BS VI c©Gabriel Nagy

Banach Spaces VI:

Vector-Valued Calculus Notions

Notes from the Functional Analysis Course (Fall 07 - Spring 08)

In this section we discuss the extension of several well known results from Calculus tofunctions that take values in a Banach space. Specifically, we consider functions f : D → Y ,where Y is a Banach space, and D is (in most cases) a subset in R or C. Although in manystatements from this section we specify that Y is a Banach space, many definitions (andresults) discussed here make sense for more general spaces, and we will do that, wheneverthe “Banach space proofs” can be easily adapted. The reader is invited to explore thepossible generalizations of those results that are stated specifically for Banach spaces, thebest candidates for such generalizations being Frechet spaces.

Convention. Throughout this note K will be one of the fields R or C, and all vectorspaces are over K.

A. Riemann Integrability

In Calculus, integration theory comes in two flavors: (s) the theory of Riemann inte-gral (which, due to Lebesgue’s Criterion for Riemann integrability, is essentially based oncontinuity), or (b) the theory of “abstract” integration (a-la Lebesgue, which is based onmeasurability). As it turns out, when one passes to the vector-valued case, the notion Rie-mann integrability can be easily generalized, and this is the subject of this sub-section.Concerning the “abstract” integration theory, this is a challenging problem technically, andwe will discuss one “naive” approach to it in sub-section B.

Definitions. Given two real numbers a < b, a partition of [a, b] is a system

∆ = (a = x0 < x1 < · · · < xn = b).

Given such a ∆, we define its size to be the positive number |∆| = maxxj − xj−1 : j =1, . . . , n.

We denote by P[a, b] the collection of all partitions of [a, b]. Given ∆ ∈ P[a, b] as above,an n-tuple T = (t1, . . . , tn) ∈ [a, b]n is said to be a tag for ∆, if xj−1 ≤ tj ≤ xj, ∀ j = 1, . . . , n.In this case, the pair (∆, T ) is referred to as a tagged partition.

The set P[a, b] carries a natural order relation ≻ defined as follows. Given two partitions∆ = (a = x0 < x1 < · · · < xn = b) and Σ = (a = y0 < y1 < · · · < ym = b), we write ∆ ≻ Σ,if x0, x1, . . . , xn ⊃ y0, y1, . . . , ym. In this case we say that ∆ is finer than Σ.

Remark 1. The ordered set (P[a, b],≻) is directed, because any two partitions have acommon refinement. The net (|∆|)∆∈P[a,b] ⊂ (0,∞) is decreasing (i.e. ∆ ≻ Σ ⇒ |∆| ≤ |Σ|)and converges to 0.

1

Definition. Suppose f : [a, b] → Y is a function which takes values in a topological vectorspace Y . Given a partition ∆ = (a = x0 < x1 < · · · < xn = b) and a tag T = (t1, . . . , tn) of ∆,we define the associated Riemann sum as the vector Rf (∆, T ) =

∑nj=1(xj − xj−1)f(tj) ∈ Y .

We say that f is Riemann integrable, if there exists some vector y ∈ Y , such that

(∗) whenever (T∆)∆∈P[a,b] a net in⊔∞

n=1 Rn, such that T∆ is a tag for ∆, for all ∆ ∈ P[a, b],it follows that the net (Rf (∆, T∆))∆∈P[a,b] ⊂ Y converges to y.

In this case, the vector y, which is unique, since Y is Hausdorff, is denoted by∫ b

af(t) dt.

Theorem 1. If Y is a complete locally convex topological vector space, then any contin-uous function f : [a, b] → Y is Riemann integrable. Moreover, the vector y =

∫ b

af(t) dt has

the following additional property:

(∗∗) for every neighborhood V of 0 in Y, there exists ε > 0, such that Rf (∆, T ) − y ∈ V,for all tagged partitions (∆, T ), with |∆| < ε.

Proof. The key step is contained in the following

Claim 1: For every neighborhood V of 0 in Y, there exists ε > 0, such that, whenever(∆, T ) and (Σ, S) are tagged partitions with Σ ≻ ∆ and |∆| < ε, it follows that the differenceRf (Σ, S) −Rf (∆, T ) belongs to V.

Since Y is locally convex, we can assume that

V = y ∈ Y : q(y) ≤ 1, (1)

where q is a continuous seminorm on Y . Since f : [a, b] → Y is continuous, an obviousuniform continuity argument shows that there exists ε > 0, such that

q(f(s) − f(t)) ≤1

b− a, for all s, t ∈ [a, b], with |s− t| < ε. (2)

Suppose now Σ = (a = x0 < x1 < · · · < xn = b) and ∆ = (a = y0 < y1 < · · · < ym = b)are two partitions, with Σ ≻ ∆ and |∆| < ε, and S = (s1, . . . , sm) and T = (t1, . . . , tn) aretags for Σ and ∆, respectively. Since Σ ≻ ∆, there are integers 0 = k0 < k1 < · · · < km = n,such that xi = yki

, ∀ i = 1, . . . ,m. Using these indices we can write

Rf (Σ, S) −Rf (∆, T ) =m−1∑

i=0

( ki+1∑

j=ki+1

(yj − yj−1)[f(sj) − f(ti+1]

)

,

so when we apply the seminorm q we get

q(

Rf (∆, T ) −Rf (Σ, S))

≤

m−1∑

i=0

( ki+1∑

j=ki+1

|xj − xj−1| · q(

f(tj) − f(si+1

)

)

. (3)

Remark now that, given 0 ≤ i ≤ m− 1 and ki + 1 ≤ j ≤ ki+1, we have xi ≤ ti+1 ≤ xi+1, aswell as:

xi = yki≤ yj−1 ≤ sj ≤ yj ≤ yki+1

= xi+1,

2

so both sj and xi+1 belong to the interval [xi, xi+1]. This forces |sj − ti+1| ≤ |∆| < ε, whichby (2) yields

q(

f(sj) − f(ti+1

)

<1

b− a,

so now if we go back to (3) we obtain

q(

Rf (Σ, S) −Rf (∆, T ))

≤1

b− a

m−1∑

i=0

( ki+1∑

j=ki+1

|yj − yj−1|

)

=

=1

b− a

m−1∑

i=0

(yki+1− yki

) =1

b− a

m−1∑

i=0

(xi+1 − xi) = 1,

which means that Rf (Σ, S) −Rf (∆, T ) indeed belongs to V .Fix now a sequence of tagged partitions (∆n, Tn), such that ∆n+1 ≻ ∆n and limn→∞ |∆n| =

0. By Claim 1, the sequence (yn)∞n=1 ⊂ Y , given by yn = Rf (∆n, Tn), is Cauchy, thus con-vergent to some (unique) y ∈ Y . It is clear that the proof of the Theorem will be complete,once we prove the following

Claim 2: The vector y has property (∗∗).

