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. 



Bounded operator on a normed space with empty spectrum
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


