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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.

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

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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

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  • $\begingroup$ This looks good, but is it obvious that that subspace really is dense? $\endgroup$
    – Geoff Robinson
    Commented Nov 28, 2014 at 13:52
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    $\begingroup$ @GeoffRobinson: $\bigcap_n \mathop{\mathrm{im}} T^n$ is dense iff $\mathop{\mathrm{im}} T$ is dense iff $\ker T^\ast = 0$. $\endgroup$
    – Alexander Shamov
    Commented Nov 28, 2014 at 14:00
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    $\begingroup$ @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 $\mathop{\mathrm{im}} T^k$. $\endgroup$
    – Alexander Shamov
    Commented Nov 28, 2014 at 14:09
  • $\begingroup$ @AlexanderShamov : Thanks for the clarification. $\endgroup$
    – Geoff Robinson
    Commented Nov 28, 2014 at 15:39
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    $\begingroup$ 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. $\endgroup$
    – Jochen Wengenroth
    Commented Dec 1, 2014 at 8:00

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