23 A theorem of Pólya on polynomials

Theorem 23.1

Let $f (z)$ be a complex polynomial of degree at least 1 and leading coefficient 1. Set $C = {z \in C : | f (z) | \leq 2}$ and let $R$ be the orthogonal projection of $C$ onto the real axis. Then there are intervals $I_{1}, \dots, I_{t}$ on the real line which together cover $R$ and satisfy

ℓ (I_{1}) + \dots + ℓ (I_{t}) \leq 4.

Proof ▶

Theorem 23.2

Let $p (x)$ be a real polynomial of degree $n \geq 1$ with leading coefficient 1, and all roots real. Then the set $P = {x \in R : | p (x) | \leq 2}$ can be covered by intervals of total length at most 4.

Proof ▶

Corollary 23.3

Let $p (x)$ be a real polynomial of degree $n \geq 1$ with leading coefficient 1, and suppose that $| p (x) | \leq 2$ for all $x$ in the interval $[a, b]$ . Then $b - a \leq 4$ .

Proof ▶

23.1 Appendix: Chebyshev’s theorem

Theorem 23.4 Chebyshev’s theorem

Let $p (x)$ be a real polynomial of degree $n \geq 1$ with leading coefficient 1. Then

max_{- 1 \leq x \leq 1} | p (x) | \geq \frac{1}{2^{n - 1}} .

Proof ▶

Theorem 23.5 Fact 1

If $b$ is a multiple root of $p^{'} (x)$ , then $b$ is also a root of $p (x)$ .

Proof ▶

Let $b_{1} < \dots < b_{r}$ be the roots of $p (x)$ with multiplicities $s_{1}, \dots, s_{r}$ , $\sum_{j = 1}^{r} s_{j} = n$ . From $p (x) = (x - b_{j})^{s_{j}} h (x)$ we infer that $b_{j}$ is a root of $p^{'} (x)$ if $s_{j} \geq 2$ , and the multiplicity of $b_{j}$ in $p^{'} (x)$ is $s_{j} - 1$ . Furthermore, there is a root of $p^{'} (x)$ between $b_{1}$ and $b_{2}$ , another root between $b_{2}$ and $b_{3}, \dots$ , and one between $b_{r - 1}$ and $b_{r}$ , and all these roots must be single roots, since $\sum_{j = 1}^{r} (s_{j} - 1) + (r - 1)$ counts already up to the degree $n - 1$ of $p^{'} (x)$ . Consequently, the multiple roots of $p^{'} (x)$ can only occur among the roots of $p (x)$ .

Theorem 23.6 Fact 2

We have $p^{'} (x)^{2} \geq p (x) p^{″} (x)$ for all $x \in R$ .

Proof ▶

If $x = a_{i}$ is a root of $p (x)$ , then there is nothing to show. Assume then $x$ is not a root. The product rule of differentiation yields

p^{'} (x) = \sum_{k = 1}^{n} \frac{p (x)}{x - a_{k}}, that is, \frac{p^{'} (x)}{p (x)} = \sum_{k = 1}^{n} \frac{1}{x - a_{k}} .

Differentiating this again we have □

\frac{p^{″} (x) p (x) - p^{'} (x)^{2}}{p (x)^{2}} = - \sum_{k = 1}^{n} \frac{1}{(x - a_{k})^{2}} < 0.