Chapter 11 Bounded H1-Calculus for Hilbert Space Operators 11.1 Bounded H1-Calculus Let D ⊆ C be some (non-empty) domain in the complex plane. We say that an operator A on a Banach space X has a bounded H1-calculus on D if there is a bounded representation Φ :H1(D) !L(X) such that its canonical extension within Mer(D) has A as its generator (in the sense that Φ(z) = A). Obviously, the case D = C is pretty uninteresting since H1(C) = C1 by Liouville's theorem. So in the following we take D 6= C as a standing assumption. In general, to say that A \has a bounded H1(D)-calculus" does not tell much. In particular, considered in this generality there is no reason to think that such a calculus must be unique. However, if we suppose even more that D 6= C, then for each λ 2 CnD one −1 1 has rλ := (λ − z) 2 H (D), and hence Φ(rλ) = R(λ, A). In this case Φ is completely determined at least on the space of rational functions bounded on D and then, by boundedness of Φ, on the closure of this space with respect to the uniform norm kfk1;D := supfjf(z)j j z 2 Dg: For certain domains D one can then use results from complex approximation 1 theory to infer that some special, explicitly constructed, H -calculus ΦA for A is bounded. (The reason for this is usually that a general function f 2 H1(D) can be approximated by a bp-convergent sequence of bounded rational functions, and the calculus ΦA is bp-continuous in a certain sense.) In such a situation, A having \a" bounded H1-calculus on D is equivalent 1 to the concretely given ΦA being bounded on H (D), and the latter is what is usually intended when one uses the former terminology. 195 196 11 Bounded H1-Calculus for Hilbert Space Operators To wit, this all holds for sectorial operators with dense domain and range, as the following result shows. Theorem 11.1. Let A be a sectorial operator with dense domain and range on a Banach space X. Then the following assertions are equivalent for ! 2 (!se(A); p) and C ≥ 0: 1 (i) The sectorial calculus for A is bounded on H (S!) with kf(A)k ≤ Ckfk1;S! (11:1) 1 for all f 2 H (S!). 1 (ii) The estimate (11.1) holds for each rational function f 2 H (S!) with f(0) = 0 = f(1). (iii) The estimate (11.1) holds for each elementary function f 2 E(S!). Sketch of Proof. The implication (i) ) (ii) is clear. Now define 1 A := ff 2 H (S!) j (11.1) holdsg: It follows from the convergence lemma (Theorem 9.10) that if A 3 fn ! f pointwise and lim infn!1 kfnk1 ≤ kfk1, then f 2 A. (ii) ) (iii): Denote by K := S! [ f1g ⊆ C1 and let R0(S!) be the algebra of rational functions f bounded on S! and with f(0) = f(1) = 0. Using [4, II, Thm.10.4] it can be shown that R0(S!) is sup-norm dense in the algebra 1 B := ff 2 C(K) \ H (S!) j f(0) = 0 = f(1)g (cf. the proof of [6, Prop.F.3]). So (ii) implies B ⊆ A. Given f 2 E(S!) define αn 1 fn := f(z ), where αn := 1 − =n, n 2 N. By Lemma 9.4.b), fn 2 B, so f 2 A. But f ! f pointwise on S and sup kf k ≤ kfk , hence n n ! n2N n 1 1 f 2 A as claimed. 1 1=n −2=n (iii) ) (i): Let f 2 H (S!) and define fn := f · z (1 + z) for n 2 N. 1 Then fn 2 E(S!) and fn ! f pointwise on S!. Moreover, kfnk1 ≤ kfk1c n , −2 where c = kz(1 + z) k1;S! . It follows that f 2 A as desired. Remark 11.2. A result analogous to Theorem 11.1 holds for operators of right half-plane type and for strong strip type operators (Exercise 11.10). See Exercise 11.1 for a similar theorem in the context of bounded operators. An estimate of the form kf(A)k ≤ Ckfk1 is usually the best one can expect since one has equality (with C = 1) for multiplication operators. To understand the situation better, suppose that A is a densely defined operator of right half-plane type 0. For such an operator one has the holomorphic calculus on each open half-plane which contains C+. But as −A need not −tz be the generator of a C0-semigroup, the operators (e )(A) need not be bounded. A fortiori, A cannot have a bounded H1-calculus on any half-plane. 