Chapter 5: the Metric Space H(G)

Chapter 5: The metric space H(G)

Course 414, 2003–04 February 22, 2004

Remark 5.1 For G ⊂ C open, we use the notation

H(G) = {f : G → C analytic} for the set (space) of all holomorphic functions on G and

C(G) = {f : G → C continuous} for the set (space) of all continuous functions on G. The main change of perspective now is that we move from looking at functions one at a time to looking at properties of the whole space of functions. It is a more simple fact that H(G) and C(G) are algebras than the fact that M(G) is an algebra. We deﬁne vector space operations on f, g ∈ C(G) and λ ∈ C via (f + g)(z) = f(z) + g(z) and (λf)(z) = λf(z)

and a multiplication on C(G) by (fg)(z) = f(z)g(z). This makes C(G) a commutative algebra over C and H(G) a subalgebra. ∞ We are interested in what it might mean for a sequence of functions (fn)n=1 (in H(G) say) to converge to a limit f ∈ H(G). Most of the ideas are the same initially for sequences in C(G) and but we will come later to properties special to holomorphic functions. The simplest way to define limn→∞ fn = f is the notion of pointwise convergence on G. This means that for each z ∈ G we have limn→∞ fn(z) = f(z) (convergence as sequences of complex numbers). Though it easy to define, it is a kind of convergence that has rather unpleasant properties. Perhaps it is more accurate to say that it fails to have many properties we would like to have linking behaviour of the functions fn to the behaviour of the limit function f. These drawbacks include (i) We can define pointwise convergence for sequences in H(G) without insisting that the limit function be in H(G) or even C(G). Then there are sequences of holomorphic functions that converge pointwise to discontinuous limits f. Such examples are hard to come by and the easiest way to show they can be found is to invoke Runges theorem (something we will come to later).

1 2 414 2003–04 R. Timoney

(ii) We can exhibit a sequence in C(C) that converges to a discontinuous limit. For example −n|z| fn(z) = e has 1 z = 0 lim fn(z) = n→∞ 0 z 6= 0

(iii) There is no distance on C(G) or H(G) so that fn → f pointwise on G is equivalent to convergence in distance, that is to

lim distance (fn, f) = 0 n→∞

This means in fact that the behaviour of pointwise convergence in H(G) or C(G) belongs in a more complicated theory than the theory of metric spaces. In metric spaces many, but not all, familiar properties of limits of sequences of scalars are still true.

The next simplest notion to consider is uniform convergence on G. Recall that fn → f uniformly on G (as n → ∞) means the following

Given ε > 0 there exists N so that

n > N, z ∈ G ⇒ |fn(z) − f(z)| < ε

With a small bit of work, we could check that this is equivalent to the following formulation

Given ε > 0 there exists N so that

n > N ⇒ sup |fn(z) − f(z)| < ε z∈G

In this way we can see a deﬁnition of distance we can use to describe this kind of convergence

distance (g, h) = dG(g, h) = sup |g(z) − h(z)| z∈G

and this looks promising. We do know that fn → f uniformly on G and each fn ∈ C(G) implies f ∈ C(G) (‘uniform limits of continuous functions are continuous’). We also know a similar fact about sequences in H(G) (see Exercises 2 where we saw that Moreras theorem could be used to show that uniform limits of sequences of holomorphic functions are holomorphic). However, there are still drawbacks:

(a) If we apply this convergence in H(C) (that is to entire functions) we ﬁnd that dC(fn, f) < ε < ∞ implies that fn(z)−f(z) is a bounded entire function, hence a constant by Liouville’s theorem. So in H(C) the sequences of functions that converge uniformly are all of the form fn(z) = f(z) + cn with cn constants converging to 0 (except for some ﬁnite number of n where fn can be anything). Thus a very restrictive notion of convergence. Chapter 5 — the metric space H(G) 3

(b) This is linked to the fact that dG(g, h) need to be ﬁnite for f, g ∈ C(G) (or even f, g ∈ H(G)). So our notion of ‘distance’ is not so great as it is not always deﬁned. It is a good notion to use if we looked only at bounded continuous functions on G (or bounded holomorphic functions, but we see that if G = C then there are only constant bounded holomorphic functions). We will not look into these spaces of bounded functions, though they have been studied extensively, as they are a rather more advanced topic.

(c) The notion of uniform convergence does not cover all the cases where we have already encountered limits of sequences of functions, especially power series. For example we have

∞ n 1 X n X j = z = lim z = lim sn(z)(|z| < 1) 1 − z n→∞ n→∞ n=0 j=1

but we don’t have uniform convergence for z ∈ D(0, 1). Similarly we have

∞ n n n z X z X z e = = lim = lim sn(z)(z ∈ C) n! n→∞ n! n→∞ n=0 j=1

but no unform convergence on C. In general for a power series with radius of convergence R > 0 ∞ n X n X j f(z) = an(z − a) = lim aj(z − a) = lim sn(z)(|z − a| < R) n→∞ n→∞ n=0 j=1

we do not have unform convergence for z ∈ D(a, R) but we do have uniform convergence for |z − a| ≤ r with any r < R. (This follows from the Weierstrass M-test.) This last point shows the way to proceed in general.

∞ Definition 5.2 If (fn)n=1 is a sequence of functions fn : G → C defined on an open G ⊂ C and f : G → C is another function, then we say the sequence converges to f uniformly on compact subsets of G if the following is satisfied

∞ for each compact subset K ⊂ G the sequence of restrictions (fn |K )n=1 converges uniformly on K to f |K . Spelling this out more it says for each compact subset K ⊂ G and any given ε > 0 we can ﬁnd N so that

n > N, z ∈ K ⇒ |fn(z) − f(z)| < ε

This can also be reformulated using a distance on each K, or on C(K) = {g : K → C continuous}, deﬁned by dK (g, h) = sup |g(z) − h(z)| z∈K 4 414 2003–04 R. Timoney

For g, h ∈ C(K) this supremum is always ﬁnite (continuous complex-valued functions on compact sets are always bounded and there is even some z ∈ K where |g(z) − h(z)| achieves its maximum). This dK distance makes C(K) a metric space, that is dK satisﬁes the following properties that seem to be natural for familiar distances:

(i) dK (g, h) ≥ 0 ∀g, h ∈ C(K)

(ii) dK (g, h) = dK (h, g) ∀g, h ∈ C(K) (iii) (triangle inequality)

dK (g1, g3) ≤ dK (g1, g2) + dK (g2, g3)(∀g1, g2, g3 ∈ C(K))

[These properties so far make C(K) with dK a pseudo-metric space (or sometimes called a semi-metric space).]

