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.