$L^2$ space of Hilbert-Schmidt operator valued functions

Question

Let $\mathscr{S}$ denote the space of all Hilbert-Schmidt operators on $L^2(\mathbb R)$. Consider the Hilbert space $L^2(\mathbb R, \mathscr S)$ of square-integrable $\mathscr S$-valued function, that is, for any $T\in L^2(\mathbb R, \mathscr S)$ we have $$\int_\mathbb{R} \|T(\lambda)\|^2_{HS} d\lambda<\infty,$$ where $\|\cdot\|_{HS}$ denotes the Hilbert-Schmidt norm. Next, consider an orthonormal basis $\{e_n\}_{n\ge 0}$ of $L^2(\mathbb R)$. For any $\lambda\in \mathbb R$ define $e^\lambda_n(x)=|\lambda|^{1/2}e_n(\lambda x)$. Note that for each $\lambda$, $\{e^\lambda_n\}_{n\ge 0}$ is an orthonormal basis in $L^2(\mathbb R)$. For each $n\ge 0$ we denote the projection operator on $\text{Span}\{e^\lambda_n\}$ by $P^\lambda_n$. Based on these projections, let us define the following subspaces of $L^2(\mathbb R, \mathscr S)$: $$H_n=\{\lambda\mapsto f(\lambda)P^\lambda_n: f\in L^2(\mathbb R)\}.$$

Question: Is it true that $$L^2(\mathbb R, \mathscr S)=\bigoplus_{n=0}^\infty H_n?$$

Answer 1

No, I don't think so. For an explicit example, for $v,u\in L^2(\mathbb{R})$, let me use the notation $v\otimes w$ for the rank $1$ operator $(v\otimes w) (f) = (w,f) v$. Then consider any fixed $g\in L^2(\mathbb{R})$ and the operator valued function $G(\lambda) = g(\lambda)e_1^\lambda \otimes e_2^\lambda$. Now let $F\in H_n$, i.e., $F(\lambda)=f(\lambda)P_n^\lambda = f(\lambda) e_n^\lambda \otimes e_n^\lambda $. Then you have that \begin{equation} tr(F(\lambda)G^*(\lambda))=f(\lambda)\overline{g(\lambda)}tr((e_n^\lambda\otimes e_n^\lambda) (e_2^\lambda \otimes e_1^\lambda)) =f(\lambda)\overline{g(\lambda)} (e_n^\lambda,e_2^\lambda)(e_n^\lambda,e_1^\lambda)=0, \forall n. \end{equation} So this $G$ is non zero and orthogonal to $ \bigoplus_{n=0}^\infty H_n$