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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!

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- Thread starter
- #1

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!

- Feb 13, 2012

- 1,704

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 $\alpha$...

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!

Kind regards

$\chi$ $\sigma$

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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?

- Feb 13, 2012

- 1,704

If You consider the function...

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?

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

$\chi$ $\sigma$

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