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Schrödinger operators on the Wiener space Ichiro SHIGEKAWA Kyoto University August 10, 2004 Beijing Normal University URL: Contents 1. Essential self-adjointness

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Schrödinger operators on the Wiener space Ichiro SHIGEKAWA Kyoto University August 10, 2004 Beijing Normal University URL: Contents 1. Essential self-adjointness 2. Domain of Schrödinger operator 3. Spectral gap of Schrödinger operator 1 1. Essential self-adjointness ffl B: a Banach space ffl μ: the Wiener measure with (B; H; μ): an abstract Wiener space ffl H: a Hilbert space ρ B B Z e p 1hx;'i μ(dx) = exp n 1 2 j'j2 H Λo; ' 2 B Λ ρ H Λ : 2 FC 1 0 f(x) = F (hx; ' 1 i; : : : ; hx; ' n i); : f : B! R such that F 2 C 1 0 (R n ); ' 1 ; : : : ; ' n 2 B Λ : L V : Schrödinger operator on L 2 (μ) L: the Ornstein-Uhlenbeck operator V : a scalar potential Question: L V essentially self-adjoin on FC 1 0? Is 3 k k 2 : L 2 -norm V + := maxfv;0g (the positive part) V := maxf V;0g (the negative part) Proposition 1.1. Assume ffl V + 2 L 2+ = Sp 2 Lp, ffl there exist 0 a 1, b 0 such that kv fk 2» aklfk 2 + bkfk 2 : Then L V is essentially selfadjoint on FC ffl (B; μ): a probability space ffl E: a Dirichlet form ffl L: the associated generator What is sufficient for kv fk 2» aklfk 2 + bkfk 2? (Defective) logarithmic Sobolev inequality B Z jfj 2 log(jfj=kfk 2 ) dμ» ffe(f; f) + fikfk 2 2 : 5 E.g. On an abstract Wiener space: We assume ffl E admits a square field operator. ffl E has a local property. Hence E has the following form B Z (1.1) E(f; g) = (f; g) dμ and has the derivation property. ffl (f; g) = rf rg, r: the gradient operator 6 Theorem 1.2. Assume B Z jfj 2 log(jfj=kfk 2 ) dμ» ffe(f; f) + fikfk 2 2 : Then, for any 0, there exist positive constants K 1, K 2 such that B Z f 2 log 2 + f dμ» ff 2 (1 + )klfk K 1 + K 2 kfk 6 2 : cf. Feissner(1975), Bakry-Meyer(1982) 7 ψ(x) = e p x+1 1 : xy» Φ(x) + Ψ(y)» x log 2 + x + 2p ye p y Hausdorff-Young inequality Set Φ(x) = x log 2 + x; ψ 1 (x) = Φ 0 (x); Define the complimentary function Ψ(x) = x 0 Z ψ(y)dy: Hausdorff-Young inequality: 8 jfj 2 log(jfj=kfk 2 ) dμ» ffe(f; f) + fikfk 2 2 Theorem 1.3. Assume the logarithmic inequality B Z and v 0, e v 2 L 2ff+ = [ L p : p 2ff Then, there exist constants 0 a 1 and b 0 such that (1.2) kvfk 2» aklfk 2 + bkfk 2 : 9 We now return to an abstract Wiener space. Gross' logarithmic Sobolev inequality B Z B Z jfj 2 log(jfj=kfk 2 ) dμ» jrfj 2 dμ ) f 2 log 2 + f dμ» (1 + )klfk2 2 + K 1 + K 2 kfk 6 2 : B Z Theorem 1.4. Assume ffl V +, e V 2 L 2+. Then L V is essentially self-adjoint on FC 1 0. cf. Segal(1969), Glimm & Jaffe(1970), Simon(1973), & Hfiegh-Krohn(1972) 10 Simon 2. Domain of Schrödinger operator A = L V + W is essentially self-adjoint on FC 1 0 ) 11 We consider a Schrödinger operator A = L V + W on an abstract Wiener space (B; H; μ). Basic assumptions (A.1) V 1, V 2 L 2+. W 0 and there exists a constant 0 ff 1 (A.2) that e W 2 L 2=ff. such The intertwining property, i.e., ffl p A = A p V : V Aim : To determine the domain, i.e., Dom(A) = Dom(L) Dom(V ) Main tools ffl The Lax-Milgram theorem. 