I was reading my lecture notes for functional analysis when I came across the following statement:
Let $(e_{n})$ be a total orthonormal sequence in a separable Hilbert space H. The right shift operator, defined as the linear operator $T: H\rightarrow{}H$ such that $Te_{n} = e_{n+1}$ for all n, is bounded.
The statement seems intuitively correct to me, but I find the proof of it quite confusing. The proof goes like this:
Proof: For $\forall x\in{}H$, since $(e_{n})$ is total, write $\displaystyle x=\lim_{n\rightarrow{\infty}}x_{n}$, where $\displaystyle x_{n}=\sum_{k=1}^{n}\left
e_{k}$ . Then we have $||Tx_{n}||^{2}=||\sum_{k=1}^{n}\leftTe_{k}||^{2}=||\sum_{k=1}^{n}\left . Therefore $||Tx||^{2}\stackrel{(\ast)}{=}\lim_{n\rightarrow{\infty}}||Tx_{n}||^{2}=\sum_{k=1}^{\infty}|\lefte_{k+1}||^{2}= \sum_{k=1}^{n}|\left |^{2}$ |^{2}=||x||^{2}$ . Thus, $T$ is bounded and isometric.
However, I think there is something fishy with the proof: In the equality $(\ast)$, I believe the proof is using that $\displaystyle ||Tx||=||T\left(\lim_{n\rightarrow\infty}x_{n}\right)||=||\lim_{n\rightarrow\infty}Tx_{n}||=\lim_{n\rightarrow{\infty}}||Tx_{n}||$. But for the second equality to hold, it is already assuming that T is indeed continuous, which implies boundedness. And that makes it a circular reasoning here...
Is my judgement about the proof right? If this proof is indeed wrong, can anybody suggest a correct way to prove the statement?