(iv) g, h ∈ C(K) and dK (g, h) = 0 ⇒ g = h.

We can say fn → f uniformly on compact subsets if and only if the following is true

for each compact K ⊂ G, limn→∞ dK (fn, f) = 0

What we have now is inﬁnitely many distances dK to consider in order to describe uniform convergence on compact sets, but we will show soon how to manage with just one distance. Meantime we show that this kind of convergence has desirable properties (and that it covers the power series situation).

∞ Proposition 5.3 Let G ⊂ C be open and (fn)n=1 a sequence in H(G) that converges uniformly on compact subsets of G to some limit function f : G → C. Then f ∈ H(G).

Proof. Analyticity of the limit is a local property, something to be shown at each point a ∈ G. We need to show that f 0(a) exists. For this ﬁx a ∈ G and choose r > 0 so that D(a, r) ⊂ G. Then pick δ with 0 < δ < r ¯ so that K = D(a, δ) ⊂ G is a compact subset. So fn → f uniformly on K and so fn → f uniformly on D(a, δ). By Exercises 2 question 3 (the one that used Moreras theorem) we have that the restriction of f to D(a, δ) is analytic. So f 0(a) exists.

∞ Proposition 5.4 If G ⊂ C is open, then we can ﬁnd a sequence (Kn)n=1 of compact subsets of G such that S∞ (i) n=1 Kn = G

◦ (ii) Kn ⊂ (Kn+1) = the interior of Kn+1 for each n = 1, 2,...

(iii) if K ⊂ G is any compact subset, then there is some n with K ⊂ Kn We call such a sequence an exhaustive sequence of compact subsets of G. Chapter 5 — the metric space H(G) 5

Proof. We already showed how to construct such a sequence in the proof of 4.21 (and see 4.22). ¯ When G = C it was easy — just put Kn = D(0, n) and for G 6= C we took 1 K = {z ∈ G : dist(z, \ G) ≥ and |z| ≤ n}. n C n Recall also [ 1 \ K = {w ∈ : |w| > n} ∪ D w, C n C n w∈C\G 1 1 and z ∈ Kn ⇒ D z, n − n+1 ⊂ Kn+1. S∞ We did not note then that the Kn satisfy (iii) but if K ⊂ G is compact then K ⊂ n=1 Kn ⊂ S∞ ◦ ◦ n=1(Kn+1) and so we have an open cover of K by the sets (Kn+1) . Hence there is a ﬁnite SN ◦ ◦ subcover K ⊂ n=1(Kn+1) = (KN+1) ⊂ KN+1. This shows (iii) holds.

Lemma 5.5 Suppose G ⊂ C is open and (Kn) is an exhaustive sequence of compact subsets of ∞ G. Let (fj)j=1 be a sequence of functions fj ∈ C(G) and f ∈ C(G). Then fj → f uniformly on compact subsets of G (as j → ∞) if and only if

∀n = 1, 2,..., lim dK (fj, f) = 0 j→∞ n

This means that instead of considering the vastly inﬁnite (uncountably inﬁnite) number of compact K ⊂ G it is enough to look only at K in the sequence Kn.

Proof. Since each Kn ⊂ G is a compact subset it is clear that if fj → f uniformly on compact subsets of G then limj→∞ dKn (fj, f) = 0 for each n.

Conversely, suppose we know limj→∞ dKn (fj, f) = 0 for each n and we take any K ⊂ G compact. Then there is some n with K ⊂ Kn and it is fairly clear then that

dK (fj, f) ≤ dKn (fj, f) → 0 as j → ∞

Example 5.6 It is now clear that in the power series examples with radius of convergence R > 0

∞ n X n X j f(z) = an(z − a) = lim aj(z − a) = lim sn(z)(|z − a| < R) n→∞ n→∞ n=0 j=1 we do have sn → f uniformly on compact subsets of D(a, R). The reason is every compact subset K ⊂ D(a, R) is contained in a subdisc D(a, r) with 0 < r < R (where we do have uniform convergence by the Weierstrass M-test argument). Another way to see this is to work out what the standard construction gives for Kn when G = D(a, R) (for n large, a sequence of closed discs of radii < R centered at a).

Remark 5.7 These dK (or dKn ) have a nice property that they are associated with a semi-norm on C(G). A semi-norm is a way of associating a length to elements of a vector space that has 6 414 2003–04 R. Timoney

all the usual properties of lengths of vectors except non-zero elements are allowed to have length = 0. (Also we are not saying that there is any inner product around.) If we deﬁne for K ⊂ G compact (here G ⊂ C is open as usual) and f ∈ C(G)

kfkK = sup |f(z)| z∈K

then we have the following characteristic semi-norm properties satisﬁed

(i) kfkK ≥ 0 for each f ∈ C(G)

(ii) (triangle inequality) kf + gkK ≤ kfkK + kgkK holds for all f, g ∈ C(G)

(iii) λ ∈ C, f ∈ C(G) ⇒ kλfkK = |λ|kfkK

The link between the semi-norm k · kK and the pseudo-metric dK is

dK (g, h) = kg − hkK and all semi-norms give rise to a pseudo-metric in this way. We will now show that there is a single metric on C(G) that describes uniform convergence on compact subsets, but we will lose the connection with a norm (or semi-norm).

