Appendix A Vector-valued Holomorphic Functions Let X be a Banach space and let Ω ⊂ C be an open set. A function f :Ω→ X is holomorphic if f(z0 + h) − f(z0) f (z0) := lim (A.1) h→0 h h∈C\{0} exists for all z0 ∈ Ω. If f is holomorphic, then f is continuous and weakly holomorphic (i.e. x∗ ◦ f ∗ ∗ is holomorphic for all x ∈ X ). If Γ := {γ(t):t ∈ [a,b]} is a finite, piecewise smooth contour in Ω, we can form the contour integral f(z) dz. This coincides Γ b with the Bochner integral a f(γ(t))γ (t) dt (see Section 1.1). Similarly we can define integrals over infinite contours when the corresponding Bochner integral is absolutely convergent. Since 1 2 f(z) dz, x∗ = f(z),x∗ dz, Γ Γ many properties of holomorphic functions and contour integrals may be extended from the scalar to the vector-valued case, by applying the Hahn-Banach theorem. For example, Cauchy’s theorem is valid, and also Cauchy’s integral formula: 1 f(z) f(w)= dz (A.2) − 2πi |z−z0|=r z w whenever f is holomorphic in Ω, the closed ball B(z0,r) is contained in Ω and w ∈ B(z0,r). As in the scalar case one deduces Taylor’s theorem from this. Proposition A.1. Let f :Ω→ X be holomorphic, where Ω ⊂ C is open. Let z0 ∈ Ω,r>0 such that B(z0,r) ⊂ Ω.Then W. Arendt et al., Vector-valued Laplace Transforms and Cauchy Problems: Second Edition, 461 Monographs in Mathematics 96, DOI 10.1007/978-3-0348-0087-7, © Springer Basel AG 2011 462 A. VECTOR-VALUED HOLOMORPHIC FUNCTIONS ∞ n f(z)= an(z − z0) n=0 converges absolutely for |z − z0| <r,where 1 f(z) an := dz. − n+1 2πi |z−z0|=r (z z0) We also mention a special form of the identity theorem. Proposition A.2 (Identity theorem for holomorphic functions). Let Y be a closed subspace of a Banach space X.LetΩ be a connected open set in C and f :Ω→ X be holomorphic. Assume that there exists a convergent sequence (zn)n∈N ⊂ Ω such that limn→∞ zn ∈ Ω and f(zn) ∈ Y for all n ∈ N.Thenf(z) ∈ Y for all z ∈ Ω. Note that for Y = {0}, we obtain the usual form of the identity theorem. ∗ 0 ∗ ∗ ∗ ∗ Proof. Let x ∈ Y := {y ∈ X : y, y =0(y ∈ Y )}.Thenx ◦ f(zn) = 0 for all n ∈ N. It follows from the scalar identity theorem that x∗ ◦ f(z) = 0 for all z ∈ Ω. Hence, f(z) ∈ Y 00 = Y for all z ∈ Ω. In the following we show that every weakly holomorphic function is holomor- phic. Actually, we will prove a slightly more general assertion which turns out to be useful. A subset N of X∗ is called norming if ∗ x 1 := sup | x, x | x∗∈N defines an equivalent norm on X. A function f :Ω→ X is called locally bounded if supK f(z) < ∞ for all compact subsets K of Ω. Proposition A.3. Let Ω ⊂ C be open and let f :Ω→ X be locally bounded such that x∗ ◦ f is holomorphic for all x∗ ∈ N,whereN is a norming subset of X∗. Then f is holomorphic. In particular, if X = L(Y,Z),whereY,Z are Banach spaces, and if f :Ω→ X is locally bounded, then the following are equivalent: (i) f is holomorphic. (ii) f(·)y is holomorphic for all y ∈ Y . (iii) f(·)y, z∗ is holomorphic for all y ∈ Y, z∗ ∈ Z∗. Proof. We can assume that x 1 = x for all x ∈ X. In order to show holomorphy at z0 ∈ Ω we can assume that z0 = 0, replacing Ω by Ω − z0 otherwise. For small h, k ∈ C\{0},let f(h) − f(0) f(k) − f(0) u(h, k):= − . h k A. VECTOR-VALUED HOLOMORPHIC FUNCTIONS 463 We have to show that for ε>0 there exists δ>0 such that u(h, k) ≤ε whenever |h|≤δ and |k|≤δ.Letr>0 such that B(0, 2r) ⊂ Ωand M := sup f(z) < ∞. z∈B(0,2r) Then by Cauchy’s integral formula, for |z| <r,|h|≤r, |k|≤r, h, k =0 ,x∗ ∈ N, 1 1 1 1 1 1 1 u(h, k),x∗ = f(z),x∗ − − − dz 2πi h z − h z k z − k z |z|=2r h − k f(z),x∗ = dz. 2πi |z|=2r z(z − h)(z − k) Hence, | u(h, k),x∗ |≤|h − k|M/r2.SinceN is norming, we deduce that M u(h, k) ≤|h − k| . r2 This proves the claim. Corollary A.4. Let Ω ⊂ C be a connected open set and Ω0 ⊂ Ω be open. Let h :Ω0 → X be holomorphic. Assume that there exists a norming subset N of ∗ ∗ X such that for all x ∈ N there exists a holomorphic extension Hx∗ :Ω→ C ∗ of x ◦ h.Ifsupx∗∈N |Hx∗ (z)| < ∞,thenh has a unique holomorphic extension z∈Ω H :Ω→ X. Proof. Again we assume that · 1 = · .Let Y := y =(yx∗ )x∗∈N ⊂ C : y ∞ := sup |yx∗ | < ∞ , x∗∈N and let H :Ω→ Y be given by H(z):=(Hx∗ (z))x∗∈N . It follows from Proposition ∗ A.3 that H is holomorphic. By x ∈ X → ( x, x )x∗∈N , one defines an isometric injection from X into Y .SinceH(z) ∈ X for z ∈ Ω0, it follows from the identity theorem (Proposition A.2) that H(z) ∈ X for all z ∈ Ω. We will extend Proposition A.3 considerably in Theorem A.7. Before that we prove Vitali’s theorem. Theorem A.5 (Vitali). Let Ω ⊂ C be open and connected. Let fn :Ω→ X be holomorphic (n ∈ N) such that sup fn(z) < ∞ n∈N z∈B(z0,r) whenever B(z0,r) ⊂ Ω. Assume that the set ' ( Ω := z ∈ Ω: lim fn(z) exists 0 n→∞ 464 A. VECTOR-VALUED HOLOMORPHIC FUNCTIONS has a limit point in Ω. Then there exists a holomorphic function f :Ω→ X such that (k) (k) f (z) = lim fn (z) n→∞ uniformly on all compact subsets of Ω for all k ∈ N0. ∞ ∞ Proof. Let l (X):={x =(xn)n∈N ⊂ X : x ∞ := sup xn < ∞}.Thenl (X) is a Banach space for the norm · ∞ and the space c(X) of all convergent sequences is a closed subspace of l∞(X). Consider the function F :Ω→ l∞(X)givenby F (z)=(fn(z))n∈N. It follows from Proposition A.3 that F is holomorphic. (One ∞ may take N to be the space of all functionals on l (X) of the form (xn)n∈N → ∗ ∗ ∗ ∗ xk,x where k ∈ N,x ∈ X , x ≤1). Since F (z) ∈ c(X) for all z ∈ Ω0, it follows from the identity theorem (Proposition A.2) that F (z) ∈ c(X) for all z ∈ Ω. Consider the mapping φ ∈L(c(X),X)givenbyφ((xn)n∈N) = limn→∞ xn. Then f = limn→∞ fn = φ ◦ F :Ω→ X is holomorphic. Finally, we prove uniform convergence on compact sets. Let B(z0,r) ⊂ Ωand k ∈ N0. It follows from (A.2) that 1 1 fn(w) f (k)(z)= dw. n − k+1 k! 2πi |w−z0|=r (w z) (k) Now the dominated convergence theorem implies that fn (z) converges uniformly (k) on B(z0,r/2) to f (z). Since every compact subset of Ω can be covered by a finite number of discs, the claim follows. If in Vitali’s theorem (fn) is a net instead of a sequence, the proof shows that f(z) = lim fn(z) exists for all z ∈ Ω and defines a holomorphic function f :Ω→ X. Next we recall a well known theorem from functional analysis. Theorem A.6 (Krein-Smulyan). Let X be a Banach space and W be a subspace of the dual space X∗. Denote by B∗ the closed unit ball of X∗.ThenW is weak* closed if and only if W ∩ B∗ is weak* closed. For a proof, see [Meg98, Theorem 2.7.11]. Now we obtain the following convenient criterion for holomorphy. Theorem A.7. Let Ω ⊂ C be open and connected, and let f :Ω→ X be a locally bounded function. Assume that W ⊂ X∗ is a separating subspace such that x∗ ◦ f is holomorphic for all x∗ ∈ W .Thenf is holomorphic. Here, W is called separating if x, x∗ =0forallx∗ ∈ W implies x =0 (x ∈ X). Proof. Let Y := {x∗ ∈ X∗ : x∗ ◦ f is holomorphic}.SinceW ⊂ Y , the subspace Y is weak* dense. It follows from Vitali’s theorem (applied to nets if X is not A. VECTOR-VALUED HOLOMORPHIC FUNCTIONS 465 separable) that Y ∩ B∗ is weak* closed. Now it follows from the Krein-Smulyan theorem that Y = X∗. Hence, f is holomorphic by Proposition A.3. Notes: Usually, Vitali’s theorem is proved with the help of Montel’s theorem which is only valid in finite dimensions. A vector-valued version is proved in the book of Hille and Phillips [HP57] by a quite complicated power-series argument going back to Liouville. The very simple proof given here is due to Arendt and Nikolski [AN00] who also proved Theorem A.7 (see also [AN06]). Appendix B Closed Operators Let X be a complex Banach space. An operator on X is a linear map A : D(A) → X,whereD(A) is a linear subspace of X, known as the domain of A.Therange Ran A,andthekernel Ker A,ofA are defined by: Ran A := {Ax : x ∈ D(A)}, Ker A := {x ∈ D(A):Ax =0}.