Let $\ell^p (\Bbb Z) = \left \{\left \{x_n\right \}_{n \in \Bbb Z}\ :\ x_n \in \Bbb C\ \text {and}\ \displaystyle {\lim\limits_{N \to \infty} \sum\limits_{n = -N}^{N} {\left \lvert x_n \right \rvert}^{p} \lt \infty} \right \}$ for $p \in [1, \infty).$ Let $\{x_n\}_{n \in \Bbb Z}$ and $\{y_n\}_{n \in \Bbb Z}$ be two elements of $\ell^1(\Bbb Z).$
(a) Prove that $\displaystyle {\lim\limits_{N \to \infty} \sum\limits_{m = -N}^{N} x_{n-m}\ y_m}$ exists for every $n \in \Bbb Z.$
(b) If $z_n = \displaystyle {\lim\limits_{N \to \infty} \sum\limits_{m = -N}^{N} x_{n-m}\ y_m,}$ then prove that $\{z_n \}_{n \in \Bbb Z} \in \ell^1(\Bbb Z).$
(c) Conclude that $\{z_n\}_{n \in \Bbb Z} \in \ell^p (\Bbb Z)$ for all $p \in (1,\infty).$
My attempt $:$ First I will show that if $\{x_m\}_{m \in \Bbb Z} \in \ell^1(\Bbb Z)$ then $\{x_{n-m}\}_{m \in \Bbb Z} \in \ell^1(\Bbb Z),$ for every $n \in \Bbb Z.$ Now for $n \geq 1$ and for $N \gt n$ we have $$\sum\limits_{m = -N}^{N} |x_{n-m}| = \sum\limits_{m = -(N-n)}^{N + n} |x_m| = \sum\limits_{m = -(N-n)}^{N-n} |x_m| + \sum\limits_{m = N-n}^{N+n} |x_m|.\ \ \ \ \ \ \ \ \ \ (1)$$ Since $ \displaystyle {\lim\limits_{N \to \infty} \sum\limits_{n = -N}^{N} {\left \lvert x_n \right \rvert} \lt \infty}$ it follows that $\lim\limits_{N \to \infty} \displaystyle \sum\limits_{m = -(N-n)}^{N-n} |x_m| = \lim\limits_{N \to \infty} \displaystyle {\sum\limits_{m = -N}^{N} |x_m| \lt \infty}$ and $\lim\limits_{N \to \infty} \displaystyle {\sum\limits_{n = 1}^{N} |x_n|} \lt \infty$ which implies $\displaystyle{\sum\limits_{n=1}^{\infty} |x_n| \lt \infty.}$ Hence by Cauchy criterion we have $\displaystyle\sum\limits_{m = N-n}^{N+n} |x_m| \longrightarrow 0$ as $N \to \infty.$ So by $(1)$ we have $\lim\limits_{N \to \infty} \displaystyle{\sum\limits_{m = - N}^{N} |x_{n-m}|} = \lim\limits_{N \to \infty} \displaystyle {\sum\limits_{n = -N}^{N} |x_m| \lt \infty,}$ for all $n \geq 1.$ Similar proof holds for $n \leq -1.$ This proves our claim.
Now by Cauchy-Schwarz's inequality we have for any $n \in \Bbb Z$ $$\left ( \displaystyle {\sum\limits_{m = -N}^{N} |x_{n-m}\ y_m|} \right )^2 \leq \left ( \displaystyle {\sum\limits_{m = -N}^{N} |x_{n-m}|^2} \right ) \left ( \displaystyle {\sum\limits_{n=-N}^{N} |y_n|^2} \right ).\ \ \ \ \ \ \ \ \ \ (2)$$ Since $\displaystyle {\lim\limits_{N \to \infty} \sum\limits_{m = -N}^{N} |x_{n-m}| \lt \infty}$ and $\displaystyle {\lim\limits_{N \to \infty} \sum\limits_{m = -N}^{N} y_m} \lt \infty$ it follows that the series $\displaystyle {\sum\limits_{m = 1}^{\infty} |x_{n-m}|}, \displaystyle {\sum\limits_{m = 0}^{\infty} |x_{n+m}|} \lt \infty$ and $\displaystyle {\sum\limits_{m = 1}^{\infty} |y_m|}, \displaystyle {\sum\limits_{m = 0}^{\infty} |y_{-m}|} \lt \infty.$ So there exists $M(n) \in \Bbb N$ such that $|x_{n-m}| , |x_{n+m}| ,|y_m|, |y_{-m}| \lt 1,$ for all $m \gt M(n).$ Let $S_{M(n)} = \displaystyle {\sum\limits_{m = - M(n)}^{M(n)} |x_{n-m}|^2 + \sum\limits_{m = -M(n)}^{M(n)} |y_m|^2}.$ Then for $N \gt M(n)$ we have by $(2)$ \begin{align*} \left ( \displaystyle {\sum\limits_{m = -N}^{N} |x_{n-m}\ y_m|} \right )^2 & \leq S_{M(n)} + \left (\sum\limits_{m = M(n)}^{N} |x_{n-m}|^2 + \sum\limits_{m=M(n)}^{N} |x_{n+m}|^2 \right ) + \left (\sum\limits_{m = M(n)}^{N} |y_{m}|^2 + \sum\limits_{m=M(n)}^{N} |y_{-m}|^2 \right ) \\ & \leq S_{M(n)} + \left (\sum\limits_{m = M(n)}^{N} |x_{n-m}| + \sum\limits_{m=M(n)}^{N} |x_{n+m}| \right ) + \left (\sum\limits_{m = M(n)}^{N} |y_m| + \sum\limits_{m=M(n)}^{N} |y_{-m}| \right ) \\ & \leq S_{M(n)} + \sum\limits_{m = -N}^{N} |x_{n-m}| + \sum\limits_{m = -N}^{N} |y_m| \end{align*} Now $S_{M(n)}$ is a finite quantity and $\lim\limits_{N \to \infty} \displaystyle {\sum\limits_{m = -N}^{N} |x_{n-m}|}, \lim\limits_{N \to \infty} \displaystyle {\sum\limits_{m = -N}^{N} |y_m|} \lt \infty$ so for all $n \in \Bbb Z$ it follows that $\lim\limits_{N \to \infty} \displaystyle {\sum\limits_{m=-N}^{N} |x_{n-m}\ y_m|} \lt \infty,$ proving (a).
How do I prove (b) and (c)? Any help in this regard will be highly appreciated.
Thanks in advance.