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Here is the question:

Considering the right shift operator $S$ on $\ell^2({\bf Z})$, what can one know about ran$(S-\lambda)$?

Here is what I thought:

  • If one wants to prove that the operator $S-\lambda$ is onto when $\lambda$ satisfies some conditions, does one have to construct the solution?

  • One needs to find the solution $(S-\lambda)x=y$ where $x,y\in \ell^2({\bf Z})$. Using the standard basis $e_n=(\delta_{nk})_{k=-\infty}^{\infty}$, one has to solve $\lambda x_{k}-x_{k-1}=y_{k}, (k\in {\bf Z})$.

  • Intuitively, when $|\lambda|=1$, there may be no way for $S-\lambda:\ell^2({\bf Z})\to \ell^2({\bf Z})$ to be onto. However, when $|\lambda|\neq 1$, can one explicitly find $x=(x_{k})_{k=-\infty}^{\infty}$?

[ADDED] Thanks to a recent editing of my question, I have learned that the right and left shift operators acting on two-sided infinite sequences are also called bilateral shifts.

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    Hint: Fourier series – Robert Israel Apr 11 '11 at 23:57
  • The basis $e_n=(\delta_{in})_{i=-\infty}^{\infty}$ seems not helpful here. I think the key point here is to solve the equation $(S-\lambda)x=y$ in $l^2({\bf Z})$. I don't understand how Fourier series work here. –  Apr 12 '11 at 01:11
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    That's not what I meant. I was referring to the unitary operator from $l^2({\mathbb Z})$ onto $L^2([0, 2 \pi))$ defined by $(U x)(t) = \sum_{k = -\infty}^{\infty} x_k e^{i k t}/\sqrt{2 \pi}$. This transforms the shift to a multiplication operator. – Robert Israel Apr 12 '11 at 02:34
  • The right shift operator is a particular unitary operator. Thus the spectral analysis of the unitary operator is useful to answer the question here. –  Apr 13 '11 at 17:27
  • @RobertIsrael : How did u come up with the idea of using Fourier series ? I would love to get some more explanation on how u came up with it . – Theorem Jan 14 '13 at 21:34
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    @Theorem: I didn't, Fourier did. But seriously, turning translation (and hence, difference or differentiation) into multiplication is one of the main reasons Fourier analysis is useful. – Robert Israel Jan 15 '13 at 04:26
  • related: http://math.stackexchange.com/q/617601/173147 – glS Jun 08 '15 at 07:11

2 Answers2

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The right shift is unitary, so its spectrum is contained in the unit circle. It has no eigenvalues, so $S-\lambda$ is always injective. The spectrum is nonempty, so there exists $\lambda_0$ with $|\lambda_0|=1$ such that $S-\lambda_0$ is not invertible. For each $\lambda$ in the unit circle, the operator $\lambda S$ is unitarily equivalent to $S$, via the diagonal unitary operator $$(\ldots,x_{-2},x_{-1},x_0,x_1,x_2,\ldots)\mapsto(\ldots,\overline{\lambda}^2x_{-2},\overline{\lambda}x_{-1},x_0,\lambda x_1,\lambda^2x_2,\ldots).$$ Thus for each $\lambda$ in the unit circle, $\sigma(S)=\sigma((\lambda\cdot\overline{\lambda_0})S)=(\lambda\cdot\overline{\lambda_0})\sigma(S)$, and therefore $\sigma(S)$ contains $\lambda$. This shows that the spectrum of $S$ is the unit circle.

For each $\lambda\in\sigma(S)$, $S-\lambda$ is not surjective, because $S-\lambda$ is injective but not invertible. However, $S-\lambda$ has dense range, which follows from the fact that the left shift $S^*$ also has no eigenvalues.

Feel free to ask for elaboration on any of the claims I've made.

