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> Let $A:D(A)\subseteq E \to E$ be a closed operator on a complex Banach lattice $E.$ Then $A$ is said to be **real** if $x+iy \in D(A) \implies x,y \in D(A)$ for all $x,y \in E\_{\mathbb R}$ and $A(D(A) \cap E\_{\mathbb R}) \subseteq E\_{\mathbb R}.$ Here $E$ is the complexification of $E\_{\mathbb R}.$
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I would like some references which talk about spectral properties of real operators on Banach lattices.



Reference request: Spectral properties of real operators
**A few preliminary remarks:**


1) Complexifications of Banach lattices are in fact a special case of the more general concept of *complexifications of real Banach spaces*.


2) Most books and articles about complex Banach lattices which contain spectral theoretic results focus on *positive* operators (or, say, generators of positive semigroups) which are, of course, real. 


3) One reason for 2) might be that real operators can actually be defined on every complexification of a real Banach space (no matter whether a lattice structure is present), so the question about "spectral properties of real operators" belongs to the realm of complexifications of real Banach spaces rather than to the realm of complex Banach lattices. 


4) While most books on Banach lattices contain a chapter on complex Banach lattices (which is, unfortunately, very brief in most cases), there does not seem to exist much literature on complexifications of real Banach spaces (although the topic often seems to be considered as some kind of "standard folklore" among functional analysts).


The only article I'm aware of which contains a rather extensive treatment of complexifications of real Banach spaces is "G. A. Muñoz, Y. Sarantopoulos, and A. Tonge: Complexifications of real Banach spaces, polynomials and multilinear maps. Studia Math., 134(1):1–33, 1999." Unfortunately, though, spectral theory is not treated there.


**References to learn about spectral properties of real operators:**


If you primarily want to learn about the spectral properties of real operators, you can find some information in the following references (I apologize for the self-advertisement, but these are really the only references I know for this topic):


[1] Appendix C in my [PhD thesis](https://dx.doi.org/10.18725/OPARU-4238) contains a thorough treatment of complexifications. Spectral theoretic aspects are discussed in Section C.3, and a few spectral theoretic properties of real operators are discussed in Proposition C.3.2 (the fact that the operators considered in this proposition are exactly the class of real operators follows from Proposition C.1.6). However, I should add that Appendix C contains almost no proofs, for the following reason (which reflects, of course, only my personal point of view): when it comes to complexfications, it is in most cases more difficult to find the correct statements rather than to prove them (proving them is very straightforward in most cases).


[2] You can also find similar information in Appendix A of the [arXiv version 1](https://arxiv.org/abs/1410.2502v1) of my paper "Spectral and Asymptotic Properties of Contractive Semigroups on Non-Hilbert Spaces" (but in version 2, which was eventually published, I removed this appendix and replaced it with a very brief discussion of complexifications at the beginning of the paper in response to a request in a referee's report).


**References to quote:**


* The above references [1] and [2] are certainly not very well-suited in case that you need a reference to quote in a paper. Indeed, [2] is only a preprint which was eventually published without the information relevant here, and [1] contains almost no proofs (moreover, one might also question whether it is a particularly good idea to refer to the appendix of a PhD thesis for information about a quite classical topic in functional analysis - even though this topic does not seem to be particularly well represented in the literature).
* Unfortunately, I do not know any other references about the spectral theory of real operators (and I have severe doubts whether such a reference exists).
* So my suggestion is: If you need spectral properties of real operators in a paper, simply state them either in the introduction or in an appendix of the paper. You may include proofs if you feel they are necessary; but in many cases those properties are such elementary that you probably won't need to include proofs.


**Additional remark (biased by my personal opinion):**


I think it is reasonable to conlcude that a survey article about complexifications of real Banach spaces which complements the aspects discussed in the article of Muñoz, Sarantopoulos and Tonge would be a very useful source of reference. Now, all that is needed is a volunteer to write it and a journal willing to publish it...




---


**Edit in response to the OP's comment:**


Here are a few facts with proofs. Throughout, let $E$ be a complexification of a real Banach space $E\_{\mathbb{R}}$.


Note that an *everywhere defined* linear operator $T$ on $E$ is real if and only if $T(E\_{\mathbb{R}}) \subseteq E\_{\mathbb{R}}$.


**Proposition 1.** Let $A: E \supseteq D(A) \to E$ be a linear operator which is real. If $z \in D(A)$, then $\overline{z}$ is also in $D(A)$ and we have $A\overline{z} = \overline{Az}$.


*Proof.* We may write $z$ as $z = x+iy$, where $x,y \in E\_{\mathbb{R}}$. As $A$ is real, we have $x,y \in D(A)$ and $Ax$ as well as $Ay$ is in $E\_{\mathbb{R}}$. Hence,
\begin{align\*}
 A\overline{z} = A(x-iy) = Ax - iAy = \overline{Ax+ iAy} = \overline{Az}.
\end{align\*}
Note that the third equality above uses the fact that both $Ax$ and $Ay$ are elements of $E\_{\mathbb{R}}$.


**Proposition 2.** Let $A: E \supseteq D(A) \to E$ be a linear operator which is real. If $\lambda$ is a real number in the resolvent set of $A$, then the resolvent $R(\lambda,A) := (\lambda - A)^{-1}$ is real, too.


