Thursday, May 15, 2014

Complex Analysis: #22 Montel`s Theorem

  • Complex Analysis: #22 Montel`s Theorem

Thinking about Weierstrass’ convergence theorem (theorem 28), let us again consider sequences of functions.

Definition 14
Let G ⊂ ℂ be a region, and for each n ∈ ℕ let fn : G → ℂ be analytic. This gives us a sequence of functions on G. We will say the sequence is locally bounded if for all z ∈ G, there exists an (open) neighborhood z ∈ U ⊂ G and an M > 0, such that |fn(w)| ≤ M, for all w ∈ U and all n ∈ ℕ.


Theorem 43
Let fn : G → ℂ be a locally bounded sequence of analytic functions. Assume there exists a dense subset T ⊂ G, such that (fn(z))n∈ℕ is a convergent sequence in ℂ for all z ∈ T. Then there exists an analytic function f : G → ℂ such that fn → f uniformly on every compact subset of G.

Proof
Begin by observing that we only need prove that for every z0 ∈ G, there exists an r > 0 such that the sequence of functions fn is uniformly convergent on B(z0 , r) (the open disc of radius r centered on z0). This follows, since given any compact subset K ⊂ G, it can be covered with finitely many such discs.

So given some z0 ∈ G, we would like to show that there exists an r > 0 such that for all ∈ > 0 there exists an N0 ∈ ℕ such that |fn(z) − fm(z)| < for all n, m ≥ N0 and |z − z0| < r. Given this, then for each such z we would have (fn(z))n∈ℕ being a Cauchy sequence, converging to a point f(z) in ℂ. Thus the sequence would be uniformly convergent in B(z0, r) to the function f.

In order to find such an r and N0, let us begin by using the property that the sequence of functions is locally bounded. Thus there is some M > 0 and an r > 0 such that |fn(z)| ≤ M for all z ∈ D(z0, 2r) = {z ∈ G : |z − z0| ≤ 2r}. Since T is dense in G, and D(z0, r) is compact, we can find some finite number of points of T in B(z0, r), call them a1, . . . , ak ∈ T, with

Complex Analysis: #22 Montel`s Theorem equation pic 1

Note that the inequality in the third line follows because the radius of the circle is 2r, and of course r is greater than both |ζ − z| and |ζ − a1|. Also the last inequality is due to the fact that we have assumed that |z − a1| < ∈ r/6M.


Theorem 44 (Montel)
Assume fn : G → ℂ (with n ∈ ℕ) is a locally bounded sequence of analytic functions. Then there exists a sub-sequence which is uniformly convergent on every compact subset of G.

Proof
Take some arbitrary sequence {a1, a2, . . . } which is dense in G. Since the sequence of points (fn(a1))n∈ℕ is bounded, there exists a convergent sub-sequence, giving a sub-sequence (f1n)n∈ℕ of the sequence of functions. Next look at the sequence of points (f1n(a1))n∈ℕ. Again, there is a convergent sub-sequence. And so forth. So for each m ∈ ℕ, we obtain a sequence of functions (fmn)n∈ℕ which is such that the sequence of points (fmn(a1))n∈ℕ converges, for all l ≤ m. Therefore the sequence of functions (fnn)n∈ℕ satisfies the conditions of theorem 43.

We can use this theorem to find a criterion for the convergence of a sequence of functions as follows.

 
Theorem 45
Again, let fn : G → ℂ be a locally bounded sequence of analytic functions. Assume there exists some z0 ∈ G such that the sequences (fn(k)(z0))n∈ℕ (that is, the sequences of k-th derivatives) converge, for all k. Then (fn)n∈ℕ is uniformly convergent on all compact subsets of G.

Proof
According to theorem 43, if (fn(z))n∈ℕ is convergent for all z ∈ G, then the sequence of functions is uniformly convergent on all compact subsets of G, and we are finished. So let’s assume that there exists some a ∈ G, such that the sequence of points (fn(a))n∈ℕ is not convergent. But at least it must be bounded, so there must be two different sub-sequences of the sequence of functions, being convergent at a to two different values, say one sub-sequence converges to the value va ∈ ℕ and the other converges to wa, where wa ≠ va. However, using Montel’s theorem, we have sub-sequences of these sub-sequences of functions, converging to two different analytic functions: f, g : G → ℂ, with f(a) = wa ≠ va = g(a). Looking at the point z0, we have

limn→∞  fn(k)(z0) = f(k)(z0) = g(k)(z0),

that is, (f − g)k(z0) = 0 for all k. Therefore, the function f − g is zero in a neighborhood of z0, but this implies that it is zero everywhere, including at the point a, which gives us a contradiction.

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