Are rank-1 operators $e_i e_j^*:\mathcal H\to\mathcal H$ dense, for a separable Hilbert space $\mathcal H$?

Question

Let $\mathcal H$ be a separable Hilbert space with orthonormal basis $\{e_i\}_{i\in \mathbb{N} }$, and consider the linear operators $E_{ij}\equiv e_i e_j^*$, defined as $$E_{ij}v=e_i \langle e_j,v\rangle.$$ In finite-dimensions, it's clear that these are $\dim(\mathcal H)^2$ operators that span the space of all linear operators on $\mathcal H$, call it $\operatorname{Lin}(\mathcal H)$.

Can anything similar be said in infinite dimensions? That is, is $\{E_{ij}\}_{i,j\in\mathbb{N} }$ a basis for $\operatorname{Lin}(\mathcal H)$, with respect to one of its "standard" topologies (norm, strong, weak, etc)? If not, could it maybe be a basis for the subspace of bounded/compact/Hilbert-Schmidt/something-else operators?

I'm aware that $\operatorname{Lin}(\mathcal H)$ isn't a Hilbert space, and the same is true for the set of bounded operators $B(\mathcal H)$, but I think my question should still be meaningful with the notion of convergence provided in a Banach space, which these spaces should be.

I've also found that $B(\mathcal H)$ is not separable. This tells me that most likely my question has negative answer with respect to bounded operators (and therefore also linear ones).

On the other hand Hilbert-Schmidt operators do form a Hilbert space, though I'm not sure whether it's separable. So it seems possible (or even likely) that the operators $\{E_{ij}\}$ are dense in the space of Hilbert space operators (and if with respect to the natural topology in the Hilbert-Schmidt space, they'd in fact be an orthonormal basis for it). In fact, this post discusses how finite-rank operators are dense in the Hilbert-Schmidts. Which is close enough, but also different (as far as I can tell) because their construction uses a basis comprised of all possible finite-rank operators, while here I'm focusing on a basis built from a specific basis for $\mathcal H$.

Answer 1

As David explained in the comments already, your answer depends on the operator space and the associated norm where you want to take the closure in.

• When considering the usual operator norm on $B(H)$, then the corresponding closure are the compact operators $K(H)$, cf., e.g., Corollary 16.4 in "Introduction to Functional Analysis" by Meise & Vogt (1997).
• When considering any Schatten-$p$ norm, then the closure yields the corresponding Schatten-$p$ class ($1\leq p<\infty$), cf, e.g., Theorem 2.3.8 in Ringrose's "Compact Non-Self-Adjoint Operators" (1971).

For either of these cases there is also a paper by Widom where, in Proposition 2.1, he proves that for strongly converging sequences $B_i\to B$, $C_i\to C$ one has $B_nAC_n^*\to BAC^*$ in $p$-norm for all $p\in[1,\infty]$. Applied to $B_i=\sum_j^ie_je_j^*=C_i$ this precisely shows that the corresponding "block approximations" $\sum_{j,k}(e_j^*Ae_k)e_je_k^*$ converge in $p$-norm to $A$ (hence, density).

• For the strong operator topology you end up with all of $B(H)$. Here I do not know of a good reference, but if I may I could plug my own work: a direct proof regarding density of $e_ie_j^*$—which even works for non-separable Hilbert spaces—can be found in the Appendix of this paper (arXiv) of mine, as Corollary 7.1. This result also contains the corresponding density results for the operator and the Schatten norms, again for arbitrary Hilbert spaces. To be fair, all I did here was to modify Widom's proof to non-separable spaces thanks to uniform boundedness of the nets $(B_i)_i,(C_i)_i$ so the underlying idea is the same.