Currently reading the Levin book on Entire functions, Taking my time practicing by trying to prove the following statements, I have wasted a couple of days fighting with these folks:
- $p_{fg} \le max(p_{f},p_{g})$
- $p_{f+g} \le max(p_{f},p_{g})$
There are similar statements for the "types" but they should be easily provable, based on statements for the "order".
$\sigma_{fg} \le \sigma_{f}+\sigma_{g})$ <- Provable by logarithm properties.
$\sigma_{f+g} \le max(\sigma_{f},\sigma_{g})$ <- proof should be based on proved 2nd stmnt.
Kindly ask for a hint and (if possible) full proof of $$p_{fg} <= max(p_{f},p_{g})$$ and $$p_{f+g} \le max(p_{f},p_{g})$$
I am very confused about these statements. The second statement will make sense(when at least one order(e.g. (both) is infinite, or one(both) of the orders is equal to zero)
Definition of the order $$ p = \left[ \limsup_{r\to\infty} \frac{log(log(M_f(r)))}{log(r)} \right] $$
Limits comparison will bring the need to compare the actual different parts of order definition.
For 1st: $$log(M_{f}(r))+log(M_{g}(r)) \le max(log(M_{f}(r)), log(M_{g}(r))$$ For 2nd: $$log(M_{f}(r)+M_{g}(r)) \le max(log(M_{f}(r), log(M_{g}(r))$$
How to prove it for any order? Should I try to do that in the Taylor series?
It is not my homework.
Thanks to Conrad, you have this proof.
Let's prove, statement 1. $p_{fg} \le max(p_{f},p_{g})$.
By utilizing the maximum modulus principle, we are having the formulas $M_{fg} \le M_f(r)*M_g(r)$, as well as $M_{f+g} \le M_f(r)+ M_g(r)$
Those equations are true for $|f|$ and $|g|$.
By definition of the order $$ p = \left[ \limsup_{r\to\infty} \frac{log(log(M_f(r)))}{log(r)} \right] $$
It is reasonable to consider only the distinct part (sum -> multiplication by log properties.). $$loglog(M_{f}(r)*M_{g}(r)) \le max(loglog(M_{f}(r)), loglog(M_{g}(r))$$
$max(log(M_{f}(r)), log(M_{g}(r))$ might be not obvious, so we will consider a much stronger condition. $$loglog(M_{f}(r)*M_{g}(r)) \le max(loglog(M^2_{f}(r)), loglog(M^2_{g}(r))$$
So, by logarithm properties, we obtain: (for ($loglog(M^2_{f}(r)$)
$$loglog(M^2_{f}(r) = log(2*log(M_{f}(r)) = log(2) + loglog(M_{f}(r))$$
The same applies to $M_g$. We can write: $$max(log(2) + loglog(M_{f}(r)),log(2) + loglog(M_{g}(r)))$$
Which equals to:
$$max(loglog(M_{f}(r)),loglog(M_{g}(r))) + log(2)$$
Let us have the final inequality written down:
parameter $ A = \limsup_{r\to\infty} \frac{loglog(M_{f}(r))}{log(r)}$
parameter $ B = \limsup_{r\to\infty} \frac{loglog(M_{g}(r))}{log(r)}$
$$\limsup_{r\to\infty} \frac{loglog(M_{fg}(r))}{log(r} \le max(A,B) + \limsup_{r\to\infty} \frac{log(2)}{log(r)}$$
So, $\limsup_{r\to\infty} \frac{log(2)}{log(r)}$ is going to be 0 as $log(2)* \frac{1}{\inf} = log(2) * 0 = 0$
Thus, $\Delta$. The statement is proven.
Statement 2 $p_{f+g} \le max(p_{f},p_{g})$
Proof is the same, as $$loglog(M_{f+g}(r)) \le max(loglog(M^2_{f}(r)), loglog(M^2_{g}(r))$$
Since $(a+b) or (a*b) \le max(a^2,b^2)$