Proof that a monic polynomial with $|F(0)| \ge 1$ has max modulus at least 2 on the unit circle

Question

Theorem: Let $F(w)$ be a monic polynomial of degree $m \ge 1$. If its constant term $F(0)$ satisfies $|F(0)| \ge 1$, then its maximum modulus on the unit circle must be at least 2. That is, $\max_{|w|=1} |F(w)| \ge 2$.

My Attempted Proof:

The proof proceeds by contradiction.

1. Let $M = \max_{|w|=1} |F(w)|$. Assume $M < 2$.
2. Choose a complex number $\lambda$ such that $M < |\lambda| < 2$.
3. Consider the polynomial $G(w) = F(w) - \lambda$. Since $|F(w)| \le M < |\lambda|$ on the unit circle $|w|=1$, $G(w)$ has no roots on the unit circle.
4. Let $N$ be the number of roots of $G(w)$ inside the unit circle, and $N^*$ be the number of roots of its reciprocal polynomial $G^*(w) = w^m \overline{G(1/\bar{w})}$ inside the circle. Since there are no roots on the circle, we have the relation $N + N^* = m$.
5. We can find $N^*$ using Rouché's Theorem on $G^*(w) = F^*(w) - \bar{\lambda}w^m$. On the unit circle, we have $|F^*(w)| = |F(w)| \le M$ and $|-\bar{\lambda}w^m| = |\lambda|$. Since $M < |\lambda|$, the term $-\bar{\lambda}w^m$ dominates. $G^*(w)$ must therefore have the same number of roots inside the circle as $-\bar{\lambda}w^m$, which is $m$. So, $N^* = m$.
6. From the counting formula, this implies $N=0$. So, all $m$ roots of $G(w)$ must lie outside the unit circle ($|w_k|>1$).
7. By Vieta's formulas, the product of these roots has magnitude $|w_1 \cdots w_m| = |F(0) - \lambda|$.
8. Since all roots are outside the circle, we must have $|w_1 \cdots w_m| > 1$. This implies the condition $|F(0) - \lambda| > 1$ must hold for any $\lambda$ in the annulus $M < |\lambda| < 2$.

Here is where I'm stuck. This conclusion must lead to a contradiction with the premise that $|F(0)| \ge 1$.

The contradiction is clear if $|F(0)|=1$, because the disk $\{\lambda : |\lambda-F(0)|\le 1\}$ contains points with modulus up to 2, so it must intersect the annulus $\{\lambda : M < |\lambda| < 2\}$.

However, if $|F(0)|$ is very large (e.g., $|F(0)|=4$), then the disk $\{\lambda : |\lambda-4|\le 1\}$ and the annulus $\{\lambda : M < |\lambda| < 2\}$ are disjoint. In this case, no contradiction seems to arise from this line of argument.

My Question:

How can this proof be completed to cover all cases where $|F(0)| \ge 1$? Is there a different, more direct way to obtain the contradiction from the statement in step 8? Or is a completely different proof of the theorem required?

Thank you!

Answer 1

Another approach with Rouche:

suppose for contradiction that $\max_{\vert w\vert =1} \big\vert F(w)\big\vert \lt 2$
$Q(w):= 2\cdot w^m - F(w)$ has all $m$ roots $\big\{\lambda_k\big\}$ in the open unit disc per Rouche yet
$1\gt \big\vert \prod_{k=1}^m \lambda_k\big\vert = \big\vert Q(0)\big\vert =\big\vert F(0)\big\vert \geq 1$, a contradiction

Answer 2

Here is a proof completely different from the Rouché's Theorem approach outlined in the question.

Let $F(w) = w^m + a_{m-1}w^{m-1} + \dots + a_1w + a_0,M = \max_{|w|=1} |F(w)|$. Our goal is to show that $M \ge 2$.

1. Construct a Symmetric Set of Points

Let $\sigma = a_0 / |a_0|$. This is the complex number on the unit circle that points in the same direction as the constant term $a_0$.

Now, consider the $m$-th roots of $\sigma$. Let $\eta_0, \eta_1, \dots, \eta_{m-1}$ be these $m$ roots. These are $m$ equally spaced points on the unit circle.

2. Sum the Polynomial Over These Points

Let's evaluate the sum of the values of $F(w)$ at these $m$ points. $$\sum_{k=0}^{m-1} F(\eta_k) = \sum_{k=0}^{m-1} \left(\eta_k^m + a_{m-1}\eta_k^{m-1} + \dots + a_1\eta_k + a_0\right)$$ By rearranging the terms, we get: $$\sum F(\eta_k) = \left(\sum \eta_k^m\right) + a_{m-1}\left(\sum \eta_k^{m-1}\right) + \dots + a_1\left(\sum \eta_k\right) + \left(\sum a_0\right)$$ 3. Simplify the Sum

The sums of the powers of these roots have a special property. Because the points $\eta_k$ are the $m$-th roots of $\sigma$:

• For $p=m$: $\eta_k^m = \sigma$. The sum $\sum_{k=0}^{m-1} \eta_k^m$ is simply $\sigma$ added to itself $m$ times, which is $m\sigma$.
• For $1 \le p < m$: The sum of the $p$-th powers of the $m$-th roots of any number is $0$.

Applying these results, every term in our main sum vanishes except for the first and the last: $$\sum F(\eta_k) = (m\sigma) + a_{m-1}(0) + \dots + a_1(0) + (ma_0) = m(\sigma + a_0)$$ 4. Apply the Triangle Inequality

Now we have an exact expression for the sum. We can take the magnitude of both sides and then apply the triangle inequality: $$\sum_{k=0}^{m-1} |F(\eta_k)| \ge m|\sigma + a_0|$$ By the definition of $M$, we know that $|F(\eta_k)| \le M$ for every $k$. Therefore, the sum of these $m$ terms is at most $mM$: $$mM \ge m|\sigma + a_0|$$ Dividing both sides by $m$, we get: $$M \ge |\sigma + a_0|=1 + |a_0| \ge 1 + 1 = 2$$ This completes the proof.