Application of Closed Graph Theorem for l2 space I have an exercise, which I am struggling with. Suppose $x\in \mathbb{R}^\mathbb{N}$ and $\forall y \in l^2$ we have $xy \in l^1$. Now I am to show that $x \in l^2$. There is a hint to use the Closed Graph Theorem.
Now I have an idea, we define a linear operator $T : D(T) \subset l^2 \mapsto \mathbb{R}$, where $T(y) = \sum_{i \in \mathbb{N}} x_i y_i$, this is well defined due to our assumption. Then if T is closed and $D(T)$ is closed, we get that T is bounded and so by Riesz Representation Theorem there is a unique $u \in D(T)$ (since a closed subset of a Hilbert Space is a Hilbert Space) such that $T( y ) = \sum_{i \in \mathbb{N}} u_i y_i$, hence we would have that $x = u \in l^2$. But of course there is the big portion I omitted of finding a closed D(T) and showing T is closed.
I'm not sure this is the right way to go about it, but I think it makes sense. I tried to take $D(T) = l^2$ directly, but I think showing the graph is closed then amounts to showing that the operator is continuous. Thanks!
Edit. $D(T)$ is some subset we define the linear operator T on i.e. domain of $T$.
Edit 2. After some thought I guess I have to try to make $D(T) = l^2$ work, since intuitively by choosing any other $D(T)$, we would have to somehow localize $x$ in $l^2$. I'm not sure if that or any of this makes sense, I'd appreciate any kind of input!
 A: There is another way to solve the problem using the Banach-Steinhaus theorem, which states:

*

*Given a family $\{T_\lambda\}_\lambda$ of linear maps $E\rightarrow L$ (with $E$ a Banach space), if $\{T_\lambda(x)\}_\lambda$ is a bounded subset of $L$ for all $x\in E$, then there is $C>0$ such that $$\|T_\lambda\|_\infty\leq C\forall\lambda$$
In particular, define $T_n:l^2\rightarrow\mathbb R$ as $$y=(y_i)_i\mapsto\sum^n_{i=0}x_iy_i$$ by hypothesis $T_n(y)$ is a convergent sequence for all $y$ (converging to $\sum_ix_iy_i$), in particular is limited. By the previous theorem there is $C>0$ such that $$\|T_n\|_\infty=\sup_{0\neq y\in l^2}\frac{|T_n(y)|}{\|y\|_2}<C$$ for all $n$.
Now, consider the sequence $x^n=(x_0,\cdots,x_n,0,\cdots,0,\cdots)$, then $$C\geq\frac{|T_n(x^n)|}{\|x^n\|_2}=\frac{|\sum_{i=0}^nx_i^2|}{\|x^n\|_2}=\frac{\|x^n\|^2_2}{\|x^n\|_2}=\|x^n\|_2$$ At this point notice that $\|x\|_2=\lim_n\|x_n\|_2\leqslant C<\infty$, so $x\in l^2$. A similar argument can be used to prove that if $x\in\mathbb R^\mathbb{N}$ satisfy $xy\in l^1$ for all $y\in l^p$, then $x\in l^q$ (where $p^{-1}+q^{-1}=1$).
A: Define $T$ a little bit differently. Let $T:\ell_2\to \ell_1$ given by
$$Ty = (y_1x_1,\,y_2x_2,\,...,y_nx_n,...),$$
for every $y=(y_i)_{i}\in \ell_2$. Then $T$ is linear. We use the closed graph theorem on this operator. For this, let $y^n=(y_i^n)_i\in \ell_2$ with $y^n\overset{\ell_2}{\rightarrow}y=(y_i)_i$ and $Ty^n=z^n$, for $z^n=(z_i^n)_i\in \ell_1$ where $z^n\overset{\ell_1}{\rightarrow}z$. We wish to show that $Ty=z$.
Since $Ty^n=z^n$ we have that $z^n_i=y^n_ix_{i}$ for all $i$. Since $y^n\overset{\ell_2}{\rightarrow}y$ and $z^n\overset{\ell_1}{\rightarrow}z$ we have that $y_i^n\to y_i$ and $z_i^n\to z_i$ as $n\to \infty,$ for all $i$. Hence, since $z_i^n=y_i^nx_i$ holds for every $n$, by letting $n\to \infty$ we get $z_i=y_ix_i$ for all $i$. In other words, $Ty=z$ and the graph is closed. Therefore, $||T||<\infty$. Now, let
$$y_N=(x_1,\,x_2,\,....,\, x_N,0,0...0)\in \ell_2.$$
Then,
$$||Ty_N||_1\leq ||T||\cdot ||y_N||_2.$$
In other words,
$$\sum_{k=1}^N|x_k|^2\leq ||T||\cdot \biggl(\sum_{k=1}^{N}|x_k|^2\biggr)^{1/2}.$$
Multiple both sides by $\biggl(\sum_{k=1}^{N}|x_k|^2\biggr)^{-1/2}$ to get
$$\biggr(\sum_{k=1}^{N}|x_k|^2\biggr)^{1/2}\leq ||T||.$$
As this holds for every $N$ on the limit we get
$$\biggl(\sum_{k=1}^{\infty}|x_k|^2\biggr)^{1/2}\leq ||T||<\infty.$$