Fix as above, a neighborhood V of 0 in Y , which we can again assume to be of the form(1). Applying Claim 1 to the neighborhood

12V = y ∈ Y : q(y) ≤ 1

2,

there exists ε > 0, such that

q(

Rf (Σ, S) −Rf (∆, T ))

≤ 12, (4)

for any two tagged partitions (∆, T ) and (Σ, S), with Σ ≻ ∆ and |∆| < ε. Fix now anarbitrary tagged partition (∆, T ) with |∆| < ε, and choose, for every n ≥ 1, a taggedpartition (Σn, Sn), such that Σn ≻ ∆,∆n. If we choose now nε ∈ N, such that |∆n| < ε,∀n ≥ nε, then by (4) we have

q(

Rf (Σn, Sn) −Rf (∆n, Tn))

≤ 12,

q(

Rf (Σn, Sn) −Rf (∆, T ))

≤ 12,

so if we add these two inequalities we obtain q(

Rf (∆, T ) −Rf (∆n, Tn))

≤ 1, that is,

q(Rf (∆, T ) − yn) ≤ 1,∀n ≥ nε. (5)

Of course, the seminorm q is continuous, to taking limn→∞ in (5) immediately yields q(Rf (∆, T )−y) ≤ 1, which by (1) means precisely that Rf (∆, T ) − y belongs to V .

Riemann integrability is compatible with compositions with linear continuous maps, asindicated in the following.

Exercise 1. Assume Y and Z are topological vector spaces, and T ∈ L(Y ,Z). Provethat, if f : [a, b] → Y is Riemann integrable, then so is the function T f : [a, b] → Z, and

furthermore one has the identity∫ b

a(T f)(t) dt = T

( ∫ b

af(t) dt

)

.

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Besides Theorem 1 and Exercise 1, Riemann integration enjoys several other nice proper-ties, analogous to those from the scalar case. These features are contained in the next groupof problems.

Exercises 2-4. Assume Y is a Banach space.

2. Prove that every Riemann integrable function f : [a, b] → Y is automatically bounded.

3. Prove that, if f : [a, b] → Y is Riemann integrable, and if the function [a, b] ∋ t 7−→‖f(t)‖ ∈ [0,∞) is Riemann integrable, then

∥

∥

∥

∥

∫ b

a

f(t) dt

∥

∥

∥

∥

≤

∫ b

a

‖f(t)‖ dt.

4. Prove that, if g : [a, b] → [c, d] is an increasing C1-diffeomorphism, and if f : [c, d] → Yis a function with the property that the function [a, b] ∋ t 7−→ g′(t)f(g(t)) ∈ Y isRiemann integrable, then f is also Riemann integrable, and furthermore one has theequality

∫ d

c

f(t) dt =

∫ b

a

g′(t)f(g(t)) dt. (6)

Remark 1. The definition of the Riemann integral∫ b

acan be extended to cover the

“wrong order” case a > b, by the usual convention:∫ b

a= −

∫ a

b. This convention can be

used in Exercise 4, to include the case of arbitrary diffeomorphisms g (which might flip theendpoints), so that the general version of (6) reads

∫ g(b)

g(a)

f(t) dt =

∫ b

a

g′(t)f(g(t)) dt.

Comment. We know that every Banach space Y can be isometrically represented asa closed linear subspace Y ⊂ CK(Ω), for some (suitably chosen) compact Hausdorff spaceΩ. Therefore it is legitimate to examine Riemann integrability, as well as continuity, forfunctions f : [a, b] → CK(Ω), and this is the subject of the next three Exercises.

Exercises 5-6. Let Ω be a compact Hausdorff space.

5.

♥ Assume T is an arbitrary topological space, Define, for every function and F : T×Ω →K, and every t ∈ T , the function

Ft : Ω ∋ ω 7−→ F (t, ω) ∈ K. (7)

Prove that, given such an F , the following are equivalent:

(i) F is continuous;

(ii) for every t ∈ T , the function Ft : Ω → K is continuous, and furthermore thefunction

f : T ∋ t 7−→ Ft ∈ CK(Ω) (8)

is continuous, when CK(Ω) is equipped with the uniform norm.

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Conclude that the correspondence F 7−→ f establishes a linear isomorphism betweenthe space CK(T × Ω) of all continuous K-valued functions on T × Ω and the spaceCCK(Ω)(T ) of all norm-continuous CK(Ω)-valued functions on T .

6. Assume f : [a, b] → CK(Ω) is a norm-continuous function, and let F ∈ CK([a, b] × Ω)be its corresponding function, which is given by

F (t, ω) = (f(t))(ω), ∀ (t, ω) ∈ [a, b] × Ω.

Prove that the function1 h : Ω ∋ ω 7−→∫ b

aF (t, ω) dt ∈ K has the following properties

(i) h is continuous, so it belongs to the Banach space CK(Ω);

(ii) h =∫ b

af(t) dt.

Comment. As expected, Riemann integrability is tied up with differentiability, exactlyas in the scalar case, as shown in Exercises 7-8 below. Differentiability for vector-valuedfunctions is defined, exactly as in the scalar case, as follows.

Definition. Given a topological vector space Y over K, a subset D ⊂ K, a functionF : D → Y , and some accumulation point z ∈ D, we say that F is K-differentiable at z, if

limw→zw∈D

(w − z)−1[F (w) − F (z)]

exists in Y . In this is the case, the limit is denoted by F ′(z) and is called the K-derivativeof F at z.

Exercise 7-8. Assume Y is a Banach space.

7. Prove that, if J ⊂ R is an arbitrary interval, and f : J → Y is continuous, then forevery a ∈ J , the function F : J → Y , defined by

F (x) =

∫ x

a

f(t) dt

is continuously R-differentiable, and satisfies the equality F ′(x) = f(x), ∀x ∈ J .

7. Prove that, if F : [a, b] → Y is R-differentiable, and the function F ′ : [a, b] → Y is

Riemann integrable, then∫ b

aF ′(t) dt = F (b) − F (a).

B. (Ultra)Weak Calculus

In this section we discuss a very useful technique that allows one to reduce vector-valuedcalculus problems to scalar-valued ones. Throughout this sub-section we will assume that Yis a fixed normed vector space. The slightly non-standard terminology (especially the use ofthe prefix “ultra”) we will adopt is as follows.

Definitions. Assume f : D → Y is some function. Suppose one is given a “calculusproperty” (P).

1 Since F is continuous, for every ω ∈ Ω, the function [a, b] ∋ t 7−→ F (t, ω) ∈ K is continuous, henceRiemann integrable.

5

A. We say that f has the weak property (P), if for every linear continuous functionalφ ∈ Y∗, the composition φ f : D → K has property (P).

B. If property (P) involves a numerical value, say V (f), we say that f has ultraweak prop-erty (P), if there exists a unique y ∈ Y , such that, for every φ ∈ Y∗, the compositionφ f has property (P), with value V (φ f) = φ(y).

Specific instances of the above terminology are listed below.

Example 1. Assume we are given a function f : [a, b] → Y .

