• Complex Analysis: #10 Analytic Continuation

We now know that an analytic function f : G → ℂ can be represented as a power series centered at any given point z₀ ∈ G. The function f is equal to the function defined by the power series in the largest possible open disc around z₀ which is contained in G. But of course G is not, in general itself an open disc. Therefore there might be parts of G where f is not given by this power series centered on z₀. Or, (thinking about the logarithm function) we might have the situation that G could be expanded to a larger region G_*, with G ⊂ G_* where the function could be defined. But then perhaps there might be different ways of “continuing” this definition of f from G to G_*.

Therefore let us consider a chain of open discs (B₁, . . ., B_n) say, with

B_j = {z ∈ ℂ : |z − p_j| < r_j} 

for a chain of points p_j which are the centers of the discs, and numbers r_j > 0, which are the radii in each case. We assume that it is a connected chain in the sense that B_j ∩ B_j+1 ≠ ∅, in each case. For each B_j let us assume that an analytic function f_j : B_j → ℂ exists, such that in the region of overlap, we have f_j(z) = f_j+1 (z) for z ∈ B_j ∩ B_j+1.

Definition 8
The functions f_j here are called function elements, and if we have f_j(z) = f_j+1 (z) for z ∈ B_j ∩ B_j+1 for all j = 1, . . . , n − 1, then we have an analytic continuation of the function elements through the chain of open discs. Or, if we consider the ordering of the discs, we can say that the final function f_n is obtained by analytic continuation of the initial function f₁ through the chain of discs.

Theorem 17 shows that if, say f₁ is given in B₁, and there exists a chain of open discs allowing some analytic continuation, then this continuation is unique.


Theorem 21
Let (B₁, . . ., B_n) be a chain of open discs with B_j ∩ B_j+1 ≠ ∅, for j = 1, . . . , n − 1. Let f₁ : B₁ → ℂ be some given analytic function. Then there exists an analytic continuation of f₁ throughout the chain [That is to say, there exists a set of function elements forming an analytic continuation, such that the first element in the chain of function elements is f₁.] if and only if there also exists an analytic continuation of f₁' : B₁ → ℂ (the derivative of f₁) throughout the chain.

Proof
“⇒” is trivial. (Just take f_j', the derivative of f_j, for each j.)

As far as “⇐” is concerned, we are assuming that there is an analytic continuation of the function . To avoid confusion, let us call this function g_j : B₁ → ℂ. i.e. g₁(z) = f₁'(z) for all z ∈ B_j. The assumption is that for each j there is an analytic function g_j : B_j → ℂ, providing an analytic continuation, starting with g₁. We now use induction on the number n. For n = 1 there is nothing to prove. So let n > 1, and assume that we have an analytic continuation of g along the chain of open discs from B₁ to B_n−1, giving g_j : B_j → ℂ such that g_j : f_j' for each j < n. According to theorem 3, g_n has an antiderivative, G_n : B_n → ℂ, with G_n' : g_n. Now in the region B_n−1 ∩ B_nwe have g_n−1 = g_n. That is, f_n−1' = G_n', or f_n−1' − G_n' = 0. Thus f_n−1 − G_n = k say, where k is a constant number. But then the function G_n + k is also an antiderivative to g_n, and we can take f_n = G_n + k.

One way to think about these chains of discs is to imagine that they are associated with a path, namely a path starting at p₁ then following a straight line to p₂, then a straight line to p₃, and so forth, finally ending at p_n. So each straight segment, say from p_j to p_j+1 is contained in the union of the two discs B_j ∪ B_j+1 .

Let’s generalize this idea in the following way. Let γ : [t₀, t₁] → ℂ be a continuous path. (It doesn’t have to be differentiable here.) Let t₀ = τ₀ < τ₁ < · · · < τ_n = t₁ be a partition of the interval [t₀, t₁]. Assume that we have a corresponding set of open discs B_j = {z ∈ ℂ : |z − γ(τ_j)| < r_j}(where r_j > 0) such that γ(t) ∈ B_j ∪ B_j+1 , for t ∈ [τ_j, τ_j+1], for all relevant j. Then we will say that we have a chain of discs along the path γ. Furthermore, if we have a sequence of analytic function elements giving an analytic continuation along this chain, then we will say that the function is analytically continued along the path. The next theorem shows that this is a property of the path, independent of the choice of discs along the path.


Theorem 22
Let γ : [t₀, t₁] → ℂ be continuous and let B = {z ∈ ℂ : |z − γ(t₀) < r} and B_* = {z ∈ ℂ : |z − γ(t₁)| < r_*} (with both r, r_* > 0) be open discs centered on γ(t₀) and γ(t₁), respectively. Let t₀ = τ₀ < τ₁ < · · · < τ_n = t₁ and t₀ = ρ₀ < ρ₁ < · · · < ρ_m = t₁ be two different partitions of the interval [t₀, t₁], giving two different chains of discs along the path, satisfying the conditions listed above, where the first and last discs are B and B_*. Assume that f : B → ℂ is an analytic function at the first disc and it has an analytic continuation with respect to the first chain of discs, finally giving the analytic function g : B_* → ℂ. Then f also has an analytic continuation with respect to the second chain of discs, and it also gives the same function g : B_* → ℂ.

Proof
If the whole path γ is completely contained within the disc B then, using theorem 9 we see that the analytic continuation is simply given by the power series representing the function.

Therefore let M ⊂ [t₀, t₁] be defined to be the set of t_* ∈ [t₀, t₁] such that the theorem is true for the interval [t₀, t_*] (with respect to this path γ). But if t ∈ M, then since γ is continuous, and since the analytic function which has been continued out to the point γ(t_*) has a power series representation in a neighborhood of γ(t_*), we must have some ∈ > 0 such that {t ∈ [t₀, t₁] ; |t − t_*| < ∈ } ⊂ M. Therefore M is an open subset of [t₀, t₁]. If [t₀, t₁] \ M ≠ ∅, then the same argument shows that M is closed in [t₀, t₁]. Yet the interval [t₀, t₁] is connected. Therefore M = [t₀, t₁].

All of these thoughts allow us to perform path integrals along continuous, but not necessarily differentiable paths. To see this, take γ to be some continuous path, allowing an analytic continuation from an open disc centered on the starting point of the path. The discs in the finite chain of open discs describing the analytic continuation have centers at various points along the path γ. But now take γ to be the piecewise linear path connecting those centers. This gives a path integral, namely ∫_γ f(z)dz, where f consists of the function elements along the path. We can now simply define ∫_γ f(z)dz to be this integral along γ. Theorem 22 then shows that the path integral for γ is well defined.