Does $\sum_{i = 1}^{\infty} |\beta_{i}|^2 < \infty$ implies that $A $ satisfies $ \sum_{i=1}^{\infty} \sum_{j=1}^{\infty} |a_{ij}|^2 < \infty $

Question

Does $\sum_{i = 1}^{\infty} |\beta_{i}|^2 < \infty$ implies that $A$ satisfies $ \sum_{i=1}^{\infty} \sum_{j=1}^{\infty} |a_{ij}|^2 < \infty $?

Here is the question I am trying to solve:

Let $A = [a_{ij}]_{i,j = 1}^{\infty}$ be an infinite matrix of real numbers and suppose that, for any $x \in \ell^2,$ the sequence $Ax$ belongs to $\ell^2.$ Prove that the operator $T,$ defined by $T(x) = Ax,$ is a bounded operator on $\ell^2.$

My trial of a solution is:

According to the givens we can define $A : \ell^2 \rightarrow \ell^2 $ by $$A x = A (\xi_{1}, \xi_{2}, …) = (\beta_{1}, \beta_{2}, …), $$ i.e., $\beta_{i} = \sum_{j=1}^{\infty} a_{ij} \xi_{j}$

The operator $T$ is bounded as the operator $A$ is bounded and this can be proved as follows:\

$|\beta_{i}| = |\sum_{j=1}^{\infty} a_{ij} \xi_{j}| \leq \sum_{j=1}^{\infty} |a_{ij} \xi_{j}| \leq (\sum_{j=1}^{\infty} |a_{ij}|^2)^{1/2} (\sum_{j=1}^{\infty} |\xi_{j}|^2)^{1/2}$ \

Which implies that for $x = (\xi_{1}, \xi_{2}, …),$

$\|Ax\|^{2} = \sum_{i=1}^{\infty} |\beta_{i}|^2 \leq \|x\|^2 (\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} |a_{ij}|^2).$

But then what? Does $\sum_{i = 1}^{\infty} |\beta_{i}|^2 < \infty$ implies that $A $ satisfies $ \sum_{i=1}^{\infty} \sum_{j=1}^{\infty} |a_{ij}|^2 < \infty $?

I do not know how to complete. could anyone help me in this please?

Answer 1

No, $A$ need not satisfy $$ \sum_{i=1}^\infty\sum_{j=1}^\infty |a_{ij}|^2<\infty\,.\tag{1} $$ As an example consider $A=(\delta_{jk})_{j,k\in\mathbb N}=\operatorname{diag}(1,1,\ldots)$ so $A$ is an "infinite identity matrix". Obviously $A$ satisfies $x\in\ell^2\Rightarrow Ax=x\in\ell^2$ but the double infinite sum (1) is equal to $\sum_{j=1}^\infty 1=\infty$. Indeed the operators (respectively: infinite matrices) which satisfy (1) are called Hilbert-Schmidt operators and are a subset of the compact operators.

As for your original quesiton the situation is even worse: as you noticed the action of $A$ and the action of $T$ is equivalent so what you are essentially asking is: "is an everywhere-defined linear Hilbert space operator automatically bounded?". The answer to that is no (see here for more detail) so your proof is in vain---although the result is of an abstract nature and such an operator cannot be constructed explicitly.