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On page 371, "Introductory Functional Analysis with Applications", written Erwin Kreyszig, the author give the following definition (Definition 7.2-1):

Let $X \neq {0}$ be a complex normed space and $T: D(T) \rightarrow X$ a linear operator with domain $D(T) \subset X$. A regular value $\lambda$ of $T$ is a complex number such that

(R1) $R_{\lambda}(T)$ exists,

(R2) $R_{\lambda}(T)$ is bounded,

(R3) $R_{\lambda}(T)$ is defined on a set which is dense in $X$.

Question: Why do we need the condition that "a set which is dense" in this case?

The resolvent set $\rho(T)$ of $T$ is the set of all regular value $\lambda$ of $T$. Its complement $\sigma(T) = \mathbb{C} - \rho(T)$ in the complex plane $\mathbb{C}$ is called the spectrum of $T$, and a $\lambda \in \sigma (T)$ is called a spectral value of $T$. Furthermore, the spectrum is partitioned into three disjoint sets as follows.

The point spectrum or discrete spectrum $\sigma_p(T)$ is the set such that $R_{\lambda} (T)$ does not exist.

Question: Why do we use the terminology "point spectrum" for this definition?

The continous spectrum $\sigma _c (T)$ is the set such that $R_{\lambda}(T)$ exists and satisfies (R3) but not (R2), that is, $R_{\lambda}(T)$ is not bounded.

Question: Why do we use the terminology "continous spectrum" for this definition? How does it make sense?

The residual soectrum $\sigma _r (T)$ is the set such that $R_{\lambda}(T)$ exists (and may be bounded or not) but does not satisfy (R3), that is, the domain of $R_{\lambda}(T)$ is not dense in $X$

Question: Why do we use the terminology "residual spectrum" for this definition? How does it make sense?

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It appears that a lot of the complication arises because of trying to allow an incomplete space $X$.

The usual definition on the completion $\hat{X}$ of $X$ involves a densely-defined linear operator $T : \mathcal{D}(A)\subseteq\hat{X}\rightarrow\hat{X}$, and the resolvent set $\rho(T)$ consists of all $\lambda\in\mathbb{C}$ such that $(T-\lambda I)$ is injective and surjective with a bounded inverse $(T-\lambda I)^{-1} : \hat{X}\rightarrow\hat{X}$. There are several subtle issues that arise in trying to define everything on a potentially incomplete space $X$, and what Kreysig has given is undoubtedly carefully considered. For example, if you have a bounded operator $A : \mathcal{D}(X)\subseteq X\rightarrow X$ defined on a dense subspace $\mathcal{D}(X)$ of $X$, then $A$ extends by continuity to a bounded operator $\hat{A} : \hat{X}\rightarrow\hat{X}$, which is almost surely part of the motivation for the definition of the extended definition of the resolvent set $\rho(A)$.

Spectrum was originally considered for symmetric or selfadjoint operators arising out differential equations of Math-Physics. For a selfadjoint operator $$ A : \mathcal{D}(A) \subseteq X\rightarrow X, $$ there are really only two kinds of spectrum: (1) point spectrum $\sigma_p(A)$ consisting of eigenvalues $\lambda$ of $A$ and (2) continuous spectrum $\sigma_c(A)$. The point spectrum was natural. For example, suppose $X=L^2[-\pi,\pi]$, and suppose $$ A = \frac{1}{i}\frac{d}{dt},\\ \mathcal{D}(A) = \{ f \in X : f \mbox{ is absolutely cont with } f' \in L^2, \; f(-\pi)=f(\pi)\}. $$ Then $0,\pm 1,\pm 2,\pm 3,\cdots$ are eigenvalues of $A$ with eigenvectors $$ A e^{inx} = n e^{inx},\;\; n=0,\pm 1,\pm 2,\cdots. $$ This is a complete orthonormal basis with respect to the inner product $$ (f,g) = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)\overline{g(t)}dt, $$ meaning that every $f\in L^2[-\pi,\pi]$ can be written as $$ f = \sum_{n=-\infty}^{\infty}(f,e^{inx})e^{inx}. $$ The operator $A$ is diagonalized with respect to this basis: $$ Af = \sum_{n=-\infty}^{\infty} n(f,e^{inx})e^{inx} $$ Discrete spectrum or point spectrum are the terms applied to this case because of how vectors are decomposed into discrete sums of Fourier components.