Lemma 5.8 Let G ⊂ C be open and K ⊂ G compact. Deﬁne ρK (f, g) for f, g ∈ C(G) by

dK (f, g) ρK (f, g) = 1 + dK (f, g)

Then

(i) ρK is a pseudo metric on C(G)

(ii) 0 ≤ ρK (f, g) < 1 for all f, g ∈ C(G)

∞ (iii) If (fn)n=1 is a sequence of functions in C(G) and f ∈ C(G), then

lim dK (fn, f) = 0 ⇐⇒ lim ρK (fn, f) = 0 n→∞ n→∞

Proof.

(i) This is the main part requiring proof. It is clear that ρK (f, g) = ρK (g, h) ≥ 0 and the part that requires an argument is that ρK satisﬁes the triangle inequality. Chapter 5 — the metric space H(G) 7

Let ρ(t) = t/(1+t) for t ≥ 0 and note that ρ0(t) = 1/(1+t)2 ≥ 0 so that ρ(t) is increasing (for t ≥ 0). Now the triangle inequality holds for dK and so we can say

ρK (f1, f3) = ρ(dK (f1, f3))

≤ ρ(dK (f1, f2) + dK (f2, f3)) d (f , f ) + d (f , f ) = K 1 2 K 2 3 1 + dK (f1, f2) + dK (f2, f3) d (f , f ) d (f , f ) = K 1 2 + K 2 3 1 + dK (f1, f2) + dK (f2, f3) 1 + dK (f1, f2) + dK (f2, f3) d (f , f ) d (f , f ) ≤ K 1 2 + K 2 3 1 + dK (f1, f2) 1 + dK (f2, f3)

= ρK (f1, f2) + ρk(f2, f3)

(ii) This is clear from the deﬁnition of ρK

(iii) If limn→∞ dK (fn, f) = 0 then limn→∞ ρK (fn, f) = 0 because the function ρ(t) = t/(1+t) is continuous at t = 0 and ρK (fn, f) = ρ(dK (fn, f)).

Conversely if limn→∞ ρK (fn, f) = 0, then limn→∞ dK (fn, f) = 0 because dK (fn, f) = ρK (fn, f)/(1 − ρK (fn, f)).

∞ Proposition 5.9 Let G ⊂ C be an open subset and (Kn)n=1 an exhaustive sequence of compact subsets of G. Then ∞ X 1 ρ(f, g) = ρ (f, g)(f, g ∈ C(G)) 2n Kn n=1 ∞ deﬁnes a metric on C(G). If (fj)j=1 is a sequence in C(G) and f ∈ C(G) then

lim ρ(fj, f) = 0 j→∞ if and only if fj → f uniformly on compact subsets of G (as j → ∞). In other words this metric ρ is such that convergence in this distance corresponds to uniform convergence on compact subsets.

Proof. To show that ρ is a pseudo metric on C(G) is not difﬁcult using Lemma 5.8. The series P∞ n deﬁning ρ(f, g) is convergent since it is smaller than n=1 1/2 = 1 and it clearly has a non- negative sum. The triangle inequality follows by applying Lemma 5.8 as does ρ(f, g) = ρ(g, f). S∞ Finally ρ(f, g) = 0 ⇒ ρKn (f, g) = 0∀n ⇒ dKn (f, g) = 0∀n ⇒ f(z) = g(z)∀z ∈ n=1 Kn = G ⇒ f = g

If fj → f uniformly on compact subsets of G, then limj→∞ dKn (fj, f) = 0 for each n and

hence limj→∞ ρKn (fj, f) = 0 (for each n). To show that limj→∞ ρ(fj, f) = 0, start with ε > 0 given and choose N large enough that

∞ X 1 ε < 2n 2 n=N+1 8 414 2003–04 R. Timoney

Then we have

∞ N ∞ N X 1 X 1 X 1 X 1 ε ρ(f , f) = ρ (f , f) ≤ ρ (f , f) + < ρ (f , f) + j 2n Kn j 2n Kn j 2n 2n Kn j 2 n=1 n=1 n=N+1 n=1

PN 1 Since limj→∞ n=1 2n ρKn (fj, f) = 0, if j is large enough (say j > j0) then this ﬁnite sum is < ε/2 and ρ(fj, f) < ε. Conversely, if limj→∞ ρ(fj, f) = 0 then notice that, for any ﬁxed n,

∞ X 1 ρ (f , f) ≤ 2n ρ (f , f) = 2nρ(f , f) → 0 as j → ∞ Kn j 2m Km j j m=1

So limj→∞ ρKn (fj, f) = 0 for each n, hence by Lemma 5.8(iii) limj→∞ dKn (fj, f) = 0 and by Lemma 5.5, fj → f uniformly on compact subsets of G as j → ∞.

Remark 5.10 From now on we will frequently consider H(G) (or C(G)) as equipped with a metric ρ of the above type. So (H(G), ρ) is now a metric space. We can define open sets in a metric space by analogy with the way we define open subsets in C. Sometimes it may help to consider a subset of H(G) as a family of analytic functions (which means the same as a set of functions). If we fix f0 ∈ H(G) and r > 0 we can define the ball about f0 of radius r as Bρ(f0, r) = {f ∈ H(G): ρ(f, f0) < r}

and then we can define a subset F ⊂ H(G) to be open if f0 ∈ F ⇒ ∃r > 0 with Bρ(f0, r) ⊂ F. We can define compactness for subsets F ⊂ H(G) by requiring that every open cover has a finite subcover. But in metric spaces, we always have an alternative way to describe compactness via sequences (as we do in C). A family F ⊂ H(G) is compact if and only if every ∞ ∞ sequence (fn)n=1 of functions fn ∈ F has a subsequence (fnj )j=1 with a limit f ∈ F. (So

limj→∞ ρ(fnj , f) = 0.)

Lemma 5.11 For G ⊂ C open and let F ⊂ H(G) be a family of holomorphic functions (or subset of H(G)). Consider two metrics ρ1, ρ2 on H(G) constructed as above (from exhaustive sequences of compact subsets of G). Then F is compact in (H(G), ρ1) if and only if it is compact in the metric space (H(G), ρ2).

Proof. As noted above compactness of subsets of a metric space can be characterised using limits of sequences. But both metrics ρ1 and ρ2 have the same convergent sequences by Proposition 5.9.

