Discrete convolution equation

Question

Let $x_1 = (x_1^k)_{k =-\infty}^{+\infty}$, $x_2 = (x_2^k)_{k=-\infty}^{+\infty}$, $x_3 = (x_3^k)_{k=-\infty}^{+\infty}$ be three sequences of real numbers such that $x_j^k = 0$ for $k < -m_j < 0$. Then the convolution $x_i * x_j$ is correctly defined: $$ [x_i * x_j]^k = \sum_{k_1 + k_2 = k} x_i^{k_1} x_j^{k_2}. $$ Suppose that $x_i^k$ are known for $k \geq 0$ and are unknown for $k < 0$ and consider the following convolution-type equation: $$ x_3 + A \cdot x_1 * x_2 + B \cdot x_1 * x_1 * x_1 = 0 $$ with known $A$ and $B$. My question is whether there is some literature considering the equations of this type? What is the natural approach to solve such equations? Of course we can consider this equation as an infinite system of the second order algebraic equations but in such a way it is too difficult for me to obtain the conditions when the solution to this problem exists and is unique.

Answer 1

One of natural approaches would be to work with the spaces $\ell_p$, $p\in [1,\infty]$, because the Young's inequality works:

$$\|x\ast y,\ell_r\|\le \|x,\ell_p\|\|y,\ell_q\|,\quad p,q,r\in [1,\infty],\quad \frac 1p+\frac 1q=1+\frac 1r.$$

For example, if you have the equation of the type (it arises in theory of difference equations, for example) $$x=B\ast x+f,$$ then for sufficiently small norm of $B$ and for all $f\in\ell_p$ you have the uniqueness of the solution $x\in\ell_p$. You just define a bounded linear operator $\mathcal B:\ell_p\to\ell_p$, $\mathcal By=B\ast y$. Clearly, $\|\mathcal B\|\le \|B,\ell_1\|$. Then if $ \|B,\ell_1\|<1$ you write $x=(Id-\mathcal B)^{-1}f$.

However, the explicit forms of solutions, their support, etc can be a much trickier question.