2

Let $T$ be a right shift operator on $l^2(\mathbb{N})$. We know that residual spectrum of $\sigma_r(T) = \{\lambda \in \mathbb{C} \mid |\lambda| < 1\}$.

Indeed, computing residual spectrum using the theory of adjoint operator by viewing $\ell^{2}$ as a Hilbert space is simple. However, this time I want to prove it directly without using the properties of adjoint. Namely, I want to construct $l^{*} \in (l^2)^*$ such that $l^{*}$ vanishes on the range of $(T-\lambda I)$ where $|\lambda| < 1$, but I am stuck on constructing such an element in the dual space.

I am pretty sure you need to use the property where $|\lambda| < 1$, and only use case I can think of per now is, in general, to use the fact that $|\lambda|^n \rightarrow 0$ as $n$ goes to infinity. I tried telescoping sum, but either I fail to achieve linearity or well-definedness (i.e. image may not be a real valued).

In this case, how could I enhance my approach and show that $\{\lambda \in \mathbb{C} \mid |\lambda| < 1\} \subset \sigma_r(T)$? Thanks in advance!

glS
  • 7,963
MHC_Class_2
  • 311
  • 1
  • 7
  • You can use the properties of the adjoint to find such $l^$, namely write $\langle (T-\lambda I)v, l^\rangle = \langle v, (T^-\overline \lambda I)l^\rangle$. Now choose $l^$ such that $(T^-\overline \lambda I)l^*=0$ – leoli1 Aug 28 '24 at 21:20
  • related: https://math.stackexchange.com/q/617601/173147 – glS Aug 30 '24 at 18:14

1 Answers1

2

Consider $\lambda \in \mathbb C$ such that $|\lambda| < 1$ and let $$l_{\lambda}(x) = \sum_{n \in \mathbb N} \lambda^n x_n$$ for $x = \{x_n\}_{n \in \mathbb N} \in l^2(\mathbb N)$. It is easy to see that $l_\lambda$ is an element of $(l^2(\mathbb N))^*$, i.e. it is a continuous linear functional. The computation $$ l_\lambda(Tx - \lambda x) = l_\lambda(Tx) - \lambda l_\lambda(x) = \sum_{n = 2}^\infty \lambda^{n}x_{n-1} - \lambda \sum_{n = 1}^\infty \lambda^n x_n = \sum_{n = 1}^\infty \lambda^{n+1} x_n - \sum_{n = 1}^\infty \lambda^{n+1} x_n =0. $$ shows that $l_\lambda$ vanishes on the range of $T - \lambda I$.

To give more context, $\lambda$ belongs to the residual spectrum of $T$ if and only if the range of $T - \lambda I$ is not dense and this holds if and only if $T^* - \overline{\lambda}I$ has nontrivial kernel. In this case every element $l \in \ker (T^* - \overline{\lambda} I)$ gives rise to a functional $x \rightarrow \langle x, l \rangle$ which vanishes on range of $T - \lambda I$ (this can be easily derived from the definition of the adjoint operator). Clearly, as $T^*$ is the left shift operator one can easily find $l \in \mathrm{ker}(T^* - \overline{\lambda} I)$ provided $|\lambda| < 1$: $l = \{\overline{\lambda^n}\}_{n \in \mathbb N} \in l^2(\mathbb N)$. Clearly $\langle x, l\rangle = \sum x_n \lambda^n = l_\lambda(x)$. That is a way to find functionals that vanish on the range explicitly using adjoint operators.

Matsmir
  • 4,159