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?
Best Answer
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.