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

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.

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).

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).