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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 .

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I have never seen this definition of resonance. In my personal experience, a resonance is a pole of the resolvent $(z-\Delta+V)^{-1}$ (either - for some authors - in the lower complex halfplane, or - for other authors - in the whole plane: that is, also eigenvalues are sometimes dubbed resonances). –  Delio Mugnolo Oct 23 '12 at 9:53
    
What is $P_C$, by the way? –  Delio Mugnolo Oct 23 '12 at 9:54
    
@ Delio Mugnolo it's the projection onto the continuous spectrum of $H$ –  user23078 Oct 23 '12 at 10:12
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In a paper of Jensen and Kato ( ams.org/mathscinet-getitem?mr=544248 ), the asymptotic expansion for low energy resolvent is computed. If there is a resonance at zero, the resolvent blows up as the energy approaches zero, though not as rapidly as if there is an eigenvalue at zero. The Schrodinger propagators are sort of Fourier transforms of the resolvent, so if the resolvent blows up at low energy, the (low frequency component of the) propagators will decay slowly at long times. –  Terry Tao Oct 24 '12 at 4:44
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@Delio Mugnolo : The Resolvent is only analytic outside the spectrum. If for instance the spectrum consists of the non negative real numbers, then it might be possible that on a subspace of the Hilbert space the resolvent can be analytically continued from the upper half plane to the lower half plane across the spectrum. If A is self adjoint then the resolvent itself can never have a pole outside the real axis, but the analytic continuation may have a pole there, and that's a resonance. And the imaginary part of the pole is interpreted as the inverse of the lifetime of the state –  jjcale Oct 24 '12 at 19:35
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