Let \begin{equation*} \bigoplus_{ \ell_2} K_n := \{ (x_1,x_2,\cdots) \in \bigoplus_{n=1}^\infty : x_n \in K_n, \sum_{n=1}^{\infty} || x_n ||^2 <\infty \} \end{equation*} where $K_n$ are Hilbert Spaces. It is clear that $\bigoplus_{ \ell_2} K_n$ is also a Hilbert space. Let be $H$ another Hilbert space, and the bounded operator \begin{equation*} \begin{split} T & : H \rightarrow \bigoplus_{ \ell_2} K_n \\ & x \mapsto (T_1 x,T_2 x, \cdots) \end{split} \end{equation*} where $T_n = \pi_n \circ T$ and $\pi_n$ are the canonical projections.
We have that $T^* = T^* \circ \pi_n^* = T^* \circ j_n$, where $j_n : K_n \rightarrow \bigoplus_{ \ell_2} K_n$ is the canonical inclusion (see Adjoint of projection onto direct sum of Hilbert spaces)
I tried to prove that \begin{equation*} T^*T = \sum_{n=1}^\infty T_n^* T_n. \end{equation*} This is my attempt:
Let $x \in H$.
We have \begin{equation*} \begin{split} \sum_{n=1}^\infty T_n^* T_n x & = \sum_{n=1}^\infty T^* j_n \pi_n Tx = T^* \left( \sum_{n=1}^\infty j_n \pi_n (T_1x,T_2x,\cdots) \right) \\ & = T^*\left(\sum_{n=1}^\infty j_n(y_n)\right) = T^* \left(\sum_{n=1}^\infty (0,\cdots,0,T_nx,0, \cdots) \right) \\ & = T^*(T_1x,T_2x,\cdots) = T^*Tx. \end{split} \end{equation*} Is my proof right? Thank you for your help.
The problem that you don't address is that of the series. A very important fact that is often overlooked, is that a series is a limit of partial sums, and a limit implies an underlying topology. But you don't say which topology.
What convergence do you consider on $\sum_n T_n^*T_n$? What about all the other sums in your reasoning? At the beginning of your reasoning your took $T^*$ out of the sum; that amounts to exchange it with the limit, so you are assuming that $T^*$ is continuous in whichever topology you are using for the sum.