# Another MVT sighting

This blog has lain fallow some three years now while I took a detour into writing for a commercial source. Now that that's done (it was great, but it was time to move on) it's time to plant some new seeds in my own backyard.

It's summer and I'm teaching Introduction to Complex Variables, a course I like very much. This week, after introducing the idea of an analytic function I set out to prove the following fact: if $$f$$ is analytic on a domain $$D$$ and if $$f'(z) = 0$$ everywhere on $$D$$, then $$f$$ is constant.

In the calculus of a single real variable the analogous statement is a consequence of the Mean Value Theorem, but of course we don't have that in the plane. Well, we sort of do, but it's not one of those things that gets taught as a rule (I don't remember it from undergraduate analysis, but then I'm a topologist and so I've let a lot of that fade from memory). Anyway, it's not obvious how we should proceed to prove what seems to be an obvious fact--the only functions with 0 derivative are the constants--so we have to think a little bit.

I have a long-running joke (with myself, mostly) that one day I'm going to write an advanced calculus text called The Real Fundamental Theorem of Calculus, by which I mean the Mean Value Theorem. It will be a Where's Waldo-style sort of thing where the reader will need to spot the MVT hiding in the text. Maybe I'll design a little cartoon representation of it to play the role of Waldo. My argument is that when you really get down to proving things about calculus you almost always need the Mean Value Theorem. In particular, the Fundamental Theorem (students' favorite part about how to evaluate definite integrals in terms of antiderivatives) is a pretty easy consequence of the MVT, hence my assertion that we should call it the Real Fundamental Theorem.

So, in complex analysis we're dealing with functions of a single complex variable. We can think of these as functions of two real variables, of course, but the real and imaginary parts are not arbitrary. A consequence of the definition of differentiability for complex functions is the Cauchy-Riemann equations: if $$f(z) = u(x,y) + i v(x,y)$$ is differentiable at $$x_0 + iy_0$$ then the partial derivatives must satisfy $$u_x = v_y$$ and $$u_y = -v_x$$. This is a serious restriction. In fact you can use them to show that the rather innocuous-looking function $$f(z) = \overline{z}$$ is not differentiable anywhere even though the real and imaginary parts have partial derivatives of all orders. Another consequence of these equations is that if $$f$$ is analytic then the derivative may be computed as $$f'(z) = u_x + iv_x$$.

Now we can prove the theorem I claimed above. If $$f'(z) = 0$$ everywhere in the connected open set $$D$$, then all the partial derivatives $$u_x, u_y, v_x, v_y$$ vanish. Suppose $$P$$ and $$P'$$ are two points in a small disc contained in $$D$$ and let $$L$$ be the line segment from $$P$$ to $$P'$$. Let $$\vec{w}$$ be a unit vector in the direction of $$L$$ and let $$s$$ be the distance along $$L$$ from $$P$$. The function $$u(x,y)$$ may be restricted to $$L$$ using the parameter $$s$$. We then compute $\frac{du}{ds} = (u_x + iu_y)\cdot \vec{w} = 0.$  But now notice that this is a function of the real variable $$s$$ and so by the Mean Value Theorem we have that $$u(x,y)$$ is constant on $$L$$. Now, given any two points in $$D$$, we can join them by a polygonal path lying entirely inside $$D$$ and this argument shows that $$u$$ is constant on each segment and hence on the whole path. Thus $$u(x,y)$$ is constant on $$D$$. A similar argument applies to $$v(x,y)$$ and so $$f(z)$$ is constant on $$D$$.

Cool. So the MVT was hiding in there after all, just as I suspected. You should hop on my bandwagon now.

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