# Spectral radius and dense subspace

Let $X$ be a Banach space, and let $E$ be a dense subspace of $X$. Let $A: X \to X$ be a bounded operator on $X$ that maps $E$ to itself. Assume that the spectral radius of $A$ restricted to $E$ is $r>0$. Is it true that the spectral radius of $A$ on $X$ is equal to $r$ (or can it "jump up")?

Edit (precisions) I am assuming that $E$ is a linear vector space, and that the spectral radius of $A$ on $E$ is equal to a finite $r>0$. I am asking if this implies that the spectral radius of $A$ on $X$ is at most $r$.

• $f = A$, I presume? – Robert Israel Feb 2 '16 at 22:53
• yes, sorry. I'll fix it – hal Feb 2 '16 at 22:57
• Suppose the $X$ in your case is an ordered Banach space, and $T$ is invariant under the cone $K=\{x\in X \colon x\geq 0\}$, then the spectral radius is in spectrum. What happens to the spectral radius of the restriction in this case? – La Rias Jun 21 '18 at 11:17

Consider the Banach space $X = C[0,1]$ of continuous functions on $[0,1]$, with the operator $A$ of multiplication by $x$ (i.e. $Af(x) = x f(x)$). This has spectrum $[0,1]$. Let $E$ be the subspace of $X$ consisting of polynomials. This is invariant under $A$. However, $A - \lambda I$ is never surjective as an operator from $E$ to $E$, e.g. there is no polynomial $g(x)$ such that $(x - \lambda) g(x) = 1$. Therefore the spectrum of the restriction of $A$ to $E$ is all of $\mathbb C$.

EDIT: In the other direction, the spectrum of $A$ is always contained in the spectrum of the restriction $A|_E$ of $A$ to $E$: if $\lambda$ is not in the spectrum of $A|_E$, i.e. $A|_E - \lambda I|_E$ has a bounded two-sided inverse $R_\lambda: E \to E$, then $R_\lambda$ extends by continuity to a bounded linear operator $\overline{R}_\lambda$ on $X$ which is a two-sided inverse of $A-\lambda I$ there, so $\lambda$ is not in the spectrum of $A$. Thus the spectral radius can never "jump up".

For an example where $A|_E$ has finite spectral radius, consider the same $A$ and $X$ as in my previous example, but let $E$ be the space of restrictions to $[0,1]$ of functions analytic in the disk $\{z: |z| < 2\}$. If $|\lambda| > 2$, $A|E - \lambda I$ has a bounded inverse on $E$, namely multiplication by $1/(x - \lambda)$. But if $|\lambda| < 2$, $A|_E - \lambda$ is not surjective. Thus the spectral radius "jumps down" from $2$ to $1$.

• interesting exemple. This proves that the spectral radius can "jump down". do you have any example where it "jumps up"? also I am assuming that the spectral radius is finite on $E$ – hal Feb 3 '16 at 18:59
• @RobertIsrael is it true/obvious that for the non-Banach space $E$ the two usual definitions of the spectral radius coincide? (ie that it is the limit of the norm of $A^n$ to the power $1/n$? – Glougloubarbaki Feb 4 '16 at 1:57

$X=R^2=Vect\{e_1,e_2\}$, $A(e_1)=e_1, A(e_2)=2e_2$, $E=R^2-Re_2$. The spectrum of $A$ restricted to $E$ is $1$ and the spectrum of $A$ is $\{1,2\}$.

• $R^2-Re_2=\{(x,y)\neq (0,y)\}$ is dense in $R^2$. – Tsemo Aristide Feb 2 '16 at 23:12
• Your $E$ is not a linear subspace. – Robert Israel Feb 3 '16 at 0:05
• Certainly, but the question requests only E to be a dense subspace – Tsemo Aristide Feb 3 '16 at 0:09
• In the context, subspaces are linear. What does "spectrum" even mean for operators on non-linear spaces? – Robert Israel Feb 3 '16 at 16:53
• For linear operators on infinite-dimensional normed spaces, spectrum is not the same thing as the set of eigenvalues. – Robert Israel Feb 3 '16 at 21:06