11.2 Plancherel's Theorem 197 But now suppose that −A generates a bounded C0-semigroup (Tt)t≥0. Then one has the Hille{Phillips calculus on C+ for A, with the estimate kf(A)k ≤ (sup kT k) kfk (f 2 LS( )): t LS(C+) C+ t≥0 The Hille{Phillips calculus is already \better" than the holomorphic half- plane calculus. However, the Laplace-Stieltjes norm is strictly stronger than the uniform norm, and the example of the shift semigroup on L1(R) shows that one cannot expect a better estimate in general (Lemma 6.9). This indicates that a supremum norm estimate for a functional calculus (and in particular a bounded H1-calculus) is quite special and one may wonder whether there are interesting non-trivial situations where such an estimate holds. We shall see soon that the answer to this question is yes. 11.2 Plancherel's Theorem The most satisfying results on H1-estimates for functional calculi hold for Hilbert space operators. The spectral theorem (Chapter 4) is the most im- pressive witness for this claim. In the remainder of this chapter we shall see what can be said when the operators in question are not normal anymore. One of the most important \gateways" to H1-boundedness results on Hilbert spaces, at least in the context of semigroups, is a vector-valued version of Plancherel's theorem. For any Banach space X the vector-valued Fourier transform is de- fined as for scalar functions by Z −it·s d 1 d (Ff)(t) := e f(s) ds (t 2 R ; f 2 L (R ; X)): d R (We shall freely use the results of Appendix A.6 on Bochner spaces from now on.) In Appendix A.9 we have collected important information about the Fourier transform for vector-valued functions. In particular, one can find a proof of the Fourier inversion theorem (Theorem A.46). Note that if H is a Hilbert space, then so is L2(Rd; H), with the scalar product being Z (f j g )L2( d;H) = (f(s) j g(s))H ds R d R for f; g 2 L2(Rd; H). For a proof of the following result, see Theorem A.49 in Appendix A.9. (Recall that S is the reflection operator, defined in Section 5.2.) Theorem 11.3 (Plancherel). Let H be a Hilbert space. Then the Fourier transform maps L1(Rd; H) \ L2(Rd; H) into L2(Rd; H) and extends to a 198 11 Bounded H1-Calculus for Hilbert Space Operators bounded operator (again denoted by F) on L2(Rd; H) in such a way that F 2 = (2p)dS and the operator − d 2 d 2 d (2p) 2 F :L (R ; H) ! L (R ; H) is unitary. Moreover, for each µ 2 M(Rd) and f 2 L2(Rd; H) F(µ ∗ f) = µb · f:b (11:2) The Gearhart{Greiner{Pr¨ussTheorem Plancherel's theorem is useful in the context of semigroup theory through the following observation. Let B be the generator of a C0-semigroup T = (Tt)t≥0 on a Hilbert space H, let ! > !0(T ), the exponential growth bound of T . Then, for s 2 R and x 2 X we have Z 1 Z 1 −(!+is)t −ist ! R(! + is; B)x = e Ttx dt = e Tt x dt: 0 0 (Recall the definition of T ! from page 92.) Hence, ! R(! + is;B)x = F(Tt x)(x 2 H): (We view T as a function on R by setting T (t) = 0 for t < 0.) Plancherel's theorem then implies that p ! 2 2 kR(! + is;B)xkL (R;H) = 2pkTt xkL (R;H) ≤ C!kxk for some C! independent of x 2 H (Exercise 11.2). The following important result of semigroup theory is a nice application. Theorem 11.4 (Gearhart{Greiner{Pr¨uss). Let B be the generator of a C0-semigroup T = (Tt)t≥0 on a Hilbert space. Then R(z;B) is not bounded on [ Re z > !0(T )]. Proof. By rescaling we may suppose without loss of generality that !0(T ) = 0 0. Suppose towards a contradiction that M := supRe z>0 kR(z; B)k < 1. For each 0 < α < ! the resolvent identity yields R(α + is;B) = I + (! − α)R(α + is;B) R(! + is;B): Hence, there is C ≥ 0 independent of α such that Z 1 −2αt 2 −1 2 e kTtxk dt = (2p) kR(α + is;B)xkL2 0 −1 0 2 2 2 2 ≤ (2p) (1 + M (! − α)) kR(! + is;B)xkL2 ≤ C kxk 11.3 Von Neumann's Inequality 199 for all x 2 H. Letting α & 0 yields Z 1 2 2 2 kTtxk dt ≤ C kxk (x 2 H): 0 !s Now fix M ≥ 0 such that kTsk ≤ Me for all s ≥ 0.