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Let $f,g$ be entire functions such that $f(g(z))=0, \forall z.$ Could anyone advise me on how to prove/disprove: either $g(z)$ is constant or $f(z) =0, \forall z \ ?$

Hints will suffice, thank you.

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Another way to make it: Assume that $g$ is not constant in particular $g'\neq 0$:
As $f\circ g=0$, then by derivation we get $g'\times (f'\circ g)=0$, but the ring of entires functions ( $\Bbb C$ connected) is an integral domain, it follow that $f'\circ g=0$, another derivation we get $g'\times (f''\circ g)=0$, hence $f''\circ g=0$, by same argument we can show that for all $n\in \Bbb N$; $f^{(n)}\circ g=0$.
Now let $a\in \Bbb C$ and $b=g(a)$, since $f$ is analytic we have : $$\forall z\in \Bbb C\ \ \ f(z)=\sum_{n=0}^{+\infty}\frac{f^{n}(b)}{n!}(z-b)^n=0$$ because $f^{(n)}(b)=f^{(n)}(g(a))=0$

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  • $\begingroup$ Your second solution is great too! I wonder why it is voted down. $\endgroup$ – Alexy Vincenzo Aug 24 '14 at 4:28
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If $g$ is not constant, then $U=g(\Bbb C)$ is an open subset, and $f |_U=0$, we get $f=0$.

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  • $\begingroup$ I like this answer better than mine, though both are useful ways of thinking about analytic continuation problems. I hope the edited syntax is reasonable. $\endgroup$ – Ian Aug 23 '14 at 17:17
  • $\begingroup$ @Hamou: Thanks for the answer. But how do I show that $U $ has an accumulation point in $\mathbb{C} \ ?$ The accumulation point can be any boundary point of $U \ ?$ $\endgroup$ – Alexy Vincenzo Aug 23 '14 at 17:30
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    $\begingroup$ The accumulation point in the result I think you are referencing is allowed to be in the set. That is, you only need that there is a sequence $\{ z_n \}_{n=1}^\infty$ which converges to some $z$ such that $f(z_n) = 0$. Whether the $z$ is in $U$ is irrelevant. With that, every point in an open subset of $\mathbb{C}$ is an accumulation point. $\endgroup$ – Ian Aug 23 '14 at 17:37
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    $\begingroup$ $U$ is open , and take $z_0\in U$, hence $\Bbb D(z_0,r)$ for some $r>0$, now remark that $z_0$ is an accumulation point in $U$. $\endgroup$ – Hamou Aug 23 '14 at 17:42
  • $\begingroup$ @Hamou: Thanks for the clarification. It seems that use of Opening mapping theorem is necessary in your proof (to show $U$ is open). Is it possible to address this problem without use of that theorem? $\endgroup$ – Alexy Vincenzo Aug 24 '14 at 0:38
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Suppose $f(z)$ isn't identically zero. Then $f(z_0) \neq 0$ for some $z_0$. The range of $g$ cannot contain $z_0$, because otherwise $f \circ g$. But $f$ is continuous at $z_0$, so it's nonzero is a small open set around $z_0$. $g$ is an entire function, so either it's constant or its image is dense in $\mathbb{C}$; the latter isn't possible (its closure cannot contain $z_0$), so it's constant.

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Suppose $g$ is not constant, $f$ is not identically zero, and $f \circ g$ is identically zero. Since $f$ is entire and not identically zero, its zero set is discrete (i.e. has no limit points). Since $g$ is not constant, we conclude that $g$ maps $\mathbb{C}$ onto a discrete set which has more than one point. Why is this impossible?

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