I'm reading an article about discrete Wiener-Hopf Operators and came across a passage I don't understand.
Let $l^p(\mathbb{Z};\omega)$ be a weighted sequence space, where $\omega = (\omega)_{n \in \mathbb{N}}$ is a Muckenhoupt weight, which means the following: $$\exists C<\infty :\quad\forall 0 \leq m \leq n \quad \frac{1}{n-m+1}(\sum_{k=m}^n\omega_k^p)^{\frac{1}{p}}(\sum_{k=m}^n\omega_k^{-q})^{\frac{1}{q}} < C \quad (\frac{1}{p}+\frac{1}{q} = 1)$$ Consider the following set: $$M_{p,\omega}:=\{ a \in L^{\infty}(\mathbb{T}) \quad|\quad \exists C_{p,\omega,a} < \infty : ||L(a)x||_{l^p(\mathbb{Z},\omega)} \leq C_{p,\omega,a}||x||_{l^p(\mathbb{Z},\omega)} \quad \forall x \in l^2(\mathbb{Z}) \cap l^p(\mathbb{Z},\omega) \}$$ Where $L(a) = (a_{n-k})_{n,k=-\infty}^{\infty}$ is called the Laurent Matrix generated by the Fourier Coefficients of $a$. $$a_n = \frac{1}{2\pi}\int_0^{2\pi}a(e^{i\theta})e^{-in\theta}d\theta$$
In the article they say that if $a \in M_{p,\omega}$ then $L(a)$ is a bounded operator and I don't really understand why. I get that it verifies the condition to be bounded for every element in $l^2(\mathbb{Z}) \cap l^p(\mathbb{Z},\omega)$, but why would that implie that it also verifies the condition for every element in $l^p(\mathbb{Z},\omega)$?
I tried searching if $l^2(\mathbb{Z}) \cap l^p(\mathbb{Z},\omega)$ is dense in $l^p(\mathbb{Z},\omega)$ but couldn't find anything about it.
Thanks in advance
So, as answered here (In what circumstances is $l^2(\mathbb{Z}) \cap l^p(\mathbb{Z},\omega)$ dense in $l^p(\mathbb{Z},\omega)$?), $l^2(\mathbb{Z})\cap l^p(\mathbb{Z},\omega)$ is dense in $l^p(\mathbb{Z},\omega)$ so given any $x \in l^p(\mathbb{Z},\omega)$ there is a sequence $(x_n)_{n\in\mathbb{N}}$ such that $x_n\to x$.
$\forall n \in \mathbb{N} \quad ||L(a)x_n||\leq C||x_n||$, and since the norm is a continuous function, we can conclude that $||L(a)x||\leq C||x||$.