Remark 5.12 Closures and closed subsets of (H(G), ρ) can also be characterised using limits of sequences. Hence they are also the same for different metrics ρ constructed as before. Open subsets of (H(G), ρ) are also the same in different such metrics ρ (since open sets are complements of closed sets). Similar remarks apply to (C(G), ρ). Chapter 5 — the metric space H(G) 9

Proposition 5.13 Let G ⊂ C be open. Then H(G) is a closed subset of C(G) (when we use a metric of the type constructed above on C(G)). (As H(G) is a vector subspace of C(G), it is commonly called a closed subspace.)

Proof. Convergent sequences of functions fn in H(G) with limn→∞ fn = f ∈ C(G) have limits f ∈ H(G) by Proposition 5.3. It follows that H(G) is closed.

Lemma 5.14 Let G ⊂ C be open and ρ a metric on C(G) constructed as before. Let K ⊂ G be compact. Then the map f 7→ kfkK : C(G) → R is continuous.

Proof. Continuity on metric spaces can be described via sequences. It sufﬁces to prove that limn→∞ ρ(fn, f) = 0 ⇒ limn→∞ kfnkK = kfkK . This follows because a version of the triangle inequality says

|kfnkK − kfkK | ≤ kf − fnkK → 0

∞ (as n → ∞) by uniform convergence of the sequence (fn)n=1 to f on K.

Deﬁnition 5.15 Let G ⊂ C be open. A subset F of C(G) (or H(G)) is called relatively compact if and only if its closure is compact. (The closure will be the set of all possible limits of convergent ∞ sequences (fn)n=1 with each fn ∈ F.) A subset F is called bounded if for each K ⊂ G compact

sup kfkK < ∞ f∈F (This could be phrased as ‘uniformly bounded on compact subsets’, but the shorter term ‘bounded’ is usually used. In a way, this is the only sensible notion of ‘bounded’. Deﬁning bounded in terms of the distance ρ is not much use as all distances are bounded by 1.)

Lemma 5.16 Let G ⊂ C be open and F ⊂ C(G) (or F ⊂ H(G)) a subset. ∞ (i) F is relatively compact if and only if every sequence (fn)n=1 in F has a convergent subse- ∞ quence (fnj )j=1 (convergent to some limit in C(G)) (ii) Relatively compact families F are bounded.

Proof. (i) (This is actually a general fact, true in any metric space.) If the closure of F is compact, ∞ then a sequence (fn)n=1 in F is also a sequence in the closure. As the closure is compact ∞ there is a subsequence (fnj )j=1 converging to a limit (in the closure, hence in C(G)). Going the other way, suppose we know every sequence in F has a convergent subsequence. ∞ ¯ Take a sequence (gn)n=1 in the closure F. Then, for each n, we can ﬁnd fn ∈ F so that 10 414 2003–04 R. Timoney

∞ ρ(gn, fn) < 1/n. By assumption, there is a subsequence (fnj )j=1 which converges to some

f ∈ C(G). But then gnj → f as j → ∞ because 1 ρ(gnj , f) ≤ ρ(gnj , fnj ) + ρ(fnj , f) ≤ + ρ(fnj , f) → 0 (as j → ∞). nj

(ii) Let F ⊂ C(G) be relatively compact. If it is not bounded, then there is some K ⊂ G compact so that sup kfkK = ∞. f∈F ∞ So for each n, we can ﬁnd fn ∈ F with kfnkK > n. A convergent subsequence (fnj )j=1 exists (with a limit f ∈ C(G)). By Lemma 5.14,

lim kfn kK = kfkK j→∞ j

but that contradicts kfnj kK > nj → ∞ as j → ∞.

Theorem 5.17 (Montels Theorem) Let G ⊂ C be open and F ⊂ H(G) a family of analytic functions. Then F is relatively compact if and only if F is bounded (in the sense of Deﬁni- tion 5.15, that is uniformly on compact subsets).

Remark 5.18 Though this theorem corresponds exactly with the situation for subsets of C and ﬁnite-dimensional vector spaces like Rn, a similar result is not usually true in inﬁnite dimensional spaces. The same statement fails in C(G). An example to show that is F = {fn : n = 1, 2,...}, fn(z) = exp(−n|z|) and G any open set containing the origin. The family is bounded because ∞ |fn(z)| = fn(z) ≤ 1, but there is no subsequence of the sequence (fn)n=1 that converges in C(G). If a subsequence did converge in (C(G), ρ) to some limit f ∈ C(G), then the subsequence would have to converge pointwise to the same limit function. This forces f(0) = 1 and f(z) = 0 for z 6= 0. So f cannot be in C(G). Our proof of Montels Theorem will require the corresponding theorem for families in C(G). That theorem (the Arzela-Ascoli Theorem) is more complicated to state and also a bit long to prove. We will relegate the proof of it to an appendix.

Deﬁnition 5.19 Let G ⊂ C be open z0 ∈ G a point and F ⊂ C(G) a family of continuous functions on G. Then the family F is called equicontinuous at z0 if for each ε > 0 there is some δ > 0 so that |z − z0| < δ, f ∈ F ⇒ |f(z) − f(z0)| < ε. (Perhaps it is worth comparing this to uniform continuity of a single function on a set E. There f is ﬁxed and z, z0 are any two points of the set E with |z − z0| < δ. Here, the δ required is as in the condition for continuity of a single function f at z0, but the same δ has to work for all f ∈ F.) Chapter 5 — the metric space H(G) 11

Theorem 5.20 (Arzela-Ascoli Theorem) Let G ⊂ C be open and F ⊂ C(G) a family of continuous functions on G. Then F is relatively compact in (C(G), ρ) if and only if it satisﬁes both of the following conditions:

(i) F is pointwise bounded on G (that is, for each point z0 ∈ G, we have supf∈F |f(z0)| < ∞)

(ii) F is equicontinuous at each point of G.

