Sunday, May 03, 2009

Nonsensical Theorem and Proof

Today I was in the mood for some Maths; and as I am home alone, I had no distractions etc. First up it was the Euler Product for the zeta function. I understood the sieve process, until the very two last lines which complete the proof. We take the sup and limits and then draw a nice black square. Well not very nice in my case, but at least I followed the main idea I consoled myself.

Next came lecture five - Derivatives of Infinite Product.

I understood the first paragraph, but then came the Theorem and its proof. Some generous lecturers offer rewards if students spot any mistakes in their notes. Although this proof and theorem is not something I originally wrote, I too will be generous and offer a reward if anyone can point out any corrections. However... to claim your reward, you have to ask for it in person! By the way I can on this occasion say that I copied exactly what was on the black board, as I remember the lecture very well (due to a small hiccup). So here goes:

Theorem
Let $f= \prod g_j$ with $g_j \in Hol(U), f \neq 0, g_j \neq 0$ in U, and $f = \prod g_{\lambda}$ (L.U). Then

$\frac{f'}{f} = \sum \frac{g_n'}{g}.$

Proof

$f_n = g_1 ... g_n \Rightarrow \frac{f_n'}{f} = \frac{g_1'}{g_1} + ... + \frac{g_n'}{g_n}.$

But $f \in$ Hol(U) and $f_n' \to f'$ by convergence lemma (2.2).

Let C be a circle in U, then $\exists \delta : | f(z)| \ge \delta$ "more than" 0,

$\exists N \text{ st } n \ge N \Rightarrow |f_n(z) | \ge \frac{\delta}{2}$.

$\Rightarrow | \frac{1}{f_n(z)} - \frac{1}{f(z)} | = \frac{|f_n(z) - f(z)|}{|f_n(z) f(z)| } \le \frac{2}{\delta^2} |f_n(z) - f(z)|.$

$\Rightarrow \frac{1}{f_n} \to \frac{1}{f}$ uniformly on C

Therefore $\frac{1}{f_n} \to \frac{1}{f}$ uniformly on each compact disc.

Therefore $\frac{1}{f_n} \to \frac{1}{f}$ (LU) $\Rightarrow \frac{f_n'}{f_n} \to \frac{f'}{f}$.
$\Box$

If you can't find any errors and understand the proof, some help would be nice! I'm being dumb, I know, but how does the conclusion in the proof help us?

Anyway, it's food time now! Erm - not fast food again... Pizza is quite healthy in my opinion, especially if it has some pineapples on it!

steve said...

Actually I don't understand the question!
but how does the conclusion in the proof help us?Help us to do what? I presume you don't mean how does fn'/fn->f'/f prove the theorem? After all, fn'/fn is just the partial sum which, by definition, tends to the infinite sum, so the infinite sum must be f'/f.

Beans said...

Hi Steve,
Many thanks for your reply. I now understand what I was misunderstanding! I got confused with the notation of the theorem, and had thought it was meant to read f_j = \prod g_j, and then f_n = g_1.g_2...g_n, which then tends to f (LU).

Once again thanks. :)

MARC said...

This question is not well posed since you do not say anything about the type of convergence of the infinite product. Do you mean normal convergence? Anyways, I think you do mean normal convergence. Everything is local and f is nonzero so you can just take the logarithm. take ln(z) and differentiate:

d/dz (ln(f(z))) = f(z)/f'(z)

Using uniform convergence on compact sets, its easy to see what happens to the left hand side.

Maybe that helps.