The following question is inspired by Problem 5.30 in The Elements of Operator Theory written by Carlos S. Kubrusly
Let $H$ be a Hilbert space and suppose $\{H_k\}_{k \in \mathbb{Z}}$ is an infinite family of non-zero pairwise orthogonal subspaces of $H$ such that $H$ = $\bigoplus_{k = -\infty}^{\infty} H_k$. Here one can show that $H$ is isometrically isomorphic to $\overline{\sum_{k = -\infty}^{\infty} H_k}$ and hence the norm in $H$ is equal to the $l_2$ norm defined in $\bigoplus_{k = -\infty}^{\infty} H_k$. (for details please look up Chapter 5 in the given textbook).
Now assume all $H_k$ has the same dimension and $U_k: H_{k-1} \rightarrow H_k$ be unitary operators for all $k \in \mathbb{Z}$. Define $S: H \rightarrow H$ to be $S(x) = \bigoplus_{k = -\infty}^{\infty} U_k x_{k-1}$ and $S^*(x) = \bigoplus_{k = -\infty}^{\infty} U_k^* x_k$ where $x_k \in H_k$.
Let $K$ be a Hilbert Space that has the same dimension as one of $H_k$. Let S' be a function defined on $\bigoplus_{k = -\infty}^{\infty} K$ (i.e. $l_2(K))$) and defined as:
$S'(\bigoplus_{k = -\infty}^{k = n-1} 0 \bigoplus x \bigoplus_{k = n+1}^{k = \infty} 0) = \bigoplus_{k = -\infty}^{k = n} 0 \bigoplus x \bigoplus_{k = n+2}^{k = \infty} 0$. Here we call $S'$ the canonical bilaterial shift. Similar to $S$, $S'|_{K_k} K_k \rightarrow K_{k+1}$ where $K = K_k$ $\forall k \in \mathbb{Z}$. Inside the textbook, it says the multiplicity of $S'$, which is the common dimension of each $H_i, (i \in \mathbb{Z})$, is the same as dim$K$
The question is: is $S$ and $S'$ unitarily equivalent? Meanwhile I wonder whether or not $lim_{k \rightarrow \infty} U_{n}U_{n-1} \ldots U_{-k}$ will converge in strong operator topology and why. If it does, then the answer will be clear.
Below is some of my attempts done previously
If $lim_{k \rightarrow \infty} U_{n}U_{n-1} \ldots U_{-k}$ exists in $SOT$ for any $n \in \mathbb{N}$, define $A_n = lim_{k \rightarrow \infty} U_{n}U_{n-1} \ldots U_{-k}$. Since $K$ has the same dimension as every $H_k$ does, if $A_n$ exists, we can let $A_n$ be defined on $K$ and maps elements in $K$ to $H_n$. $A_n$ will also be unitary because $SOT$ limit is preserved by composition.
Pick an element $x \in \bigoplus_{k = -\infty}^{\infty} K$ and say $x = (\dots, x_{-2}, x_{-1}, x_{0}, \dots)$. Although each $x_j \in K (j \in \mathbb{Z})$ but we can fix the bijection here so that $x_{-2} \in H_{-2}, x_{-1} \in H_{-1}, x_{0} \in H_{0}, \dots$.
Denote $U$ by ($ \dots A_{-1}, A_{0}, A_{1}, A_{2}, \dots$) such that $U: \bigoplus_{k = -\infty}^{\infty} K \rightarrow \bigoplus_{k = -\infty}^{\infty} K$ and let $Ux =$ ($\dots, A_{-1}x_{-2}, A_{0}x_{-1}, A_{1}x_{0}, A_{2}x_{1}, \dots$). In this case, $A_n$ maps elements from $H_{n-1}$ to $H_{n}$. Note $A_{-1}x_{-2} \in H_{-1}, A_{0}x_{-1} \in H_{0}, A_{1}x_{0} \in H_{1}$. As a result, each $A_{n}^*$ maps the opposite direction.
$SUx = ( \dots, U_0A_{-1}x_{-2}, U_1A_{0}x_{-1}, U_2A_{1}x_{0}, \dots) = (\dots, A_0x_{-2}, A_1x_{-1}, A_2x_{0}, \dots)$. Here $A_0x_{-2} \in H_0, A_1x_{-1} \in H_1$ and $A_2x_{0} \in H_2$.
$U^* = ( \dots, A_{-1}^*, A_{0}^*, A_{1}^*, \dots)$. $U^*SUx = (\dots, A_0^*A_0x_{-2}, A_1^*A_1x_{-1}, A_2^*A_2x_{0}, \dots) = (\dots, x_{-2}, x_{-1}, x_{0}, \dots)$. By definition of $A_{n}^*$, $A_{n}^*A_nx_{n-2} \in H_{n-1}$. Originally the $x_{-2}$ is on the $H_{-2}$ coordinate but now it is on the $H_{-1}$ coordinate and then we can conclude $U*SU$ is the canonical bilaterial shift.
I am not very sure about the part related to $A_n^*$ and problem related to coordinate of $x \in \bigoplus_{k = -\infty}^{\infty} K$. Please feel free to point out any mistakes and any responses will be appreciated.
For $k<n$ let $F_{n,k}:= U_{n}\cdots U_{k+2}U_{k+1} : H_{k}\to H_n$ which is unitary as a composition of unitaries. For $k> n$ let $F_{n,k}:= F_{k,n}^{-1}$ as well as $F_{k,k}=\Bbb1_{H_k}$. Note that $F_{n,k}\circ F_{k,l} = F_{n,l}$ and that $F_{n,n-1}=U_n$. Now fix a number $n$ as well as a unitary isomorphism $G: H_n\to K$ and for $k\in\Bbb Z$ let $$F_k := G\circ F_{n,k}: H_k\to K$$ which is unitary as a composition of unitaries. Let $F:\bigoplus_k H_k \to \bigoplus_k K$ be given by $F=\bigoplus_k F_k$, this is unitary.
For $x\in H_k$ note that $$S(x) = U_k(x)= F_{k,k-1}(x) = (F_{k,n}\circ F_{n,k-1})(x)=(F_{n,k}^{-1}\circ F_{n,k-1})(x) = (F_{k}^{-1}\circ F_{k-1})(x)=(F^{-1}\circ S'\circ F)(x).$$ This is the step that implies $S = F\circ S'\circ F^{-1}$ and $S$ and $S'$ are unitarily equivalent.
You are also asking about $\lim_k U_nU_{n-1}\cdots U_{-k}$. But note that as $k$ changes the domain of this map changes, and as such it doesn't make much sense to compare them, unless with $U_k$ you mean $\bigoplus_{l<k}0\oplus U_k \oplus\bigoplus_{l>k}0$. In that case the that product will go to $0$ in the strong topology.