Let $(X,||\cdot||)$ be a Banach space on $\mathbb{C}$ and $\mathcal{L}(X)$ the set of linear, continuous functions from $X$ to itself. For $T\in\mathcal{L}(X)$, define the norm $||T||_{\mathcal{L}(X)}:=\sup_{v\neq 0}\frac{||T(v)||}{||v||}$ and the continuous spectrum $\sigma(T):=\{\lambda\in\mathbb{C}\mid A-\lambda I\,\text{ has no inverse}\}$. Define the sets: $$GL(X):=\{T\in\mathcal{L}(X)\mid T\text{ is invertible}\}$$ $$\mathcal{H}(X):=\{T\in GL(X)\mid T\text{ is hyperbolic}\}$$ (where hyperbolic means $|\lambda|\neq 1$ for all $\lambda\in\sigma(T)$)

Prove that: (1) $GL(X)\subset\mathcal{L}(X)$ is open and (2) $\mathcal{H}(X)$ is open, dense in $GL(X)$.

(1) Take $A\in GL(X)$ and the open ball $B_r(A)$, where $r:=1/||A^{-1}||$. If $B\in B_r(A)$, then $||I-BA^{-1}||\leq||A-B||\cdot||A^{-1}||<r||A^{-1}||=1$. The series $\sum_{n=0}^\infty(I-BA^{-1})^n$ is therefore convergent, so $BA^{-1}$ is invertible with $(BA^{-1})^{-1}=(I-(I-BA^{-1}))^{-1}=\sum_{n=0}^\infty(I-BA^{-1})^n$. Therefore, $B$ is invertible $\Rightarrow B_r(A)\subset GL(X)$. $_\blacksquare$

(2) We prove $GL(X)\setminus\mathcal{H}(X)$ is closed in $GL(X)$. If $\{A_n\}_{n\in\mathbb{N}}\subset GL(X)\setminus\mathcal{H}(X)$ a Cauchy sequence converging to $A\in GL(X)$, we need to prove $A\notin \mathcal{H}(X)$. By definition, for all $n$ there is $\lambda_n\in \mathbb{C}$ such that $|\lambda_n|=1$ and $A_n-\lambda_nI\notin GL(X)$. Since $\mathbb{S}^1:=\{\lambda\in\mathbb{C}\mid |\lambda|=1\}$ is compact, there is a convergent subsequence $\{\lambda_{i_n}\}_{n\in\mathbb{N}}$ converging to $\lambda\in\mathbb{S}^1$, so $\lim_{n\to\infty} A_{i_n}-\lambda_{i_n}I=$ $A-\lambda I$. Since $A_{i_n}-\lambda_{i_n} I\notin GL(X)$ (which is open), $A-\lambda I\notin GL(X)$, so $A\in GL(X)\setminus\mathcal{H}(X)$.

To prove $\mathcal{H}(X)$ is dense in $GL(X)$, my idea is this: let $A\in GL(X)\setminus\mathcal{H}(X)$ and $\epsilon>0$ arbitrary, then there exist $z\in\mathbb{C}$ with $|z|<\epsilon$ such that $A+zI\in\mathcal{H}(X)$. I feel like this could work, but I don't know how to guarantee $\sigma(A+zI)\cap\mathbb{S}^1=\emptyset$.

  • 4
    $\begingroup$ Adding a small multiple of identity is not going to work. For example, the spectrum of shift operator $S$ is the closed unit disk. Let $T=S + (3/2)I$; this shifts the spectrum by $3/2$, so $T$ is invertible. But $T$, and any $T+zI$ with $|z|<1/2$, still has the spectrum intersecting the unit circle. $\endgroup$
    – user357151
    May 28, 2018 at 6:24
  • 2
    $\begingroup$ Out of curiosity, where did this problem come from? $\endgroup$
    – Wraith1995
    Jun 1, 2018 at 20:22

2 Answers 2


The statement that for $X$ an $\mathbb{F}$-Banach space ($\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$), the set $H(X)$ of hyperbolic automorphisms is dense in $GL(X)$ is true when $\dim(X)<\infty$, and not necessarily true in general (as opposed to openness, which holds without the finite dimension assumption).

(I'm assuming the adjective "continuous" of spectrum above is a typo.)

