We have a set of scalars $(A_{ij}\mid i,j\in\mathbb N)$, which are supposed to be the coefficients of a continuous linear operator $A$ on a real Hilbert space, with respect to an orthonormal basis $(e_i\mid i\in\mathbb N)$.

$$x=\sum_ix_ie_i\mapsto Ax=\sum_{i,j}A_{ij}x_je_i$$

$$A_{ij}=e_i\cdot Ae_j$$

When is this possible? How can we tell from $A_{ij}$ whether this produces a well-defined continuous operator?

Here are some necessary results of continuity:


$$\forall j,\quad\lVert Ae_j\rVert^2=\sum_iA_{ij}^2<\infty$$

$$\sup_j\lVert Ae_j\rVert^2=\sup_j\sum_iA_{ij}^2<\infty$$

Here are some sufficient conditions:

$$\max\{i+j\mid A_{ij}\neq0\}<\infty$$




Is any such expression necessary and sufficient? Of course we need precisely

$$\lVert A\rVert^2=\sup_{x\neq0}\frac{\lVert Ax\rVert^2}{\lVert x\rVert^2}=\sup_{x\neq0}\frac{\sum_i\left(\sum_jA_{ij}x_j\right)^2}{\sum_ix_i^2}<\infty$$

and we can take $x$ to be in the countable dense set ($\cong c_{00}\cap\mathbb Q^\mathbb N\subset\ell^2$) of finite, rational, linear combinations of $e_i$. But this seems difficult to work with.

The adjoint $A^*$ has the same operator norm $\lVert A^*\rVert=\lVert A\rVert$, so $A$ is bounded if and only if $A^*$ is bounded. And $\lVert A^*A\rVert=\lVert A\rVert^2$, so $A$ is bounded if and only if $A^*A$ is bounded. Thus, we need only to consider symmetric positive-semidefinite operators.

$$S=A^*A;\quad S_{ij}=\sum_kA_{ki}A_{kj}$$



The answer is probably no, mainly because of what happens in the finite dimensional case... as an example, asking that $\sum_{ij} A_{ij}^2<+\infty$ is surely enough, but in a $n$-dimensional vector space the following holds (and it is sharp): $$\sup_{\|x\|=1} \|Ax\|^2\leq \sum_{ij} A_{ij}^2\leq n \sup_{\|x\|=1} \|Ax\|^2$$ and, as the constant is linear in $n$, you cannot expect to have anything like that in infinite dimension. Just take, for every $k>0$, a $k\times k$ matrix $A^{(k)}$ which reaches the upper bound in $\mathbb{R}^k$, such that $$\sup_{\|x\|=1} \|A^{(k)}x\|^2=1$$ and construct a block-diagonal infinite matrix $A$ with blocks $A^{(1)}$, $A^{(2)}$ and so on. Then, given $x$ in your Hilbert space $H$, write $$x=x^{(1)}\oplus x^{(2)}\oplus \cdots$$ (which is an orthogonal decomposition, if you started with a o.n. basis) and compute $$\|Ax\|^2=\|(A^{(1)}x^{(1)}\oplus A^{(2)}x^{(2)}\oplus\cdots\|^2=\sum_{j}\|A^{(j)}x^{(j)}\|^2\leq \sum_{j}\|x^{(j)}\|^2=\|x\|^2\;.$$ So, the matrix $A$ induces a bounded operator on $H$, but $\sum_{ij}A_{ij}^2=+\infty$.

The same is true for other matrix norms, such as $\sum_{ij} |A_{ij}|$.

On the other hand, even with a condition such as $\sum_{i}A_{ij}^2\leq C$ and $\sum_{j}A_{ij}^2\leq C$, you cannot obtain the continuity of the operator induced by $A$, again by the same kind of reasoninig.

Can we find, in $\mathbb{R}^n$, a symmetric matrix $A^{(n)}$ such that the images of the canonical basis are all bounded by $1$, but there is another vector of norm $1$ which is mapped into a vector whose norm is an unbounded function of $n$?

Yes: we take a matrix that sends the whole space to one line, which is equally inclined with respect to every element of the canonical basis, such that the kernel of the matrix is the hyperplane orthogonal to that line. I.e., we take $$u=(1,1,\ldots, 1)$$ and set $$A^{(n)}=\frac{1}{\sqrt{n}}u^tu\;.$$ For every $e_j$ (column vector) in the canonical basis, $A^{(n)}e_j=\frac{1}{\sqrt{n}}u^t$, so $\|A^{(n)}e_j\|^2=1$, but $$\left\|A^{(n)}\left(\frac{1}{\sqrt{n}}u^t\right)\right\|^2=\left\|\frac{1}{n}u^tuu^t\right\|^2=n$$ and $n^{-1/2}u^t$ is a unit vector.

Therefore, the norm of every column and row of $A^{(n)}$ is $1$, but its operator norm is $\sqrt{n}$. Now, we perform the same construction as before, with a block-diagonal infinite matrix and we get the desired counterexample.

I am no expert, but I am not aware of any easy condition such that $A_{ij}$ represents a bounded linear operator on a Hilbert space with respect to some o.n. basis. This is probably due (but this is just a speculation) to the fact that the $2$-norm is not basis-dependent (as long as you only consider o.n. bases). For example, if you consider instead $\ell^1(\mathbb{R})$ or $\ell^{\infty}(\mathbb{R})$, it is much easier to give a necessary and sufficient condition such that $A_{ij}$ represents a bounded linear operator.

