Polynomials such that $|p(z)|\leq p(|z|)$

Question

Let $p(x)=1+p_1x+p_2x^2+\cdots+p_nx^n$ be a polynomial with real coefficients and no positive zeros. Define $$\mu(x)=\frac{xp'(x)}{p(x)}, \hspace{3ex} \sigma(x)=x \mu'(x).$$ Many years ago, as part of my PhD dissertation (in 1990 or 1991) I proved this result: For each $r>0$, $\sigma(r)>0$ if and only if there is a neighborhood $V$ of $r$ in ${\mathbb C}\setminus \lbrace 0\rbrace $ such that $|p(z)|\leq p(|z|)$ for all $z\in V$. My proof was elementary but rather complicated. Now I am revisiting this material and I wonder if this result is found somewhere else, and if it can be proved in a simple way, perhaps within a context that I do not know about.

Answer 1

The claim holds with $p$ replaced by any positive real analytic function on $(0,+\infty)$ that is not a monomial function $z \mapsto C z^\alpha$ (which also satisfies $|p(z)| \leq p(|z|)$, but for which $\sigma$ vanishes identically).

Fix $r>0$. The appearance of the scaling vector field $x \frac{d}{dx}$ indicates that an logarithmic change of variables will be useful. Accordingly, write $p(r e^u) = e^{f(u)}$, then $f$ is real analytic on ${\bf R}$ with $\mu(r e^u) = f'(u)$ and $\sigma(r e^u) = f''(u)$. We can then extend $f$ holomorphically to a neighborhood of the origin. $f$ is not linear, since otherwise $p$ would be a monomial function.

If $f''(0) = \sigma(r)$ is positive, then from the Taylor expansion $f(u) = f(0) + f'(0) u + \frac{1}{2} f''(0) u^2 + O(|u|^3)$ for small $u$ we have $$ \frac{d}{db} \mathrm{Re} f(a+ib)|_{b=0} = 0; \quad \frac{d^2}{db^2} \mathrm{Re} f(a+ib)|_{b=0} < 0$$ for all small $a$, hence $$ \mathrm{Re} f(u) \leq f( \mathrm{Re} u ) \tag{1}$$ for small $u$, which is equivalent to $|p(z)| \leq p(|z|)$ for $z$ near $r$. If $f''(0)$ is negative, the same argument shows that the estimate $|p(z)| \leq p(|z|)$ fails. The remaining case is if $f''$ vanishes to some positive order $m$ at the origin, then Taylor expansion gives $$ f(u) = f(0) + f'(0) u + c u^{2+m} + O( |u|^{3+m})$$ for some real non-zero $c$. Setting $u = \varepsilon e^{i\theta}$ in (1) and extracting out the $\varepsilon^{2+m}$ component then gives $$ c\cos( (2+m) \theta ) \leq c\cos^{2+m}(\theta),$$ which is absurd by setting $\theta = \frac{2\pi}{2+m} \in (0,\pi)$ (if $c$ is positive) or $\theta = \frac{\pi}{2+m} \in (0,\pi)$ (if $c$ is negative).

Answer 2

The result proved in the answer of @Terry Tao is actually due to Teichmuller, see, for example

L. Ahlfors, Conformal invariants, section 3-4.

For an interesting related result, see

MR0344465 Boĭčuk, V. S.; Golʹdberg, A. A. On the three lines theorem. (Russian) Mat. Zametki 15 (1974), 45–53.

And references there.

English translation: Math. Notes 15, 26-30 (1974). (Free review in Zbl 0289.30034).