A sequence with uniformly bounded second-variation Let $\left(a_k\right)_{k\in\mathbb{Z}}$ be a bounded bi-sequence of nonnegative real numbers, indexed by the integers $\mathbb{Z}$. Assume that for all $n\in\mathbb{Z}$,
$$v_{n}:=\left(a_{k-n}-a_{k}\right)_{k\in\mathbb{Z}}\in l^{2}\left(\mathbb{Z}\right)$$
and, moreover,
$$\sup_{n\in\mathbb{Z}}\left\Vert v_{n}\right\Vert _{l^{2}\left(\mathbb{Z}\right)}<\infty.$$
My question is

Does it follow that $a:=\lim_{\left|k\right|\to\infty}a_k$ exists and further $\sum_{k\in\mathbb{Z}}\left|a_{k}-a\right|^{2}<\infty$?

Let me make two remarks:

*

*By $\lim_{\left|k\right|\to\infty}a_k$ I mean that both limits $\lim_{k\to\infty}a_{k}$ and $\lim_{k\to-\infty}a_{k}$ exist and are equal. In case that both $\lim_{k\to\infty}a_{k}$ and $\lim_{k\to-\infty}a_{k}$ exist and are different, I think that it is not hard to see that $\sup_{n\in\mathbb{Z}}\left\Vert v_{n}\right\Vert _{l^{2}\left(\mathbb{Z}\right)}=\infty.$ Thus, the assumption in the question rules out this case. The bigger problem for me is to rule out cases when at least one of these limits does not exist.

*I do not have any idea what approach should be taken toward such question, either to prove or to disprove. Yet, let me put this question in a context that might be useful.
It is a basic fact that a sequence $\left(a_k\right)_{k\in\mathbb{Z}}$ of a bounded variation, that is $\sum_{k\in\mathbb{Z}}\left|a_{k}-a_{k-1}\right|<\infty$, is a Cauchy sequence so it converges. On the other hand, it is not hard to see that if we only have that $\sum_{k\in\mathbb{Z}}\left|a_{k}-a_{k-1}\right|^2<\infty$ then this fails and the sequence can diverge. In fact, in this case one can see that for each $n\in\mathbb{Z}$ it holds that
$$\left\Vert \left(a_{k-n}-a_{k}\right)_{k\in\mathbb{Z}}\right\Vert _{l^{2}\left(\mathbb{Z}\right)}^{2}\leq\left|n\right|\cdot\left\Vert \left(a_{k-1}-a_{k}\right)_{k\in\mathbb{Z}}\right\Vert _{l^{2}\left(\mathbb{Z}\right)}^{2}<\infty.$$
But, of course, we see that this quantity may not be uniformly bounded. My question regards the more restrictive assumption when $\left\Vert \left(a_{k-n}-a_{k}\right)_{k\in\mathbb{Z}}\right\Vert _{l^{2}\left(\mathbb{Z}\right)}^{2}$ is uniformly bounded in $n\in\mathbb{Z}$.

 A: Indeed such a sequence must be of the form $a + b_k$ with $(b_k) \in \ell^2(\mathbb{Z})$. We don't need the non-negativity assumption, $(a_k)$ can be any complex bi-sequence whose $n^{\text{th}}$ differences form a bounded set in $\ell^2(\mathbb{Z})$.
On the space of all complex bi-sequences define the translation operator $\tau$ by $(\tau a)_k = a_{k-1}$. We then have $v_n = \tau^na - a$ and
$$v_n = \sum_{\nu = 0}^{n-1} \tau^{\nu}v_1 \tag{1}$$
for $n > 0$ (we need only consider positive $n$, but since $\tau^n v_{-n} = - v_n$ it is the same to demand that $\{v_n : n \in \mathbb{Z}\}$ is bounded in $\ell^2(\mathbb{Z})$ as to demand that $\{v_n : n > 0\}$ is).
The $v_n$ belong to $\ell^2(\mathbb{Z})$, hence each is the bi-sequence of the Fourier coefficients of some $V_n \in L^2([-\pi,\pi])$. For such bi-sequences, the translation $\tau$ corresponds to multiplying the function with $e^{it}$, thus $(1)$ becomes
$$V_n(t) = \sum_{\nu = 0}^{n-1} e^{i\nu t}V_1(t) = \frac{e^{int} - 1}{e^{it}-1}\cdot V_1(t) \tag{2}$$
(almost everywhere) for every $n > 0$. Since $\lVert \hat{f}\rVert_{\ell^2(\mathbb{Z})} = \lVert f\rVert_{L^2([-\pi,\pi])}$, the family $\{V_n : n > 0\}$ is bounded, i.e. there is a $C \in [0, + \infty)$ such that
$$\int_{-\pi}^{\pi} \biggl\lvert \frac{e^{int}-1}{e^{it}-1}\biggr\rvert^2\cdot \lvert V_1(t)\rvert^2\,dt \leqslant C \tag{3}$$
for all $n$. Now $\lvert e^{ix} - 1\rvert = 2\lvert \sin \frac{x}{2}\rvert$ for $x \in \mathbb{R}$, hence we can write $(3)$ as
$$\int_{-\pi}^{\pi} \frac{\lvert V_1(t)\rvert^2}{\sin^2 \frac{t}{2}}\cdot \sin^2 \biggl(\frac{nt}{2}\biggr)\,dt \leqslant C \tag{4}$$
for all $n > 0$. A fortiori, for every $\delta \in (0,\pi)$ we have
$$\int_{\delta \leqslant \lvert t\rvert \leqslant \pi}\frac{\lvert V_1(t)\rvert^2}{\sin^2 \frac{t}{2}}\cdot \sin^2 \biggl(\frac{nt}{2}\biggr)\,dt \leqslant C\,.$$
Using $\sin^2 x = \frac{1}{2}(1 - \cos (2x))$ and the fact that $\frac{\lvert V_1(t)\rvert^2}{\sin^2 \frac{t}{2}}$ is integrable on $[-\pi, -\delta] \cup [\delta, \pi]$ since $\frac{1}{\sin^2 \frac{t}{2}}$ is bounded there, the Riemann-Lebesgue lemma yields
$$\int_{\delta \leqslant \lvert t\rvert \leqslant \pi} \frac{\lvert V_1(t)\rvert^2}{\sin^2 \frac{t}{2}}\,dt \leqslant 2C$$
for all $n > 0$ and all $\delta \in (0,\pi)$. Thus, by monotone convergence it follows that $\frac{V_1(t)}{\sin \frac{t}{2}} \in L^2([-\pi,\pi])$. Multiplying with the measurable function $\frac{1}{2ie^{it/2}}$ of constant (non-zero) modulus we see that
$$f(t) = \frac{V_1(t)}{e^{it}-1} \in L^2([-\pi,\pi])\,.$$
Now computing the Fourier coefficients of $V_1(t) = (e^{it}-1)f(t)$ yields
$$a_{k-1} - a_k = (v_1)_k = \hat{f}(k-1) - \hat{f}(k)\,. \tag{5}$$
But rerranging $(5)$ shows that $a_k - \hat{f}(k)$ is constant, call its value $a$. Thus indeed
$$a_k = a + \hat{f}(k)$$
for all $k \in \mathbb{Z}$.
