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Relationship between two entire functions

Bingk

New member
Jan 26, 2012
16
Hello, this was another question on the exam which I wasn't sure about:

Let [TEX]f[/TEX] and [TEX]g[/TEX] be entire such that [TEX]|f(z)| \leq |g(z)| \ \forall z \in \mathbb{C}[/TEX]. Find a relationship between [TEX]f[/TEX] and [TEX]g[/TEX].

I'm kinda lost on this one...

Thanks!
 

chisigma

Well-known member
Feb 13, 2012
1,704
Hello, this was another question on the exam which I wasn't sure about:

Let [TEX]f[/TEX] and [TEX]g[/TEX] be entire such that [TEX]|f(z)| \leq |g(z)| \ \forall z \in \mathbb{C}[/TEX]. Find a relationship between [TEX]f[/TEX] and [TEX]g[/TEX].

I'm kinda lost on this one...

Thanks!
A consequence of the Liouville's theorem is that if $\displaystyle |f(z)|\le |g(z)| \forall z \in \mathbb{C}$, then it must be $f(z)=\alpha\ g(z)$ for some complex $\alpha$...

Kind regards

$\chi$ $\sigma$
 

Bingk

New member
Jan 26, 2012
16
Hi, that's actually what I got, but I'm pretty sure I got it the wrong way. I don't remember exactly what I did (the exam wasn't returned to us), but my method involved the Maximum Modulus Principle (sort of like applying Liouville's on open balls).

How exactly is that a consequence of Liouville's theorem?
I can see that for the case of g being constant, then f should be constant, so g will be a multiple of f. What about for when g is non-constant?

From what I can recall, what I think I did was I said that in an open ball, f and g attain their maximum modulus on the boundary and the modulus of g will be greater than that of f at that point in the boundary. This will happen for any open balls. So, if we consider two balls with the same center, and let the radius of one approach the other, and see what happens to the modulus of f and g at that boundary, it should turn out that g is a multiple of f. Is this sort of right?
 

chisigma

Well-known member
Feb 13, 2012
1,704
Hi, that's actually what I got, but I'm pretty sure I got it the wrong way. I don't remember exactly what I did (the exam wasn't returned to us), but my method involved the Maximum Modulus Principle (sort of like applying Liouville's on open balls).

How exactly is that a consequence of Liouville's theorem?
I can see that for the case of g being constant, then f should be constant, so g will be a multiple of f. What about for when g is non-constant?

From what I can recall, what I think I did was I said that in an open ball, f and g attain their maximum modulus on the boundary and the modulus of g will be greater than that of f at that point in the boundary. This will happen for any open balls. So, if we consider two balls with the same center, and let the radius of one approach the other, and see what happens to the modulus of f and g at that boundary, it should turn out that g is a multiple of f. Is this sort of right?
If You consider the function...

$\displaystyle h(z)=\frac{f(z)}{g(z)}$ (1)

... then, because is $\displaystyle |f(z)| \le |g(z)|$ h(*) is bounded. That means that all the zeroes of g(*) must be also zeroes of f(*) and h(*) is entire, so that h(*) for the Liouville's theorem must be a constant that we call $\alpha$...

Kind regards

$\chi$ $\sigma$
 

Bingk

New member
Jan 26, 2012
16
Thank you! I didn't think to consider the function h ... but it should've occured to me.