Jonas Meyer
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    Thanks for your answer. First of all, why the the spectrum of the unitary operator is "contained in the unit circle"? Second, I think it is because $S-\lambda$ is always injective that it has no eigenvalues, correct? What's more, most part of your argument based on the property of the unitary operator. Would you please recommend any reference books for this topic? Some books for the functional analysis do not elaborate the spectral theory topic. –  Apr 12 '11 at 01:57
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    @Jack: For your first question, $|S|=1$ and $|S^-1|=1$, so the spectral radii of $S$ and of $S^{-1}$ are less than or equal to $1$. Since $\sigma(S^{-1})={\lambda^{-1}:\lambda\in\sigma(S)}$, the spectrum consists of complex numbers of modulus less than $1$ whose reciprocals also have norm less than $1$, and therefore it is contained in the unit circle. For your second questoin, yes, the statement that $S$ has no eigenvalues is the same as the statement that $S-\lambda$ is injective for all $\lambda$. – Jonas Meyer Apr 12 '11 at 02:03
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    For reference, any good text on functional analysis that includes a little theory of operators on Hilbert space should be helpful. There are many, not all of which I'm very familiar with, so I'm hesitant to recommend particular choices, but I like J.B. Conway's A course in functional analysis for one. I do highly recommend Halmos's Hilbert space problem book, but this is not a traditional text. You wouldn't need an in depth treatment for just this purpose; my claims are either not very deep, or completely standard in the sense that they should be in most books on functional analysis. – Jonas Meyer Apr 12 '11 at 02:08
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    For example, non-emptiness of the spectrum holds for all operators on Banach space, and this is a theorem of Gelfand. Similar operators have equal spectrum, and this is a purely algebraic fact, as is $\sigma(aT)=a\sigma(T)$ when $a$ is a scalar. And of course, you're welcome to ask for further elaboration. – Jonas Meyer Apr 12 '11 at 02:15
  • @Jonas Meyer: I still don't quite understand how did you deduce that $S-\lambda$ has dense range from the property of $S^*$. –  Apr 13 '11 at 06:58
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    @Jack: If $T$ is an operator on a Hilbert space $H$ and $T^$ is its adjoint, then $(TH)^\perp=\mathrm{ker}T^$. So for all $\lambda\in\mathbb{C}$, $((S-\lambda)\ell^2(\mathbb{Z}))^\perp=\mathrm{ker}(S^*-\overline{\lambda})={0}$, which implies that $S-\lambda$ has dense range. – Jonas Meyer Apr 13 '11 at 11:54
  • Excuse me, could you please tell me why $\sigma(S) = \sigma(\lambda \bar{\lambda}_0 S)$? – Ryze Jun 28 '20 at 01:53
  • @Fyhswdsxjj: Unitarily equivalent operators have the same spectrum. – Jonas Meyer Jun 28 '20 at 13:59
  • @JonasMeyer Thanks! But what wrote is “$\lambda S$ is unitarily equivalent to $S$”, not “ $\lambda \bar{\lambda_0}$ is unitarily equivalent to $S$”. – Ryze Jun 28 '20 at 14:13
  • @Fyhswdsxjj: The first $\lambda$ you quoted is an arbitrary point in the circle, for the statement of the general fact. Where it is applied, the fact that $\lambda \overline {\lambda_0}$ is a point in the circle is used. – Jonas Meyer Jun 28 '20 at 16:27
  • @JonasMeyer Thank you very much! – Ryze Jun 29 '20 at 00:39
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To address the final part of your question: you can write down the inverse of $S-\lambda$ for $|\lambda|\ne1$, so in this sense you can "explicitly find $x$". (Although the spectral argument is a more efficient way to answer the range question).

If $\|T\|<1$ then $1-T$ is invertible, and $$ (1-T)^{-1}=\sum_{k\ge0}T^k.$$ Wikipedia calls this the Neumann series of $T$.

Since $S^{-1}$ is the backward shift, we have $\|S^{-1}\|=1$. So if $0<|\lambda|<1$ then $\|\lambda S^{-1}\|<1$ and $$(S-\lambda)^{-1}=-S^{-1}(1-\lambda S^{-1})^{-1}=S^{-1}\sum_{k\ge0}(\lambda S^{-1})^k.$$

Since $\|S\|=1$, if $|\lambda|>1$ then $\|\lambda^{-1}S\|<1$ so $$ (S-\lambda)^{-1}=-\lambda^{-1}(1-\lambda^{-1} S)^{-1}=-\lambda^{-1}\sum_{k\geq 0}(\lambda^{-1}S)^k.$$

mac
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