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I have a fundamental doubt about a concrete step in this Corollary (I've copied the statement and its proof from Stein's textbook):

Corollary to Liouville's Theorem Every non-constant polynomial $P(z)=a_nz^n+...+a_0$ with complex coefficients has a root in $\mathbb C$.

Proof If $P$ has no roots, then $\dfrac{1}{P(z)}$ is a bounded holomorphic function. To see this, we can of course assume that $a_n \neq 0$, and write $$\dfrac{P(z)}{z^n}=a_n+(\dfrac{a_{n-1}}{z}+...+\dfrac{a_0}{z^n})$$ whenever $z \neq 0$. Since each term in the parentheses goes to $0$ as $|z| \to \infty$, we conclude that there exists $R>0$ so that if $c=\dfrac{|a_n|}{2}$, then $$|P(z)|\geq c|z|^n \space \text{whenever}\ \space |z|>R$$.

In particular, $P$ is bounded from below when $|z|>R$. Since $P$ is continuous and has no roots in the disc $|z|\leq R$, it is bounded from below in that disc as well, thereby proving our claim.

By Liouville's theorem we then conclude that $\dfrac{1}{P}$ is constant. This contradicts our assumption that $P$ is non-constant and proves the corollary.

My doubt My question is more related to a more general analysis matter rather than a complex-analysis one, I don't understand this excerpt of the proof:

"Since $P$ is continuous and has no roots in the disc $|z|\leq R$, it is bounded from below in that disc as well...". I can't understand why $P(z)$ it's bounded from below in $|z|\leq R$ from the condition that it is continuous; it would be greatly appreciated if someone could give me a detailed proof or a well explained justification of this part.

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    $\begingroup$ If it were not bounded below 0 would be a limit. $\endgroup$ – Rene Schipperus Jun 15 '14 at 1:15
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    $\begingroup$ The disc $K = \{|z|\leq R\}$ is compact, so $|P|$ attains its minimum there. Since $P$ is nonzero on $K$, that minimum must be positive. $\endgroup$ – anomaly Jun 15 '14 at 1:29
  • $\begingroup$ Do you mean "$1/P$ is bounded from below when $|z|>R$" instead of "$P$ is bounded from below when $|z|>R$"? $\endgroup$ – mfl Jun 15 '14 at 1:29
  • $\begingroup$ @nfl, well, the final conclusion would be that $1/P$ is bounded (from below and from above), but answering your question: no, I exactly mean P is bounded from below when $|z|>R$. $\endgroup$ – user100106 Jun 15 '14 at 2:14
  • $\begingroup$ @anomaly that's exactly what I was looking for, now I understand it, thank you very much. $\endgroup$ – user100106 Jun 15 '14 at 2:15
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The disk $K = \{|z| \leq R\}$ is compact, so $|P|$ attains its minimum there. Since $P$ is nonzero on $K$, that minimum must be positive.

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