Let $H_n$ be a separable complex Hilbert space for each $n\in \mathbb{N}$. And
denote
$$H= \bigoplus_{n=1}^\infty H_n,$$
its direct sum in the category of separable complex Hilbert spaces.
Suppose for a family $(F_n)_n$ of Hilbert spaces we have the exactly same conditions and notation. I wonder how I can calculate the Hilbert-Schmidt norm of an operator
$$A:F \longrightarrow H.$$
I found something like
$$\|A\|^2_{HS}=\sum^{\infty}_{n=1}\|A|_{F_n}\|_{HS}^2.$$
But I dont know why this is the case, usually (by definition) for an arbitrary orthonormal basis $(e_n)_n$ of $F$, it should be
$$\|A\|^2_{HS}=\sum^\infty_{n=1}\|Ae_n\|^2_{H}.$$
Could some one please give me a conection between these two equalities.
Best Answer
One can view $F=\oplus_{n=1}^\infty F_n$ as the collection of infinite(countable) tuples which are square summable such that entries at $n^{th}$ co-ordinate is from $F_n$. Just choose an orthonormal basis $\{e^{(F_n)}_i\}_{i\in \mathbb N}$ for each Hilbert space $F_n$. Let $\delta_{n,i}\in F$ be the tuple whose $n^{th}$ co-ordinate is $e^{(F_n)}_i$. Then $$E=\{\delta_{n,i}: n,i\in \mathbb Z\}$$ is an orthonormal basis for $F=\oplus_{n=1}^\infty F_n$. Thus for any bounded linear operator $A:F\to H$, using the your definition for HS-norm \begin{align*} ||A||^2_{HS}&=\sum_{n,i\in \mathbb N}||A\delta_{n,i}||^2\\&=\sum_{n\in \mathbb N}\sum_{i\in \mathbb N}||A\delta_{n,i}||^2 \end{align*} But $\sum_{i\in \mathbb N}||A\delta_{n,i}||^2=\sum_{i\in \mathbb N}||A_{\restriction F_n}(e^{F_n}_i)||^2=||A_{\restriction F_n}||^2_{HS}$. Thus , from previous computations, we observe that $$||A||^2_{HS}=\sum_{n\in \mathbb N}||A_{\restriction F_n}||^2_{HS}$$