Sequences - Convolution

Question

We got the following setting:

Let $A$ the set of all complex sequences $x=(x_n)_{n \in \mathbb N_0}$ with \begin{align*} \Vert x \Vert := \sum_{n = 0}^\infty \vert x_n \vert e^{-n^2} < \infty. \end{align*} Consider the convolution $(xy)_n = \sum_{k = 0}^n x_k y_{n - k}$ as multiplication from $A$. Then $A$ is a commutative unital Banach Algebra.

Now what I have so far:

1. A is obviously a normed space, this is rather easy to show.
2. The convolution is welldefined and submultiplicative, because $\begin{eqnarray*} \Vert x y \Vert &=& \sum_{n = 0}^\infty \vert \sum_{k = 0}^n x_k y_{n - k} \vert e^{-n^2} \leq \sum_{n = 0}^\infty \sum_{k = 0}^n \vert x_k \vert \vert y_{n - k} \vert e^{-n^2} \leq \sum_{n = 0}^\infty \vert y_n \vert \sum_{k = 0}^\infty \vert x_k \vert e^{-(n + k)^2} \\ &=& \sum_{n = 0}^\infty \vert y_n \vert e^{-n^2} \sum_{k = 0}^\infty \vert x_k \vert e^{-k^2} e^{-2nk} \leq \sum_{n = 0}^\infty \vert y_n \vert e^{-n^2} \sum_{n = 0}^\infty \vert x_n \vert e^{-n^2} = \Vert x\Vert \Vert y \Vert < \infty \end{eqnarray*}$
3. A is commutative because $(xy)_n = \sum_{k = 0}^n x_k y_{n - k} = \sum_{k = 0}^n y_k x_{n - k} = (yx)_n$ for all $n \in \mathbb N_0$.
4. A is unital because $e = (1, 0, \dots) \in A$ and $(xe)_n = \sum_{k = 0}^n x_k e_{n - k} = x_n$ for all $n \in \mathbb N_0$.

Now to my problem: How can I show that $A$ is a Banach Space? I allready tried the following:

Let $(x^{(k)})_{k \in \mathbb N}$ a Cauchy sequence in $A$. Let $\epsilon > 0$. Then there exists a $N \in \mathbb N$ with \begin{align*} \Vert x^{(k)} - x^{(l)} \Vert = \sum_{n = 0}^\infty \vert x^{(k)}_n - x^{(l)}_n \vert e^{-n^2} \leq \epsilon & & \text{ for all } l, k \geq N \end{align*} Hence for all $n \in \mathbb N_0$ it follows that $\vert x^{(k)}_n - x^{(l)}_n \vert \leq e^{n^2} \epsilon$. So the sequence $(x^{(k)}_n)_{k \in \mathbb N}$ is a Cauchy sequence for all $n \in \mathbb N_0$ in $\mathbb C$ and thus converges to a $x_n \in \mathbb C$. Set $x =: (x_n)_{n \in \mathbb N_0}$. We need to show that $x \in A$. Now I would like to have something like that: \begin{eqnarray*} \Vert x \Vert = \sum_{n = 0}^\infty \vert x_n \vert e^{-n^2} &\leq& \sum_{n = 0}^\infty \vert x_n - x^{(k)}_n \vert e^{-n^2} + \sum_{n = 0}^\infty \vert x^{(k)}_n \vert e^{-n^2} \\ &\leq& \sum_{n = 0}^\infty \vert x^{(k)}_n \vert e^{-n^2} + \Vert x^{(k)} \Vert < \infty \end{eqnarray*} But I couldnt't show that $\sum_{n = 0}^\infty \vert x^{(k)}_n \vert e^{-n^2} < \infty$. I would be grateful for some help. Have a nice day!

Answer 1

Since the function $w(n)= e^{-n^2}$ is positive, the formula $$\mu(A) = \sum_{n\in A} e^{-n^2}\quad (A\subseteq \mathbb{N})$$ defines a measure on the power-set of $\mathbb{N}$. Then $A = L_1(\mu)$, so it is a Banach space.