# Complex Analysis: #22 Montel`s Theorem

**Deﬁnition 14**

Let G ⊂ ℂ be a region, and for each n ∈ ℕ let f

_{n}: 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 |f

_{n}(w)| ≤ M, for all w ∈ U and all n ∈ ℕ.

**Theorem 43**Let f

_{n}: G → ℂ be a locally bounded sequence of analytic functions. Assume there exists a dense subset T ⊂ G, such that (f

_{n}(z))

_{n∈ℕ}is a convergent sequence in ℂ for all z ∈ T. Then there exists an analytic function f : G → ℂ such that f

_{n}→ f uniformly on every compact subset of G.

*Proof*

Begin by observing that we only need prove that for every z

_{0}∈ G, there exists an r > 0 such that the sequence of functions f

_{n}is uniformly convergent on B(z

_{0}, r) (the open disc of radius r centered on z

_{0}). This follows, since given any compact subset K ⊂ G, it can be covered with ﬁnitely many such discs.

So given some z

_{0}∈ G, we would like to show that there exists an r > 0 such that for all ∈ > 0 there exists an N

_{0}∈ ℕ such that |f

_{n}(z) − f

_{m}(z)| < for all n, m ≥ N

_{0}and |z − z

_{0}| < r. Given this, then for each such z we would have (f

_{n}(z))

_{n∈ℕ}being a Cauchy sequence, converging to a point f(z) in ℂ. Thus the sequence would be uniformly convergent in B(z

_{0}, r) to the function f.

In order to ﬁnd such an r and N

_{0}, 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 |f

_{n}(z)| ≤ M for all z ∈ D(z

_{0}, 2r) = {z ∈ G : |z − z

_{0}| ≤ 2r}. Since T is dense in G, and D(z

_{0}, r) is compact, we can ﬁnd some ﬁnite number of points of T in B(z

_{0}, r), call them a

_{1}, . . . , a

_{k}∈ T, with

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 − a

_{1}|. Also the last inequality is due to the fact that we have assumed that |z − a

_{1}| < ∈ r/6M.

**Theorem 44 (Montel)**Assume f

_{n}: 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 {a

_{1}, a

_{2}, . . . } which is dense in G. Since the sequence of points (f

_{n}(a

_{1}))

_{n∈ℕ}is bounded, there exists a convergent sub-sequence, giving a sub-sequence (f

_{1n})

_{n∈ℕ}of the sequence of functions. Next look at the sequence of points (f

_{1n}(a

_{1}))

_{n∈ℕ}. Again, there is a convergent sub-sequence. And so forth. So for each m ∈ ℕ, we obtain a sequence of functions (f

_{mn})

_{n∈ℕ}which is such that the sequence of points (f

_{mn}(a

_{1}))

_{n∈ℕ}converges, for all l ≤ m. Therefore the sequence of functions (f

_{nn})

_{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 f

_{n}: G → ℂ be a locally bounded sequence of analytic functions. Assume there exists some z

_{0}∈ G such that the sequences (f

_{n}

^{(k)}(z

_{0}))

_{n∈ℕ}(that is, the sequences of k-th derivatives) converge, for all k. Then (f

_{n})

_{n∈ℕ}is uniformly convergent on all compact subsets of G.

*Proof*

According to theorem 43, if (f

_{n}(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 (f

_{n}(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 v

_{a}∈ ℕ and the other converges to w

_{a}, where w

_{a}≠ v

_{a}. 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) = w

_{a}≠ v

_{a}= g(a). Looking at the point z

_{0}, we have

lim

_{n→∞}f_{n}^{(k)}(z_{0}) = f^{(k)}(z_{0}) = g^{(k)}(z_{0}),that is, (f − g)

^{k}(z

_{0}) = 0 for all k. Therefore, the function f − g is zero in a neighborhood of z

_{0}, but this implies that it is zero everywhere, including at the point a, which gives us a contradiction.

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