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Estimates for the Resolvent and Spectral Gaps for Non Self-Adjoint Operators

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Estimates for the Resolvent and Spectral Gaps for Non Self-Adjoint Operators Estimates for the resolvent and spectral gaps for non self-adjoint operators Vesselin Petkov University Bordeaux Mathematics Days in Sofia, July 9, 2014 Vesselin Petkov (University BordeauxEstimates) for the resolvent and spectral gaps for non self-adjoint operators 1 / 29 Outline 1. Holomorphic functions without zeros in a strip. 2. Estimates for (I + B(z))−1. 3. The cut-off resolvent and the scattering operator. 4. Spectral gap for generators of semigroupes. 5. Dynamical zeta function. Vesselin Petkov (University BordeauxEstimates) for the resolvent and spectral gaps for non self-adjoint operators 2 / 29 1. Holomorphic functions without zeros in a strip 1. Holomorphic functions without zeros in a strip We start with the following question. Let f (z) be a holomorphic function in + C = {z ∈ C : Im z > 0} such that for some α ≥ 0, C > 0, M ∈ N we have M α Im z |f (z)| ≤ C(1 + |z|) e , z ∈ C+. (1) i Assume that f (z) 6= 0 in the strip 0 < Im z < 1 and consider z = x + 2 . What is the 1 optimal estimate for |f (x+i/2)| . Theorem 1 (Borichev, -P.) + Let f (z) be a holomorphic function in C satisfying (1). Assume that f (z) 6= 0 for 0 < Im z < 1. Then log |f (x + i/2)| lim = 0. (2) |x|→∞ x 2 It is clear that (2) implies for every 0 < 1 the estimate 1 2 ≤ C e|x| , x ∈ . |f (x + i/2)| R Vesselin Petkov (University BordeauxEstimates) for the resolvent and spectral gaps for non self-adjoint operators 3 / 29 1. Holomorphic functions without zeros in a strip f (z)eiαz Considering the function F (z) = C(z+i)M , we reduce the proof to the case when |f (z)| ≤ 1 in C+. Then G(z) = log(1/|f (z)|) is a positive harmonic function for 0 < Im z < 1. If for some x > 1 we have G(x + i/2) ≥ cx 2, c > 0, then by Harnack inequality we get 1 1 G(x + t + i/2) ≥ c cx 2, − ≤ t ≤ , c > 0. 1 4 4 1 Thus if log |f (x + i/2)| lim inf = −c < 0, |x|→∞ x 2 we can find a sequence of points xn ∈ R, |xn+1| > |xn| + 1, n ≥ 0 so that Z xn+1/4 1 dy log 2 ≥ c2c > 0 xn−1/4 |f (y + i/2)| 1 + y and then Z +∞ 1 dy log 2 = +∞. −∞ |f (y + i/2)| 1 + y ∞ + This contradicts the standard theorem for functions in H (C ) and we obtain the result. Vesselin Petkov (University BordeauxEstimates) for the resolvent and spectral gaps for non self-adjoint operators 4 / 29 1. Holomorphic functions without zeros in a strip Remark 1 + The assertion of Theorem 1 holds for holomorphic functions f (z) in C for which we have f (z) 6= 0 for 0 < Im z < 1 and M |f (x)| ≤ C(1 + |x|) , ∀x ∈ R, α|z| |f (z)| ≤ Ce , α ≥ 0, ∀z ∈ C+. In fact we can consider the function f (z)eiαz g(z) = (z + i)M and apply the Phragm´en-Lindel¨ofprinciple in the first and the the second quadrant of C to conclude that g(z) is bounded in C+. Vesselin Petkov (University BordeauxEstimates) for the resolvent and spectral gaps for non self-adjoint operators 5 / 29 1. Holomorphic functions without zeros in a strip To verify that the result of Theorem 1 is rather sharp, we establish the following Proposition 1 (Borichev, -P.) Let ρ(x) be a positive function such that limx→∞ ρ(x) = 0. Then there exists a + holomorphic function B(z) in C without zeros in the domain {z ∈ C : 0 < Im z < 1} such that log |B(x + i/2)| lim inf < 0. x→∞ ρ(x)x 2 This estimate means that there exists α > 0 so that for some sequence xn, |xn| → ∞ we have 1 2 ≥ Ceαρ(xn)|xn| . |B(xn + i/2)| Vesselin Petkov (University BordeauxEstimates) for the resolvent and spectral gaps for non self-adjoint operators 6 / 29 1. Holomorphic functions without zeros in a strip Proof. We choose two sequences xn → ∞, xn ≥ 1 and kn ∈ N, n ≥ 1 so that 2 kn ≥ ρ(xn)xn , n ≥ 1, (3) X kn 2 < ∞. (4) xn n≥1 Next we set zn = xn + i, n ≥ 1 and consider Blaschke product 2 k Y“ |zn + 1| z − zn ” n B(z) = 2 · . zn + 1 z − z¯n n≥1 The condition (4) guarantees the convergence of the infinite product. On the other hand, using (3) we get ˛ (x + i/2) − (x + i) ˛kn 2 n n −kn −ρ(xn)xn |B(xn + i/2)| ≤ ˛ ˛ = 3 ≤ e . ˛ (xn + i/2) − (xn − i) ˛ Vesselin Petkov (University