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This is my second attempt at calculating the spectrum of the left shift operator on a Hilbert space. I got stuck again and I would be grateful if someone could help. (You can find my previous (failed) attempt here).

Let $H$ be a Hilbert space with orthonormal base $e_i$ and let $L\in B(H)$ be the left shift operator, that is, $L(e_i) = e_{i-1}$.

In the following I will assume that if there exists a sequence $x_n\in H$ with $\|x_n\|=1$ and $Tx_n \to 0$ then $T$ is not invertible.

First note that $L$ is not invertible hence $0 \in \sigma (L)$.

Let $\lambda \in \mathbb C$ be non-zero.

I am stuck trying to find $x_n$ with $\|x_n\|=1$ and $(T-\lambda I)x_n \to 0$.

Edit

After doing some calculations I think that $L - \lambda I$ is never invertible for any $\lambda \in \mathbb C$ but I can't find $x \neq 0$ such that $(L-\lambda I)x = 0$. Could someone help me find such $x$ please? I'm sure that $\sigma(L) = \mathbb C$.

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1 Answers1

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Since $\lVert L\rVert = 1$, it follows that $L-\lambda I$ is invertible for all $\lambda$ with $\lvert\lambda\rvert > 1$, we can see that using the Neumann series. For $\lvert \lambda\rvert > \lVert L\rVert$, the series

$$\sum_{n=0}^\infty \lambda^{-n}L^n$$

is absolutely convergent, and since $B(H)$ is a Banach space, it is convergent. One computes

$$(I-\lambda^{-1}L)\sum_{n=0}^\infty \lambda^{-n}L^n = \lim_{N\to\infty}(I-\lambda^{-1}L)\sum_{n=0}^N \lambda^{-n}L^n = \lim_{N\to \infty} I - \lambda^{-N-1}L^{N+1} = I,$$

and ditto for $\biggl(\sum\limits_{n=0}^\infty \lambda^{-n}L^n\biggr)(I-\lambda^{-1}L)$, and from that one sees that $L-\lambda I$ is invertible when $\lvert\lambda\rvert > \lVert L\rVert$, and the inverse is given by the series

$$(L-\lambda I)^{-1} = -\sum_{n=0}^\infty \lambda^{-n-1}L^n$$

then.

This argument uses no specific property of $L$, it works for all bounded linear operators on Banach spaces. The spectrum of a bounded linear operator $T$ on a Banach space is always conatined in the closed disk with radius $\lVert T\rVert$ and centre $0$.

For the left shift operator $L$, we have - using the separable Hilbert space $\ell^2(\mathbb{N})$ to get a convenient notation -

$$(L-\lambda I)x = (x_1 -\lambda x_0, x_2 - \lambda x_1, x_3 - \lambda x_2,\dotsc).$$

Thus $(L-\lambda I)x = 0$ if and only if we have $x_1 = \lambda x_0$, $x_2 = \lambda x_1$, and so on, which becomes $x_n = \lambda^n x_0$. For $\lvert \lambda\rvert < 1$, that defines a one-dimensional subspace of $H$, spanned by

$$\nu_\lambda = \sum_{n=0}^\infty \lambda^n\cdot e_n.$$

Thus we have $\ker (L-\lambda I) \neq \{0\}$ for $\lvert\lambda\rvert < 1$, and hence $\sigma(L)$ contains the open unit disk.

Since the spectrum of a bounded linear operator is compact, and we saw above that $\sigma(L)$ is contained in the closed unit disk, it follows that

$$\sigma(L) = \overline{\mathbb{D}} = \{ \lambda\in\mathbb{C} : \lvert \lambda\rvert \leqslant 1\}.$$

Daniel Fischer
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  • Regarding the calculation of $|L|$: could you check if I'm doing it right? Here is how I do it: $$\$$

    (1) $|L|$ is $\le 1$ because $|Lx|\le|x|$ for all $x \in H$. $$\$$

    (2) $|L|$ is $\ge 1$ because $x=(0,1,0,\dots)$ has norm $1$ and so does $Lx$.