Proof. (of Theorem 5.17 using Theorem 5.20) One direction is already covered by Lemma 5.16 (i). If F ⊂ H(G) is relatively compact then it must be bounded. For the other direction, we use Theorem 5.20. Assume F ⊂ H(G) is bounded. Then F is certainly pointwise bounded since points z0 ∈ G make singleton compact sets K = {z0} (and F is uniformly bounded on K). Next we claim F is equicontinuous at each z0 ∈ G. Fix z0 ∈ G and ε > 0. We can ﬁnd r > 0 with D(z0, r) ⊂ G. If 0 < δ0 < r/2, then the closed disk K = ¯ D(z0, 2δ0) ⊂ D(z0, r) ⊂ G is a compact subset of G. So

M = sup kfkK < ∞. f∈F

By the Cauchy integral formula, for z ∈ D(z0, δ0) we have Z 0 1 f(ζ) f (z) = 2 dζ 2πi |ζ−z0|=2δ0 (ζ − z)

and so we can estimate

0 1 M 2M |f (z)| ≤ (2π(2δ0)) 2 = = M1 (say) 2π δ0 δ0

(using |ζ − z| = |(ζ − z0) − (z − z0)| ≥ |ζ − z0| − |z − z0| = 2δ0 − |z − z0| ≥ 2δ0 − δ0 = δ0). Thus for |z − z0| < δ0 we have

Z z 0 |f(z) − f(z0)| = f (ζ) dζ ≤ |z − z0|M1 z0

Thus if we take δ = min(δ0, ε/M1) (which is independent of f ∈ F), we have

|z − z0| < δ, f ∈ F ⇒ |f(z) − f(z0)| < ε

Thus, by the Ascoli theorem 5.20, we know that F is relatively compact if we view it as a family of continuous functions (F ⊂ H(G) ⊂ C(G)). In other words its closure in C(G) is compact. But, since H(G) is closed in C(G), the closure of F in C(G) is actually contained in H(G) and so is the same as its closure in H(G). Hence F is relatively compact in H(G). 12 414 2003–04 R. Timoney

Theorem 5.21 Suppose G ⊂ C is open and F ⊂ H(G) a bounded subset. If a sequence ∞ (fn)n=1 in F is pointwise convergent to a function f : G → C, then the sequence is automatically uniformly convergent on compact subsets (and the limit f ∈ H(G)). (In other words, for bounded sequences of analytic functions, pointwise convergence is after all the same as uniform convergence on bounded sets. It follows that if a sequence of analytic functions fn : G → C converges pointwise to a limit f : G → C and if f fails to be differentiable or continuous at a point z0 ∈ G, then the sequence cannot be bounded (uniformly on compact subsets of G). In fact we can replace G be a small disc D(z0, δ) ⊂ G and say that the sequence could not be bounded on any such disc. So

sup(sup{|fn(z)| : |z − z0| < δ}) = ∞ n

for each δ > 0 (small enough that D(z0, δ) ⊂ G). This makes it quite hard to ﬁnd such sequences fn. As stated before, we can use Runges theorem to show there are such sequences, once we ﬁnd out about Runges theorem.)

∞ Proof. By Montels Theorem 5.17, the sequence has a convergent subsequence (fnj )j=1. So the subsequence converges in (H(G), ρ) (or equivalently, uniformly on compact subsets of G) to some limit g ∈ H(G). As singleton subsets K = {z} ⊂ G are compact it follows that the subsequence converges pointwise to g. That is

g(z) = lim fn (z) = lim fn(z) = f(z)(each z ∈ G) j→∞ j n→∞

and so f = g ∈ H(G). ∞ To show that the sequence (fn)n=1 converges, uniformly on compact subsets of G, to f we need to know that for each K ⊂ G compact

lim kfn − fkK = 0. n→∞

If that fails to be so, it fails for some K ⊂ G compact. Failing for K means we can ﬁnd ∞ ε > 0 so that kfn − fkK ≥ ε for inﬁnitely many n. That means a subsequence (fnk )k=1 where

kfnk − fkK ≥ ε for all k. Using Montels theorem, we can ﬁnd a subsequence (f )∞ of the subsequence which con- nkj j=1 ∞ verges in (H(G), ρ). We can rename this sub-subsequence as (fmj )j=1. Repeating the argument at the beginning of the proof we can see that this subsequence must converge to f and so

lim kfm − fkK = 0 j→∞ j

contradicting kf − fk = kf − fk ≥ ε > 0 (∀j). mj K nkj K

Thus we must have fn → f in (H(G), ρ). Chapter 5 — the metric space H(G) 13

∞ Theorem 5.22 (Osgoods theorem) Let G ⊂ C be open and (fn)n=1 a sequence of analytic functions fn ∈ H(G) that converges pointwise to a limit function f : G → C. Then there is a dense open subset G0 ⊂ G so that the restriction of f to G0 is analytic and ∞ the restriction of the sequence (fn)n=1 converges uniformly on compact subsets of G0 to f |G0 .

The proof of this relies on the previous theorem (5.21) and the Baire category theorem. The idea is to take

∞ \ Sm = {z ∈ G : sup |fn(z)| ≤ m} = {z ∈ G : |fn(z)| ≤ m} n n=1

◦ S∞ Gm = (Sm) the interior of Sm and G0 = m=1 Gm. Clearly G0 is open. The Baire Category theorem can be used to show that Gm is not empty for m big enough, and in fact that G0 is dense in G. This is the main part of the proof. S∞ Once these facts are established, we can see that every compact subset K ⊂ G0 = m=1 Gm Sm0 and so the Gm form an open cover of K. Hence K is contained in a ﬁnite union m=1 Gm = Gm0 and so the sequence fn is uniformly bounded on K (by m0). From Theorem 5.21, the rest of the result follows. We will leave the rest of the details to an appendix.

A Proof of Arzela-Ascoli Theorem

First, the ‘easy’ direction of the theorem.

Proposition A.1 Let G ⊂ C be open and F ⊂ C(G) a family of continuous functions on G. Suppose that F is a relatively compact in (C(G), ρ). Then it satisﬁes both of the following conditions:

(i) F is pointwise bounded on G

(ii) F is equicontinuous at each point of G.

