Let $X$ be the space of all weakly measurable functions $\gamma:\mathbb{R}^n \to L^2(\mathbb{R}^n)$ (modulo functions that are 0 almost everywhere) for which $$\|\gamma\|_X^2 := \sup_{\|g\|_{L^2}=1} \int_{\mathbb{R}^n} |\langle g, \gamma(x) \rangle|^2 \mathrm{d} x < \infty.$$ Some authors call these functions bounded Carleman functions (e.g. Halmos, Sunder: Bounded Integral Operators on $L^2$ Spaces, p. 62). I need to determine the dual space $X'$ of the Banach space $(X,\|\cdot\|_X)$. I have tried the following:
It is easy to see that the algebraic tensor product $L^2(\mathbb{R}^n) \otimes_{\mathrm{alg}} L^2(\mathbb{R}^n)$ is contained in $X'$ because simple tensors $f\otimes g$ induce functionals on $X$ via $$(f\otimes g, \gamma) := \int_{\mathbb{R}^n} f(x) \langle \overline{g},\gamma(x) \rangle \mathrm{d}x.$$ In fact, $$|(f\otimes g, \gamma)| \leq \|f\|_{L^2} \|g\|_{L^2} \|\gamma\|_{X}.$$ Thus, also elements of the projective tensor product $L^2(\mathbb{R}^n) \hat{\otimes}_{\mathrm{\pi}} L^2(\mathbb{R}^n)$ define continuous functionals on $X$. However, I am sure that $X'$ is larger than $L^2(\mathbb{R}^n) \hat{\otimes}_{\mathrm{\pi}} L^2(\mathbb{R}^n)$.
My second approach was to consider $X$ as a closed subspace of $\mathfrak{B}(L^2(\mathbb{R}^n))$, the bounded operators on $L^2(\mathbb{R}^n)$, via the isometric embedding $$\begin{align}\Phi: X &\hookrightarrow \mathfrak{B}(L^2(\mathbb{R}^n)), \\ \gamma &\mapsto (g \mapsto \langle g,\gamma(\cdot)\rangle).\end{align}$$ Then we obtain $$X' = \mathfrak{B}(L^2(\mathbb{R}^n))' / X^\perp,$$ where $X^\perp$ is the annihilator of $X$ (i.e. all $\varphi \in \mathfrak{B}(L^2(\mathbb{R}^n))'$ for which $\varphi(\gamma) = 0$ for all $\gamma \in X$). However, the dual space $\mathfrak{B}(L^2(\mathbb{R}^n))' = (L^2(\mathbb{R}^n) \hat{\otimes}_\pi L^2(\mathbb{R}^n))''$ seems to be quite large and I do not see a way to simplify the quotient $\mathfrak{B}(L^2(\mathbb{R}^n))' / X^\perp$.
