In general, my question is whether the operator norm of a (bounded) Laurent operator is equal to the operator norm of its corresponding Toeplitz operator. That is, the Toeplitz operator is obtained by truncating the Laurent operator to its bottom-right corner.
Motivation and more details: I have a mapping given by a state-space model: \begin{align} x_{t+1}&= Ax_t + Bu_t\\ y_t &= Cx_t. \end{align} where $\{u_t\}$ is the input sequence and $\{y_t\}$ is the output sequence. I am trying to understand whether there is a difference if we compute the operator norm when the time horizon is semi-infinite ($t\in\mathbb{N}$) or doubly infinite ($t\in\mathbb{Z}$). The matrix $A$ is assumed to be stable so that the mapping is linear and bounded.
My attempt: The mapping from inputs to outputs can be described with a doubly-infinite (block) Toeplitz operator with the Markov parameters $F_i = CA^iB$ for $i>0$. That is, \begin{align} y &= Fu, \end{align} where the $i$th diagonal of $F$ is given by $CA^iB$.
My question is now simple: are the norms of $F$ and the truncation of $F$ to the semi-infinite horizon equal? if not, are they related?
Let $T:\ell^2(\mathbb{Z})\to \ell^2(\mathbb{Z})$ be the bounded Laurent operator corresponding to the sequence $\{a_n\}_{n=-\infty}^\infty.$ In particular the columns of $T$ are square summable, i.e. $\sum|a_n|^2<\infty.$ The operator $T$ is unitarily equivalent to the operator $\widetilde{T}:L^2(0,2\pi)\to L^2(0,2\pi)$ acting by $\widetilde{T}(f)=af,$ where $a(t)=\sum_{n=-\infty}^\infty a_ne^{int}.$ It is not hard to see that $\widetilde{T}$ is bounded iff $a\in L^\infty(0,2\pi)$ and $\|T\|=\|\widetilde{T}\|=\|a\|_\infty.$
Consider the Toeplitz operator acting on $\ell^2(\mathbb{N}_0)$ by $T_{\rm tr}x=P_0Tx,$ where $P_0$ denotes the orthogonal projection from $\ell^2(\mathbb{Z})$ onto $\ell^2(0,1,2,\ldots).$ Clearly $\|T_{\rm tr}\|\le \|T\|.$ The norm of $T_{\rm tr}$ is the same as the norm of the operator $T_0=P_0TP_0$ acting on $\ell^2(\mathbb{Z}).$ As $T$ commutes with translations, i.e. $TS=ST,$ where $(Sx)_n=x_{n+1},$ the operator $T_0$ is unitarily equivalent to the operator $T_n=P_nTP_n,$ where $P_n$ is the orthogonal projection onto the subspace $\ell^2(-n,-n+1,-n+2,\ldots).$ As $P_n\to I$ strongly on $\ell^2(\mathbb{Z}),$ then $T_n\to T$ strongly. Therefore $\|T\|\le \liminf_{n\to \infty} \|T_n\|.$ However $\|T_n\|=\|T_0\|$ for every $n,$ hence $\|T\|\le \|T_0\|.$ Summarizing $\|T_{\rm tr}\|=\|T_0\|=\|T\|=\|a\|_\infty.$