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I would just like a brief sanity check for a question in Brezis's functional analysis book so I can be sure I understand the basics of spectral theory.

Take $l^p(\mathbb{R})$, for $1\leq p \leq \infty$. For a fixed, bounded, real sequence $\lambda_n$ define the multiplication operator $M: l^p \to l^p$ by: $$ M(x_1,x_2,x_3, \cdots ) = (\lambda_1 x_1 , \lambda_2x_2, \lambda_3x_3 ,\cdots) $$ i.e, a pseudo-dot product type thing. The question asks me to determine the Eigenvalues of $M$, and then the spectrum, $\sigma(M)$.

First, if we have an eigenvalue $\mu$, we have: $$ \mu x_1 = \lambda_1 x_1 \\ \mu x_2 = \lambda_2 x_2 \\ \mu x_3 = \lambda_3x_3 \\ \vdots $$ But $\lambda_n$ need not be a constant sequence, so this operator in general does not have an eigenvalue. For the Spectrum, we have that: $$ \sigma(M) = \{\mu \in \mathbb{R} : (M - \mu I) \text{ is not invertible}\} $$ In general, if $M - \mu I$ has an inverse, it is reasonable to assume its action on some $y$ is given by: $$ \left(\frac{1}{\lambda_1 - \mu}y_1, \frac{1}{\lambda_2- \mu}y_2,\frac{1}{\lambda_3 - \mu}y_3 , \cdots \right) $$ Thus, the operator is not invertible when $\mu = \lambda_k$ for some $\lambda_k$. Thus, the spectrum is given by $\{\lambda_n : n \in \mathbb{N}\}$.

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    $\begingroup$ How about the unit vectors $e_n$? Shouldn't they be eigenvectors with corresponding eigenvalues $\lambda_n$? Regarding the spectrum, what if $\mu-\lambda_n$ can get arbitrarily close to zero? Can you invert the operator $M-\mu I$ then? $\endgroup$
    – MSDG
    Commented Dec 8, 2018 at 17:34
  • $\begingroup$ Thank you, I see now the flaw in the first part of the question. However, I am not understanding what you mean by $\mu - \lambda_n$ approaching $0$. It seems that even if that did happen (knowing nothing about $\lambda_n$ here) the quotients $(\mu - \lambda_n)^{-1}$ would still exist and hence we would still have an inverse. $\endgroup$ Commented Dec 8, 2018 at 17:42
  • $\begingroup$ Assuming that it was invertible, it would give a bijection between $\ell^p$ and $\ell^p$. In particular, you should be surjective, which would allow you to take a sequence whose inverse image is not in $\ell^p$, which is absurd. So to achieve invertibility, you should require the existence of positive constants $\epsilon, \delta$ such that $\epsilon < |\lambda_n-\mu| < \delta$. Do you see what the spectrum must be then? $\endgroup$
    – MSDG
    Commented Dec 8, 2018 at 18:01
  • $\begingroup$ I believe I understand your point now. If $\lambda_n \to \mu$ then we may have the issue that $\frac{1}{\lambda_n - \mu}$ explodes, so that the $l^p$ norm of the product does not converge. Is this what you mean? If this is the case it would also have to be the case that the $\lambda_n$ converge to $\mu$ rapidly enough to make the $l^p$ sum diverge, although I am not quite sure how that would be formalized. $\endgroup$ Commented Dec 8, 2018 at 18:43
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    $\begingroup$ Yes, that was the idea idea. The newly posted answer formalizes this quite elegantly in my opinion. $\endgroup$
    – MSDG
    Commented Dec 8, 2018 at 18:43

1 Answer 1

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Since we are discussing the spectrum of an operator on a normed space, we consider the operator norm, which in the case of a multiplication operator happens to be the infinity norm. This is relevant because when discussing the spectrum, invertibility means the existence of a bounded inverse.

You will have eigenvectors wherever the sequence $\lambda$ is constant on some subset. In particular, on single points. Namely, for each set $E_m=\{n:\ \lambda_n=\lambda_m\}$, the sequence $x=1_{E_m}$, that is the sequence with $x_n=1$ if $n\in E_m$ and zero otherwise, satisfies $Mx=\lambda_m x$.

Thus $\{\lambda_n:\ n\in\mathbb N\}\subset\sigma(M)$. As the spectrum is always closed, $\overline{\{\lambda_n:\ n\in\mathbb N\}}\subset\sigma(M)$.

Conversely, if $\mu\not\in\overline{\{\lambda_n:\ n\in\mathbb N\}}$, there exists $\delta>0$ such that $|\mu-\lambda_n|>\delta>0$ for all $n$. Then $$\left|\frac1{\mu-\lambda_n}\right|<\frac1\delta$$ for all $n$, which gives that the sequence $\{1/(\mu-\lambda_n)\}_n$ is bounded, and so $\mu I-M$ is invertible. This shows that $\overline{\{\lambda_n:\ n\in\mathbb N\}}\supset\sigma(M)$ and so $$ \sigma(M)=\overline{\{\lambda_n:\ n\in\mathbb N\}} $$

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  • $\begingroup$ Thank you for the answer, I am having some issues understanding your sequencing argument however. I understand that if we assume $\mu$ is not in the closure, there is no sequence of $\lambda_n$ s converging to $\mu$, hence there exist infinitely many $m$ such that $\frac{1}{\lambda_m - \mu} < \frac{1}{\delta}$. However, how can we choose $N$ such that the above is true for all $n \geq N$? It seems that all we know is the existence of at least one such for value for every natural number $N$. $\endgroup$ Commented Dec 8, 2018 at 19:06
  • $\begingroup$ If $K$ is closed and $\mu\not\in K$, this means that $\operatorname{dist}(\mu,K)>\delta$ for some $\delta$. That is, $|\mu-\lambda|>\delta$ for all $\lambda\in K$. $\endgroup$ Commented Dec 8, 2018 at 19:14
  • $\begingroup$ That makes sense. Thank you. $\endgroup$ Commented Dec 8, 2018 at 19:16
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    $\begingroup$ Why does $\{1/(\mu-\lambda_n)\}$ being bounded imply $\mu I - M$ invertible? I am trying to work through this as well as the comments in the OP, but even if we guarantee the terms don't blow up, it does not mean anything. $\endgroup$
    – user760286
    Commented Apr 16, 2020 at 6:58
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    $\begingroup$ Multiplication by $\{\tfrac1{\mu-\lambda_n}\}$ is precisely the inverse of $\mu I-M$. $\endgroup$ Commented Apr 16, 2020 at 12:58

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