Problem: Let $x: \mathbb{N} \rightarrow \mathbb{C}$ be a sequence with the property that $xy \in l^1 (\mathbb{N})$ (coordinate-wise multiplication) whenever $y \in l^2 (\mathbb{N})$. Use the closed graph theorem to show that $x \in l^2 (\mathbb{N})$.
Attempt: I'm not sure how to solve this problem. I was thinking of defining a continuous operator $l^1(\mathbb{N}) \rightarrow l^2 (\mathbb{N})$ and then use the closed graph theorem somehow to show the statement. But I don't know how this operator should be defined.
Recall: the closed graph theorem states that a linear operator $T: X \rightarrow Y$ between Banach spaces is continuous (bounded) iff $T$ has closed graph.
An equivalent statement: if $(x_n) \to x$ in $X$, and $T(x_n) \to y $ in $Y$, then $T(x) = y$.
you can define the operator $T: \ell^2\to\ell^1$ defined by $T(y) = xy$ is well-defined by hypothesis (is linear between banach spaces), if you want to apply closed graph th. you must take a sequence $(y^n)_{n\in\mathbb{N}} \subset \ell^2$ converging to $y\in\ell^2$ such that $T(y^n)\to z\in \ell^1$ in $\ell^1$ and demonstrate that $T(y) = z$. Hint for this: prove that $\forall k,\forall n: |x_ky_k - z_k|\leq |x_k|\| y^n - y \|_2 + \| xy^n - z \|_1$.
edit: one hint for the last part, you take the sequences $y^n = (x_1, x_2,...,x_n,0,0,....)\in\ell^2$