Finite dimensional Case: In this case the already mentioned argument of "adding a small multiple of identity" works (moving finitely many points on the circle in one direction on the complex plane so that none are on the circle is not hard). One can even be very explicit as to how small of a multiple one ought to use; see Palis and de Melo's Geometric Theory of Dynamical Systems, p.50, Prop.2.11 for details (for the $\mathbb{F}=\mathbb{C}$ case). (An analogous infinitesimal statement is also presented there: put $\mathfrak{h}(X):=\{A\in \operatorname{End}(X)\,|\, \operatorname{spec}(A)\cap i\mathbb{R}=\emptyset\}$ (these are also called hyperbolic, or, elsewhere, elementary). Then $\mathfrak{h}(X)$ is also open and dense (in $\operatorname{End}(X)$). The key observation is the continuity of $\exp: \mathfrak{h}(X)\to H(X)$. It is interesting to note that $\exp: \mathfrak{h}(X)\to H(X)$ is analogously well-defined in the general case by Dunford's Spectral Mapping Theorem.)

General Case: Here is an explicit example due to M. Monge (see https://www.fc.up.pt/pessoas/jfalves/pub/senegal.pdf (pp.14-15, Ex.5.3) or https://arxiv.org/abs/1510.05831 for details):

Put $D:=\{z\in\mathbb{C}\,||, |z-1|< 1/2\}$ and take $X$ to be the Banach space of all bounded analytic functions $D\to \mathbb{C}$ (with $C^0$ norm). Then

$$T: X\to X,\,\, f\mapsto [z\mapsto zf(z)]$$

is an automorphism with $\operatorname{spec}(T)=\overline{D}\ni 1$. For $S\in GL(X)$, $\Vert T-S\Vert<1/8$, $1\in\operatorname{spec}(S)$. The proof is by elementary complex analysis.

From a more dynamical point of view, Eisenberg and (J.H.) Hedlund proved that any hyperbolic automorphism is uniformly expansive ($T$ is uniformly expansive if $\exists n\in\mathbb{Z}_{>0}, \forall x\in X: |x|=1$ $\implies |T^n(x)|\geq 2 \text{ or } |T^{-n}(x)|\geq2$), and if $X$ is Hilbert then hyperbolic (Banach) automorphisms are not dense among uniformly expansive (Banach) automorphisms. This is a consequence of a result by Halmos and Lumer regarding the semicontinuity of the complement of the approximate point spectrum in the spectrum. (See Eisenberg & Hedlund's "Expansive automorphisms of Banach spaces" and Hedlund's follow-up "Expansive automorphisms of Banach spaces. II" for details.)

  • $\begingroup$ when you use the word 'automorphism', you mean continuous linear bijection? Because a linear map is a homomorphism of vector spaces, I would just use automorphism if in addition it was bijective, not necessarily continuous. $\endgroup$
    – NotaChoice
    Jan 3, 2022 at 0:07
  • $\begingroup$ I used "Banach automorphism" if both the map and the inverse are linear and continuous. More precisely I'm referring to automorphisms in the category of topological vector spaces (or automorphisms in the full subcategory of Banach spaces). $\endgroup$
    – Alp Uzman
    Jan 3, 2022 at 2:57

I didn't take a course about the spectral theorem yet but let me try to answer your question. I appreciate to be corrected if I am mistaken.

Your idea can work : To prove $\mathcal{H}(X)$ is dense in $GL(X)$, let $A\in GL(X)\setminus \mathcal{H}(X)$ and $\epsilon> 0$ be arbitraries, then $\exists \lambda_0\in \sigma(A): |\lambda_0|= 1$ , we are looking for a $T\in \mathcal{H}(X)$ ( i.e. $\forall \lambda\in \sigma(T):\ |\lambda|\neq 1$ ) , with $\|A-T\|< \epsilon$

