You say Holomorphic, I say Analytic.

November 9, 2010

In another post, I noted something a bit strange at first reading: some authors use the word holomorphic to describe a power series expansion and reserve analytic for complex differentiable while other authors swap those terms.  I then noted that “this doesn’t matter.”  Well, why not?  I mean, definitions are pretty important in mathematics!  My reasoning is: these are really the same thing.  If $f$ is holomorphic then it is also analytic, and vice versa.  I’ve been putting off doing this proof for way too long, so let’s just get it over with.  It’s not hard, it’s just an analysis proof, which means that it’s extremely easy to describe (“take a little ball and do something in it”) but extremely tedious to work out (“take an epsilon such that this epsilon is less than the sum of the minimum of the supremums of…”) but I’m going to try to have a complete proof and motivate every step.

After this, I’ll give a short proof that if a function is complex differentiable (holomorpic, to me.) once, then it is complex differentiable infinitely many times.  It’s not a direct corollary, but it’s a nice fact to know.

First, let’s formally say what it means for a function to be holomorphic and analytic.

A function $f$ is holomorphic at a point $z_{0}$ if it is complex differentiable in a neighborhood of $z_{0}$.  A function $f$ is analytic at a point $z_{0}$ if there is a power series expansion about some neighborhood of $z_{0}$; ie, that

$\displaystyle f = \sum_{i=1}^{\infty} a_{i}(z-z_{0})^{i}$

and a function is said to be entire if it is holomorphic or analytic everywhere on the complex plane.

Note that familiar functions like $f(z) = z^{2} + 2z + 1$ and $g(z) = e^{z}$ are analytic everywhere = entire.  It comes up quite a bit that $e^{z}$ is entire, so that’s a good one to keep in the back of your mind.

Now, let’s start proving the theorem.  It is non-trivial to say, but kind of a pain to prove that every complex power series is complex differentiable, but it’s easy to see this: just do it term by term.  In fact, the radius of convergence even stays the same, so nice things are happening all over the place.  Therefore, we can hand-wave and say that all analytic functions are holomorphic.  It now suffices to prove that holomorphic functions are analytic, which is (if you think about what this means) non-trivial to even think about how to prove this!

In fact, the proof that I will follow (Lang’s Complex Analysis.) uses the local Cauchy formula to prove this.  The other complex books have a similar proof, but use much more “analysis”-type stuff.  Therefore, let us state (but not prove) the local cauchy formula.

Theorem (Local Cauchy Formula): If $\bar{D}$ is a closed disk of positive radius and $f$ is holomorphic on the closed disk (meaning that $f$ is holomorphic on the open disk $D$), then let $\gamma$ be a simple closed curve which is the boundary of $\bar{D}$ (ie, let $\gamma$ be the circle on the outside of the disk); then for each $z_{0}\in D$ we have

$\displaystyle f(z_{0}) = \frac{1}{2\pi i}\int_{\gamma}\frac{f(z)}{z-z_{0}}dz.$

So, this is a pretty nice formula.  Applying this to things is the topic of another post, but it essentially says that if $f$ is holomorphic (= analytic) and $g$ is just $f$ divided by $z-z_{0}$, then we can integrate g around some simple closed curve, and the solution is given by this formula above.  You’d be surprised at how many times you need to divide out by such a value in complex analysis.  Either way, let’s go onwards.

Theorem (Holomorphic implies Analytic): If $f$ is holomorphic on an open set $U$ containing $z_{0}$, then $f$ is analytic on $U$.

Proof. Since we’re all so good at topology, we note that an open set will be the union of small open balls.  Thus, it suffices to prove this theorem for an open ball.  Thus, let’s say we’re given a small ball $D_{R}$ of radius $R$ such that $f$ is holomorphic on $D_{R}$.  Let $C_{R}$ be the circle bounding $D_{R}$.  Also note that we can easily prove that the values of $f$ must be bounded by values on $C_{R}$, but this is more of a formality that we don’t get any messy stuff going to infinity in our circle.  Let’s just assume this is true (if you are genuinely curious about it, it is done in almost every complex analysis book I know of).  By the Local Cauchy Formula, we note that

$\displaystyle f(z) = \frac{1}{2\pi i}\int_{C_{R}}\frac{f(\alpha)}{(\alpha - z_{0})}d\alpha$

where $\alpha$ is just some dummy variable.  So what are we doing here?  We’re integrating writing $f$ in a way that uses an integral; this is kind of strange at first, right?  The right side seems much more complex than the left side, but it will turn out to make nice things happen for us exactly because that denominator term is there.  Consider a small ball about $z_{0}$ and notice that

