|=1} \max(0,|1-z|-\epsilon)(1-z)^{-2} dz$ won't converge as $\epsilon\to 0.$ $\endgroup$
    – Dap
    Dec 31, 2017 at 9:37
  • $\begingroup$ Thanks for the rewording. I'm confused by the second part of your comment though. What I was saying is that if $\int z^{-m}(1-z)^{-n} d\mu_k \to \int z^{-m}(1-z)^{-n}d\mu$ as $k \to \infty$ for all $n,m$, then this is weak* convergence (by Stone Weierstrass probably). So if we have solutions $(a_{n,m}^{(k)})_{n,m}$ for $k=1,2,\dots$ with corresponding measures $\mu_1,\mu_2,\dots,\mu_k,\dots$, then pointwise convergence of $a_{n,m}^{(k)} \to a_{n,m}$ corresponds to weak* convergence of $\mu_k$ to $\mu$, where $\mu$ is the measure associated to $(a_{n,m})_{n,m}$. $\endgroup$ Dec 31, 2017 at 11:05

4 Answers 4

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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.
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  • $\begingroup$ nice cutdown! :) $\endgroup$ Feb 18, 2020 at 1:49
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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 $(2)$.

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    $\begingroup$ Great solution! Everything looks right, though I haven't checked the Lemma yet. $\endgroup$ Feb 5, 2018 at 18:38
  • $\begingroup$ @mathworker21 Thanks ! I will post a proof of it soon $\endgroup$
    – charmd
    Feb 5, 2018 at 20:42
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    $\begingroup$ @EwanDelanoy I think that part of his solution is fine. Can you be more precise what you mean by the upper left quarter? As I see it, all rows $n < 0$ are fine as he said, and for rows $n \ge 0$ and columns $m < 0$, you get $0$. And the nontrivial case is $n \ge 0$ and $m \ge 0$, which is handled by the Lemma. $\endgroup$ Feb 6, 2018 at 6:51
  • $\begingroup$ @mathworker21 You're right, my mistake. Thx for the explanation $\endgroup$ Feb 6, 2018 at 8:00
  • $\begingroup$ This is a correct answer, but I chose to accept Josse van Dobben de Bruyn's answer because I found their counterexamples particularly instructive. $\endgroup$
    – Dap
    Feb 9, 2018 at 8:41
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As a by-product of my other answer, I found a neat topological argument why there must be more than conjectured by OP.

Prerequisites and notation.

  • Let $\mathbb{N} = \{0,1,2,\ldots\}$ denote the natural numbers with zero.
  • Once again let $\mathcal S_{\mathbb{R}}$ denote the space of all double sequences satisfying (1.) and (2.), and let $\tau_{\mathcal S}$ be the locally convex topology given by the family of seminorms $$ \left\lVert \{a_{n,m}\}_{n,m\in\mathbb{Z}} \right\rVert_k \: := \: \sup_{m\in\mathbb{Z}} |a_{k,m}|\tag*{($k\in\mathbb{Z}$).} $$ As I pointed out in my other answer, this topology $\tau_{\mathcal S}$ is compatible with a complete invariant metric, for instance $$ d\big(\{a_{n,m}\},\{b_{n,m}\}\big) \: := \: \sum_{k\in\mathbb{Z}} \frac{1}{2^{|k|}} \cdot \frac{\lVert \{a_{n,m}\} - \{b_{n,m}\}\rVert_k}{1 + \lVert \{a_{n,m}\} - \{b_{n,m}\}\rVert_k}. $$ As such, $(\mathcal S_{\mathbb{R}},\tau_{\mathcal S})$ is a Fréchet space.
  • Like in mathworker21's answer, for $\varepsilon \in (0,\pi)$ we let $\mathbb{T}_\varepsilon\subseteq \mathbb{T}$ denote the arc $$ \mathbb{T}_\varepsilon \: := \: \{e^{\theta i} \: : \: \theta \in [\varepsilon,2\pi-\varepsilon]\}. $$
  • Like in my other answer, let $M(\mathbb{T})$ denote the space of all complex Borel measures on $\mathbb{T}$.
  • Let $X \subseteq M(\mathbb{T})$ denote the set of all measures $\mu \in M(\mathbb{T})$ supported on a subset of some $\mathbb{T}_\varepsilon$ and such that the double sequence defined by OP's formula (x) is real-valued.
  • Let $A : X \to \mathcal S_{\mathbb{R}}$ denote the map defined by OP's formula (x). (Warning: this is not exactly the same as the map $A$ from my other answer. The domain is smaller, and therefore the range is also smaller.)

Solution.