We can consider $T= A+D$ where $D$ is a 'diagonal' operator with respect to a specific basis $\{b_i\}_{i\in I}$ of $X$ , where $Db_i= z_ib_i$ for each $i\in I$ and to satisfy the second condition we can choose the $z_i\in \mathbb{C}$ such that $|z_i|< \frac{\epsilon}{2} \ \forall i\in I$
For an obvious reason in what follows, we need $\{b_i\}_{i\in I}$ to be taken exactly as the basis ( not sure if we need a Hamel basis or a Schauder basis here? ) formed by extending the set of 'eigenvectors' of $A$ such that $Ab_j= \lambda_jb_j \ \forall \lambda_j\in \sigma(A)$ so here $j\in J\subset I$ , whereas $Ab_l= b_l \ \forall l\in I\setminus J$ .
We note that this extension of a linearly independent set to a basis is always possible.
Now to guarantee $\sigma(A+D)\cap S^1= \emptyset$ , we have $$\sigma(A+D)= \{\lambda'_i+z_i\big|\ i\in I ,\ \ \lambda'_i= \lambda_i \ \ \text{if }\ i\in J,\ \ \ \lambda'_i=1 \text{ otherwise}\}$$ For each $i\in I$ to guarantee that $|\lambda'_i+z_i|\neq 1$ we can choose each $z_i$ to have the same argument as $\lambda'_i$ if $|\lambda'_i|>1$ , otherwise if $|\lambda'_i|<1$ we pick $z_i$ to have the 'opposite' argument, that is $arg(z_i)= arg(\lambda'_i)+\pi$ , so to stay away from $S^1$, and we can check that when $\lambda'_i= 1$ then any $z_i\in \mathbb{R}\setminus \{0\}$ works.

Edit : (Thanks to Wraith1995)

We will use $\sigma(T)=\overline{\sigma_p(T)}$, where the $\sigma_p(T)$ is the point spectrum, to 'avoid' an open set containing $S^1$, say the circular sector $S= \{z\in \mathbb{C}:\ 1-\frac{\epsilon}{3}<|z|<1+\frac{\epsilon}{3}\}$

$\{b_i\}_{i\in I}$ is taken exactly as the basis formed by extending the set of 'eigenvectors' of $A$ such that $Ab_j= \lambda_jb_j \ \forall \lambda_j\in \sigma(A)$ and $j\in J\subset I$
$D$ is a 'diagonal' operator with respect to $\{b_i\}_{i\in I}$ of $X$ , where $Db_i= z_ib_i$ for each $i\in J$ and $Db_l= 0 \ \forall l\in I\setminus J$

To guarantee $\sigma(A+D)\cap S^1= \emptyset$ , we have $\sigma_p(A+D)= \{\lambda_i+z_i\big|\ i\in J \}$ and $$\sigma(A+D)= \overline{\{\lambda_i+z_i\big|\ i\in J \}}$$ For each $i\in J$ to guarantee that $|\lambda_i+z_i|\notin S$ we can choose each $z_i$ to have a module $|z_i|>\frac{\epsilon}{3}$ and the same argument as $\lambda_i$ if $|\lambda_i|>1$ , otherwise when $|\lambda'_i|<1$ we pick $z_i$ to have the 'opposite' argument that is $arg(z_i)= arg(\lambda'_i)+\pi$ , so that the closure doesn't intersect $S^1$ as desired.

Of course $\epsilon$ is taken to be very small, say $<\frac{1}{7}$

  • $\begingroup$ I think your calculation for $\sigma(A+D)$ is wrong or at least not justified enough; the continuous spectrum is where $A+D-\lambda I$ has no continuous inverse, but is still injective (so has no associated eigenvectors!). You've not really show why $A+D-(\lambda^{\prime}_{i} + z_{i})I$ has no continuous inverse. The use of eigenvectors to analyze spectrum only applies to point spectrum in infinite dimensions. $\endgroup$
    – Wraith1995
    Dec 29, 2021 at 16:49
  • $\begingroup$ I don't understand all what you are saying, but if you look at the infinite matrix representation of A+D? In the particular basis, the eigenvalues aren't necessarily unique and independent of the basis? So we can't claim if they are exactly $\lambda_i + z_i$ in this basis then they are also for any basis? $\endgroup$
    – NotaChoice
    Dec 29, 2021 at 19:45
  • $\begingroup$ I am saying that $\sigma$ is not the set of eigenvalues! That is the point spectrum. The continuous spectrum, $\sigma(A)$ is simply the place where although $A-\lambda I$ is injective (meaning no eigenvector is associated to any value in $A-\lambda I$) that $A-\lambda I$ is not surjective (so has no continuous inverse). Reasoning about eigenvectors associtaed to eigenvalues in $\sigma$ makes no sense. $\endgroup$
    – Wraith1995
    Dec 29, 2021 at 21:30
  • $\begingroup$ I got your point and corrected the argument $\endgroup$
    – NotaChoice
    Dec 30, 2021 at 23:01
  • $\begingroup$ The update still does not work; the continuous spectrum is not the closure of the point spectrum (in fact, point spectrum and continuous spectrum are disjoint). I am not sure, but I also think the continuous spectrum need not even be a subset of the closure of the point spectrum. $\endgroup$
    – Wraith1995
    Dec 31, 2021 at 0:20

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