BordeauxEstimates) for the resolvent and spectral gaps for non self-adjoint operators 7 / 29 1. Holomorphic functions without zeros in a strip + If we have a holomorphic function in C with growth condition |z|β + |f (z)| ≤ Ce , 1 < β < 2, z ∈ C , and if f (z) 6= 0 for 0 < Im z < 1, then we can prove that log |f (x + i/2)| lim inf > −∞. |x|→∞ x β+1 This result is also sharp but we will not discuss this class of functions in this talk. Vesselin Petkov (University BordeauxEstimates) for the resolvent and spectral gaps for non self-adjoint operators 8 / 29 2. Estimates for (I + B(z))−1 2. Estimates for (I + B(z))−1 Let H be Hilbert space with norm k.k. We denote also by k.k the norm of the operators + in H and by L the space of bounded linear operators on H. Let B(z): C 3 z −→ L(H) + be an operator valued holomorphic function in C . Theorem 2 (Borichev, -P.) + Let B(z) be holomorphic in C and such that for some constants α ≥ 0,C > 0,M ≥ 0 we have M α Im z kB(z)k ≤ C(1 + |z|) e , z ∈ C+. Assume that (I + B(z))−1 ∈ L(H) for 0 < Im z < 1 and let Image B(z) ⊂ V , V being a finite dimensional space of H independent of z ∈ C+. Then for every > 0 we have −1 |x|2 k(I + B(x + i/2)) k ≤ Ce . In particular, we cover the cases of matrix-valued operators. The argument is based on an estimate of the determinant Vesselin Petkov (University BordeauxEstimates) for the resolvent and spectral gaps for non self-adjoint operators 9 / 29 2. Estimates for (I + B(z))−1 Now consider an operator valued holomorphic function + A(z): z ∈ C → T1, where T1 denotes the space of trace class operators in H with the norm k · k1. A compact P operator B is trace class if j |λj (B)| < ∞, λj (B) being the eigenvalues of B. Recall that for every B ∈ T1 we can define the determinant Y det(I + B) = (1 + λj (B)), j and | det(I + B)| ≤ ekBk1 . Moreover, given B(z) ∈ T1, we may consider the function ˆ ˜ −1 FB (z) = det(I + B(z)) (I + B(z)) which extends to an entire operator valued function in C such that kB(z)k1|z| kFB (z)k ≤ e , µ ∈ C. −1 If kB(z)k1 ≤ C|z|, an estimate for |(det(I + B(z))) | implies an estimate for the resolvent k(I + B(z))−1k, but for infinite dimensional spaces H this estimate in general is not optimal. It is an open problem to extend the above results for trace class operators + B(z) holomorphic in C . Vesselin Petkov (University BordeauxEstimates) for the resolvent and spectral gaps for non self-adjoint operators 10 / 29 3. The cut-off resolvent and the scattering operator 3. The cut-off resolvent and the scattering operator n Consider a bounded connected domain K ⊂ R , n ≥ 2, with smooth boundary ∂K. Set n Ω = R \ K¯ and consider the Dirichlet problem 8 (∂2 − ∆)u = 0 in × Ω, <> t Rt u = 0 on Rt × ∂K, (5) > :u(0, x) = f0(x), ut (0, x) = f1(x). 2 Let −∆D be the Dirichlet Laplacian in Ω which is a self-adjoint operator on H = L (Ω) 1 2 2 −1 with domain D = H0 (Ω) ∩ H (Ω). For Im λ > 0 the resolvent (−∆D − λ ) is a ∞ bounded operator from H to D and, for all ϕ ∈ C0 (Ω), the cut-off resolvent 2 −1 − RD (λ) = ϕ(−∆D − λ ) ϕ admits a meromorphic continuation in C for n odd and in − − − C \ iR for n even with poles in C = {λ ∈ C : Im λ < 0} The poles of this continuation are independent on the choice of ϕ and they are called scattering resonances. Vesselin Petkov (University BordeauxEstimates) for the resolvent and spectral gaps for non self-adjoint operators 11 / 29 3. The cut-off resolvent and the scattering operator The estimates of the cut-off resolvent in the poles free regions play a crucial role for the decay of local energy and many problems in mathematical physics. In several important examples there exists a strip Uδ = {λ ∈ C : −δ ≤ Im λ ≤ 0, | Re λ| ≥ C0 > 0}, δ > 0, where RD (λ) has no poles. In this case Uδ is called spectral gap. This is the case for non-trapping obstacles, where RD (λ) is holomorphic in the region {λ ∈ C : | Im λ| ≤ C(log | Re λ| + 1), | Re λ| > C0}. Moreover, in this region for non trapping obstacles we have a bound M kRD (λ)k ≤ C(1 + |λ|) . The case of trapping obstacles is much more difficult since we can have different type trapping rays and the geometry of the obstacle can be rather complicated.
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