By definition, $\tau_S$ is the smallest topology making all of the seminorms $\lVert \:\cdot\: \rVert_k$ continuous. Hence, for any sequence $\{C_k\}_{k=0}^\infty$ of positive real numbers, the set $$ \mathcal V\left(\{C_k\}_{k=0}^\infty\right) \: := \: \big\{\{a_{n,m}\}_{n,m\in\mathbb{Z}} \in \mathcal S_{\mathbb{R}} \: : \: \lVert \{a_{n,m}\}_{n,m\in\mathbb{Z}} \rVert_k \leq C_k \ \ \text{for all $k\in\mathbb{N}$} \big\} $$ is closed, being the intersection over all $k\in\mathbb{N}$ of the inverse image of the closed interval $[-C_k,C_k]$ under the continuous map $\lVert\:\cdot\:\rVert_k$. We shall be interested in the sequence $\{W_n\}_{n=1}^\infty$ of closed sets given by $$ W_n \: := \: \mathcal V\left(\left\{n\cdot \left|\frac{1}{1 - e^{i/n}}\right|^k\right\}_{k=0}^\infty\right). $$


Claim 1. We have $$ \text{ran}(A) \: \subseteq \: \bigcup_{n=1}^\infty W_n. $$ Proof. Let $\mu \in X$ be given. Suppose that $\mu$ is supported on $\mathbb{T}_\varepsilon$, then by Hölder's inequality we have $$ \hspace{-18mm}\lVert A(\mu)\rVert_k \: \leq \: \left\lVert \left(\frac{1}{1 - z}\right)^k \right\rVert_{\mathscr{L}^1(|\mu|)} \: \leq \: \lVert 1\rVert_{\mathscr{L}^1(|\mu|)} \cdot \left\lVert \left(\frac{1}{1 - z}\right)^k\right\rVert_{\mathscr{L}^\infty(|\mu|)} \: \leq \: \lVert \mu\rVert \cdot \left|\frac{1}{1 - e^{i\varepsilon}}\right|^k\tag*{($k\in\mathbb{N}$).} $$ Thus, if we take $n := \left\lceil \max(\lVert \mu\rVert,\frac{1}{\varepsilon}) \right\rceil$, then we certainly have $A(\mu) \in W_n$.$\quad\Box$


Claim 2. Every $W_n$ has empty interior.

Proof. The interior of a convex, balanced set is again convex and balanced. Hence, if $W_n$ should have non-empty interior, then in particular $0$ will be an interior point. This is however not the case: pick some $\varepsilon \in (0,\frac{1}{n})$ and let $\mu_\varepsilon \in X \subseteq M(\mathbb{T})$ be the measure defined by $$ \mu_\varepsilon\left(\{e^{i\varepsilon}\}\right) = \tfrac{1}{2}, \qquad \mu_\varepsilon\left(\{e^{-i\varepsilon}\}\right) = \tfrac{1}{2},\qquad \mu_\varepsilon\left(\mathbb{T} \setminus \{e^{i\varepsilon},e^{-i\varepsilon}\}\right) = 0. $$ The sequence $\{\frac{1}{k} A(\mu_\varepsilon)\}_{k=1}^\infty$ converges to zero in $\mathcal S_{\mathbb{R}}$, but no element of this sequence lies in $W_n$.$\quad\Box$


Conclusion. Now the result follows from the Baire category theorem: each $W_n$ is closed and has empty interior, so their union also has empty interior. It follows that $\text{ran}(A)$ has empty interior as well, so in particular it is not the whole space $\mathcal S_{\mathbb{R}}$.


(Observe the important role played by completeness in both my solutions. This comes to show some of the power of the theory of complete metric spaces.)

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I wanted to post an answer just containing what is needed to answer the OP's question. This answer is due to Josse van Dobben de Bruyn and not me.

Yes, there is a solution not of the form (x). Define the measure $\mu$ on $\mathbb{T} := \{z \in \mathbb{C} : |z| = 1\}$ by $d\mu = e^{-\frac{1}{|z-1|}}dz$, and let $a_{n,m} = \int_{\mathbb{T}} z^{-m}(1-z)^{-n}d\mu(z)$. Since $\mu$ decays rapidly enough near $1$, $a_{n,m}$ is defined for each $n,m$; in fact, the rapid decay implies $\sup_m |a_{n,m}| \le \int_{\mathbb{T}} |1-z|^{-n}d\mu(z) < \infty$ for each $n$, verifying (2). Also, of course (1) is satisfied. It remains to show $(a_{n,m})_{n,m}$ is not of the form (x). If it were, then we'd have $\int_{\mathbb{T}} z^{-m}d\nu(z) = \int_{\mathbb{T}} z^{-m}d\mu(z)$ for each $m \in \mathbb{Z}$, for some finite, signed $\nu$ supported away from $1$. This means $\nu$ and $\mu$ have the same fourier coefficients and are thus equal, but $\mu$ is supported at/near $1$.

This question seems very trivial in hindsight...

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  • $\begingroup$ Haha, fair enough, I guess my answer is much longer than needed. I've been meaning to trim it down a bit — thanks for reminding me. ☺ $\endgroup$ Feb 17, 2020 at 18:08

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