The Fourier transform was the other significant example of a selfadjoint operator. Here, one considers $$ A = \frac{1}{i}\frac{d}{dt}, \\ \mathcal{D}(A) = \{ f \in L^2 : f \mbox{ is absolutely continuous, with } f'\in L^2 \}. $$ You would like to say that $e^{ist}$ is an eigenvector with eigenvalue $s$ because $$ \frac{1}{i}\frac{d}{dt} e^{ist} = s e^{ist}, $$ but the problem is that $e^{ist}\notin L^2$. By analogy to the "discrete spectrum" of the first example $A$, this operator $A$ has "continuous spectrum," and elements may be represented as "continuous" sums: $$ f = \int_{-\infty}^{\infty} "(f,e^{ist})" e^{ist} ds, \\ (f,e^{ist}) = \frac{1}{2\pi}\int_{-\infty}^{\infty}f(t)e^{-ist}dt, $$ and $A$ is diagonalized using this "continuous basis" $\{ e^{ist} \}_{s=\infty}^{\infty}$. $$ Af = \int_{-\infty}^{\infty} s(f,e^{ist})e^{ist}ds. $$ The problem with dealing with the Fourier transform in this way is that $e^{ist}$ is not an eigenvector because $e^{ist} \notin L^2$. However, $$ e_{s,\epsilon}(t)=\int_{s-\epsilon}^{s+\epsilon}e^{i\lambda t}d\lambda $$ is in the space, and the Plancherel theorem gives \begin{align} \|A e_{s,\epsilon}- se_{s,\epsilon} \|^2 & = \left\|\int_{s-\epsilon}^{s+\epsilon}(\lambda-s)e^{i\lambda t}dt\right\|^2 \\ & = 2\pi\int_{s-\epsilon}^{s+\epsilon}(\lambda-s)^2ds \\ & \le \epsilon^2 \| e_{s,\epsilon}\|^2. \end{align} In other words, $e_{s,\epsilon}$ is an approximate eigenvector with $$ \|(A-sI)e_{s,\epsilon} \| \le \epsilon \|e_{s,\epsilon}\|. $$ This is the "approximate point spectrum" or "continuous spectrum," the first term being more descriptive, and the second term referring to an integral decomposition instead of a discrete one.

The term "residual spectrum" is literally the "left over spectrum" that occurs when studying non-selfadjoint operators. For selfadjoint operators, $$ \mathcal{N}(A-\lambda I) = \mathcal{R}(A-\lambda I)^{\perp},\;\;\; \lambda\in\mathbb{R}. $$ For more general types of operators, you can have situations where $\mathcal{R}(A-\lambda I)$ is not dense, even though $\mathcal{N}(A-\lambda I)\ne \{0\}$. This type of spectrum is the "residual spectrum."

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  • $\begingroup$ Thank you so much. Your explanation is very clear and helpful for me. I really respect you. Thank you. $\endgroup$ – Trần Linh Apr 19 '17 at 13:51
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    $\begingroup$ You're welcome. History always helped me put things into perspective. Some people like this sort of thing, and others don't. $\endgroup$ – DisintegratingByParts Apr 19 '17 at 14:07
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    $\begingroup$ @TrầnLinh If you are satisfied with the answer, you should accept it by clicking the tick. $\endgroup$ – Hua May 18 '17 at 11:16
  • $\begingroup$ Thank you so much. I did it! $\endgroup$ – Trần Linh May 18 '17 at 15:38

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