Resonance of Schrödinger operator

2012-10-23T09:15:23Z

Consider the dispersive estimates for the Schrödinger 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 .

Resonance of Schrödinger operator - MathOverflow

Resonance of Schrödinger operator