A. We say that f is weakly Riemann integrable, if φf : [a, b] → K is Riemann integrable,for every φ ∈ Y∗.

B. We say that f is ultraweakly Riemann integrable, if there exists y ∈ Y , such that

(i) f is weakly Riemann integrable, but also

(ii)∫ b

a(φ f)(t) dt = φ(y), ∀φ ∈ Y∗.

In this case the vector y, which we denote by w-∫ b

af(t) dt, is referred to as the weak

Riemann integral of f .

Example 2. Assume we are given D ⊂ K, a function f : D → Y , and an accumulationpoint z ∈ D.

A. We say that f is weakly K-differentiable at z, if φ f : D → K is K-differentiable at z,for every φ ∈ Y∗.

B. We say that f is ultraweakly K-differentiable at z, if there exists y ∈ Y , such that

(i) f is weakly K-differentiable at z, but also

(ii) (φ f)′(z) = φ(y), ∀φ ∈ Y∗.

In this case, the vector y, which we denote by f ′w(z), is referred to as the weak K-derivative of f at z.

Comment. In both Examples 1 and 2, the uniqueness of y was not stipulated. Theuniqueness follows from the fact that functionals in Y∗ separate the points in Y , whichmeans that the condition φ(y) = φ(y′), ∀φ ∈ Y∗, forces y = y′.

Example 3. If D is a topological space, and z ∈ D, then when we discuss the aboveversions of “continuity at z” for a function f : D → Y , we only address the “weak” version.

Remark 2. It is pretty clear that if (P) is any one of the above three “calculus proper-ties,” all of which have “strong” versions (for instance “norm” versions, when Y is a Banachspace), then one has the implications:

“strong (P)” ⇒ “ultraweak (P)” ⇒ “weak (P)”.

Moreover, following Examples 1 and 2, we have the following implications

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(i) if f is Riemann integrable, then∫ b

af(t) dt = w-

∫ b

af(t) dt;

(ii) if f is K-differentiable at z, then f ′(z) = f ′w(z).

Comment. The benefit of the use of normed vector spaces is illustrated perfectly byExample 4, Proposition 1 and Exercise 9 below.

Example 4. If D is an arbitrary set, then, exactly as in Example 3, when we discussthe versions of “boundedness” for a function f : D → Y , we again only talk about “weak”version. As it turns out, f is weakly bounded, if and only if it is norm-bounded. Thisfollows from the Uniform Boundedness Principle, as discussed in BS II. This means that the“strong” and the “weak” notions of boundedness coincide.

Proposition 1. Assume D ⊂ K, z ∈ D is an accumulation point, and f : D → Y isweakly K-differentiable at z. Then f is norm-continuous at z.

Proof. Fix some sequence (wn)∞n=1 ⊂ D r z, and let us prove that

limn→∞

‖f(wn) − f(z)‖ = 0. (9)

Consider the function g : N ∋ n 7−→ (wn − z)−1[f(wn) − f(z)] ∈ K, and remark that, sincef is weakly K-differentiable at z, it follows that

limn→∞

(φ g)(n) = limn→∞

(wn − z)−1φ(

f(wn) − f(z))

= (φ f)′(z), ∀φ ∈ Y∗.

In particular, this means that, for every φ ∈ Y∗, the function φ g : N → K is bounded,which in turn means that g : N → Y is weakly bounded. By Example 4, it follows that gis norm-bounded, so there exists some M > 0, such that ‖g(n)‖ ≤ M , ∀n. Of course, thismeans that

M ≥ ‖(wn − z)−1[f(wn) − f(z)]‖ =‖f(wn) − f(z)‖

|wn − z|, ∀n ∈ N. (10)

Of course, (10) gives ‖f(wn) − f(z)‖ ≤M |wn − z|, ∀n ∈ N, which clearly implies (9)

Exercise 9.

♥ Prove that, if f : [a, b] → Y is weakly Riemann integrable, then f isnorm-bounded.

Exercise 10. Prove that if (P) is any of the “calculus properties” discussed above,then both “ultraweak (P)” and “weak (P)” are compatible with compositions with linearcontinuous maps T : Y → Z, as in Exercise 1.

Exercises 11-12. Let Y be a Banach space, and let S be an arbitrary set (for now).Consider the Banach space ℓ∞Y (S) of all norm-bounded functions f : S → Y , which isa special case of the vector-valued ℓ∞-space introduced2 in BS II. The norm on ℓ∞Y (S) isdefined as ‖f‖sup = sups∈S ‖f(s)‖.

2 In BS II a slightly more complicated notation was used, in order to accommodate an S-tuple (Xs)s∈S

of Banach space. Here all the Xs’s are equal to Y.

7

11.

♥ Prove that, if S is a compact Hausdorff space, then the space

CY(S) = f : S → Y : f norm-continuous

is a closed linear subspace in ℓ∞Y (S), therefore (CY(S), ‖ . ‖sup) is a Banach space.

12.

♥ Assume S = [a, b], and consider the space

RY [a, b] = f : [a, b] → Y : f Riemann integrable.

By Exercise 9 we know that RY [a, b] is a linear subspace of ℓ∞Y ([a, b]). Prove that:

(i) RY [a, b] is closed in ℓ∞Y ([a, b]), therefore (RY [a, b], ‖ . ‖sup) is a Banach space;

(ii) The integration map RY [a, b] ∋ f 7−→∫ b

af(t) dt ∈ Y defines a linear continuous

operator.

(Hint: Prove (ii) first, by showing that∣

∣φ( ∫ b

af(t) dt

)∣

∣ ≤ (b− a) · ‖φ‖ · ‖f‖sup, ∀φ ∈Y∗, f ∈ RY [a, b]. To prove (i), start with some sequence (fn)∞n=1 ⊂ RY [a, b], which

converges in the norm ‖ . ‖sup to some f ∈ ℓ∞Y ([a, b]), and let yn =∫ b

afn(t) dt. By part

(ii), the sequence (yn)∞n=1 ⊂ Y is Cauchy, so it converges to some y ∈ Y . Prove that f

is Riemann integrable with∫ b

af(t) dt = y.)

We conclude this sub-section with a discussion of another important calculus property:“abstract” integrability. Since “abstract” integrability is always naturally linked to theconcept of measurability, it is very difficult3 to define a “strong” notion, so our discussionwill be limited to the (ultra)weak versions.

Definitions. Assume (X,A, µ) is a measure space, Y is a Banach space, and f : X → Yis some function.

A. We say that f is weakly µ-integrable, if φ f : X → K is µ-integrable, i.e. φ f ∈L

1K(X,A, µ), for every φ ∈ Y∗.

B. We say that f is ultraweakly µ-integrable, if there exists y ∈ Y , such that

(i) f is weakly µ-integrable, but also

(ii)∫

Xφ f dµ = φ(y), ∀φ ∈ Y∗.

