Chapter 8 The Holomorphic Functional Calculus Approach to Operator Semigroups In Chapter 6 we have associated with a given C0-semigroup an operator, called its generator, and a functional calculus, the Hille{Phillips calculus. Here we address the converse problem: which operators are generators? The classical answer to this question and one of our goals in this chapter is the so- called Hille{Yosida theorem, which gives a characterization of the generator property in terms of resolvent estimates. However, instead of following the standard texts like [1], we shall approach the problem via a suitable functional calculus construction. Our method is generic, in that we learn here how to use Cauchy integrals over infinite con- tours in order to define functional calculi for unbounded operators satisfying resolvent estimates. (The same approach shall be taken later for defining holomorphic functional calculi for sectorial and strip-type operators.) This chapter is heavily based on the first part of [2], which in turn goes back to the unpublished note [3]. 8.1 Operators of Half-Plane Type Recall from Chapter 6 that if B generates a C0-semigroup of type (M; !), then s(B) is contained in the closed half-plane [ Re z ≤ ! ] and M kR(λ, B)k ≤ (Re λ > !): Re λ − ! In particular, if ! < α then R(·;A) is uniformly bounded for Re λ ≥ α. This \spectral data" is our model for what we shall call below an operator of \left half-plane type". Recall, however, that the associated Hille-Phillips calculus for the semi- group generated by B is actually a calculus for the negative generator 135 136 8 The Holomorphic Functional Calculus Approach to Operator Semigroups A := −B. So in order to stay in line with the Hille{Phillips picture we shall construct our new functional calculus for operators of right half-plane type. Let us turn to the precise definitions. For ! 2 [−∞; 1] we let L! := fz 2 C j Re z < !g and R! := fz 2 C j Re z > !g be the open left and right half-plane at the abscissa !. (So R! = C++! in the terminology of Chapter 6.) An operator A on a Banach space X is said to be of right half-plane type ! 2 (−∞; 1] if s(A) ⊆ R! and M(A; α) := supfkR(z; A)k j Re z ≤ αg < 1 for every α < !. We say that A is of right half-plane type if A is of right half-plane type ! for some !. For an operator A of right half-plane type we call !hp(A) := sup α sup kR(z; A)k < 1 Re z≤α the abscissa of uniform boundedness of its resolvent. An operator B is called of left half-plane type ! if A = −B is of right half-plane type −!. The abscissa of uniform boundedness of the resolvent of such an operator is denoted by s0(B) := inf α sup kR(λ, B)k < 1 2 R [ {−∞} Re λ≥α as it is common in semigroup theory. (See [4, p.342] where this number is called the pseudo-spectral bound of B.) Remark 8.1. The definition of a half-plane type operator may appear com- plicated and unintuitive at first glance. Why do we not define A to be of right half-plane type ! simply by s(A) ⊆ R!; sup kR(z; A)k < 1 ? (8:1) Re z<! The answer is that, with our definition, A is of right half-plane type !hp(A), whereas this would be false with the alternative definition. See Exercise 8.1. In the following, when we speak of half-plane type operators, it has to be read as right half-plane type, unless explicitly noted otherwise. 8.2 Functional Calculus on Half-Planes As already announced, we are aiming at defining a functional calculus for operators A of half-plane type. There is a \cheap" way of doing this: pick 8.2 Functional Calculus on Half-Planes 137 some λ 2 r(A) and define Aλ := R(λ, A). Then set up the Dunford-Riesz calculus for Aλ and play it back to A. (See Exercise 8.2 for details.) This approach, elegant as it appears, does not use at all the resolvent estimates which are characteristic for operators of half-plane type. Moreover, it is principally unsuitable to reach our goal, because the functions e−tz have an essential singularity at 1 and hence are not accessible