Nash Type Inequalities for Fractional Powers of Non-Negative Self-adjoint Operators ( Wroclaw 006) P.Maheux (Orléans. France) joint work with A.Bendikov. European Network (HARP) (to appear in T.A.M.S) 1
Classical example: Heat semigroup on n R with dx the Lebesgue measure: T t f(x) = h t f(x) = h t (x, y) = 1 (4πt) n/ exp R n h t(x, y)f(y) dy ( x y 4t ) Af = f, f C 0 (Rn ) D T t 1 := sup{ T t f, f 1 = 1} (Ult) T t 1 = h t (0, 0) = 1 (4πt) n/, t > 0 (Sob) f n/n c( f, f), (Nash) f +4/n c( f, f) f 4/n 1,
Let (T t ) t>0 be a symmetric submarkovian semigroup acting on L (X, µ) with µ a σ-finite measure on X and let A be its generator with domain D. f D T t+s = T t T s t, s > 0 lim T tf = f (in L ) t 0 + 0 f 1 0 T t f 1 t > 0 T t f f Af = lim t 0 + t L T t = e ta acts on L p (1 p + ) space as a contraction semigroup: T t f p f p, t > 0 3
Theorem[V,C-K-S] Let ν >. The following conditions are equivalent : (Sob) f ν/ν c (Af, f), f D (Nash) f +4/ν c (Af, f) f 4/ν 1, f D L 1 (Ult) T t 1 c t ν/, t > 0 4
Fact: Nash inequality for A implies Nash inequality for all α (0, 1) (by subordination): f +4α/ν c (A α f, f) f 4α/ν 1 Nash-type (or generalized) inequality(def) Let B : [0, + [ [0, + [ be a non-decreasing function. A satisfies a Nash-type inequality if f B( f ) (Af, f), f D f 1 1 Classical case in R d (Nash) c f ( f ) /n (Af, f) with B(x) = cx /n 5
Theorem [T.Coulhon. J.F.A 141] (Ult) T t 1 m(t), t > 0. implies Nash type inequality f B( f ) (Af, f), f D, f 1 1 with B(x) = sup s>0 [ s log x sg(s)] G(s) = log m(1/s) 6
Classical example: Polynomial decay: Under the general assumptions above on the semigroup: T t 1 c =: m(t) tν/ then B(x) = cx /ν Another example: One exponential decay Proposition Let γ > 0. Assume that T t 1 c exp(1/t γ ) t > 0 Then for all f D with f 1 1, f Example : On T, [ log+ ( f )] 1+1/γ (Af, f) A = + k=1 k 1 γ k
7 Another example: Double exponential decay Proposition Let β > 0. Assume that T t 1 c e e1/tβ t > 0 Then for all f D with f 1 1, f log + ( f ) [ log+ ( f )] 1/β (Af, f) Example : On T, A = + k=1 (ln k) γ k 8
An example of Nash-type inequality without Ultracontractivity: The Ornstein-Uhlenbeck operator A = +x on L n (R, γ n ) with the Gaussian measure: γ n (x) = 1 ( ) x exp dx (π) n/ Proposition f log + ( f ) (Af, f), f 1 1 then Θ(x) = x log + (x) But (T t ) is not ultracontractive! Proof. Deduce from Log-Sobolev inequality of Gross (but slightly weaker). 9
Spectral Theory Let A = A be a non negative self-adjoint operator (unbounded) on L (X, µ) with µ a positive σ-finite Spectral theory: A = + 0 λ de λ (E λ spectral measure). Functional calculus: Let f : [0, + [ R be a Borel function, Examples: f(a) = + 0 f(λ) de λ (t > 0) f t (λ) = e tλ : f(λ) = λ α, 0 < α < + : T t = e ta semi-group A α fractional power f(λ) = log(1 + λ) : log(i + A) 10
General Question: Assume a Nash type inequality holds for A: f B( f ) (Af, f), f D, f 1 1 Is it true that g(a) satisfies a Nash type inequality for some function g 0? If yes: describe the function B g s.t: f B g( f ) (g(a)f, f) f D g, f 1 1 In particular with: g(x) = x α, 0 < α < 1 : Fractional powers g(x) = log(1 + x) : log -semigroup 11
Why are we so interested by g(a)? If g : [0, + [ [ is real and g 0 then g(a) is also non negative self adjoint operator (defined by spectral theory). Then, we can define the associated semigroup T g t = e tg(a) 1
Locally Compact Abelian Groups Setting Let G be a l.c.a group with dual group Ĝ. ( g is Bernstein function) g is a Bernstein function if g :]0, + [ [0, + [, g is C and ( 1) p f (p) 0 p N {0}. Examples: g(x) = 1 e sx, s > 0 : Poisson sgr. with jump s g(x) = x α, 0 < α < 1 : Fractional powers g(x) = log(1 + x) : Γ-semigroup Why are we so interested by g(a)? 1a
Let µ t be a convolution semigroup on G given by ˆµ t (y) = e tψ(y), y Ĝ with ψ the associated continuous negative definite function. Then (goψ) is a continuous negative definite function ( µ g t )ˆ(y) = e t (goψ)(y), y Ĝ is also a convolution semigroup! Representation formula for Berstein function g(x) = a + bx + + 0 (1 e xs ) dµ(s) with µ a positive measure on [0, + [ 1 + s dµ(s) < dµ(s) < 0 1 1b
Examples: g(x) = x α (0 < α < 1) then g(a) = A α is the fractional operator. g(x) = log(1 + x) i.e g(a) = log(i + A). The operator g(a) generates a Markov semigroup. Is it true that g(a) satisfies a Nash type inequality? The answer is yes with A α. Open problem: Nash inequality for log(i + A) (Partial results in R n and also in the abstract setting) 13
The Assumptions: Let (X, µ) be a measure space with σ-finite measure µ. Let A be a non-negative self-adjoint operator with domain D(A) L (X, µ). Suppose that the semigroup T t = e ta acts as a contraction on L 1 (X, µ) Let B : [0, + [ [0, + [ be a non-decreasing function which tends to infinity at infinity. 14
Theorem(A.Bendikov,P.M) If the operator A satisfies f B ( f ) (Af, f) f D(A), f 1 1 Then, for any α > 0, f [ B ( f )] α (A α f, f) f D(A α ), f 1 1. Picture : Operators: A A α Functions: > B(x) > B α (x) 15
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