12 How to define an operator A? We define a vector field b by = b = 1 r rv log V: 2 2V and a bilinear form E A by E A (f; g) = (rf;rg) + (b rf; g) (f; b rg) + ((V W jbj 2 )f; g): By a formal computation, the associated generator is given by (2.1) A = L 2b r + (r Λ b V + W + jbj 2 ): 13 A (f; g) = ^E A (f; g) :::::::::::: E symmetric»e A (f; g) :::::::::::: + skew-symmetric ^E A (f; g) = (rf;rg) + ((V W jbj 2 )f; g);»e A (f; g) = (b rf; g) (f; b rg): ^E A (f; g) = ^E A (f; g) + (f; g): L V (f; g) = (rf;rg) + (V f; g): E 14 Decompose E A as where Moreover, we set The bilinear form associated to L V is e W +jbj2 2 L 2=ff Clearly Dom(E L V ) = Dom(r) Dom( p V ): We will show that Dom( ^E A ) = Dom(E L V ). Additional assumptions We assume either (B.1) or there exists a constant C 0 such that jbj 2» ffv + C: (B.2) 15 (W + jbj 2 f; f)» ff :: 2.1. Assume (A.1), (A.2) and one of (B.1) Proposition (B.2). Then there exists a constant fi such that and appeared in (A.2) E L V (f; f) + fi(f; f) and hence (1 ff)e L V (f; f)» ^E A (f; f) + fi(f; f)» (1 + ff)e L V (f; f) + fi(f; f): Therefore ^E A ) = Dom(E L V ) = Dom(r) Dom( p V ): Dom( 16 Estimate of»e A 2.2. Assume (A.1), (A.2) and one of (B.1) Proposition (B.2). Then, for sufficiently large, there exists a and constant K 0 such that j»e A (f; g)j» K ^E A (f; f) 1=2 ^E A (g; g) 1=2 : Therefore E A satisfies the sector condition. E A = ^E A +»E A is a closed bilinear form. 17 p V A = A p V ; Intertwining property Instead of we show (2.2) E A (f; p V g) = E A ( p V f; g): 2.3. Assume (A.1), (A.2) and one of (B.1) Proposition (B.2). Then (2.2) holds for f,g 2 FC 1 0. Moreover, and we have, for f 2 Dom(A), g 2 Dom(A Λ ), (2.3) (Af; p V g) = ( p V f; A Λ g): 18 K 1 k(a )fk 2» klfk 2 + kv fk 2 K 1, K 2 depend only on constants in (A.1), Remark. (B1), (B.2). (A.2), Domain of the Schrödinger operator 2.4. Assume (A.1), (A.2) and one of (B.1) Theorem (B.2). Then Dom(A) = Dom(L) Dom(V ). and Moreover, for sufficiently large, there exist positive constants K 1, K 2 such that» K 2 k(a )fk 2 : 19 3. Spectral gap of Schrödinger operator it consists of point spectrums of finite multiplicity. i.e., 20 A Schrödinger operator A = L V + W on an abstract Wiener space (B; H; μ). ff(a): the spectrum of A = L V + W. Bounded potential 3.1. Assume V is bounded and W = 0. Theorem l = sup ff(a) is a point spectrum of multiplicity Then one and the associated eigenfunction can be chosen to be positive. Moreover, the spectrum is discrete on (l 1; l], General potential 3.2. Assume (A.1), (A.2) and one of (B.1) Theorem (B.2). Then l = sup ff(a) is a point spectrum of and multiplicity one and the associated eigenfunction can be chosen to be positive. Moreover, the spectrum is discrete on (l 1; l], i.e., it consists of point spectrums of finite multiplicity. 21 ff(l Vn ) is discrete on ( (V n ) 1; (V n )] )( (V n ) = sup ff(l V n ). where Proof of Theorem 3.1 Approximation method f' i g 1 i=1 BΛ : a c.o.n.s of H Λ. F n := ff(' 1 ; ' 2 ; : : : ; ' n ). V n = E[V jf n ]. 22 We set G (n) = ( L + V n ) 1 ; G = ( L + V ) 1 : claim: G (n)! G in norm sense G G (n) = G (n) (V V n )G: We show k(v V n )Gk op! 0. the logarithmic Sobolev inequality and the By inequality xy» x log x x + e y Hausdorff-Young 23 k(v V n )Gfk 2 2» 1 N f2e[jrgfj2 ] + kgfk 2 2 log kgfk2 2 = E[(V V n ) 2 (Gf) 2 ] = 1 N E[N(V V n) 2 (Gf) 2 ]» 1 N E[(Gf)2 log(gf) 2 (Gf) 2 + e N (V V n) 2 ] kgfk E[eN (V V n) 2 ]g: 24 k(v V n )Gfk 2 2 Now replacing f with f=kgfk 2,» 1 N f2e[jrgfj2 ] + E[e N (V V n) 2 1]kGfk 2 2 g» 1 N fe[f 2 ] + E[(Gf) 2 ] + E[jV j(gf) 2 ] + E[e N (V V n) 2 1]kfk 2 2 g» 1 N f(2 + kv k 1)kfk E[eN (V V n) 2 1]kfk 2 2 g: Hence k(v V n )Gk 2 op» 1 N f2 + kv k 1 + E[e N (V V n) 2 1]g: 25 lim k(v V n)gk op = 0: n!1 Now letting n! 1 and then letting N! 1, we have This completes the proof. 26

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