Consider the dispersive estimates for the SchrÃ¶dinger flow $$ e^{itH}P_{c},\quad H=-\Delta+V \quad \text{on}\quad \mathbb{R}^n,n\ge 1 $$ where $P_{c}$ is the projection onto the continuous spectrum of $H$, and we will be most concerned with whether it has the form $$ \|e^{itH}P_{c}\|_{L^1\to L^{\infty}}\leq C |t|^{-\frac{n}{2}} $$ In order to get this estimate, some decay and regularity condition must be put on the potential $V$, an important assumption is that zero is neither an eigenvalue nor a resonance.

If $0$ is a eigenvalue, then it's easy to see that the above estimates may fail. My question is then if zero is a resonance but not an eigenvalue, why will the estimates above go wrong?

Zero is said to be a resonance in the sense that if the operator $(I-V\Delta^{-1})^{-1}$ is bounded on $L^1$(why not on $L^2$ ?),see the paper of Vodev,I found this is less illuminating for me, so I want to know if there are some better understanding of this definition to make it more intuitive.

**Edit**
As Terry and Delio have commented,the key point is the asymptotic expansions of the resolvents around the zeero energy.for odd dimension,with $\Im z>0$,one can write
$$
(-\Delta+V-z)^{-1}=\frac{A_{-1}}{z}+\frac{A_{-\frac12}}{z^{\frac12}}+A_{0}+O(z)
$$
(for even dimension,the $log z$ terms are included)where $A_{-1}$ is the projection onto the eigenspace of $H$,and $A_{-\frac12}$ is related to both eigenspace and resonance functions. So in order to get the optimal decay ($t^{-\frac{n}{2}}$)for large t,one need $A_{-1}=A_{-\frac12}=0$,that is zero is neither an eigenvalue nor a resonance .