Harmonic Growth as Related to Complex Analytic Growth

Let’s change gears a bit. This post will be on something I haven’t talked about in probably a year…that’s right, analysis. Since the last post was short, I’ll do another quick one. The past few days have had varying efforts to solve a problem of the form if f is an analytic function and we know that |Ref(z)|\leq M|z|^k (for large |z| say), do we actually know something like |f(z)|\leq M|z|^k?

Let’s rephrase this a bit. Essentially we’re talking about growth. It would be sufficient to show something along the lines of: if u is harmonic, and grows at some rate, then v the harmonic conjugate also must grow at a related rate. But all of this growth talk is vague. What does this even mean?

One measure of growth would be |\nabla u|=\sqrt{\left(\frac{\partial u}{\partial x}\right)^2+\left(\frac{\partial u}{\partial y}\right)^2}. In fact, gradient points in the direction of greater change, so this is in some sense an upper bound on the growth. Another is f'(z). Does this help? Well, first off, if this is our notion of growth, then by the Cauchy-Riemann equations, we immediately get that the harmonic conjugate grows exactly the same: |\nabla u|=|\nabla v|. Let’s check how useful this is in recovering growth of f.

Since I haven’t talked about complex analysis much, note that the derivative operator for complex functions is \frac{1}{2}\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right).

Now f'(z)=\frac{1}{2}\left(\frac{\partial(u+iv)}{\partial x}-i\frac{\partial(u+iv)}{\partial y}\right)
= \frac{1}{2}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+i\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)\right)
= \frac{\partial u}{\partial x}-i\frac{\partial u}{\partial y} by Cauchy-Riemann

Thus |f'(z)|=|\nabla u|.

Did this solve our original problem? Yes. Since if we work out the partial derivatives we get that if $|u|\leq M(x^2+y^2)^k/2$, then |\nabla u(z)|\leq Mk|z|^{k-1}.

In particular, |f'(z)|\leq Mk|z|^{k-1}. So we wanted to show that f was a polynomial of degree at most k, and we can now use Cauchy estimates to get that.

If any of what I just wrote is true, then there is some really obvious way of doing it that isn’t messy like this at all. I mean, the result is |f'(z)|=|\nabla u|. Is this for real? Am I horribly mistaken? I can’t find this in any book…


3 thoughts on “Harmonic Growth as Related to Complex Analytic Growth

  1. That seems right to me. I remember working out something similar in the proof of the Hadamard factorization theorem. In that case you only have that exp(f(z)) is bounded by exp(M|z|^n). i.e. if Re(f(z)) is bounded above by M|z|^n then f is a degree n polynomial.

  2. This is pretty cool. I learned the Schwarz formula specifically to equip myself to solve this problem if it appears, and now I will happily forget it.

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