Proof. We know from Lemma 5.16 (i) that F is bounded (uniformly on compact subsets) and so it is pointwise bounded (because singleton subsets K = {z0} ⊂ G are compact). To show it must be equicontinuous at each point of G, ﬁx z0 ∈ G and suppose F fails to be equicontinuous at z0. Then there is some ε > 0 for which no δ > 0 works. Fix such an ε > 0 and r > 0 with D(z0, r) ⊂ G. Consider the fact that δ = r/n does not work. This means there exists zn with |zn − z0| < r/n (hence zn ∈ G) and fn ∈ F so that |fn(zn) − fn(z0)| ≥ ε. ∞ As F is relatively compact in C(G), there is a subsequence (fnj )j=1 which converges in (C(G), ρ) (equivalently on compact subsets of G) to some f ∈ C(G). As f is continuous at z0 we know there exists δ0 > 0 so that ε |z − z | < δ ⇒ |f(z) − f(z )| < 0 0 0 3 14 414 2003–04 R. Timoney

¯ We can assume that δ0 < r so that the closed disc D(z0, δ0) ⊂ G is a compact subset of G. Thus ¯ fnj → f uniformly on K = D(z0, δ0) and if j is large enough

sup |fnj (z) − f(z)| < ε/3 z∈D¯(z0,δ0)

This contradiction shows that F must be equicontinuous at z0.

Deﬁnition A.2 Let G ⊂ C be open and F ⊂ C(G) a family of continuous function on G. For E ⊂ G a subset, we say that F on E if it satisﬁes:

Given ε > 0 there exists δ > 0 so that

f ∈ F, z0 ∈ E, z ∈ G, |z − z0| < δ ⇒ |f(z) − f(z0)| < ε

(One might describe this better as ‘equicontinuous uniformly at all points of E’.)

Lemma A.3 Let G ⊂ C be open and F ⊂ C(G) a family of continuous function on G. If F is equicontinuous at each point of G and if K ⊂ G is compact, then F is equicontinuous on K. (So equicontinuity at each point implies equicontinuity on compact subsets.)

Proof. Fix K. For each ζ ∈ K, by equicontinuity at ζ we know there exists δζ > 0 so that ε f ∈ F, |z − ζ| < δ ⇒ |f(z) − f(ζ)| < ζ 2

n δζ o Now D ζ, 2 : ζ ∈ K is an open cover of K and so it has a ﬁnite subcover

δ δ δ K ⊂ D z , z1 ∪ D z , z2 ∪ · · · ∪ D z , zn 1 2 2 2 n 2

Now put δ = min(δz /2, δz /2, . . . , δz /2). 1 2 n δzi Take now z0 ∈ K and z with |z − z0| < δ. Then z0 ∈ D zi, 2 for some i (1 ≤ i ≤ n) and then δ |z − z | ≤ |z − z | + |z − z | ≤ δ + zi ≤ δ ⇒ z ∈ D(z , δ ) ⊂ G. i 0 0 i 2 zi i zi

As the same δ works for all z0 ∈ K, we have the result. Chapter 5 — the metric space H(G) 15

Proof. (of Theorem 5.20) One direction is already done in Proposition A.1 and so what remains is to show that if F ⊂ C(G) is both pointwise bounded and equicontinuous at each point of G, then F is relatively compact. ∞ Take F satisfying the two conditions and (fn)n=1 a sequence in F. To ﬁnd the appropriate subsequence of functions we use what is known as a diagonal argument. First we pick a countable dense subset of points of G. For example S = {z ∈ G :

sup |f(s1)| < ∞ ⇒ {f(s1): f ∈ F} ⊂ C is relatively compact f∈F

Thus {f0,n(s1): n = 1, 2,...} is a relatively compact (or bounded) subset of C. Hence the ∞ ∞ sequence (f0,n(s1))n=1 has a subsequence (f1,j(s1))j=1 that converges to some limit in C. So

∃ lim f1,j(s1) ∈ j→∞ C

∞ Next {f1,n(s2): n = 1, 2,...} is relatively compact in C and so the sequence (f1,n(s2))n=1 ∞ has a subsequence (f2,j(s2))j=1 that converges to some limit in C. So

∃ lim f2,j(s2) ∈ j→∞ C

∞ ∞ Continuing in this way, once we have (fn,j)j=1, we choose a subsequence (fn+1,k)k=1 so that

∃ lim fn+1,k(sn+1) ∈ C k→∞

∞ The diagonal argument is now to choose gn = fn,n. Then (gj)j=n is a subsequence of ∞ (fn,k)k=1 and so lim gj(sn) = lim fn,k(sn) exists in C j→∞ k→∞

We claim that f(z) = limj→∞ gj(z) exists for all z ∈ G, that f ∈ C(G) and that gj → f in (C(G), ρ) as j → ∞. ∞ Fix z ∈ G and we claim that (gj(z))j=1 is a Cauchy sequence in C, hence convergent. Let ε > 0. Then, using the assumption that F is equicontinuous at z, there exists δ > 0 so that f ∈ F, |ζ − z| < δ ⇒ |f(ζ) − f(z)| < ε/3. Note that this applies with f = gj since gj = fj,j is one of the fn ∈ F. By density, there exists sn ∈ S with |sn − z| < δ. Then, since

∃ lim gj(sn) ∈ j→∞ C 16 414 2003–04 R. Timoney

∞ the sequence (gj(sn))j=1 is Cauchy in C and so there is j0 so that

j, k ≥ j0 ⇒ |gj(sn) − gk(sn)| < ε/3

Thus

j, k ≥ j0 ⇒

∞ We now have (gj(z))j=1 a Cauchy sequence in C for each z ∈ G and so we can deﬁne f : G → C by

f(z) = lim gj(z) j→∞

To show gj → f uniformly on compact subsets of G, ﬁx K ⊂ G compact and ε > 0. ◦ Then there is a compact subset K1 ⊂ G so that K ⊂ (K1) . (For example, using an exhaustive sequence of compact subsets of G we can show this.)

