# 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…

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

1. George Lowther says:

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.

3. Actually, I tried to go through your steps and got stuck on the third-to-last paragraph. Why does the hypothesis on |u| imply that inequality?