In Arveson's book "A Short Course on Spectral Theory", the concept of a unilateral weighted shift operator is introduced in one of the excercises in the Section 1.6.
Given a bounded sequence $(a_n)$ of complex numbers and a Hilbert space $\mathcal{H}$ with orthonormal basis $\{e_i\}$, we want to show that there is a ($\textbf{necessarily unique}$) bounded operator $A \in B(\mathcal{H})$ satisfying $Ae_n = a_n e_{n+1}$ for $n = 1,2,…$. Such an operator is called a unilateral weighted shift.
I can see that this weighted shift is the product of a shift ($Se_n = e_{n+1}$) and a compatible diagonal operator ($Pe_n = a_ne_n$). The derivation of the spectrum $\sigma(A)$ is also fairly straightforward. However, I do not see why it is necessarily unique? Why there is a bounded operator satisfying the above? I feel like I am missing something rather obvious. Any guiding in the darkness would be appreciated.
Best Answer
$\|A(\sum c_ne_n)\|=\|\sum a_nc_n e_n\|=\sqrt {\sum |a_nc_n|^{2}}\leq M \|\sum c_ne_n\|$ where $M=\sup_n |a_n|$. Hence, $A$ is a bounded opeartor with norm at most $M$. Uniqueness is because $A(\sum c_ne_n)=\sum a_nc_n e_n$.
[Recall that $\sum c_ne_n$ converges if and only if $\sum |c_n|^{2} <\infty$].