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 is an analytic function and we know that (for large say), do we actually know something like ?

Let’s rephrase this a bit. Essentially we’re talking about growth. It would be sufficient to show something along the lines of: if is harmonic, and grows at some rate, then 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 . In fact, gradient points in the direction of greater change, so this is in some sense an upper bound on the growth. Another is . 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: . Let’s check how useful this is in recovering growth of .

Since I haven’t talked about complex analysis much, note that the derivative operator for complex functions is .

Now

by Cauchy-Riemann

Thus .

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 .

In particular, . So we wanted to show that was a polynomial of degree at most , 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 . Is this for real? Am I horribly mistaken? I can’t find this in any book…

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I am a mathematics graduate student fascinated in how all my interests fit together.

September 14, 2009 at 5:17 pm

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.

September 15, 2009 at 10:29 am

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.

September 15, 2009 at 11:14 am

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?