The following Lemma is used in a proof for the Fundamental Theorem of Algebra, but I cannot understand some steps.
Lemma: Let $f:\mathbb{C} \to \mathbb{C}$ be any polynomial function.
Then there exists a point $z_0 \in \mathbb{C}$ where the function
$|f|$ attains its minimum value in $\mathbb{R}$.
The proof starts like this:
If $f$ is a constant polynomial function, then the statement of the
Lemma is trivially true since $|f|$ attains its minimum value at
every point in $\mathbb{C}$ . So choose, e.g., $z_0=0$.If $f$ is not constant, then the degree of the polynomial defining
$f$ is at least one. In this case, we can denote $f$ explicitly by$f(z)=a_n z^n + … + a_1 z + a_0$
with $a_n \neq 0$. Now, assume $z \neq 0$ , and set
$A=max\{|a_0|,…,|a_{n−1}|\}$ . We can obtain a lower bound for
$|f(z)|$ as follows:$f(z)=|a_n| |z|^n |1+\frac{a_{n-1}}{a_n} \frac{1}{z}+…+\frac{a_0}{a_n} \frac{1}{z^n}| \leq |a_n| |z|^n (1 – \frac{A}{|a_n|} \sum \limits_{k=1}^{\infty} \frac{1}{|z|^k})=|a_n| |z|^n (1 – \frac{A}{|a_n|} \frac{1}{|z|-1})$.
I don't see how these two steps follow. We can distribute the absolute value over all the factors, but how does one turn the second factor from $1+…$ into a $1-…$ using an infinite sum? After that it seems like an application of a geometric series, but how do we know that $|z|>1$?
I've been stuck for quite some while now with no notable progress. Am I missing anything obvious here?
Edit:
The next step of the proof is a follows (so it justifies the use of the formula for the geometric series):
For all $z \in \mathbb{C}$ such that $|z| \geq 2$, we can further
simplify this expression and obtain$ (*) |f(z)| \geq |a_n| |z|^n (1 – \frac{2A}{|a_n| |z|})$.
It follows from this inequality that there is an $R > 0$ such that $|f(z)| > |f(0)|$, for all $z \in \mathbb{C}$ satisfying $|z| > R$.
Now this confuses me once again. We have $|f(0)|=a_0$, so how does this follow from the inequality given above? I think it ought to have something to do with the definiton of $A$, but I can't quite figure it out.
Could anyone please provide a hint or an explanation?
Best Answer
We don't need geometric series, etc., in order to prove the Lemma. As you remark, we can assume $$f(z)=a_nz^n+a_{n-1}z^{n-1}+\ldots+a_0,\qquad n\geq1,\qquad a_n\ne0\ .$$ This implies$${f(z)\over z^n}=a_n+{a_{n-1}\over z}+\ldots+{a_0\over z^n}\to a_n\quad\bigl(|z|\to\infty\bigr)\ ,$$ so that there is an $R>0$ with $$\left|{f(z)\over z^n}\right|>{|a_n|\over2}\qquad\bigl(|z|>R\bigr)\ .$$ From this we can infer that there is an $R'\geq R$ with $$\bigl|f(z)\bigr|>|z|^n{|a_n|\over2}>|a_0|\qquad\bigl(|z|>R'\bigr)\ .\tag{1}$$ On the compact set $\overline{D_{R'}}$ the continuous function $z\mapsto\bigl|f(z)\bigr|$ assumes a global minimum $\>\leq|f(0)|=|a_0|$. Due to $(1)$ this minimum value is not assumed when $|z|>R'$, hence is the global minimum of $|f|$ on ${\mathbb C}$.