by that calculus nor by its canonical extension. So one has to take a different route. The idea is to mimic the construction of the Dunford{Riesz calculus but with contours leading into the possible singularity, which here is the point at infinity. Since the spectrum of an oper- ator A of half-plane type may be a whole right half-plane, a straightforward choice for such contours are vertical straight lines, oriented from top to down. (Picture this on the Riemann sphere and you will see that this comes as close as possible to \surrounding once the spectrum counterclockwise".) s(A) ! δ !hp(A) R γ Fig. 8.1 The path γ runs from δ + i1 to δ − i1, where ! < δ < !hp(A). In the following, we shall work out this plan. Let A be an operator of half- plane type on a Banach space X and fix ! < δ < !hp(A). We want to define 1 Z Φ(f) := f(z)R(z; A) dz 2pi γ 138 8 The Holomorphic Functional Calculus Approach to Operator Semigroups for suitable functions f 2 Hol(R!), where γ is the top-to-down parameriza- tion of the vertical line [ Re z = δ ], see Figure 8.2. (We use the alternative expressions 1 Z −1 Z δ+i1 f(z)R(z; A) dz and f(z)R(z; A) dz 2pi Re z=δ 2pi δ−i1 to denote this integral. Observe that the orientation of the contour is implicit in the first of these.) Of course, as the contour is infinite, not every holomorphic function f on R! will be suitable. Rather, the properties of f and R(z;A) should match in order to guarantee that the integral is convergent. As the resolvent is assumed to be bounded, it is reasonable to require that f is integrable on vertical lines. This is the reason for the following definition. For ! 2 R we let1 1 1 E(R!) := f 2 H (R!) j 8 δ > ! : f 2 L (δ + iR) and call it the algebra of elementary functions on the half-plane R!. By writing f 2 L1(δ + iR) we mean that the function f(δ + ix) is in L1(R), i.e., that Z jf(δ + ix)j dx < 1: R 1 It is easily checked that E(R!) is a (non-unital) subalgebra of H (R!). For elementary functions we have the following version of Cauchy's integral the- orem. Lemma 8.2. Let f 2 E(R!), δ > !, and a 2 C with Re a 6= δ. Then ( 1 Z f(z) f(a) if δ < Re a; dz = (8:2) 2pi Re z=δ z − a 0 if δ > Re a: Moreover, f 2 C0(Rδ). Proof. Consider first the case Re a < δ. Define Z f(z) g(a) := dz (Re a < δ): Re z=δ z − a Then, by some standard arguments, g is holomorphic and Z 0 f(z) g (a) = 2 dz: Re z=δ (z − a) We claim that for fixed a this integral does not depend on δ > Re a. Indeed, 1 For an open set O ⊆ C we denote by H1(O) the space of bounded and holomorphic functions. It is a unital Banach algebra with respect to the sup-norm. 8.2 Functional Calculus on Half-Planes 139 1 Z f(z) 0 = 2 dz; 2pi γR (z − a) by Cauchy's theorem, where γR is the positively oriented boundary of the rectangle [δ; δ0]×[−R; R] with δ0 > δ and R > 0. Since f is uniformly bounded, the integrals over the horizontal line segments tend to zero as R ! 1, resulting in 1 Z f(z) 1 Z f(z) 0 = 2 dz − 2 dz: 2pi Re z=δ (z − a) 2pi Re z=δ0 (z − a) But now we can let δ0 ! 1 and, again by the boundedness of f, we conclude that g0(a) = 0. Hence, g is constant. On the other hand, g(a) ! 0 as Re a ! −∞, which yields g = 0. Consider now the case Re a > δ. Again we employ Cauchy's theorem with 0 the contour γR as above, but now with δ < Re a < δ and R > jIm aj. This yields 1 Z f(z) f(a) = dz: 2pi γR z − a Again, the integrals over the horizontal line segments tend to zero as R ! 1. It follows that 1 Z f(z) 1 Z f(z) f(a) = dz − dz: 2pi Re z=δ z − a 2pi Re z=δ0 z − a By what we have proved first, the second integral is equal to zero, and we are done. For the last assertion, fix δ0 > δ. Then it follows from (8.2) and the dominated 0 convergence theorem that f(a) ! 0 as a ! 1 within Rδ0 . As δ > δ > ! were arbitrary, the claim follows. For f 2 E(R!) we now let (as announced above) 1 Z ΦA(f) := f(z)R(z; A) dz; (8:3) 2pi Re z=δ where A is the given half-plane type operator and ! < δ < !hp(A), see also Figure 8.2.