     –  Dec 24 '14 at 02:55
  • I have another question about your answer, if I may: you write "using the separable Hilbert space $\ell^2$ to get convenient notation" -- does this mean that all Hilbert spaces with a countable orthonormal basis are isomorphic? –  Dec 24 '14 at 04:37
  • And there is a third thing troubling me: in a previous question of mine it was pointed out to me that the eigenvalues are not the only points in the spectrum. But in your answer you are only computing the eigenvalues ($\lambda$ with $(L-\lambda I)x=0$) if I understand correctly. So normally this method you use would not work but in the special case of the shift operator it happens to work nicely? –  Dec 24 '14 at 04:41
  • Are my questions stupid? –  Dec 25 '14 at 00:18
  • Don't think so, I have missed the notifications. Give me a few minutes to read your questions. – Daniel Fischer Dec 25 '14 at 00:19
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    Confirmed, none of them is stupid. 1. Your calculation of $\lVert L\rVert$ is correct and pretty much the best way to do it (the "best way" is non-unique here). 2. Yes, even more general, all Hilbert spaces with orthonormal bases of the same cardinality are isometrically isomorphic. If you have an orthonormal basis indexed by $A$, your Hilbert space is isometrically isomorphic to $\ell^2(A)$ by Parseval's identity, and $\ell^2(A) \cong \ell^2(B)$ if and only if there is a bijection between $A$ and $B$. So every Hilbert space with a countable (infinite) orthonormal basis is isometrically ... – Daniel Fischer Dec 25 '14 at 00:26
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    ... isomorphic to $\ell^2(N)$. 3. Yes, the spectrum generally does not consist only of eigenvalues (but it can of course, consider the identity or $0$ for trivial examples). That is also the case here. But since we know that the spectrum is contained in the closed disk of radius $\lVert L\rVert$ and is closed, and here every point in the open disk with radius $\lVert L\rVert$ is an eigenvalue, we know that $\sigma(L)$ is a closed set with $${ z : \lvert z\rvert < \lVert L\rVert} \subseteq \sigma(L)\subseteq { z : \lvert z\rvert\leqslant \lVert L\rVert},$$ hence the latter is an equality. – Daniel Fischer Dec 25 '14 at 00:31
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    So in this case, to determine the spectrum, it is sufficient to find the eigenvalues. That is a nice property of the left-shift operator. In general, determining the spectrum of an operator can be much more difficult since one also must find the values $\lambda$ such that $\lambda I - T$ is not surjective, which tends to be harder than finding the eigenvalues. – Daniel Fischer Dec 25 '14 at 00:34
  • @student: The left shift also has spectral values which are not eigenvalues, namely all values on the unit circle are so: $\mathbb{S}=\partial\sigma(L)$ – C-star-W-star Dec 26 '14 at 18:18
  • But I don't really need to know that all Hilbert spaces with countable orthonormal basis are isomorphic to $\ell^2$ in order to calculate this spectrum, right? I could argue that because the spectrum is contained in the unit disk the only candidates for the spectrum are $\lambda$ with $|\lambda |\le 1$. Then I can argue that $(L-\lambda)x = 0$ if $x=(\lambda, \lambda^2, \lambda^3 , \dots)$ hence for all $\lambda$ in the closed unit disk, $L-\lambda$ is not invertible. ...Right? –  Jan 03 '15 at 02:34
  • (I have to additionally note that it is clear that $L$ is not invertible and therefore $0$ is also in the spectrum...) –  Jan 03 '15 at 02:35
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    @student Right. You don't need to know that all separable Hilbert spaces are isometrically isomorphic to $\ell^2(\mathbb{N})$ to calculate the spectrum, I just find the notation nicer for $\ell^2$. If you take the eigenvector of $L-\lambda I$ as $x = (1,\lambda,\lambda^2,\lambda^3,\dotsc)$, you don't even need to special-case $\lambda = 0$. That tells us every $\lambda$ with $\lvert\lambda\rvert < 1$ is an eigenvalue, and hence every $\lambda$ with $\lvert\lambda\rvert\leqslant 1$ is a spectral value, since the resolvent set is open. – Daniel Fischer Jan 06 '15 at 17:07