14.1: Analytic Continuation
- Page ID
- 6557
If we have an function which is analytic on a region \(A\), we can sometimes extend the function to be analytic on a bigger region. This is called analytic continuation.
Consider the function
\[F(z) = \int_{0}^{\infty} e^{3t} e^{-zt} \ dt. \nonumber \]
We recognize this as the Laplace transform of \(f(t) = e^{3t}\) (though we switched the variable from \(s\) to \(z\)). The integral converges absolutely and \(F\) is analytic in the region \(A = \{\text{Re} (z) > 3\}\).
Can we extend \(F(z)\) to be analytic on a bigger region \(B\)? That is, can we find a region \(B\) a function \(\hat{F} (z)\) such that
- \(B\) contains \(A\)
- \(\hat{F} (z)\) is analytic on \(B\)
- \(\hat{F} (z)\) agrees with \(F\) on \(A\), i.e. \(\hat{F} (z) = F(z)\) for \(z \in A\).
Solution
Yes! We know that \(F(z) = \dfrac{1}{z - 3}\) -valid for any \(z\) in \(A\). So we can define \(\hat{F} (z) = \dfrac{1}{z - 3}\) for any \(z\) in \(B = C - \{3\}\).
We say that we have analytically continued \(F\) on \(A\) to \(\hat{F}\) on \(B\).
Usually we don’t rename the function. We would just say \(F(z)\) defined by Equation 14.2.1 can be continued to \(F(z) = \dfrac{1}{z - 3}\) on \(B\).
Suppose \(f(z)\) is analytic on a region \(A\). Suppose also that \(A\) is contained in a region \(B\). We say that \(f\) can be analytically continued from \(A\) to \(B\) if there is a function \(\hat{f} (z)\) such that
- \(\hat{f} (z)\) is analytic on \(B\).
- \(\hat{f} (z) = f(z)\) for all \(z\) in \(A\).
As noted above, we usually just use the same symbol \(f\) for the function on \(A\) and its continuation to \(B\).
The region \(A = \text{Re} (z) > 0\) is contained in \(B = \text{Re} (z) > -1\).
We used analytic continuation implicitly in, for example, the Laplace inversion formula involving residues of \(F(s) = \mathcal{L} (f;s)\). Recall that we wrote that for \(f(t) = e^{3t}\), \(F(s) = \dfrac{1}{s - 3}\) and
\[f(t) = \sum \text{ residues of } F. \nonumber \]
As an integral, \(F(s)\) was defined for \(\text{Re} (s) > 3\), but the residue formula relies on its analytic continuation to \(C - \{3\}\).
Analytic Continuation is Unique
Suppose \(f, g\) are analytic on a connected region \(A\). If \(f = g\) on an open subset of \(A\) then \(f = g\) on all of \(A\).
- Proof
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Let \(h = f - g\). By hypothesis \(h (z) = 0\) on an open set in \(A\). Clearly this means that the zeros of \(h\) are not isolated. Back in Topic 7 we showed that for analytic \(h\) on a connected region \(A\) either the zeros are isolated or else \(h\) is identically zero on \(A\). Thus, \(h\) is identically 0, which implies \(f = g\) on \(A\).
There is at most one way to analytically continue a function from a region \(A\) to a connected region \(B\).
- Proof
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Two analytic continuations would agree on \(A\) and therefore must be the same.
Extension. Since the proof of the theorem uses the fact that zeros are isolated, we actually have the stronger statement: if \(f\) and \(g\) agree on a nondiscrete subset of \(A\) then they are equal. In particular, if \(f\) and \(g\) are two analytic functions on \(A\) and they agree on a line or ray in \(A\) then they are equal.
Here is an example that shows why we need \(A\) to be connected in Theorem \(\PageIndex{1}\).
Suppose \(A\) is the plane minus the real axis. Define two functions on \(A\) as follows.
\[f(z) = \begin{cases} 1 \text{ for } z \text{ in the upper half-plane} \\ 0 \text{ for } z \text{ in the lower half-plane} \end{cases} \nonumber \]
\[g(z) = \begin{cases} 1 \text{ for } z \text{ in the upper half-plane} \\ 1 \text{ for } z \text{ in the lower half-plane} \end{cases} \nonumber \]
Both \(f\) and \(g\) are analytic on \(A\) and agree on an open set (the upper half-plane), but they are not the same function.
Here is an example that shows a little care must be taken in applying the corollary.
Suppose we define \(f\) and \(g\) as follows
\[f(z) = \log (z) \text{ with } 0 < \theta < 2\pi \nonumber \]
\[g(z) = \log (z) \text{ with } -\pi < \theta < \pi \nonumber \]
Clearly \(f\) and \(g\) agree on the first quadrant. But we can’t use the theorem to conclude that \(f = g\) everywhere. The problem is that the regions where they are defined are different. \(f\) is defined on \(C\) minus the positive real axis, and \(g\) is defined on \(C\) minus the negative real axis. The region where they are both defined is \(C\) minus the real axis, which is not connected.
Because they are both defined on the upper half-plane, we can conclude that they are the same there. (It’s easy to see this is true.) But (in this case) being equal in the first quadrant doesn’t imply they are the same in the lower half-plane.