In this case the vector y (which is unique, since Y∗ separates the points of Y), isdenoted by w-

∫

Xf dµ, and is referred to as the weak µ-integral of f , so that the above

identity reads:

φ

(

w-

∫

X

f dµ

)

=

∫

X

(φ f) dµ, ∀φ ∈ Y∗ (11)

3 One successful program, which develops both “strong” measurability and integration theory, is due toBochner. This approach is beyond the scope of these notes.

8

Example 5. Consider the measure space ([a, b],MLeb[a, b],Leb), where MLeb[a, b] de-notes the σ-algebra of Lebesgue measurable subsets of [a, b], and Leb denotes the Lebesguemeasure. Since in the K-valued case Riemann integrability implies Lebesgue integrability, itfollows that Riemann integrable function f : [a, b] → Y is ultraweakly Lebesgue integrable,and furthermore one has the equality

w-

∫

[a,b]

f dLeb =

∫ b

a

f(t) dt. (12)

Our next result is the ultraweak analogue of Exercise 6.

Proposition 2. Assume X is a locally compact Hausdorff space, Ω is compact Hausdorffspace, and µ is a positive Radon measure on X. Assume also f : X → CK(Ω) is a norm-continuous function, with compact support, i.e. there exists a compact subset T ⊂ X, suchthat f(x) = 0, ∀x ∈ X r T . Let F : X × Ω → K be its corresponding function (which iscontinuous, as discussed in Exercise 5), which is given by

F (x, ω) = (f(x))(ω), ∀ (x, ω) ∈ X × Ω.

The function h : Ω ∋ ω 7−→∫

XF (x, ω) dµ(x) ∈ K is continuous, so it belongs to the Banach

space CK(Ω). Moreover, one has4

∫

X

(φ f) dµ = φ(h), ∀φ ∈ CK(Ω)∗, (13)

so in particular, when one works with the measure space (X,Bor(X), µ), the function f isultraweakly µ-integrable, and furthermore, one has the equality: w-

∫

Xf dµ = h.

Proof. Before we proceed with the proof, let us first notice that, since F (x, ω) = 0, ∀x ∈X r T, ω ∈ Ω, the function h is in fact defined by

h(ω) =

∫

T

F (x, ω) dµ(x), ∀ω ∈ Ω. (14)

By Exercise 5 (applied to the product space T × Ω with the roles reversed), it follows that

(a) for every ω ∈ Ω, the function Gω : T ∋ x 7−→ F (x, ω) ∈ K is continuous, and

(b) the map g : Ω ∋ ω 7−→ Gω ∈ CK(T ) is norm-continuous.

In particular, whenever (ωλ)λ∈Λ ⊂ Ω is a net which is converges to some point ω ∈ Ω, itfollows that limλ∈Λ

[

supx∈T |F (x, ωλ) − F (x, ω)|]

= 0, which clearly implies

limλ∈Λ

∫

T

F (x, ωλ) dµ(x) =

∫

T

F (x, ω) dµ(x),

which by (14) means that limλ∈Λ h(ωλ) = h(ω), thus proving that h is indeed continuous.

4 For every φ ∈ CK(Ω)∗, the function φ f : Ω → K is continuous with compact support, hence µ-integrable, so the left-hand side of (13) makes sense.

9

In order to prove the equality (13) we first remark that, since f(x) = 0, ∀x ∈ X r T , inthe left-hand side of (13) we can replace X with T . Since every linear continuous functionalon CK(Ω) is a linear combination of positive ones, it suffices to prove (13) when φ is positive,which means that φ is defined by means of a positive Radon measure ν on Ω, by

φ(H) =

∫

Ω

H dν, H ∈ CK(Ω).

Using this presentation, the desired equality now reads:

∫

T

[∫

Ω

F (x, ω) dν(ω)

]

dµ(x) =

∫

Ω

[∫

T

F (x, ω) dµ(x)

]

dν(ω),

and everything follows from Fubini’s Theorem. (All measures involved here are finite, andthe product σ-algebra is the Borel σ-algebra on T × Ω.)

Corollary 1. If X is a locally compact Hausdorff space, µ is a positive Radon measureon X, Y is a Banach space, then every norm-continuous function f : X → Y with compactsupport is ultraweakly µ-integrable.

Proof. Fix some compact Hausdorff space Ω, such that Y is (isometrically isomorphic to) aclosed linear subspace in CK(Ω), and consider the annihilator space

Yann = φ ∈ CK(Ω)∗ : ψ∣

∣

Y= 0,

so that by the bi-annihilator Theorem (see DT I), we can represent

Y = g ∈ CK(Ω) : ψ(g) = 0, ∀ψ ∈ Yann. (15)

When regarding f as a norm-continuous CK(Ω)-valued function, we know that f is ultra-weakly µ-integrable, so we can consider the function h = w-

∫

Xf dµ ∈ CK(Ω), which satisfies

(13). In order to finish the proof, it suffices to show that

(i) h belongs to Y ;

(ii) φ(h) =∫

Xφ f dµ, ∀φ ∈ Y∗.

In order to prove (i) we use (15), by which it suffices to show that

ψ(h) = 0, ∀ψ ∈ Yann. (16)

But this is pretty obvious, since by the definition of Yann, combined with the fact that ftakes values in Y , we have ψ f = 0, ∀ψ ∈ Yann, and then the equality (16) follows from(13).

To check condition (ii) we use again (13) combined with the Hahn-Banach Theorem,which gives fact that for any φ ∈ Y∗ there exists φ′ ∈ CK(Ω)∗, with φ = φ′

∣

∣

Y.

Corollary 2 (“Easy” Fubini Theorem). Assume X1, X2 are locally compact spaces, µ1,µ2 are Radon measures on X1 and X2 respectively, Y is a Banach space, and f : X1×X2 → Yis a norm-continuous function with compact support.

10

(i) The function

g : X1 ∋ x1 7−→ w-

∫

X2

f(x1, x2) dµ2(x2) ∈ Y

is norm-continuous, with compact support.

(ii) If we equip the product space X1 ×X2 with the product Radon measure µ1 ⊠ µ2, then

w-

∫

X1×X2

f d(µ1 ⊠ µ2) = w-

∫

X1

g dµ1. (17)

Proof. (i). Since we are concerned with norm-continuity, we can assume Y = CK(Ω) forsome compact space Ω, so, by Exercise 5, there exists some continuous function F : X1 ×X2 × Ω → K, with compact support, such that f(x1, x2)(ω) = F (x1, x2, ω), ∀ (x1, x2, ω) ∈X1 × X2 × Ω. If we define the function G : X1 × Ω ∋ (x1, ω) 7−→

∫

X2F (x1, x2, ω) dµ2(x2),

then by Proposition 2 it follows that G is continuous, so by Exercise 5, the function g, whichis obviously defined by g(x1)(ω) = G(x1, ω), is also norm-continuous.

