Let $T$ be the operator on $\ell_1$ such that $$T(x_1, x_2, \ldots) = (x_2, x_3, \ldots).$$
I am going through an example in Reed & Simon's book on functional analysis which makes the following claims:
- The adjoint of $T$, $T'$, acts on $\ell_\infty$ by $$T'(x_1, x_2, \ldots) = (0, x_1, x_2). \tag{1}$$
- $\|T\| = \|T'\| = 1$
- All $\lambda$ with $|\lambda| > 1$ are in the resolvent set of $T$ and $T'$, denoted $\rho(T)$ and $\rho(T')$, respectively.
I would like to go through these examples to make sure I have understood all the definitions presented so far.
The adjoint of a Banach operator $T: X \rightarrow Y$ between two Banach spaces is defined as the operator $T': Y' \rightarrow X^*$ defined as $$(T\ell)(x) = \ell(Tx), \quad \forall \ell \in Y^*, \forall x \in X.$$ In the example above, $X = Y = \ell_1$ and so $X' = Y' = \ell_\infty$. Thus we are looking for an operator such that $$(T' \ell)(x_1, x_2, \ldots) = \ell(x_2, x_3, \ldots)$$ for any $\ell \in \ell_\infty$. How does (1) follow from this?
It is obvious that $\|T'\| = 1$, but how is $$\|T\|_{\ell_1} = \sup_{\|(x_1, x_2, \ldots)\|_{\ell_1} = 1} \frac{\|(x_2, x_3, \ldots)\|_{\ell_1}}{\|(x_2, x_3, \ldots)\|_{\ell_1}} = 1?$$
Why is the assumption $|\lambda| > 1$ necessary here?