1
$\begingroup$

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 $r$.

$\endgroup$
  • $\begingroup$ $f = A$, I presume? $\endgroup$ – Robert Israel Feb 2 '16 at 22:53
  • $\begingroup$ yes, sorry. I'll fix it $\endgroup$ – hal Feb 2 '16 at 22:57
  • $\begingroup$ 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? $\endgroup$ – La Rias Jun 21 '18 at 11:17
1
$\begingroup$

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 $1$.

$\endgroup$
  • $\begingroup$ 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 $E$ $\endgroup$ – hal Feb 3 '16 at 18:59
  • 2
    $\begingroup$ @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 $1/n$? $\endgroup$ – Glougloubarbaki Feb 4 '16 at 1:57
-1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ $R^2-Re_2=\{(x,y)\neq (0,y)\}$ is dense in $R^2$. $\endgroup$ – Tsemo Aristide Feb 2 '16 at 23:12
  • $\begingroup$ Your $E$ is not a linear subspace. $\endgroup$ – Robert Israel Feb 3 '16 at 0:05
  • $\begingroup$ Certainly, but the question requests only E to be a dense subspace $\endgroup$ – Tsemo Aristide Feb 3 '16 at 0:09
  • 1
    $\begingroup$ In the context, subspaces are linear. What does "spectrum" even mean for operators on non-linear spaces? $\endgroup$ – Robert Israel Feb 3 '16 at 16:53
  • 2
    $\begingroup$ For linear operators on infinite-dimensional normed spaces, spectrum is not the same thing as the set of eigenvalues. $\endgroup$ – Robert Israel Feb 3 '16 at 21:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.