(ii). Having prove that g is norm-continuous (with compact support), the identity (17)is equivalent to

∫

X1×X2

φ f d(µ1 ⊠ µ2) =

∫

X1

φ g dµ1, ∀φ ∈ Y∗. (18)

By definition, the right-hand side of (18) is

∫

X1

[

φ(

w-

∫

X2

f(x1, x2) dµ2(x2))

]

dµ(x1) =

∫

X1

[∫

X2

(φ f)(x1, x2) dµ2(x2)

]

dµ1(x1),

which is clearly equal to the left-hand side of (18) by the (scalar) Fubini Theorem. (Asdiscussed in the proof of Proposition 2, we basically integrate over compact sets, where themeasures are finite.)

Comment. Corollary 1 can also be proved using another approach, which is a firststep in the development of Bochner’s “strong” integral. Given X a locally compact spaceand f : X → Y a norm-continuous function with compact support T , one can construct asequence (fn)∞n=1 of elementary function, i.e. so that for each n ∈ N, there exist disjoint setsBn1, . . . , Bnkn

∈ Bor(X), and vectors yn1, . . . , ynkn∈ Y , such that

(a) fn(x) =∑kn

j=1 χBnj(x)ynj, ∀x ∈ X;

(b) Bn1 ∪ · · · ∪Bnkn= T ;

(c) limn→∞ fn = f , uniformly, i.e. in ℓ∞Y (X), which means that

limn→∞

[

supx∈X

‖fn(x) − f(x)‖]

= 0.

(Specifically, one uses the fact that Range f is compact, so it can be covered with a finitecollection of balls B1/n(ynj) of radius 1/n and centers ynj, j = 1, . . . , kn. Taking the preimages

Anj = f−1(

B1/n(ynj))

, we define Bn1 = An1 ∩ T , and Bnj = T ∩Anj r⋃j−1

k=1Ank, for j > 1.)

11

It is pretty clear that all fn’s are ultraweakly integrable5, and w-∫

Xfn dµ =

∑kn

j=1 µ(Bnj)ynj.

Notice that, if we denote w-∫

Xfn dµ simply by yn, then:

|φ(ym − yn)| =

∣

∣

∣

∣

∫

X

φ (fm − fn) dµ

∣

∣

∣

∣

≤ µ(T ) · supx∈X

|(φ (fm − fn))(x)| ≤

≤ µ(T ) · ‖φ‖ · supx∈X

‖fm(x) − fn(x)‖, ∀m,n ∈ N, φ ∈ Y∗

which then yields the inequalities

‖ym − yn‖ ≤ µ(T ) · supx∈X

‖fm(x) − fn(x)‖, ∀m,n ∈ N.

By condition (c) above, it follows that (yn)∞n=1 is Cauchy, thus convergent to some y ∈ Y .Again by conditions (a)-(c) it is pretty obvious that limn→∞

∫

Xφ fn dµ =

∫

Xφ f dµ,

which yields φ(y) =∫

Xφ f dµ, ∀φ ∈ Y∗.

C. Power Series

In this sub-section we are going to fix a Banach space Y , and we are going to investigateconvergence properties of power series with coefficients in Y , which are understood as formalexpressions of the form

S(t) =∞

∑

n=0

tnyn, (19)

where (yn)∞n=0 is a sequence of vectors in Y . At this point, we think t as a formal parameter,so exactly as in the scalar case, we can define the K-vector space Y [[t]] of all (formal) powerseries (19). As it turns out, in the scalar case, we can multiply two formal power series withcoefficients in K, so that K[[t]] becomes a K-algebra. Likewise we can multiply formally anypower series with coefficients in Y with a formal power series with coefficients in K, so Y [[t]]becomes a K[[t]]-module. Inspired from the scalar case, as well as from the discussion fromthe preceding sub-section, we introduce the following terminology.

Definitions. Fix a formal power series (19), and some ζ ∈ K.

A. We say that (19) is norm-convergent at t = ζ, if the series∑∞

n=0 ζnyn is convergent in

the Banach space Y , in the norm topology, that is, there exists y ∈ Y , such that

limN→∞

∥

∥y −N

∑

n=0

ζnyn

∥

∥ = 0.

In this case, the vector y is denoted by∑∞

n=0 ζnyn, or simply by S(ζ).

B. We say that (19) is weakly convergent at t = ζ, if the (scalar) series (φ S)(t) =∑∞

n=0 ζnφ(yn) is convergent (in K), for every linear continuous functional φ ∈ Y∗.

5 In the theory of Bochner Integral, functions g : X → Y of the form g(x) =∑k

j=1χBj

(x)yj , withµ(Bj) < ∞ are declared elementary integrable, and the “strongly” integrable functions are those that canbe suitably approximated by elementary integrable ones.

12

C. We say that (19) is ultraweakly convergent at t = ζ, if there exists y ∈ Y , such that,for every φ ∈ Y∗, the scalar series (φ S)(ζ) =

∑∞

n=0 ζnφ(yn) is convergent to φ(y). In

this case, the vector y will be denoted by w-∑∞

n=0 ζnyn, or simply by Sw(ζ).

Remark 3. Exactly as with the other “calculus properties,” for a power series (19), onehas the implications: “norm-convergent at t = ζ” ⇒ “ultraweakly convergent at t = ζ” ⇒“weakly convergent at t = ζ,” and furthermore, if S(t) is norm-convergent at t = ζ, thenS(ζ) = Sw(ζ).

Proposition-Definition 3. Assume Y is a Banach space over K, and S(t) =∑∞

n=0 tnyn

is a power series with coefficients in Y. Define the set

DK(S) = ζ ∈ K : S(t) is convergent at t = ζ.

(i) If we define the “number” RS ∈ [0,∞] by

RS =1

lim supn→∞n√

‖yn‖,

then the set such DK(S) satisfies the inclusions:

ζ ∈ K : |ζ| < RS ⊂ DK(S) ⊂ ζ ∈ K : |ζ| ≤ RS. (20)

(ii) Moreover, for any ρ ∈ [0, RS), the series S(ζ) =∑∞

n=0 ζnyn converges uniformly, on

the closed ball Bρ,K = ζ ∈ K : |ζ| ≤ ρ, in the sense that

limN→∞

[

supζ∈Bρ,K

∥

∥S(ζ) −N

∑

n=0

ζnyn

∥

∥

]

= 0. (21)

The set DK(S) is called the (K-)domain of convergence of S(t), and the “number” RS iscalled the radius of convergence of S(t).

Proof. We are going to prove both (i) and (ii) together, and we notice that the first inclusionin (20) clearly follows from (ii). To prove the second inclusion in (20), it suffices to show that,if we start with some ζ ∈ K, with |ζ| > RS, then the series

∑∞

n=0 ζnyn is not convergent.

For the series to be non-convergent, it suffices to show that the sequence ‖ζnyn‖ = |ζ|n · ‖yn‖does not converge to 0. But this is quite clear, since the inequality |ζ| > RS means that1/|ζ| < lim supn→∞

n√

‖yn‖. By the definition of lim sup, there exists some sequence of

integers 1 ≤ n1 < n2 < . . . , such that 1/|ζ| < nk

√

‖ynk‖, ∀ k, and this forces the inequalities

|ζ|nk · ‖ynk‖ > 1, ∀ k ∈ N,

which show that the sequence (‖ζnyn‖)∞n=1 cannot converge to 0.

