In QFT, one usually talks about operator-valued distributions. But let's take, for instance, $L^{2}(\mathbb{R}^{3})$ and its associated Fock space $\mathcal{F} = \bigoplus_{n=0}^{\infty}L^{2}(\mathbb{R}^{3n})$. Creation and annihilation operators, respectivelly denoted by $a^{*}(h)$ and $a(h)$, $h \in L^{2}(\mathbb{R}^{3})$ are treated as operator-valued distributions on $\mathcal{F}$ as follows. First, one restricts $a^{*}(h)$ and $a(h)$ to a dense subspace $\mathcal{D}_{\mathscr{S}}$, given by:
$$\mathcal{D}_{\mathscr{S}}:= \{\psi \in D_{0}: \psi_{n} \in \mathscr{S}(\mathbb{R}^{3n}), \forall n\}$$
with $D_{0}=\{\psi \in \mathcal{F}: \mbox{$\exists N$ such that $\psi_{n} = 0$ if $n\ge N$}\}$ and, since $\mathscr{S}(\mathbb{R}^{3}) \subset L^{2}(\mathbb{R}^{3})$, one thinks of $a^{*}(h)$ and $a(h)$ as functions $\mathscr{S}(\mathbb{R}^{3}) \ni h \mapsto a^{\#}(h)$. However, a distribution is characterized as being a *linear* map and $a(h)$ is an anti-linear in $h$. We could, however, define $h \mapsto a(h^{*})$, where $h^{*}$ is the complex conjugate of $h$, but this seems to mess up all results about $a(h)$, since the theory is described in terms of $a(h)$ instead of $a(h^{*})$. So, what is the problem here? Shouldn't an operator-valued distribution be linear? Is the use of the term "operator-valued distribution" conceptually misleading when considering $a(h)$ and we should treat this operator-valued function as an anti-linear map? Or should we, in fact, consider $a(h^{*})$ instead of $a(h)$ in this context? It doesn't seem what physicists do, however. 

**ADD:** As discussed in the comments, if $a(h)$ is anti-linear in $h$, there's nothing wrong on defining $X(h) := a(h^{*})$. But let me stress two poits which seems to be troublesome:

*(1)* Because the Fock space is $\bigoplus_{n=0}^{\infty}L^{2}(\mathbb{R}^{3n})$, the application of $a(h)$ to an element $\psi \in L^{2}(\mathbb{R}^{3n})$ should lead to:
\begin{eqnarray}
(a(h)\psi)_{n} \equiv (a(h)\psi)(x_{1},...,x_{n-1}) = \sqrt{n}\int_{\mathbb{R}^{3}}\overline{h(x)}\psi(x,x_{1},...,x_{n-1})dx \tag{1}\label{1}
\end{eqnarray}

*(2)* To turn the theory into a field theory, one usually write:
\begin{eqnarray}
a(h) = \int \overline{h(x)}a(x)dx \tag{2}\label{2}
\end{eqnarray}
Here, $a(x)$ is formally treated as $a(\delta_{x})$, an operator on $\mathcal{F}$. 

For clarity, these two topics were based on [Sigal & Gustafson][1]. But here is the problem: on [Reed & Simon's][2] book, the authors state that (\ref{2}) should be treated as an equality in terms of quadratic forms, i.e. (\ref{2}) holds in the sense of:
\begin{eqnarray}
\langle \psi, a(h)\varphi \rangle = \int_{\mathbb{R}^{3}}\overline{h(x)}\langle \psi, a(x)\varphi \rangle \tag{3}\label{3}
\end{eqnarray}
If, however, we introduce $X(h) = a(h^{*})$ and use it instead of $a(h)$, then:
$$X(h) = a(h^{*}) = \int \overline{h(x)}a(x)dx$$
is not consistent with (\ref{1}) anymore, since $\overline{h(x)}$ in (\ref{1}) becomes $h(x)$. Moral: it seems that $a(h)$ itself is treated as an operator-valued distribution, not $X(h)$, but $a(h)$ is not linear in $h$. Am I getting something wrong? 



  [1]: https://www.springer.com/gp/book/9783642218651
  [2]: https://www.amazon.com/Fourier-Analysis-Self-Adjointness-Methods-Mathematical/dp/0125850026