# The norm, resolvent set, and adjoint of the operator $T(x_1, x_2, \ldots) = (x_2, x_3, \ldots)$ on $\ell_1$

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:

1. The adjoint of $$T$$, $$T'$$, acts on $$\ell_\infty$$ by $$T'(x_1, x_2, \ldots) = (0, x_1, x_2). \tag{1}$$
2. $$\|T\| = \|T'\| = 1$$
3. 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.

1. 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?

2. It is obvious that $$\|T'\| = 1$$, but how is $$\|T\|_{\ell_1} = \sup_{\|(x_1, x_2, \ldots)\|_{\ell_1} = 1} \frac{\|(x_2, x_3, \ldots)\|_{\ell_1}}{\|(x_2, x_3, \ldots)\|_{\ell_1}} = 1?$$

3. Why is the assumption $$|\lambda| > 1$$ necessary here?

1. Write $$\ell = (y_1, y_2, \ldots)$$. Then \begin{align*}(T'(y_1,y_2,\ldots))x &= (T'\ell)x \\ &= \ell(Tx) \\ &= y_1x_2 + y_2x_3 + \cdots \\ &= 0x_1 + y_1x_2 + y_2x_3 + \cdots \\ &= (0,y_1,y_2,\ldots)x,\end{align*} so that $$T'(y_1,y_2,\ldots) = (0,y_1,y_2,\ldots)$$.

2. Note that $$\|Tx\|_1 = \|x\|_1 - |x_1| \leq \|x\|_1$$ and if $$x_2 = 1$$ and $$x_i = 0$$ for every $$i\neq 2$$, then $$\|Tx\|_1 = \|x\|_1 = 1$$.

3. At least in the case of $$T$$, the necessity of $$|\lambda| > 1$$ can be seen by the fact that any $$\lambda$$ such that $$|\lambda| < 1$$ is in the point spectrum of $$T$$. This can be seen by setting $$x_i = \lambda^i$$ and noting that $$x \in \ell_1$$ when $$|\lambda| < 1$$ but also $$Tx = \lambda x$$. Then just note the interior of the set corresponding to $$|\lambda| \geq 1$$ is $$|\lambda| > 1$$. This of course doesn't answer the possibly more interesting question about the sufficiency of $$|\lambda| > 1$$, which is what your quoted text seems to actually imply.

We can visualize the situation in the following commuting diagram: $$\require{AMScd} \begin{CD} (\ell^1)^\prime @>{T^\prime}>>(\ell^1)^\prime\\ @AAJA @AAJA \\ \ell^\infty @>{\tilde{T}^\prime}>> \ell^\infty \end{CD}$$

Here $$J: \ell^\infty \to (\ell^1)^\prime$$ is the natural isomorphism given by $$J(y) = \big(x \mapsto \sum_{i=1}^\infty x(i) y(i)\big)$$ and $$\tilde{T}^\prime$$ is the induced operator $$\tilde{T}^\prime = J^{-1} T^\prime J$$ (the action of $$T^\prime$$ on $$\ell^\infty)$$.

So the assertion in equation 1 is, that $$\tilde{T}^\prime (y_1, y_2, \dots ) = (0, y_1, y_2, \dots)$$.

To show this let $$y \in \ell^\infty$$ and $$x \in \ell^1$$. Then we have:

$$T^\prime(J(y)) (x) = J(y) \circ T (x) =\sum_{i=1}^\infty y(i) (Tx)(i) = \sum_{i=2}^\infty y(i-1) x(i) = J((0, y_1, y_2, \dots)) (x).$$

And so $$\tilde{T}^\prime = J^{-1}T^\prime(J(y)) = (0,y_1, y_2, \dots )$$.