11
$\begingroup$

A bounded operator acting on a complex Banach space has non-empty spectrum, and the proof of this fact uses the completeness of the space.

Is there any example of bounded operator acting on a complex non-complete normed space with empty spectrum?

I understand that the spectrum of an operator $T$ is the set of all complex numbers $\lambda$ such that $T-\lambda I$ is not bijective.

$\endgroup$
1
  • 1
    $\begingroup$ Your clarification of "spectrum" was essential, since every bounded operator on a normed space has an approximate eigenvalue. $\endgroup$ Nov 28, 2014 at 16:41

1 Answer 1

14
$\begingroup$

Take an operator on a Banach space whose image is dense, whose spectrum is $\{0\}$ but that has no kernel, for example $$ T(f)(x)=\int_x^1f(y)dy $$ acting on $H:=L^2([0,1])$.

Then its restriction to the dense subspace $D:=\bigcap_n Im(T^n)$ should have the property you desire.
First of all, $D$ is dense in $H$ -- see the comments below by Alexander Shamov.

Clearly, $T|_D$ is invertible.

So all that remains to be checks is that for $\lambda\not=0$, the operator $(T-\lambda)$ is invertible on $D$. Recall that $(T-\lambda)^{-1}$ makes sense on $H$. The trick is to note that $(T-\lambda)^{-1}$ preserves $D$. Indeed it preserves each subspace $Im(T^n)$: If $x\in Im(T^n)$ write it as $x=T^ny$ and then we have $(T-\lambda)^{-1}x=(T-\lambda)^{-1}T^ny=T^n(T-\lambda)^{-1}y\in Im(T^n)$. QED

$\endgroup$
6
  • $\begingroup$ This looks good, but is it obvious that that subspace really is dense? $\endgroup$ Nov 28, 2014 at 13:52
  • 1
    $\begingroup$ @GeoffRobinson: $\bigcap_n \mathop{\mathrm{im}} T^n$ is dense iff $\mathop{\mathrm{im}} T$ is dense iff $\ker T^\ast = 0$. $\endgroup$ Nov 28, 2014 at 14:00
  • 1
    $\begingroup$ @GeoffRobinson: It's actually a general fact that if we have a sequence of spaces $B_0 \supset B_1 \supset B_2 \supset \dots$ that are complete, respectively, w.r.t. norms $\Vert \cdot \Vert_0 \le \Vert \cdot \Vert_1 \le \Vert \cdot \Vert_2 \le \dots$ and all $B_{k+1}$ are dense in $(B_k, \Vert \cdot \Vert_k)$ then $\bigcap_k B_k$ is dense in $B_0$. In this case take the norms $\Vert T^{-k} (\cdot) \Vert$ on $\mathop{\mathrm{im}} T^k$. $\endgroup$ Nov 28, 2014 at 14:09
  • $\begingroup$ @AlexanderShamov : Thanks for the clarification. $\endgroup$ Nov 28, 2014 at 15:39
  • 1
    $\begingroup$ The general fact Alexander is referring to is sometimes called "abstract Mittag-Leffler theorem" (a version of this is e.g. in Bourbaki). The first explicit appearance however is in an article of R. Arens from 1958. $\endgroup$ Dec 1, 2014 at 8:00

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.