To prove (ii), we assume, of course, R > 0, and we fix ρ ∈ [0, RS), as well as somer ∈ [0, 1), such that ρ < rRS. Since r/ρ > 1/RS = lim supn→∞

n√

‖yn‖, by the definition of

lim sup, there exists some N0 ∈ N, such that r/ρ ≥ n√

‖yn‖, ∀n ≥ N0, which by taking thenth power, and multiplying by ρn, yields:

ρn‖yn‖ ≤ rn, ∀n ≥ N0. (22)

13

Fix, for the moment some ζ ∈ Bρ,K. On the one hand, since ‖ζnyn‖ = |ζ|n · ‖yn‖ ≤ ρn‖yn‖and r ∈ [0, 1), by (22) it follows that

∞∑

n=0

‖ζnyn‖ =

N0−1∑

n=0

‖ζnyn‖ +∞

∑

n=N0

‖ζnyn‖ ≤

N0−1∑

n=0

‖ζnyn‖ +∞

∑

n=N0

rn <∞,

so by the Summability Test (see BS II) it follows that the series∑∞

n=0 ζnyn is indeed con-

vergent. On the other hand, as we denote the sum of this series by S(ζ), again by (22) wehave the inequalities

∥

∥S(ζ) −N

∑

n=0

ζnyn

∥

∥ =∥

∥

∞∑

n=N+1

ζnyn

∥

∥ ≤∞

∑

n=N+1

‖ζnyn‖ ≤∞

∑

n=N+1

rn =rN+1

1 − r, ∀N ≥ N0.

This proves that

supζ∈Bρ,K

∥

∥S(ζ) −N

∑

n=0

ζnyn

∥

∥ ≤rN+1

1 − r, ∀N ≥ N0,

from which (21) follows immediately.

Comment. In Proposition 3 we only discussed the domain and radius of norm-convergence.It is, of course, possible to define the domain and the radius of ultraweak convergence, butas the next result shows, this more-or-less coincides with the one above.

Proposition 4. Assume Y is a Banach space over K, and S(t) =∑∞

n=0 tnyn is a power

series with coefficients in Y. If we denote, for every φ ∈ Y∗, by RφS the radius of convergence

of the power series (φ S)(t) =∑∞

n=0 ζnφ(yn), then the radius of convergence of S(t) is also

given byRS = inf

φ∈Y∗

RφS. (23)

Proof. Let RS denote the right-hand side of (23). Since “norm-convergent” ⇒ “weaklyconvergent,” is is clear that RS ≤ Rφ

S, ∀φ ∈ Y∗, so one has the inequality RS ≤ RS. Toprove the other inequality, we will assume RS > 0 (otherwise, there is nothing to prove),Fix, for the moment, some ρ ∈ (0, RS). For every φ ∈ Y∗, we know that

ρ < RφS =

1

lim supn→∞n√

|φ(yn)|,

which means that 1/ρ > lim supn→∞n√

|φ(yn)|, which by the definition of lim sup gives the

existence of some Nφ ∈ N, such that 1/ρ ≥ n√

|φ(yn)|, ∀n ≥ Nφ, which in turn gives (bytaking nth power, and multiplying by ρn):

|φ(ρnyn)| ≤ 1, ∀n ≥ Nφ. (24)

Of course, by (24) it follows that

supn≥0

|φ(ρnyn)| <∞, ∀φ ∈ Y∗,

14

and then, by the Uniform Boundedness Principle, it follows that

supn≥0

‖ρnyn‖ <∞.

If we denote this supremum by M , we obtain the inequality ρn‖yn‖ ≤M , which yields

n√

‖yn‖ ≤ ρ−1M1/n, ∀n ≥ 0,

so taking lim sup, we get

lim supn→∞

n√

‖yn‖ ≤ lim supn→∞

(q−1M1/n) = ρ−1,

thus proving the inequality

RS =1

lim supn→∞n√

‖yn‖≥ ρ.

Since the inequality RS ≥ ρ holds for all ρ ∈ (0, RS), it clearly implies RS ≥ RS.

D. Vector-valued Complex AnalysisThroughout this sub-section, Y is assumed to be a Banach space over C. In the devel-

opment of Complex Analysis for Y-valued functions, one uses (as in the scalar case) threetools:

(i) C-differentiability, as discussed in sub-section A);

(ii) Taylor series, which are a special type of the power series we discussed in sub-sectionC;

(iii) line integrals, which we introduce below.

Definitions. Assume Ω ⊂ C is an open set.

A. A piece-wise C1-curve in Ω is a continuous function γ : [a, b] → Ω, for which thereexists a partition π = (a = x0 < x1 < · · · < xn = b), such that γ

∣

∣

[xj−1,xj ]is continuously

differentiable, for each j. In this case, we call π a C1-partition for γ.

B. Given a piece-wise C1-curve in Ω, and a continuous function f : Ω → Y , we define theline integral

∮

γ

f(ζ) dζ =n

∑

j=1

∫ xj

xj−1

γ′(t)f(

γ(t))

dt,

where π = (a = x0 < x1 < · · · < xn = b) is any C1-partition of γ. One can show thatthe vector

∮

γf(ζ) dζ is independent of π.

15

Remark 4. If (fλ)λ∈Λ is a net of norm-continuous Ω → Y , which converges to a functionf : Ω → Y , uniformly on compact sets, in the sense that

limλ∈Λ

[

supζ∈K

‖fλ(ζ) − f(ζ)‖]

= 0,

for every compact subset K ⊂ Ω, then using Exercises 11 and 12, it follows that f is alsonorm-continuous, and

limλ∈Λ

∮

γ

fλ(ζ) dζ =

∮

γ

f(ζ) dζ,

for all piece-wise C1-curves γ in Ω.

Comment. (Line) integration along curves can be extended to so-called chains of piece-wise C1-curves, which are finite sets γ = γk : [ak, bk] → Cm

k=1 consisting of piece-wiseC1-curves with disjoint ranges (in Ω). For such a γ, and f as above, one defines

∮

γ

f(ζ) dζ =m

∑

k=1

∮

γk

f(ζ) dζ.

As a special case, if γ is a chain of piece-wise C1-loops, i.e. γk(ak) = γk(bk), ∀ k = 1, . . . ,m,then given a point z “not sitting on the chain,” i.e. z 6∈

⋃mk=1 Range γk, one defines the

quantity

indγ(z) =1

2πi

∮

γ

1

ζ − zdz,

which is referred to as the index of γ relative to z. It is well known that the index is alwaysan integer.