Use equicontinuity of F on K1 to ﬁnd δ > 0 so that

ε z ∈ G, z ∈ K , f ∈ F, |z − z | < δ ⇒ |f(z) − f(z )| < 0 1 0 0 4

◦ Now, for each z ∈ K, since S is dense in G, there is some s ∈ S ∩ (Bd(z, δ) ∩ (K1) ). ◦ Turning this around z ∈ Bd(s, δ) for some s ∈ S ∩ (K1) . We can say then that

◦ {Bd(s, δ): s ∈ S ∩ (K1) }

is an open cover of K. Thus there is a ﬁnite subcover

n0 [ 0 K ⊂ Bd(sn, δ) n=1

0 0 0 ◦ for some s1, s2, . . . , sn0 ∈ S ∩ (K1) . 0 0 0 Since f(sn) = limj→∞ gj(sn) for each sn, there is a j0 so that

0 0 ε j > j0 ⇒ max |gj(sn) − f(sn)| < 1≤n≤n0 4 Chapter 5 — the metric space H(G) 17

j > j0 ⇒ sup |gj(z) − f(z)| = dK (gj, f) < ε z∈K

This means gj → f uniformly on K, for each K. Hence f ∈ C(G) and limj→∞ gj = f in (C(G), ρ).

B Baire Category Theorem and Proof of Osgoods Theorem

The Baire category theorem is usually stated for complete metric spaces. In our case, we can get by with using it only for compact metric spaces (which are automatically complete).

Deﬁnition B.1 A metric space (X, d) is a set X together with a function (which is commonly called a distance function) d: X × X → R satisfying

(i) d(x1, x2) ≥ 0 (∀x1, x2 ∈ X)

(ii) d(x1, x2) = d(x2, x1)(∀x1, x2 ∈ X)

(iii) (triangle inequality) d(x1, x3) ≤ d(x1, x2) + d(x2, x3)(∀x1, x2, x3 ∈ X)

(iv) x1, x2 ∈ X, d(x1, x2) = 0 ⇒ x1 = x2

∞ Deﬁnition B.2 A sequence (xn)n=1 is a metric space (X, d) (so xn ∈ X∀n) is called convergent to a limit x ∈ X (and we write limn→∞ xn = x) if

lim d(xn, x) = 0 n→∞

∞ Deﬁnition B.3 A sequence (xn)n=1 is a metric space (X, d) is called a Cauchy sequence if 18 414 2003–04 R. Timoney

gien any ε > 0 there exists N ∈ N so that

n, m > N ⇒ d(xn, xm) < ε

(This means all the terms of the sequence are close to each other, except for some at the start.)

∞ Lemma B.4 A convergent sequence (xn)n=1 in a metric space (X, d) is always a Cauchy sequence.

Proof. Exercise

Deﬁnition B.5 A metric space (X, d) is called complete if every Cauchy sequence in X is convergent (to some limit in X).

Example B.6 R with the usual absolute value distance is complete. Q is not.

Deﬁnition B.7 If (X, d) is a metric space x0 ∈ X and r > 0 then the (open) ball of radius r about x0 is

Bd(x0, r) = {x ∈ X : d(x, x0) < r} If S ⊂ X is a subset and s ∈ S, then s is called an interior point of S if there is some ball Dd(s, r) ⊂ S of positive radius r > 0 about s contained in S. The interior S◦ of a subset S ⊂ X is the set of all its interior points. A subset U ⊂ S is called open if U ◦ = U (all its points are interior points). A subset E ⊂ S is called closed if its complement S \ E is open.

Proposition B.8 Let (X, d) be a metric space. S (i) arbitrary unions i∈I Ui of open subsets Ui ⊂ X (i ∈ I = any index set) are open.

(ii) the interior S◦ of any subset S ⊂ X is open

(iii) the interior S◦ of any subset S ⊂ X coincides with [ {U : U ⊂ S,U open in X}

(iv) the interior S◦ of any subset S ⊂ X is the largest open subset of X that is contained in S T (v) arbitrary intersections i∈I Ei of closed subsets Ei ⊂ X (i ∈ I = any index set) are closed.

(vi) the empty subset ∅ and X are both open and closed. Chapter 5 — the metric space H(G) 19

(vii) For each subset S ⊂ X, there is a smallest closed subset S¯ ⊂ X containing S. It is called the closure of S and can be given as \ S¯ = {E : E ⊂ S,E closed in X}

or as the complement of the interior of the complement:

S¯ = X \ (X \ S)◦

Proof. Exercise.

Definition B.9 Let (X, d) be a metric space and K ⊂ X a subset. An open cover of K is any S family U = {Ui : i ∈ I} of open subsets Ui ⊂ X such that K ⊂ i∈I Ui (that is, K is contained in their union). A subcover of a cover U of K is a smaller family V ⊂ V so that K ⊂ S V = S{U : U ∈ V}. A finite subcover is a subcover V that has only finitely many sets in it. A subset K ⊂ X is called compact if every open cover of K has a finite subcover.

Proposition B.10 For subsets of a metric space (X, d) closure, closedness and compactness can be characterised via limits of sequences:

¯ ∞ (i) if S ⊂ X, then x ∈ S ⇐⇒ there is a sequence (sn)n=1 of points sn ∈ S with limn→∞ sn = x.

∞ (ii) if S ⊂ X, then S is closed if and only if every sequence (sn)n=1 of points sn ∈ S that converges in (X, d) has its limit in S.

∞ (iii) if K ⊂ S, then K is compact if and only if every sequence (xn) of points xn ∈ K has a ∞ subsequence (xnj )j=1 which converges to a limit in K.

Proof. Omitted.

Proposition B.11 Let (X, d) be a metric space. Then (X, d) is complete if and only if each Cauchy sequence in X has a convergent subsequence.

Proof. Exercise. It is not hard to show that for a Cauchy sequence with a convergent subsequence, the whole sequence must converge (to the same limit as the subsequence).

Corollary B.12 Compact metric spaces are complete.

