Bounded, well-defined linear operators on complex matrices in $l^2$ - necessary conditions for the matrix.

Question

Suppose $A = [a_{ij}]_{i,j=1,2,\dots}$ is a matrix of complex numbers and lets define an operator $(T_{A}x)_i = \sum_{j=1}^{\infty}a_{ij}x_j$ for $i = 1,2,\dots$ and $x = (x_j)_{j\geq1} \in l^2$. What rules must A obey for T to be a well defined bounded operator on $l^2$?

Well, this was asked in $l^1$, here but with assumption of well-defined operator and here in a slight another context but none of them gave me a straight answer.

I see that the operator basically takes the sequence $x \in l^2$ and creates a new sequence where i-th element is made from a "dot product" of i-th row and the sequence.

For a $x \in l^2$ $$ \|Tx\|_2 = \sqrt{\sum_{i=1}^{\infty}\left(\sum_{j=1}^{\infty}a_{ij}x{j}\right)^2} \leq \sqrt{\sum_{i=1}^{\infty}\left(\sum_{j=1}^{\infty}a_{ij}^2\right)\left(\sum_{j=1}^{\infty}x_{j}^2\right)} \leq $$ by Cauchy Schwarz. Then: $$ \leq \sum_{i=1}^{\infty}\left(\sqrt{\sum_{j=1}^{\infty}a_{ij}^2}\sqrt{\sum_{j=1}^{\infty}x_{j}^2}\right) = \sum_{i=1}^{\infty}\sqrt{\sum_{j=1}^{\infty}a_{ij}^2}\|x\|_2 \leq \\ \sum_{i=1}^{\infty}\sum_{j=1}^{\infty}|a_{ij}|\|x\|_2 $$ since $\|x\|_2 < \infty$ the condition is that modules of entire matrix must be summable (i.e. $\forall_i \sum_{i,j=1}^\infty|a_{ij}| < \infty$).

Is that correct? How do we prove that the operator is well defined or what are the necessary matrix conditions?

Answer 1

For well-definedness of $T_A$, it is necessary that $(a_{ij})_{j\ge1}\in\ell^\infty$ for all $i$ because $\lim_{j\to\infty}a_{ij}x_j$ must vanish for each $i$ for the series to converge. Now, let $\alpha_i:=(a_{ij})_{j\ge1}\in\ell^\infty$ and let $\langle\cdot,\cdot\rangle\colon\ell^2\times\ell^\infty\to\mathbb{C}$ be a partial pairing defined as $\langle x,y\rangle=\sum_jx_jy_j :=\lim_{n\to\infty}\sum_{j=1}^nx_jy_j $, whenever the limit exists. Note that $T_A(x)=(\langle x\,,\alpha_i\rangle)_{i\ge1}$. Now, we have the following:

$\alpha_i\in\ell^1$ for all $i$ is sufficient for $T_A$ to be well-defined (and for continuity, $(\|\alpha_i\|_2)_{i\ge 1}\in\ell^2$ suffices.

Indeed, if $\alpha_i\in\ell^1$, then by the Triangle Inequality, $$|(T_Ax)_i|=|\langle x\,,\alpha_i\rangle|\le\|x\|_\infty\|\alpha_i\|_1<\infty$$ and hence $T_A$ is well-defined once $\alpha_i\in\ell^1$ for each $i$. For the continuity, this follows from Cauchy-Schwarz Inequality, that is $$\|T_Ax-T_Ay\|_2^2=\sum_i|\langle x-y\,,\alpha_i\rangle|^2\le\|x-y\|_2^2\sum_i\|\alpha_i\|_2^2\,.$$

$\alpha_i\in\ell^2$ for all $i$ is necessary for $T_A$ to be well-defined and continuous.

To see this, observe that functional $$f_i\colon\ell^2\to\mathbb{C}\,,\,~\,~\,~x\mapsto(T_A(x))_i$$ is linear and continuous; indeed, if $x_n\to x$ then $$\lim_{n\to\infty}f_i(x_n)= \lim_{n\to\infty}(T_A(x_n))_i=(T_A(x))_i=f_i(x)\,,$$ thereby establishing continuity. Linearity follows equally, that is for scalars $\beta,\gamma$, we have $$f_i(\beta x+\gamma y)=(T_A(\beta x+\gamma y))_i= \beta (T_A(x))_i+\gamma(T_A(y))_i= \beta f_i(x)+\gamma f_i(y)\,.$$ Since $(\ell^2)^*=\ell^2$, it follows that $ f_i(\cdot)=\sum_ja_{ij}(\cdot)\in\ell^2$; that is, $\alpha_i\in\ell^2$.