*Proof.* Let $x \in E\_{\mathbb{R}}$ and set $y := R(\lambda,A)x$. Then $y$ is an element of $D(A)$ and it can be written as $y = a + ib$ for unique $a,b \in E\_{\mathbb{R}}$. Since $A$ is real, we conclude that $a,b \in D(A)$.
Clearly, $x = (\lambda - A)y = (\lambda-A)a + i (\lambda - A)b$. Since $A$ is a real operator and $\lambda$ is a real number, both vectors $(\lambda - A)a$ and $(\lambda - A)b$ are elements of $E\_{\mathbb{R}}$. As the decomposition of each vector in $E$ into its real and imaginary part is unqie, we conclude that $x = (\lambda - A)a$. Hence, $R(\lambda,A)x = a \in E\_{\mathbb{R}}$.
This shows that $R(\lambda,A)$ is indeed real.


**Proposition 3.** Let $A: E \supseteq D(A) \to E$ be a linear operator which is real.


(a) If $\lambda \in \mathbb{C}$ is an eigenvalue of $A$ with eigenvector $z \in E$, then $\overline{\lambda}$ is an eigenvalue of $A$ with eigenvector $\overline{z}$.


(b) If $\lambda \in \mathbb{C}$ is a spectral value of $A$, then so is $\overline{\lambda}$.


*Proof.* (a) According to Proposition 1 we have $\overline{z} \in D(A)$ and $A\overline{z} = \overline{Az} = \overline{\lambda z} = \overline{\lambda} \overline{z}$.


(b) Assume that $\overline{\lambda}$ is not a spectral value of $A$. Then $A$ is closed, so we only have to show that $\lambda - A$ is a bijective mapping from $D(A)$ to $E$. 
Clearly $\lambda - A$ is injective, since otherwise it would follow from (a) that $\overline{\lambda} - A$ was not injective.
In order to show that $\lambda - A$ is surjective, let $z \in E$. Since $\overline{\lambda} - A$ is surjective, we can find a vector $c \in E$ such that $(\overline{\lambda} - A)c = \overline{z}$. Hence, we obtain
\begin{align\*}
 \overline{(\lambda - A)\overline{c}} = \overline{\lambda} c - Ac = (\overline{\lambda} - A) c = \overline{z}
\end{align\*}
(where we used Proposition 1 for the first equality), so $(\lambda - A)\overline{c} = z$. This proves that $\lambda - A$ is indeed surjective.


**Remark 4.** Let $A: E \supseteq D(A) \to E$ be a linear operator which is real and let $\lambda$ be a complex number. Then the following assertions hold:


(a) The number $\lambda$ is an eigenvalue of $A$ if and only if $\overline{\lambda}$ is an eigenvalue of $A$.


(b) The number $\lambda$ is a spectral value of $A$ if and only if $\overline{\lambda}$ is a spectral value of $A$.


*Proof.* The "only if" implications are the content of Proposition 3, and the converse implications also follow from Proposition 3 by applying the proposition to the number $\overline{\lambda}$ instead of $\lambda$.


**Remark 5.** The arguments in the proof of Proposition 3 actually show that we have
\begin{align\*}
 \overline{R(\lambda,A)z} = R(\overline{\lambda},A) \overline{z}
\end{align\*}
for each $\lambda$ in the resolvent set of $A$ and each $z \in E$. This can be considered as a generalisation of Proposition 2.


**Theorem 6.** Let $A: E \supseteq D(A) \to E$ be a linear operator which is real and let $\sigma$ be a compact subset of the spectrum $\sigma(A)$ such that $\sigma(A) \setminus \sigma$ is closed. Let $P$ denote the spectral projection associated to $\sigma$.


If $\sigma$ is invariant under complex conjugation (i.e. $\overline{\lambda} \in \sigma$ for all $\lambda \in \sigma$), then the operator $P$ is real.


*Sketch of the proof.* By Proposition 3 (or Remark 4) the entire spectrum $\sigma(A)$ is conjugation invariant. Since $\sigma$ is compact and $\sigma(A) \setminus \sigma$ is closed, we can find a closed (and smooth) cycle $\gamma$ in $\mathbb{C}$ which circumvents no element of $\sigma(A) \setminus \sigma$ but each element of $\sigma $ exactly once. As $\sigma$ and $\sigma(A) \setminus \sigma$ are conjugation invariant, we can choose $\gamma$ to be conjugation invariant, too, meaning that walking along $\overline{\gamma}$ is the same as walking along $\gamma$ in the converse direction. By definition of the spectral projection we have
\begin{align\*}
 P = \frac{1}{2\pi i} \int\_{\gamma} R(\mu,A) \, d\mu
\end{align\*}
Applying this to a vector $z \in E$ and employing Remark 5 we conclude that
\begin{align\*}
 \overline{Pz} = - \frac{1}{2\pi i} \int\_\gamma R(\overline{\mu}, A) \overline{z} \, \overline{d\mu} = \frac{1}{2\pi i} \int\_\gamma R(\mu,A) \overline{z} \, d\mu = P\overline{z}.
\end{align\*}
If $z \in E\_{\mathbb{R}}$ this implies that $\overline{Pz} = P\overline{z} = Pz$, so $Pz \in E\_{\mathbb{R}}$. Hence, $P$ is real.


Of course, one can prove many related results, but I hope that the above arguments suffice to give you an idea of how things usually work in the spectral theory of real operators.