$\displaystyle \frac{1}{\alpha - z} = \frac{1}{\alpha - z_{0} - (z - z_{0})}$

by adding and subtracting $z_{0}$, and then by some clever factoring, we get:

$= \displaystyle \left(\frac{1}{\alpha - z_{0}}\right)\left(\frac{1}{1 - \frac{z - z{0}}{\alpha - z_{0}}}\right)$

and if you’ve worked with geometric series, you might recall seeing things like the factor on the right.  In fact, we have that

$\displaystyle \sum_{n=1}^{\infty}x^{n} = \frac{1}{1 - x}$

for $|x| < 1$, and this is exactly the power series expansion we’re going to do for that right-hand factor above.  Specifically, we note

$= \displaystyle \left(\frac{1}{\alpha - z_{0}}\right)\left(\frac{1}{1 - \frac{z - z{0}}{\alpha - z_{0}}}\right)$

$\displaystyle = \frac{1}{\alpha - z_{0}}\left(1 + \frac{z - z_{0}}{\alpha - z_{0}} + \left(\frac{z - z_{0}}{\alpha - z_{0}}\right)^{2} + \cdots \right)$

which is a geometric series which converges uniformly if the ball is of a sufficiently small radius (!).  So, let’s just recount what we’ve done here: we’ve taken $f$ and written it as an integral, but we had to divide by something inside of the integral.  But we made the thing we’re dividing by in the integral into a power series!  This is going to change the integral into something nicer.  Because convergence of this power series is uniform, we can write the power series above as a sum in the integral and integrate this whole thing term by term.  Specifically,

$\displaystyle f(z) = \sum_{n = 0}^{\infty}\frac{1}{2\pi i}\int_{C_{R}}\frac{f(\alpha)}{(\alpha - z_{0})^{n+1}}\, d\alpha \cdot (z - z_{0})^{n}$

Now, notice that at this point the $(z - z_{0})^{n}$ is a not dependent on $\alpha$ for each $n$, and so we can actually write this whole integral as:

$\displaystyle = \sum_{n = 0}^{\infty}a_{n}(z-z_{0})^{n}$

for the $a_{n}$ given by the local cauchy formula:

$\displaystyle a_{n} = \frac{1}{2\pi i}\int_{C_{R}}\frac{f(\alpha)}{(\alpha - z_{0})^{n+1}}\, d\alpha$

which explicitly gives us a power series for $f$.  Finally.  This shows that $f$ is analytic.  $\Box$

Great, okay, so now we have analytic if and only if holomorphic.  Now, let’s give a short intuitive proof about complex differentiability.

Theorem: If a function $f$ has a first complex derivative, it has complex derivatives of all orders.

Proof. Once we note that a power series has a derivative that you can take term-by-term, and the derivative has the same radius of convergence as the original function, this proof becomes really nice.  If $f$ has a first complex derivative, it is holomorphic.  By the preceding theorem, it is analytic.  Therefore, take its power series expansion.  This is infinitely differentiable (why?) and therefore so is our function $f$.  Neat.

Good.  So I can now officially refer to all of my holomorphic functions as analytic and my analytic functions as holomorphic, and make the claim that “once a complex differentiable function, always a complex differentiable function,” with no one whining to me about how I didn’t prove it yet.

One thing you might want to think about is the following: what is so special about integrating $\frac{1}{z}$?  Note that if a function has an anti-derivative everywhere, then our integral about any closed path is zero.  So what’s the anti-derivative of $\frac{1}{z}$ and why is it so bad?  Any calculus student worth his beans will tell you that the integral of this is simply $Log(z)$, but in the complex plane things start getting ugly when you introduce logs…

4 Responses to “You say Holomorphic, I say Analytic.”

1. jacqui kotzee said

wow. that was really comprehensive, maybe you could help me: i need to prove that any f: D -> C(complex) which is holomorphic in D subset C is continuous in D. I think I need to use the open ball characteristic and then sequences to prove it continuous i.e. x -> x(0) => f(x) -> f(x(0))

2. jacqui said

thanks :)

3. spensi said

Getting here late but thanks for this post. I really needed something to help me make a clear distinction between holomorphic and analytic.