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For an entire function $f(z)$, Which of the following condition implies that $f(z)$ is a constant?

(A) If $f(z)=u+iv$ and $u^2\leq v^2+2004$

(B) If $f(z)=f(2z)\forall z\in \mathbb C$

(C)$|f(z)|\leq 2|log(z)|+3 \forall z\in \mathbb C$

(D)$|f(z)|\leq 10 \forall z\in \{z\in \mathbb C: Re(z)>0\}$

My attempt:-

(A) $u^2-v^2\leq 2004$ will be a region in $u-v$ plane between hyperbola. which is unbounded. I am not able to draw anything else from this.

(B) For any $z\in \mathbb C$, $f(z)=f(\frac{z}{2^n})$. So, By continuity of $f$ we get $f(z)=f(0)$, which is a constant.

(C)$|f(z)|\leq 2|log(z)|+3 \implies |f(z)|-2|\log(z)|\leq 3$ The Left hand part in the inequality isneed not be analytic. Hence, we can not apply Lioville Theorem.

(D) Here $f(z)$ Bounded in First and Fourth Quadrant. If it would be bounded in $\mathbb C$, I could use Lioville Theorem.

Math geek
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  • I think C gives you that $f$ is constant as well: https://math.stackexchange.com/questions/439909/showing-that-an-entire-function-with-the-following-inequality-is-constant – Cameron Williams May 19 '19 at 14:06

2 Answers2

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For first one, consider $e^{f^2}$. Now consider $\vert e^{f^2} \vert$ and use hypothesis and then use Liouvilles theorem

Chinnapparaj R
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For (A), note that an entire function cannot miss a nonempty open set $U$ (otherwise $\frac1{f(z)-p}$ is bounded entire, where $p\in U$).

For (C), use Cauchy representation formula for derivatives and the usual estimates.

For (D), $\exp(-z)$ is such an example.

user10354138
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  • Can you explain (C) bit more? – Math geek May 19 '19 at 14:49
  • You get the $n$-th derivative at $0$ is (up to a factor involving $n!$ and $2\pi i$) $\int_C f(z)z^{-n-1},\mathrm{d}z$. Now estimate this when $C$ is a circle of radius $R$, and take limit $R\to\infty$ for $n\geq 1$, see this answer where the bound on $f$ is slightly different. – user10354138 May 19 '19 at 14:55