Convention. An open set D ⊂ C is said to be “tame,” if its closure D is compact, andthere exists a chain of piece-wise C1-loops γ = γk : [ak, bk] → Cm

k=1, such that:

• all γk are simple, i.e. the restrictions γk

∣

∣

[ak,bk)are

injective;

• the boundary ofD is given as ∂D =⋃m

k=1 Range γk;

• indγ(z) = 1, ∀ z ∈ D;

• indγ(z) = 0, ∀ z ∈ C rD.

(The simplest examples of “tame” open sets are those with polygonal boundary, such as theone depicted above. The arrows on the three loops that make up the boundary indicate thecorrect orientation.)

Given now a norm-continuous function f : Ω → Y , defined on some open set Ω thatcontains D, we denote the line integral

∮

γf(ζ) dζ by

∮

∂Df(ζ) dζ. Implicit in this notation

is the fact that the integral∮

γf(ζ) dζ is independent on the chain γ, as long as it has the

above properties.

Comment. Theorem 2 below provides an important characterization of an importantclass of vector-valued functions. The equivalence (i) ⇔ (ii) is known to hold in the scalar

16

case Y = C (when (i) ⇔ (i′) holds trivially), where (25) is (and will also be from this pointon) referred to as the Cauchy Integral Formula.

Theorem-Definition 2. Assume Ω ⊂ C is an open set. For a function f : Ω → Y, thefollowing conditions are equivalent.

(i) f is C-differentiable at all points z ∈ Ω.

(i’) f is weakly C-differentiable at all points z ∈ Ω.

(ii) f is norm-continuous, and for every “tame” open set D with D ⊂ Ω, one has theformula

f(z) =1

2πi

∫

∂D

(ζ − z)−1f(ζ) dζ, ∀ z ∈ D. (25)

Exactly as in the scalar case, a function, satisfying the above equivalent conditions, issaid to be holomorphic.

Proof. The implication (i) ⇒ (i′) is trivial.(i′) ⇒ (ii). Assume f is weakly C-differentiable at every point z ∈ Ω. By Proposition

1, we know that f is norm-continuous. Fix D a “tame” open set with D ⊂ Ω, and considerthe function

g : D ∋ z 7−→1

2πi

∮

∂D

1

ζ − zf(ζ) dζ ∈ Y .

(It is the norm-continuity of f that makes it possible for the line integral to be defined.)Fix for the moment a linear continuous functional φ ∈ Y∗. On the one hand, since the

composition φ f : Ω → C is C-differentiable at all z ∈ Ω, by the scalar Cauchy IntegralFormula, it follows that

(φ f)(z) =1

2πi

∮

∂D

(φ f)(ζ)

ζ − zdζ, ∀ z ∈ D. (26)

On the other hand, by the continuity properties of the Riemann integral (Exercise 1), wealso know that

1

2πi

∮

∂D

(φ f)(ζ)

ζ − zdζ = φ

(

1

2πi

∮

∂D

1

ζ − zf(ζ) dζ

)

= φ(

g(z))

, ∀ z ∈ D,

so if we fix z ∈ D, by (26), we now have

φ(

f(z))

= φ(

g(z))

, ∀φ ∈ Y∗.

Since Y∗ separates the points in Y , the above equality forces f(z) = g(z), which by thedefinition of g is precisely (25).

(ii) ⇒ (i). Assume f satisfies condition (ii), fix z ∈ Ω, and let us prove that f isC-differentiable at z.

17

Let first remark that, if D is any “tame” open set, with z ∈ D ⊂ D ⊂ Ω, then, using(25), we have

f(w) − f(z) =1

2πi

[∮

∂D

1

ζ − wf(ζ) dζ −

∮

∂D

1

ζ − zf(ζ) dζ

]

=

=1

2πi

∮

∂D

[

1

ζ − w−

1

ζ − z

]

f(ζ) dζ =

=1

2πi

∮

∂D

w − z

(ζ − w)(ζ − z)f(ζ), ∀w ∈ D,

so if we divide by w − z, we obtain the identity

(w − z)−1[f(w) − f(z)] =1

2πi

∮

∂D

1

(ζ − w)(ζ − z)f(ζ) dζ, ∀w ∈ D r z. (27)

Let us now specialize the set D as follows. Fix some positive r < dist(z,C r Ω), so thatthe open disk Dr(z) = ζ ∈ C : |ζ − z| < r is a “tame” open set, whose closure Dr(z) iscontained in Ω. The boundary ∂Dr(z) is (correctly) parametrized using the curve γ : [0, 2π] ∋t 7−→ reit+z ∈ ∂Dr(z), so for any open set E ⊃ ∂Dr(z), and every norm-continuous functionF : E → Y , one has

∮

∂Dr(z)

F (ζ) dζ =

∫ 2π

0

ireitF (reit + z) dt. (28)

Fix for the moment some w ∈ Dr(z). If we specialize (28) to the open set E = Ω r z, w,and to the function F (ζ) = 1

(ζ−w)(ζ−w)f(ζ), we get

∮

∂D

1

(ζ − w)(ζ − z)f(ζ) dζ =

∫ 2π

0

rieit

reit(reit + z − w)f(reit) dt =

∫ 2π

0

i

reit + z − wf(reit) dt,

so if we go back to (27), we now obtain

(w − z)−1[f(w) − f(z)] =

∫ 2π

0

1

reit + z − wf(reit) dt, ∀w ∈ Dr(z) r z. (29)

Remark now that, for w ∈ Dr(z) one has |reit + z − w| ≥ |reit| − |w − z| = r − |w − z|, sowe also have

∣

∣

∣

∣

1

reit + z − w−

1

reit

∣

∣

∣

∣

=

∣

∣

∣

∣

w − z

reit(reit + z − w)

∣

∣

∣

∣

≤|w − z|

r(r − |w − z|), ∀ t ∈ [0, 2π]. (30)

Fix now a sequence (wn)∞n=1 ⊂ Dr(z) r z, with limn→∞wn = z, and notice that by (30)it follows that the sequence (hn)∞n=1 of continuous functions hn : [0, 2π] ∋ t 7−→ 1/(reit +z − wn) ∈ C, converges uniformly to the function h : [0, 2π] ∋ t 7−→ 1/(reit) ∈ C, so ifwe consider the continuous function F : [0, 2π] ∋ t 7−→ f(reit) ∈ Y , then the sequenceof continuous functions hnF converges uniformly to hF . By Exercise 12, it follows thatlimn→∞

∫ 2π

0hn(t)F (t) dt =

∫ 2π

0h(t)F (t) dt (in norm). Going back to (29) it now follows that

limn→∞

(wn − z)−1[f(wn) − f(z)] =1

2π

∫ 2π

0

1

reitf(reit + z) dt,

with the limit computed in norm. Since the limit is independent of the sequence, this showsthat f is C-differentiable at z.

18

Comment. Corollary 3 below is known to hold in the C-valued case. The formulas (31)are referred to as the higher order Cauchy Integral Formulas.

