Let there be a separable Hilbert space $\mathcal{H}$ and let $\hat\phi(t,\vec{x})$ be a bounded linear operator on $\mathcal{H}$ for every $t\in\mathbb{R},\vec{x}\in\mathbb{R}^3$. There exists a Hilber basis $\{|n\rangle\}_{n=0}^\infty$ that allows me to write $$\hat\phi(t,\vec{x}) = \sum_n\sum_m\phi^n_m(t,\vec{x})\,|n\rangle\langle m|,$$ and these real-valued functions are at least well-behaved enough to have a Fourier Transform, by hypothesis.
I am told in Quantum Field theory that I can define the Fourier Transform of $\hat\phi(t,\vec{x})$, which I will call $\hat{\tilde{\phi}}(t,\vec{p})$, by doing
$$\hat\phi(t,\vec{x}) = \sum_{n,m}\phi^n_m(t,\vec{x})\,|n\rangle\langle m| = \sum_{n,m}\left(\iiint_{\mathbb{R}^3}\mathrm{e}^{\mathrm{i}\vec{p}\cdot\vec{x}}\tilde\phi{}^n_m(t,\vec{p})\,\mathrm{d}^3p\right)|n\rangle\langle m| = \iiint_{\mathbb{R}^3}\mathrm{e}^{\mathrm{i}\vec{p}\cdot\vec{x}}\underbrace{\sum_{n,m}\tilde\phi{}^n_m(t,\vec{p})\,|n\rangle\langle m|}_{\hat{\tilde\phi}(t,\vec{p})}\,\mathrm{d}^3p.$$
Is this actually rigorous? Does the expression in components for $\hat{\tilde\phi}(t,\vec{p})$ always converge by virtue of some theorem (Paserval's, Riezs' or some other one, I tried but didn't succeed) or is this definition only valid for a subclass of all possible field operators?
So, if you are looking at the special case of the $L^2$ theory Fourier transform, this is very compatible with Hilbert-Schmidt operators. The time variable plays no role in this question so I drop it. One can write an $x$ dependent Hilbert-Schmidt operator in the form (see e.g. Simon, Trace Ideals and their Applications, Th. 1.4) $$ \Phi(x) = \sum_{n} \phi_n(x)\,|\varphi_n\rangle\langle\psi_n| $$ where $\{\varphi_n\}$ and $\{\psi_n\}$ are two orthonormal sets. Then the Hilbert-Schmidt norm of the operator for every fixed $x\in\Bbb R^3$ is just $$ \|\Phi(x)\|_2 = {\rm Tr}(\Phi(x)^*\Phi(x))^{1/2} = \left(\sum_{n} |\phi_n(x)|^2\right)^{1/2} $$ In the case of a finite rank operator, the family $\{\psi_n\}$ is finite and this is just the classical finite dimensional Euclidian norm and these finite rank operators from a dense subset of general Hilbert-Schmidt operators.
Now we want to take the Fourier transform. One can directly use the classical theory of $L^2$ Fourier transform, and by the Parceval inequality, one gets $$ \sum_n \int_{\Bbb R^3} |\psi_n(x)|^2\,\mathrm dx = \|\|\Phi(x)\|_2\|_{L^2_x}^2 \\ = \|\|\mathcal F_x\Phi(x)\|_2\|_{L^2_x}^2 = \int_{\Bbb R^3} \sum_n|\mathcal F_x\psi_n(x)|^2\,\mathrm dx $$ where the sum and integrals can be exchanged without any problem as a property of the Lebesgue integral of positive functions.
In particular, the operator $$ \mathcal F\Phi(p) = \sum_n \mathcal F\phi_n(p)\,|\varphi_n\rangle\langle\psi_n| $$ is a $L^2$ function of $p$ with value to Hilbert-Schmidt operators (i.e. it is in the space $L^2(\Bbb R^3, \mathrm{HS})$) and as such is in particular defined for almost every $p\in\Bbb R^3$.