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I am stuck at a problem which says :

" Let $f$ and $g$ are entire functions such that $e^f, e^g$ and $1$ are linearly dependant over $\mathbb{C}$. Show that $f$, $g$ and 1 are also linearly dependant."

So basically we have constants $C_1$, $C_2$, $C_3$ in $\mathbb{C}$, not all zero, such that $C_1 e^f + C_2 e^g + C_3=0$ . It becomes easy if one of the $C_i$ 's is zero. But I am stuck in the case where all of them are non-zero.

I can understand that it suffices to show that $|Af + Bg|$ is bounded on $\mathbb{C}$ for some complex constants $A, B$, not both of them zero. I tried using the inequality $|e^z|\leq e^{|z|}$, but it is not working.

Please help with any suggestion as soon as possible :( Thanks in advance !

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Why "as soon as possible"? Sounds like it is a "hand-in/homework" problem. – AD. Jan 25 '13 at 8:23
If none of $C_i$ is zero, both $f$ and $g$ are constant by Little Picard's theorem. – achille hui Jan 25 '13 at 10:12
I obviously over thinking for a while before I realize this... – Bombyx mori Jan 25 '13 at 10:36
@AD. I wrote that because it is bothering me for quite a few days, and now I have started to feel that may be my approach is wrong. – user44349 Jan 25 '13 at 18:14

2 Answers 2

up vote 1 down vote accepted

Suppose we have $c_{1} e^{f}+c_{2} e^{g}+ c_{3}=0$ and $c_1\ne0$ then we have $$e^g= \frac{c_3-c_1 e^f}{c_1}$$ and also we have $$ c_{1}f' e^{f}+c_{2}g'e^{g}=0$$ which implies $$\frac{f'}{g'}=\frac{-c_2 e^{g}}{c_1 e^f}$$ Thus we have $$\frac{f'}{g'}=\frac{-c_2(c_3-c_1 e^f)}{c_1^2 e^f}=\frac{b}{e^f}-c$$ for some constants $b$ and $c$. This implies that $\frac{f'}{g'}$ is bounded, Thus, using Liouville's theorem we get that $\frac{f'}{g'}$ is constant that is $$f'=kg'$$ Now by integrating both sides we get that $$f=kg+l$$ which implies that $\{f,g,1\}$ are linearly dependent.

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Thanks a lot ! But how did you infer that $f'/g'$ is bounded ? For example, it can be that $f(z) =-z$ for all $z$. – user44349 Feb 7 '13 at 22:50
@CB_Student I assumed $f$ is not bounded which implies that $e^f$ is not bounded, Thus, $\left|\frac{f'}{g'}\right|\le\frac{|b|}{\left|e^f\right|}+\left|c\right|$ but we know that $\left|e^f \right|\longrightarrow\infty$ as $z\longrightarrow \infty$ so we can have a big enough circle such that $\frac{1}{|e^f|}$ is bounded outside it and it bounded insided it by analyticity (continuouity ) of $\frac{1}{e^f}$ thus $\frac{f'}{g'}$ is bounded. – i.a.m Feb 7 '13 at 23:26
@CB_Student is there something wrong with the answer?! – i.a.m Feb 13 '13 at 22:00
No, absolutely not ! Thanks a lot for taking time to write it all down ! :) – user44349 Feb 14 '13 at 0:34
@CB_Student you are welcome – i.a.m Feb 14 '13 at 1:02

Assume all the constants are not zero, then we may assume without loss of generality that $e^f=e^g+1$, then you use the Picard theorem, because the range of $e^f$ doesn't contain 0 and 1(because $e^g$ can't be 0). So you get the conclusion that f and g must be constants.

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Thanks! I get it, but actually little Picard's theorem is not covered in the class till now. So I was trying to do it without using it. Is that possible ? – user44349 Jan 25 '13 at 18:06
I've been thinking about that for a while. But I haven't got another solution yet. – lee Jan 30 '13 at 2:41
I think there is a necessary and sufficient condition for a function to be exponential of another entire function. – lee Jan 31 '13 at 5:21

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