Formal Book

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 \(C = \{ z \in \mathbb {C} : |f(z)| \leq 2 \} \) and let \(\mathcal{R}\) 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 \(\mathcal{R}\) and satisfy

\[ \ell (I_1) + \cdots + \ell (I_t) \leq 4. \]
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 \(\mathcal{P} = \{ x \in \mathbb {R} : |p(x)| \leq 2\} \) 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

TODO

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

TODO

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 \(b_1 {\lt} \cdots {\lt} b_r\) be the roots of \(p(x)\) with multiplicities \(s_1, \ldots , s_r\), \(\sum _{j=1}^{r} s_j = n\). 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 \(b_3, \ldots \), 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 \(x \in \mathbb {R}\).

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

\[ p'(x) = \sum _{k=1}^{n} \frac{p(x)}{x - a_k}, \quad \text{that is,} \quad \frac{p'(x)}{p(x)} = \sum _{k=1}^{n} \frac{1}{x - a_k}. \]

Differentiating this again we have

\[ \frac{p''(x)p(x) - p'(x)^2}{p(x)^2} = - \sum _{k=1}^{n} \frac{1}{(x - a_k)^2} {\lt} 0. \]