Suppose $f$ and $g$ are entire functions, and $|f(z)| \leq |g(z)|$ for all $z \in \mathbb{C}$, Prove that $f(z)=cg(z)$.

Question

My try :

I consider $h(z)=\frac{f(z)}{g(z)}$. If I prove that $h(z)$ is entire, using the fact that $|h(z)|\leq 1$, the result follows immediately from Liouville's theorem.

To prove that $h(z)$ is entire, I have to prove that $h(z)$ has removable singularities at possible zeros of $g(z)$.

Suppose $g(z_0)=0 \Rightarrow |g(z_0)|=0 \Rightarrow |f(z_0)|=0$, BUT what if $z_0$ be a root of of order $k$ of $g(z)$ (i.e. $f^{(n)}(z_0) = 0 $ for all $ 0 \leq n \leq k-1$) and be a zero of order $m$ of $f(z)$ and $k>m$ ?

And if that's not the case, how to rigorously show that $z_0$ is a removable singularity ?

Thank you all in advance.

possible duplicate of Some proofs in complex related to F.T of algebra and Cauchy's inequality maybe — Ayman Hourieh, Dec 20 '13 at 01:27

Andrés E. Caicedo · Accepted Answer · 2013-12-20T01:53:51.480

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Suppose that $g$ has a zero at $a$. If $f$ is identically zero, we are done, as $0=0g$, so we may assume this is not the case, which implies that $g$ is not identically zero either, and therefore we may assume instead that it has a zero of order precisely $k$ at $a$, so $g(z)=(z-a)^kh(z)$, where $h$ is analytic in a neighborhood of $a$, and $h(a)\ne 0$.

Since $|f|\le|g|$, then $f(a)=0$, so $f(z)=(z-a)^mj(z)$ where $j(a)\ne0$ and $j$ is analytic in a neighborhood of $a$. For $z\ne a$ near $a$, $|f|\le|g|$ means that $$\displaystyle\left|\frac{(z-a)^{m-k}j(z)}{h(z)}\right|\le 1.$$

Since $j$ and $h$ are not zero near $a$, by continuity there is a constant $K$ such that $|j/h|\ge 1/K$ in a neighborhood of $a$, so $$\left|\frac{(z-a)^{m-k}j(z)}{h(z)}\right|\ge\frac{|z-a|^{m-k}}K,$$ or $|z-a|^{m-k}\le K$ for $z\ne a$ near $a$. But this is impossible if $m<k$.

edited Dec 20 '13 at 01:53

answered Dec 20 '13 at 01:34

Andrés E. Caicedo

79,201

Thanks . I think it must be $|z-a|^{m-k} \leq \frac{1}{K}$ , Am I right ? – the8thone Dec 20 '13 at 01:48
@Roozbeh-unity Ah, well spotted, thanks. There was a detail pending. – Andrés E. Caicedo Dec 20 '13 at 01:54
May I know why your line says it is impossible for $m<k$? – nan Sep 13 '18 at 10:43
@ilovewt Because the function $|z-a|^{m-k}$ would be unbounded in any neighborhood of $a$ in that case. – Andrés E. Caicedo Sep 13 '18 at 11:36
Hi sorry if I sound a bit unclear, but if $m<k$ then the $|z-a|^{m-k}$ would be of $\frac{1}{|z-a|^{k-m}}$ but why is this unbounded. Sorry I’m just keen to know. – nan Sep 13 '18 at 12:08

score 9 · Answer 2 · answered Dec 20 '13 at 01:24

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If $z_0$ is a singularity of $h$, and $h$ is bounded in a neighbourhood of $z_0$, then $z_0$ is a removable singularity of $h$. Because if it isn't removable, then it is a pole (and not bounded) or essential (which is also not bounded, but actually this cannot happen).

answered Dec 20 '13 at 01:24

Stephen Montgomery-Smith

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Thanks, But the problem is I still do "not" know that $h(z)$ is bounded, unless I know that I am not dividing by zeros of higher order of $g$ (relative to the order of that particular zero of $f$) – the8thone Dec 20 '13 at 01:28
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@Roozbeh-unity The key here is that $h$ is bounded in a deleted neighborhood of every zero of $g$. Thus the singularity is removable. See the answer I linked to above. – Ayman Hourieh Dec 20 '13 at 01:31
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You know that whenever $g(z) \ne 0$ that $|h(z)| \le 1$. And the zeros of $g$ are isolated. – Stephen Montgomery-Smith Dec 20 '13 at 01:32
@AymanHourieh Thanks for that link , I read that, Are we using the continuity of $h(z)$ at all points other than the zeros of $g$ ? So it is impossible that at all other points $h(z)$ is bounded an at exactly one point becomes $\infty$ ? Am I right ? – the8thone Dec 20 '13 at 01:36
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There is a theorem that isolated singularities are only of three types. Removable, poles, and essential. If it is bounded, then it has to be removable. – Stephen Montgomery-Smith Dec 20 '13 at 01:41
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@Roozbeh-unity The proof is slightly more involved (and more rigorous) than this. But it's a standard theorem found in complex analysis courses. – Ayman Hourieh Dec 20 '13 at 01:43

score 7 · Answer 3 · answered Aug 22 '14 at 18:51

@Shubhodip Mondal, @Andres Caicedo: My proof appears different from those of the above replies. Hope it is not too far off?

Consider $h:=\dfrac{f}{g},$ so singularties of $h$ are zeroes of $g.$

1.) Suppose $g$ has zero of order $n$ at $z_o$ and $f(z_o)\neq 0,$ then $\lim _{\ z\to z_o}h(z)= \infty.$ This contradicts $|h| \leq 1,$ in a neighbourhood of $z_o.$

2.) Suppose $f,g$ have zero of order $m,n$ at $z_o.$ Then $h= \dfrac{\phi(z)(z-z_o)^m}{\beta(z)(z-z_o)^n},$ for some $\phi, \beta$ analytic at $z_o$ and $\phi(z_o), \beta(z_o) \neq 0.$

If $m < n,$ then $ h(z)= \dfrac{\phi(z)}{\beta(z)(z-z_o)^{\ n-m}}$ and hence $ \lim _{\ z\to z_o}h(z)= \infty.$ (Contradiction)

So, $m \geq n$ and $ \ h(z)=\dfrac{\phi(z)(z-z_o)^{m-n}}{\beta(z)}.$ Hence $z_o$ is removable singularity of $h,$ whence $h$ can be extended to an entire function. Also, by continuity $, |h(z)| \leq 1,$ where $z$ is removable singularity of $h.$ The result follows from Lioville's theorem.

score -1 · Answer 4 · answered Dec 21 '13 at 08:51

Let $z_0$ be a zero of $g(z)$.Unless $g$ is identically zero, it can not be zero at points arbitrarily close to $z_0$. (By "identity principle".) And in that case $\displaystyle \lim_{z \to z_0} (z-z_0)\frac{f(z)}{g(z)} = 0$, (Since $|f(z)| \le |g(z)|$ ) which tells you that the singularity indeed is removable.

Suppose $f$ and $g$ are entire functions, and $|f(z)| \leq |g(z)|$ for all $z \in \mathbb{C}$, Prove that $f(z)=cg(z)$.

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