# Determine, by its action on an orthonormal basis, whether a linear operator can be continuous

*Solution:*

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.