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    $\begingroup$ This is helpful. But I'll wait to see if anyone else answers. $\endgroup$ – mr_e_man Dec 6 '19 at 0:18
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    $\begingroup$ Easier/Different argument for the first part of your answer: If $\sum_{i,j} |A_{i,j}|^2 <\infty$, then the operator $A$ is Hilbert-Schmidt, and hence compact. Thus, this condition is certainly not necessary (also, it does not even hold for the identity operator). $\endgroup$ – PhoemueX Dec 6 '19 at 19:28
  • $\begingroup$ You are absolutely right! My argument is not the easiest/quickest way to show the non-necessity (does this word exist?). The reason why I gave that explanation was to exemplify the idea of passing to the limit form finite-dim when (sharp) estimates are involved, with constants that blow up in $n$. $\endgroup$ – wisefool Dec 7 '19 at 14:29
  • $\begingroup$ You could try to find an expert's opinion to support our suspicion that there's no easy solution. Or just include examples of stronger necessary conditions and weaker sufficient conditions, to tighten the difference between them. $\endgroup$ – mr_e_man Dec 9 '19 at 3:49
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    $\begingroup$ @mr_e_man the illogicalness of your last comment caused me mental pain. you can come on stack exchange to ask for help on a particular question but still google things on your own... $\endgroup$ – mathworker21 Dec 14 '19 at 13:13

You ask several questions, amongst them "Is any such expression necessary and sufficient?" The answer is negative, and given by example: the Hilbert matrix $$(h_{ij})\:=\:\left[\frac 1{i+j-1}\right]_{i,j=1}^\infty$$ defines a bounded linear operator. Thus $(h_{ij})$ fulfils the necessary conditions, but it does not satisfy any of the sufficient conditions.

Let's first consider the necessary stuff (and $\mathbb R$ or $\mathbb C$ does not really matter):
Given an infinite matrix $(A_{ij})$, we assume that an $A\in\mathscr L(\mathsf H)$ exists and for some $\{\text{O$\mathbb N$B}\;e_n\}\subset\mathsf H\,$ one has $A_{ij}=\langle Ae_i|e_j\rangle\,$ for all $i,j$. Then by Parseval's identity $$\sum_{n=1}^\infty|A_{in}|^2\:=\:\|Ae_i\|^2 \quad\text{and}\quad\sum_{n=1}^\infty|A_{nj}|^2\:=\:\|A^*e_j\|^2\,,$$ expressing that each column and each row of $(A_{ij})$ is square-summable. Furthermore, $$|\langle Ae_i|e_j\rangle|\:\leqslant\:\|Ae_i\|\:\leqslant\:\|A\|$$ holds, and all the preceding resumes the necessary conditions given in the OP.
They are seen to be fulfilled for the Hilbert matrix, note that each column or row in it is a truncated harmonic sequence, hence square-summable.

Now let's go through the sufficient conditions:
1) says that only finitely many entries in the matrix under consideration are non-zero.
Not true in the Hilbert matrix.

2) Consider $$\sum_{i+j\leqslant n+1}|h_{ij}|^2\:=\: \sum_{k=1}^n\sum_{i+j=k+1}(i+j-1)^{-2}\:=\: \sum_{k=1}^n\frac 1k$$ which diverges.
If $\sum_{i,j=1}^\infty|A_{ij}|^2 < \infty$ holds true, then $A$ turns out to be a Hilbert-Schmidt operator, with Hilbert-Schmidt norm $\,\|A\|_2 = \sqrt{\operatorname{trace}(A^*\!A)} = \sqrt{\sum_{i,j=1}^\infty|A_{ij}|^2}$
(which is also known as Frobenius matrix norm if finite matrices are considered).

3) is a stronger condition than 2), hence not satisfied by $(h_{ij})$.

4) Both $\sup$-expressions are infinite, since they contain the harmonic series.

To your initial question "How can we tell from $A_{ij}$ whether this produces a well-defined continuous operator?" there is, according to current knowledge, no criterion or answer like "Yes, if & only if $(A_{ij})$ fulfills this & that."

Condition 4) refers to Schur's test dating back to 1911, which is an important sufficiency criterion in this context. See here for a statement plus proof on math.SE.
It may be applied to the Hilbert matrix to prove it belongs to a bounded operator (choice of the weights $w_n=(n-1/2)^{-1/2}$ will do). By the way, the corresponding operator is non-compact and its operator norm equals $\pi$. You may consult section 3 in Contributions of Issai Schur to Analysis.

Another application of Schur's test in a concrete case is here.

And a reference for an infinite matrix satisfying 'the' necessary conditions which is not representable by a bounded operator.

Math history notice
In the very beginning the (spectral) theory was developed by David Hilbert in terms of quadratic and bilinear forms, thus often drowning in indices. This was one reason for John von Neumann's success in consistently using the operator concept. It avoided the difficult ground at which your questions target.

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  • $\begingroup$ Sorry, by "any such expression" I did not mean to restrict to the ones shown; I meant "any expression similar to these". $\endgroup$ – mr_e_man Dec 15 '19 at 19:46
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    $\begingroup$ Ah, I see @mr_e_man. Then your question gets much broader, but the general answer does not change, cf the text following the separation bar. Notice that in the very beginning the (spectral) theory was developed by Hilbert in terms of quadratic and bilinear forms, thus often drowning in indices. This was one reason for von Neumann's success in consistently using the operator concept. It avoided the difficult ground at which your question targets. $\endgroup$ – Hanno Dec 15 '19 at 20:33

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