# 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 $\displaystyle x_{n}=\sum_{k=1}^{n}<x,e_{k}>e_{k}$. Then we have $||Tx_{n}||^{2}=||\sum_{k=1}^{n}<x,e_{k}>Te_{k}||^{2}=||\sum_{k=1}^{n}<x,e_{k}>e_{k+1}||^{2}= \sum_{k=1}^{n}|<x,e_{k}>|^{2}$. Therefore $||Tx||^{2}\stackrel{(\ast)}{=}\lim_{n\rightarrow{\infty}}||Tx_{n}||^{2}=\sum_{k=1}^{\infty}|<x,e_{k}>|^{2}=||x||^{2}$. 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 $\displaystyle ||Tx||=||T(\lim_{n\rightarrow\infty}x_{n})||=||\lim_{n\rightarrow\infty}Tx_{n}||=\lim_{n\rightarrow{\infty}}||Tx_{n}||$. 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?

Thank you!!

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There is a mistake: $Te_n = e_{n+1}$. – Siminore Apr 28 '12 at 14:53
@Siminore Oops :) – Vokram Apr 28 '12 at 14:54
A question: how do you extend the definition of $T$ on the basis to any element of $H$? I mean: what is $Tx$ if $x \in H$? – Siminore Apr 28 '12 at 14:56
@Siminore Yeah your question got me... It seems without continuity you can't really define $Tx$ for arbitrary $x$... Hmm so are you suggesting that if we ever want to define the right shift operator, we should assume already that it is continuous and bounded? – Vokram Apr 28 '12 at 15:01
I think this proof is not so meaningless because with assumptions that $T$ is bounded it is also proves that $T$ is isometric. – Norbert Apr 28 '12 at 15:05
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## 1 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.

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Thank you so much Nate, it is a enlightening explanation. – Vokram Apr 28 '12 at 19:38
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