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I want to prove this function has a modulus greater of equal than 1 : Find all entire functions $f$ such that for all $z\in \mathbb{C}$, $|f(z)|\ge \frac{1}{|z|+1}$

One answer says that “clearly” $\vert f(z) \vert \geq 1 $ for all $z$. But its not clear to me. How can I prove it? I am thinking a lot about this but I dont see it.

UDAC
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    I would guess that answerer did the mental arithmetic with the inequalities too quickly. A-priori, It is not clear at all that $|f(z)|\geq 1$ for all $z$. The best bet is to follow the accepted answer: $\frac{1}{f}$ is an entire function which never vanishes, and which is bounded above by $1+|z|$. Hence (a modification of the proof of Liouville's theorem), $\frac{1}{f}$ is a polynomial (of degree $\leq 1$) and hence is constant (because you have degree $\leq 1$, this is obvious; more generally you can appeal to FTA). Hence, $f$ is constant (it is clear now that the constant is $\geq 1$). – peek-a-boo Feb 09 '23 at 01:03
  • @peek-a-boo How do I know that f(z) is never zero? – UDAC Feb 09 '23 at 01:08
  • I said $\frac{1}{f}$ is never zero; this is extremely obvious simply because $f$ is complex-valued. The better question is why $\frac{1}{f}$ is entire, and the answer is that $f$ never vanishes (by the inequality) so its reciprocal is well-defined and entire. – peek-a-boo Feb 09 '23 at 01:09
  • If $|f(z)|\ge \frac{1}{|z|+1}$ what happens if $\vert z\vert $ goes to infinite? We can say that $\vert f(z) \vert \geq 0 $, but how we can discard the equal to zero? – UDAC Feb 09 '23 at 01:13
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    why are you taking limits? For each $z\in\Bbb{C}$, $|f(z)|\geq \frac{1}{|z|+1}>0$, and hence $f(z)\neq 0$. This is literally what it means for $f$ to be nowhere vanishing. Note $\infty\notin\Bbb{C}$. If you're still not convinced, do the following simple exercise: sketch the function $\frac{1}{1+t}$ for $t\in [0,\infty)$. Does this graph ever intersect the horizontal axis? (you had better not say 'it intersects it at infinity':) – peek-a-boo Feb 09 '23 at 01:14
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    And if that is still not convincing, try a proof by contradiction. Suppose there is some $z\in\Bbb{C}$ such that $f(z)=0$. Then, $0=|f(z)|\geq \frac{1}{|z|+1}>0$, i.e $0>0$. How crazy is that? – peek-a-boo Feb 09 '23 at 01:22
  • @peek-a-boo My reasoning was based in the fact that, if we take an example from real functions , if we have $|f(x)|\geq \frac{1}{|x| +1}$, we can not rule out the posibility that f goes to zero as $\vert x\vert $goes to infinity growths. Suppose this is true (f goes to zero). Then 1/f goes to infinity as x growths. – UDAC Feb 09 '23 at 01:28
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    the LIMIT (if it exists) could be zero, yes. I'm not disputing that at all. But the inequality instantly tells us that the function itself is never equal to zero, and this is all that matters to us. – peek-a-boo Feb 09 '23 at 01:29
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    Because the function is entire, if the limit of $|f|$ exists at $\infty$ and is $< \infty$, the function has to be constant. If it diverges to $\infty$ at $\infty$, the function has to be a polynomial. All other entire functions have essential singularities at $\infty$. But this is of no consequence for what peek-a-boo has been saying, which follows from the properties of the function on $\Bbb C$, and again $\infty \notin \Bbb C$. – Paul Sinclair Feb 09 '23 at 21:24

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