Is the right shift operator bounded? 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
  $\displaystyle x_{n}=\sum_{k=1}^{n}\left<x,e_{k}\right>e_{k}$. Then we have
  $||Tx_{n}||^{2}=||\sum_{k=1}^{n}\left<x,e_{k}\right>Te_{k}||^{2}=||\sum_{k=1}^{n}\left<x,e_{k}\right>e_{k+1}||^{2}= \sum_{k=1}^{n}|\left<x,e_{k}\right>|^{2}$. Therefore
  $||Tx||^{2}\stackrel{(\ast)}{=}\lim_{n\rightarrow{\infty}}||Tx_{n}||^{2}=\sum_{k=1}^{\infty}|\left<x,e_{k}\right>|^{2}=||x||^{2}$.
  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?
 A: You are right that there is circularity here.
The problem is in your definition of the right shift operator as "the" linear operator such that $T e_n = e_{n+1}$.  In fact, there are many such linear operators.  (Using Zorn's lemma, we can extend $\{e_n\}$ to a Hamel basis for $H$ by adding some additional vectors $\{u_\alpha\}$.  Then we can define an operator $T$ by setting $T e_n = e_{n+1}$ and setting $T_{u_\alpha}$ to be whatever we want, and this uniquely defines a linear operator.)
So the statement that $T x = \lim T x_n$ will have to be part of the definition of $T$.  Following your approach, given $x \in H$, let $x_n = \sum_{k=1}^n \langle x,e_k \rangle e_k$.  Then $T x_n$ is unambiguously given by $\sum_{k=1}^n \langle x, e_k \rangle e_{k+1}$.  Show that the sequence $\{T x_n\}$ is Cauchy and hence converges to some $y \in H$.  Then we can define $Tx$ to be $y$.
Now that $T$ is well defined, one can go ahead and check that $T$ is linear, bounded, and an isometry.
The moral is that defining a linear operator on a total orthonormal set is only well defined if the operator is assumed to be bounded.
