0
$\begingroup$

Problem: Let $\lambda \in l^{\infty} (\mathbb{Z})$. Define the operator $$ T: l^2 (\mathbb{Z}) \rightarrow l^2 (\mathbb{Z}): (Tx)(n) = \lambda(n) x(n+1). $$ What is $T^{*}$?

Attempt: The adjoint satisfies $\langle Tx, y \rangle = \langle x, T^{*} y \rangle. $

I have $$ \langle Tx, y \rangle = \sum_{ n \geq 0} (Tx)(n) \overline{y(n)} = \sum_{ n \geq 0} \lambda (n) x(n+1) \overline{y(n)} = \sum_{k\geq 1} \lambda(k-1) x(k) \overline{y(k-1)} \\ = \sum_{k \geq 1} x(k) \overline{ \overline{\lambda(k-1)} y(k-1)} = \langle x, T^{*} y \rangle. $$ So I would say the adjoint is $(T^{*}x)(n) = \overline{\lambda(n)} x(n) $ if $n \geq 1$, and zero otherwise.

Is my reasoning correct? Help appreciated.

|cite|improve this question
$\endgroup$
  • $\begingroup$ Your approach is correct, but there are a few mistakes. For example, at the top you have $\ell^2(\mathbb{Z})$, but then $n \geq 0$ in the sum. Also, at the end you should have $x(n-1)$ and $\lambda(n-1)$. $\endgroup$ – Klaus Nov 15 at 14:23
  • $\begingroup$ So $(T^{*} x)(n) = \overline{\lambda (n-1)} x(n-1)$? $\endgroup$ – Kamil Nov 15 at 15:15
  • 2
    $\begingroup$ Yes. An easier way of thinking about this is that $T = MS$, where $M$ is the multiplication with $\lambda$ and $S$ is the shift. Hence you get $T^* = S^*M^*$, where $S^*$ is the backward shift and $M^*$ is the multiplication with $\overline{\lambda}$. $\endgroup$ – Klaus Nov 15 at 15:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.