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?


2 Answers 2

  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.

  • $\begingroup$ Thank you for your answer! $\endgroup$
    – CBBAM
    Jun 22 at 7:03

To answer the first question:

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 ) $.

The other two questions are already answered in detail in the answer by Brian Moehring.


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