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In Hilbert spaces, the spectrum of any bounded (= continuous) operator is nonempty.

  • Could someone give me a reference on this? $~$Preferably, a simple text available to a physicist.

  • Is the spectrum of a bounded operator always nonempty also in Banach spaces?

Michael_1812
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    Yes. More than that, the spectrum of an element in any complex Banach algebra is not empty. For bounded operators over a Banach space it is the special case where we consider the algebra of bounded operators. Not sure if there is a different proof in the special case of operators. The proof of the general case can be found in any text about Banach algebras. For example, in the first chapter here: http://www.math.nagoya-u.ac.jp/~richard/teaching/s2014/Course_Wilde.pdf – Mark Oct 16 '23 at 21:29
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    Yes, it is non-empty. See here https://math.stackexchange.com/questions/3652590/proof-that-the-spectrum-in-non-empty It is essentially an application of Liouville's theorem (saying bounded holomorphic functions are constant). – Severin Schraven Oct 16 '23 at 21:29
  • @Mark Your answer was about Banach algebras, i.e. complex Banach spaces which are also algebras. My question, however, was about ordinary Banach spaces, not algebras. – Michael_1812 Oct 16 '23 at 22:40
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    @Michael_1812 My point is that if $X$ is a Banach space (doesn't have to be an algebra), then the space $L(X)$ of bounded linear operators $X\to X$ is a Banach algebra. So the spectrum of a bounded operator is a spectrum of an element in the algebra $L(X)$. – Mark Oct 16 '23 at 22:53
  • @Mark ...and is this spectrum always nonempty? (Sorry of my question is silly. I am not a mathematician.) – Michael_1812 Oct 16 '23 at 23:05
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    @Michael_1812 Yes, it is nonempty, as I wrote in the first comment. If you are interested in a proof, it appears in the link which I put in that comment. (on pages $6-7$) The space $L(X)$ of bounded linear operators on a Banach space $X$ is just a special case of a Banach algebra. – Mark Oct 16 '23 at 23:18
  • @Mark I am now reading the proof on page 7 of the lecture notes to which you kindly referred me. (I am interested specifically in the nonemptinness of the spectrum.) In the proof of this item, where exactly are we using the fact that the operators are on a Banach space -- i.e., normed and complete? – Michael_1812 Oct 21 '23 at 07:32
  • @Mark I guess it is important that the employed space of continuous linear functionals is dual to the space $A$ of the operators $x$. Is it dual in the sense that each such functional is acting on an operator to produce a $\mathbb C$ number? And then, again, why is it necessary here that the space of operators is Banach? – Michael_1812 Oct 21 '23 at 07:37
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    @Michael_1812 As I remember, they use that the spectrum is compact. This follows from the fact that the invertible elements of $A$ is an open set, and this requires completeness. (you need the fact that an absolutely convergent series is convergent) Note however that if $A$ is a normed algebra, it has a completion $B$ which will be a Banach algebra, as it can be proved. For $a\in A$, if $a-\lambda 1$ is not invertible in $B$ then in particular it is not invertible in $A$. So the spectrum of $a$ in $A$ is also not empty. So it follows that the statement itself is true in any normed algebra. – Mark Oct 21 '23 at 18:22
  • @Mark Did you really mean that "an absolutely convergent series is convergent"? Or, perhaps, "a convergent series is absolutely convergent" ? – Michael_1812 Oct 21 '23 at 18:59
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    @Michael_1812 I really mean that every absolutely convergent series is convergent. This is equivalent to a normed space $X$ being a Banach space. Without completeness, this is false. – Mark Oct 21 '23 at 19:47
  • @Mark "Without completeness, this is false." Wow... I need to cogitate about this. Not a mathematician, I easily get embarrassed by such counterintuitive things. :) – Michael_1812 Oct 21 '23 at 20:01
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    @Michael_1812 Well, if you remember how absolute convergence implies convergence for series of real (or complex) numbers, you can see that absolute convergence actually only implies that the sequence of partial sums of the original series is a Cauchy sequence. To deduce it converges we need completeness. – Mark Oct 21 '23 at 20:21
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    @Michael_1812 I mean to deduce the original series converges. (given that the series of norms is convergent) – Mark Oct 21 '23 at 20:24
  • @Mark I see a proof of this fact here. Is this what you imply? – Michael_1812 Oct 21 '23 at 20:26
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    @Michael_1812 No, I mean convergence of series in normed spaces. It's not about operator sequences. See here, for example: https://math.stackexchange.com/questions/999121/if-sum-n-1-infty-x-n-lt-infty-then-lim-k-to-infty-sum-n-1k-x?noredirect=1&lq=1 – Mark Oct 21 '23 at 20:30
  • @Mark Got it now. Your kind help is greatly appreciated! – Michael_1812 Oct 21 '23 at 20:32

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