I was dealing with this exercise of my functional analysis course :
Let $\mathcal{H} = l^2(\mathbb{Z})$, $\hspace{2mm}$ $U:\mathcal{H} \rightarrow \mathcal{H}$ such that $(Ux)_k = x_{k+1}$
a)Find $U^{*}$ and show that $U$ is unitary.
b)Show that $\forall x,y \in \mathcal{H}$ $\hspace{2mm}$ $(x,U^{n}y) \rightarrow 0$
For a) I know that if $U^{*}$ is the adjoint of $U$ it should be such that \begin{equation} (x,Uy) = (U^{*}x,y) \end{equation} and by the scalar product in $l^2$, I think I should do \begin{equation} (x,Uy) =\sum_{k \in \mathbb{Z}}x_k(Uy)_{k} = \sum_{k \in \mathbb{Z}}x_ky_{k+1} \end{equation} \begin{equation} (Ux,y) = \sum_{k \in \mathbb{Z}}(Ux)_ky_k = \sum_{k\in \mathbb{Z}}x_{k+1}y_k \end{equation}
So they must be the same, and I was thinking to change variable setting for example $k+1 = s$ to get in the first equation \begin{equation} \sum_{s \in \mathbb{Z}}x_{s-1}y_{s} \end{equation}
So I could say that $U^{*}$ is $(Ux)_k = x_{k-1}$? And How to show that it is Unitary?
For b) I really don't know where to start.. just thinking something like this \begin{equation} (x,U^ny) = \sum_{k \in \mathbb{Z}}x_k(U^{(n)}y)_k = \sum_{k \in \mathbb{Z}}x_ky^{(n)}_{k+1} \end{equation}
Any help is much appreciated! Thank you
For part a) your work is correct. To check that $U$ is a unitary, you have to see that it is a surjective isometry. "Surjective" is trivial, and so is "isometry": $$ \|Ux\|^2=\sum_n|x_{n-1}|^2=\sum_n|x_n|^2=\|x\|^2. $$ Equally easier is to notice directly that $U^*U=UU^*=I$.
For part b), given $\varepsilon>0$ there exists $k_0$ such that $\sum_{|k|>k_0}|x_k|^2<\varepsilon^2/\|x\|^2$, $\sum_{|k|>k_0}|y_k|^2<\varepsilon^2\|y\|^2$. Then, for $n>2k_0$, \begin{align} |(x,U^ny)|&=\sum_k\overline{x_k}y_{k+n}=\sum_{k\geq -k_0}\overline{x_k}y_{k+n}+\sum_{k<-k_0}\overline{x_k}y_{k+n}\\ \ \\ &\leq\|x\|\,\left(\sum_{k\geq- k_0}|y_{n+k}|^2\right)^{1/2} +\|y\|\,\left(\sum_{k<- k_0}|x_k|^2\right)^{1/2}\\ \ \\ &=\|x\|\,\left(\sum_{k\geq n- k_0}|y_{k}|^2\right)^{1/2} +\|y\|\,\left(\sum_{k<- k_0}|x_k|^2\right)^{1/2}\\ \ \\ &\leq\|x\|\,\left(\sum_{k\geq k_0}|y_{k}|^2\right)^{1/2} +\|y\|\,\left(\sum_{k<- k_0}|x_k|^2\right)^{1/2}\\ \ \\ &\leq\frac\varepsilon2+\frac\varepsilon2=\varepsilon. \end{align}