Corollary 3. If Ω ⊂ C is open and f : Ω → Y is holomorphic, then f is infinitely manytimes C-differentiable at all points in Ω, thus all higher order derivatives f (n) : Ω → Y,n ≥ 1, are also holomorphic. Moreover, id D is a “tame” open set, with D ⊂ Ω, then

f (n)(z) =n!

2πi

∫

∂D

(ζ − z)−n−1f(ζ) dζ, ∀ z ∈ D, n ∈ N. (31)

Proof. To prove the first statement, it suffices to show (by an obvious inductive argument)that the first derivative f ′ : Ω → Y is holomorphic. By Theorem 2, it suffices to show that,for every φ ∈ Y∗, the composition φ f ′ : Ω → C is holomorphic. This follows immediatelyfrom the identity (see sub-section B):

(φ f ′)(z) = (φ f)′(z), ∀ z ∈ Ω, (32)

and the scalar case, which implies that (φ f)′ is holomorphic.To prove the formulas (31) we use again (32) inductively to obtain the indentities

(φ f (n))(z) = (φ f)(n)(z), ∀φ ∈ Y∗, z ∈ Ω, n ∈ N,

so using the scalar version, as well as Exercise 1, we have

φ(

f (n)(z))

= (φ f)(n)(z) =n!

2πi

∮

∂D

(ζ − z)−n−1(φ f)(ζ) dζ =

=n!

2πi

∮

∂D

φ(

(ζ − z)−n−1f(ζ))

dζ =

= φ

(

n!

2πi

∮

∂D

(ζ − z)−n−1f(ζ) dζ

)

, ∀φ ∈ Y∗, z ∈ D, n ∈ N.

Arguing exactly as in the proof of Theorem 2, the above identities clearly imply (31).

Comment. Corollary 4 below is also known to hold in the C-valued case.

Corollary 4 (Taylor Series Expansion.) Let Ω ⊂ C be an open set. For a functionf : Ω → Y, the following are equivalent:

(i) f is holomorphic;

(ii) For every z ∈ Ω, there exists a power series Sz(t) =∑∞

n=0 tnyn(z), with coefficients in

Y, such that6∞

∑

n=0

ζnyn(z) = f(ζ + z), (33)

for all ζ ∈ C, with |ζ| < dist(z,C r Ω). In particular, the radius of convergence RSz

of Sz(t) satisfies the inequality RSz≥ dist(z,C r Ω).

6 If |ζ| < dist(z, C r Ω), then clearly ζ + z ∈ Ω.

19

Moreover, in this case, for every z ∈ Ω, the coefficients of the power series Sz(t) are yn(z) =(n!)−1f (n)(z), ∀n ≥ 0.

Proof. (i) ⇒ (ii). Assume f is holomorphic, fix z ∈ Ω, and consider the power seriesSz(t) =

∑∞

n=0 tn(n!)−1f (n)(z). Since φf is holomorphic, for every φ ∈ Y∗, and (φf)(n)(z) =

φ(

f (n)(z))

, it follows that the power series φ Sz is given by

(φ Sz)(t) =∞

∑

n=0

tn(n!)−1(φ f)(n)(z).

Since we know that the C-valued case is true, it follows that

(a) the radius of convergence RφSz

of the power series φ Sz satisfies the inequality RφSz

≥dist(z,C r Ω), and furthermore,

(b) for any ζ that satisfies |ζ| < dist(z,C r Ω)(≤ RφSz

), one has

(φ Sz)(ζ) = (φ f)(ζ + z). (34)

By Proposition 4, we know that the radius of convergence RSzof the power series Sz is

given by RS = infφ∈Y∗ RφSz

, so by (a) it follows that RSz≥ dist(z,C r Ω). In particular,

if |ζ| < dist(z,C r Ω)(≤ RSz), then Sz(t) is norm-convergent at t = ζ, and furthermore by

(b), one has

φ(

∞∑

n=0

ζn(n!)−1f (n)(z))

= (φ Sz) = φ(

f(ζ + z))

, ∀φ ∈ Y∗,

so again using the fact that Y∗ separates the point in Y , it follows that∞

∑

n=0

ζn(n!)−1f (n)(z) = (φ Sz) = f(ζ + z).

(ii) ⇒ (i). Assume, for each z ∈ Ω, there exist vectors(

yn(z))∞

n=0⊂ Y , satisfying

condition (ii), and let us show that f is holomorphic. By Theorem 2, it suffices to show thatφ f : Ω → C is holomorphic, for every φ ∈ Y∗. But this property is obvious, since for eachφ ∈ Y∗, the coefficients

(

φ(

yn(z)))∞

n=0⊂ C satisfy condition (ii) for the function φ f , so by

the scalar version, this forces φ f to be holomorphic.

We conclude this sub-section with the vector-valued versions of three other well knowncomplex analysis results. Exercise 14 contains the vector-valued version of Liouville’s Theo-rem. Exercise 15 deals with “half” of the Maximum Modulus Principle.

Exercise 13.

♥Assume Ω ⊂ C is an open set, and (fλ)λ∈Λ is a net of holomorphic functionsfλ : Ω → Y . which is uniformly Cauchy on compact sets, i.e.

limλ∈Λ

[

supσ,ν≻λz∈K

‖fσ(z) − fν(z)‖

]

= 0,

for all compact sets K ⊂ Ω.Prove that, for each z ∈ Ω, the net

(

fλ(z))

λ∈Λ⊂ Y is norm-convergent to some vector

f(z) ∈ Y , and furthermore,

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(i) the function f : Ω → Y is holomorphic;

(ii) for every n ≥ 0, the net (f(n)λ )λ∈Λ is uniformly convergent on compact sets to f (n), i.e.

limλ∈Λ

[

supz∈K

‖f(n)λ (z) − f (n)(z)‖

]

= 0,

for all compact sets K ⊂ Ω.

Exercise 14.

♥Prove that a bounded holomorphic function f : C → Y is constant. (Hint:For every φ ∈ Y∗, the function φf : C → C is holomorphic and bounded, so by the C-valuedLiouville’s Theorem, it is constant.)

Exercise 15.

♥ Assume Ω ⊂ C is a bounded open set (so that the closure Ω is compact),and f : Ω → Y is a continuous function, such that f

∣

∣

Ω: Ω → Y is holomorphic. Prove that

maxz∈Ω

‖f(z)‖ = maxz∈∂Ω

‖f(z)‖.

(Hint; We know that the statement is true in the C-valued case, by the Maximum ModulusPrinciple. Let M = maxz∈∂Ω ‖f(z)‖. For every linear continuous functional φ ∈ Y∗, bythe Maximum Modulus Principle, we know that, for every w ∈ Ω, we have

∣

∣φ(

f(w))∣

∣ ≤maxz∈∂Ω

∣

∣φ(

f(z))∣

∣, so by the definition of M , it follows that∣

∣φ(

f(w))∣

∣ ≤ ‖φ‖ ·M , ∀φ ∈ Y∗,which forces ‖f(w)‖ ≤M .)

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