Generating functions over $\mathbb{Z}$ Let $(a_n)_{n \in \mathbb{Z}}$ be a sequence such that both limits $\lim_{n \to \infty} a_n$ and $\lim_{n \to -\infty} a_n$ exists. Consider recursive relation
$$
2b_n - \frac{1}{2}(b_{n-1} + b_{n+1}) = a_n, \qquad n \in \mathbb{Z}.
$$
Using the Banach fixed point theorem I can show that there exists unique sequence $(b_n)_{n \in \mathbb{Z}}$ satisfying these equations (and in addition with limits at $\pm\infty$), but how can I find ''explicit'' (probably in the form $\sum_\mathbb{Z} \alpha_n a_n$) formula for $b_n$?
Typically, I would use generating functions method, but I am not sure how to do this in case sequences are both-side infinite.
Of course by the Banach theorem I know that
$$
(b_n)_n = \sum_{i = 0}^\infty T^i(a_n)_n,
$$
where $T(a_n)_n = \frac{1}{4} (a_{n-1} + a_{n+1})_n$, but I do not know how to go any further.
 A: You could use a formal series to get a generating function relationship:
$$A(x) = \sum_{n=-\infty}^\infty a_n x^n,\ \ B(x) = \sum_{n=-\infty}^\infty b_n x^n$$
Then $$A(x) = \sum_{n=-\infty}^\infty \left(2b_n - \frac12 (b_{n-1}+b_{n+1})\right) x^n = 2B(x)-\frac12 xB(x) - \frac12 x^{-1}B(x) = B(x) \left(2 - \frac{x}{2} - \frac{1}{2x}\right) = B(x) \left( \frac{-x^2+ 4x - 1}{2x} \right)$$
Here we have $A(x) = B(x) H(x)$ where $H$ is the transfer function associated with the transformation from $(a_n)$ to $(b_n)$.
From here we may write $$B(x) = \frac{2x}{-x^2 + 4x - 1}A(x)$$
Note that the roots of the denominator are $2\pm \sqrt{3}$ so we can find (via partial fraction decomposition) $R$ and $S$ for which:
$$B(x) = \left( \frac{R}{\left( 1 - \frac{x}{2-\sqrt{3}} \right)} + \frac{S}{\left( 1 - \frac{x}{2+\sqrt{3}} \right)}\right)A(x)$$ $$= \left[\sum_{n=0}^\infty \left( R\left( \frac{1}{2-\sqrt{3}}\right)^n + S \left( \frac{1}{2+\sqrt{3}} \right)^n \right)x^n \right]A(x)$$
Finally we use the convolution product to find $b_k$:
$$b_k = \sum_{n=0}^\infty \left( R\left( \frac{1}{2-\sqrt{3}}\right)^n + S \left( \frac{1}{2+\sqrt{3}} \right)^n \right) a_{k-n}.$$