Definition B.13 A subset S ⊂ X of a metric space (X, d) is called nowhere dense if the interior of its closure is empty, (S¯)◦ = ∅. A subset E ⊂ X is called of first category if it is a countable union of nowhere dense subsets, or equivalently, the union E = S∞ S of a sequence of nowhere dense sets ((S¯ )◦ = ∅∀n). n=1 n n A subset Y ⊂ X is called of second category if it fails to be of first category. 20 414 2003–04 R. Timoney

Example B.14 (a) If a singleton subset S = {s} ⊂ X fails to be nowhere dense, then the interior of its closure is not empty. The closure S¯ = S = {s} and if that has any interior it means it contains a ball of some positive radius r > 0. So

Bd(s, r) = {x ∈ X : d(x, s) < r} = {s}

and this means that s is an isolated point of X (no points closer to it than r). An example where this is possible would be X = Z with the usual distance (so B(n, 1) = {n}) and S any singleton subset. Another example is X = D(2, 1) ∪ {0} ⊂ C (with the distance on X being the same as the usual distance between points in C) and S = {0}. (b) In many cases, there are no isolated points in X, and then a one point set is nowhere dense. So is a countable subset is then of first category (S = {s1, s2,...} where the elements can be listed as a finite or infinite sequence). For example S = Z is of first category as a subset of R, though it of second category as a subset of itself. S = Q is of first category both in R and in itself (because it is countable and points are not isolated). The idea is that first countable means ‘small’ in some sense, while second category is ‘not small’ in the same sense. While it is often not hard to see that a set is of first category, it is harder to see that it fails to be of first category. One has to consider all possible ways of writing the set as a union of a sequence of subsets.

Theorem B.15 (Baire Category) Let (X, d) be a complete metric space. Then the whole space S = X is of second category in itself.

S∞ Proof. If not, then X is of ﬁrst category and that means X = n=1 Sn where each Sn is a ¯ ◦ nowhere dense subset Sn ⊂ X (with (Sn) = ∅). The argument may be simpliﬁed if we assume each Sn is closed (which we can do if we ¯ replace Sn by its closure) but we will just continue with Sn. ¯ Since Sn has empty interior, its complement is a dense open set. That is

¯ ¯ ◦ X \ Sn = X \ (Sn) = X ¯ Thus if we take any ball Bd(x, r) in X, there is a point y ∈ (X \ Sn) ∩ Bd(x, r) and then because ¯ ¯ X \ Sn is open there is a (smaller) δ > 0 with Bd(y, δ) ⊂ (X \ Sn) ∩ Bd(x, r). Start with x0 ∈ X any point and r0 = 1. Then, by the above reasoning there is a ball ¯ Bd(x1, r1) ⊂ (X \ S1) ∩ Bd(x0, r0). In fact, making r1 smaller if necessary, we can ensure that the closed ball ¯ ¯ Bd(x1, r1) = {x ∈ X : d(x, x1) ≤ r1} ⊂ (X \ S1) ∩ Bd(x0, r0) and r1 < 1. We can then ﬁnd x2 and r2 ≤ r1/2 < 1/2 so that ¯ ¯ Bd(x2, r2) ⊂ (X \ S2) ∩ Bd(x1, r1) Chapter 5 — the metric space H(G) 21

and we can continue this process to select x1, x2,... and r1, r2,... with 1 0 < r ≤ r /2 < , B¯ (x , r ) ⊂ (X \ S¯ ) ∩ B (x , r )(n = 1, 2,...) n n−1 2n−1 d n n n d n−1 n−1

∞ We claim the sequence (xn)n=1 is a Cauchy sequence in X. This is because m ≥ n ⇒ xm ∈ n Bd(xn, rn) ⇒ d(xm, xn) < rn < 1/2 . So, if n, m are both large

1 1 d(x , x ) < min , m n 2n 2m is small. ¯ By completeness, x∞ = limn→∞ xn exists in X. Since the closed ball Bd(xn, rn) is a ¯ closed set in X and contains all xm for m ≥ n, it follows that x ∈ Bd(xn, rn) for each n. ¯ ¯ ¯ But Bd(xn, rn) ⊂ X \ Sn and so x∈ / Sn. This is true for all n and so we have the contradiction

∞ [ ¯ x∈ / Sn = X n=1

Thus X cannot be a union of a sequence of nowhere dense subsets.

Corollary B.16 Let (X, d) be a compact metric space. Then the whole space S = X is of second category in itself.

Proof. Compact metric spaces are complete. So this follows from the theorem. Proof. (of Osgoods Theorem 5.22) As outlined previously, take

∞ \ Sm = {z ∈ G : sup |fn(z)| ≤ m} = {z ∈ G : |fn(z)| ≤ m} n n=1

◦ S∞ Gm = (Sm) the interior of Sm and G0 = m=1 Gm. Clearly G0 is open. ∞ Note that z ∈ G ⇒ limn→∞ fn(z) = f(z) ∈ C and so the sequence (fn(z))n=1 must be S∞ bounded. If m is big enough z ∈ Sm and so we have m=1 Sm = G. To show that G0 is dense in G, ﬁx z ∈ G and a disc D(z, r) about z of small enough radius that its closure D¯(z, r) ⊂ G. Then D¯(z, r) is a compact metric space and

∞ ∞ ¯ [ ¯ [ ¯ D(z, r) ⊂ G = Sm ⇒ D(z, r) = Sm ∩ D(z, r) m=1 m=1

Applying the Baire category theorem to the compact metric space D¯(z, r) we ﬁnd there is m ¯ ¯ ¯ so that Sm ∩ D(z, r) is not nowhere dense in D(z, r). As Sm ∩ D(z, r) is closed, that means it 22 414 2003–04 R. Timoney

has nonempty interior as a subset of D¯(z, r). There is therefore a ball center w and radius δ > 0 ¯ ¯ in the metric space D(z, r) that is contained in Sm ∩ D(z, r). This ball is in fact the intersection

D(w, δ) ∩ D¯(z, r)

of an open and a closed disc. Thus D(w, δ) ∩ D(z, r) is not empty, open and is contained in ¯ ◦ Sm ∩ D(z, r). So D(w, δ) ∩ D(z, r) ⊂ (Sm) = Gm ⊂ G0 and we have

D(z, r) ∩ G0 6= ∅

This shows that G0 is dense in G. The rest of the proof of Osgoods theorem was given earlier.

Richard M. Timoney February 22, 2004