Is the particle-hole transform rigorous in infinite-dims?

$DeclareMathOperator{tr}{tr}$Let me only consider fermions, i.e., fermionic Fock space $mathscr{F}$ of a single-particle Hilbert space $mathscr{H}$. Let us assume that $dim mathscr{H}=aleph_0$ (infinitely countable).

Based on my understanding, the particle-hole transform (in the most general sense) is either a unitary or anti-unitary operator $W$ on the Fock space which takes
$$
W c^*(x)W^*=c(x)
$$
where $x$ is some orthonormal basis of $mathscr{H}$, or more generally,
$$
W c^*(varphi) W^*=c(varphi)quad text{or} quad c(bar{varphi})
$$
where $varphiin mathscr{H}$, the bar notation is the anti-linear complex conjugation $C$ with respect to the basis $x$ and the "or" is depending on whether $W$ is unitary or anti-unitary.

Now I know that if $dim {mathscr{H}} <infty$ is finite-dim, I can construct such a unitary/anti-unitary operator basically by the Bogoliubov transform (Theorem 2.2 in Bach, V., Lieb, E. H., & Solovej, J. P. (1994). Generalized Hartree-Fock theory and the Hubbard model. Journal of statistical physics, 76(1), 3-89.). Indeed, the theorem tells us that if $u,v$ are (special, but not important to this question) bounded linear operators on $mathscr{H}$ and $tr {v^*v} <infty$, then there exists a unitary operator $W$ on the Fock space such that
$$
Wc^*(varphi) W^*=c^*(uvarphi)+c(vbar{varphi})
$$
Taking $v=I$ and $u=0$ and we obtain a unitary particle-hole transform $W$. Moreover, if $K$ is the second-quantization of complex conjugation $C$ on the Fock space, i.e., $K=bigoplus Cotimes cdots otimes C$, then $KW$ is the anti-unitary particle-transform operator. Either way, the essential thing is that $mathscr{H}$ is finite-dim so that $tr {v^*v} <infty$. In Solovej’s Many Body Quantum Mechanics, it’s also proven that $tr {v^*v} <infty$ (Shale-Stinespring condition) is a necessary condition for such a Bogoliubov unitary transform to exist (though I haven’t checked the proof).

Knowing all this, I can’t understand why such a particle-hole transform would exist in an infinite-dim space, despite seeing it often mentioned in such context. Maybe I’m overlooking something, or maybe it’s just that we only care about finite-dim system which are taken to the "thermodynamic limit" so that our systems are always finite-dim (albeit very large).

Side note. Notice that my particle-hole transform transforms all particles to holes and vice-versa, even though the usual way is to transform particles with energy $<0$ to holes and keep the particles with energy $ge 0$ as particles (as in the Fermi sea). In this case, as long as the number of states wth energy $<0$ is finite, then the aforementioned particle-hole transform is well-defined, even if the entire single-particle space is infinite-dim. What I’m basically asking is what happens if the number of states with energy $<0$ is infinite too.

EDIT. I realized that we probably never actually need to consider a case where the number of states with energy $<0$ is infinite. Because if we do, this implies that the Fermi sea ground state has an an energy of $ = -infty$. What those texts probably mean is that if were to take the thermodynamic limit, then the number of states with energy $<0$ would $to infty$.