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.