# Analytic Combinatorics: upper bound for sum of absolute values of two complex functions: $|z f'(z)| + |2 f(z) - zf'(z)| \leq 2f(|z|)$

Let $f \colon \mathbb C \to \mathbb C$ be a complex-valued analytic function with non-negative coefficients of Taylor series at 0 (suppose that radius of convergence is $+\infty$ for simplicity): $$f(z) = c_0 + c_1 z + c_2 z^2 + \ldots \enspace ,$$ $c_k \geq 0$ for $k = 0, 1, 2, \ldots$, and also $c_1 > 0$.

It is well-known that for such functions their characteristic function $\phi_0(z) = \dfrac{z f'(z)}{f(z)}$ is nondecreasing when $z \in \mathbb R_+$. We also introduce characteristic function of the derivative $\phi_1(z) = \dfrac{z f''(z)}{f'(z)}$ which is also nondecreasing when $z \in \mathbb R_+$. So, there exists unique real positive solution $\widehat z$ of the equation $\phi_1(\widehat z) = 1$, since $c_1 > 0$ and $\phi_1(0) = 0$, under some mild conditions (there should be at least one nonzero coefficient among $c_3, c_4, \ldots$).

Conjecture. If $0 < |z| < \widehat z$, then it holds $$|\phi_0(z)| + |2 - \phi_0(z)| \leq 2 \dfrac{f(|z|)}{|f(z)|} \enspace .$$ Moreover, if the function $f(z)$ is aperiodic in the sense $$\mathrm{GCD}\{i - j \colon c_i \neq 0, \, c_j \neq 0\} = 1$$ then equality holds only iff $z$ is a real positive, i.e. $z = |z|$.

Hints. I should mention some of my previous attempts on this conjecture. It is important to mention that if $z = |z| < \widehat z$ then $\phi_0(z) \leq 2$. The equality is attained only for some degenerate functions $f(z)$. The sketch of the proof is following. The equation $\phi_1(\widehat z) = 1$ is equivalent to $$c_1 = 3 c_3 z^2 + 8 c_4 z^3 + \ldots + (k-1)(k+1) c_{k+1}z^k + \ldots \enspace ,$$ whereas the inequality $\phi_0(z) \leqslant 2$ is equivalent to $$2 c_0 + c_1 z \geq c_3 z^3 + 2 c_4 z^4 + \ldots + (k-2) c_k z^k + \ldots \enspace ,$$ which is true because all the coefficients are nonnegative, and $(k-1)(k+1) \geq k-2$.

Some obvious tricks with triangle inequalities didn't work in my case, because there is an obvious lower bound $$2 \leq |\phi_0(z)| + |2 - \phi_0(z)| \enspace .$$ I also tried to prove equivalent inequality $$|z f'(z)| + |2 f(z) - z f'(z)| \leq 2 f(z_0) \enspace ,$$ which seems easier, because of explicit representation for $2f(z) - zf'(z)$. If we denote $z = z_0 e^{i \theta}$, then the function $$\Phi(\theta) = |z f'(z)| + |2f(z) - zf'(z)|$$ can have a plot with large number of local maxima and minima. I plot first summand (1), second summand (2) and lower bound $|2f(z)|$ (3). The sum (1+2) is also depicted. We can see that the behavior is "very tight".

Picture: Separate summands of $\Phi(\theta)$.

Every strategy I tried, didn't seem to give hopeful result, though I tried to simplify and consider particular cases like $f(z) = e^z$, $f(z) = \sinh z$. It is complicated to work with absolute value, since it is not analytic. The idea of $|\omega| = \sqrt{\omega \overline \omega}$ also didn't lead me to the solution, however maybe I was not patient enough to finish that, i mean consider square root of infinite sums. I also tried to simulate the inequality in ipython for many dfferent sequences, and it always turned out valid.

Motivation. This lemma is important in graph enumeration, because the conjecture leads to some saddle-point estimates for asymptotic number of graphs with certain structure. In particular, the generating function of graph with $n$ unrooted trees, after some transformation, takes the form $g(z)e^{nh(z)}$, and this is the contour integral we want to calculate. Usually, $\Re h(z)$ on a circle $z = z_0 e^{i\theta}$ has maximal value at $z = z_0$, i.e. $\theta = 0$.

Any helpful ideas are appreciated!

## 1 Answer

I think, it is true and follows from the following elementary

Lemma. If $\alpha>\beta>0$ and $1/\beta-1/\alpha\geqslant 2$, then $|1+\alpha z|+|1-\beta z|\leqslant 2+\alpha-\beta$ for any vector $z$ with $|z|=1$.

Proof of the lemma. Some nice argument must exist, but you may simply denote $z=e^{it}$ and differentiate LHS in $t$. For real $z$ the inequality is clear, else in a point where derivative equals 0 you get $\alpha/|1+\alpha z|=\beta/|1-\beta z|$, $|z+1/\alpha|=|z-1/\beta|$, but perpendicular bisector to the segment $[-1/\alpha,1/\beta]$ does not contain non-real points of a unit circle provided that $1/\beta-1/\alpha\geqslant 2$.

Now we have $c_1\geqslant \sum_{k\geqslant 2} (k^2-1)c_{k+1}|z|^k$ and need to prove $$|c_1z+2c_2z^2+3c_3z^3+\dots |+|2c_0+c_1z-c_3z^3-2c_4z^4-\dots|\leqslant\\ 2c_0+2c_1|z|+2c_2|z|^2+\dots,$$ for proving that at first we use triangle inequality for removing terms with $c_0$ and $c_2$, then divide by $|z|$, we reduce our inequality to $$|c_1+3c_3z^2+\dots |+|c_1-c_3z^2-2c_4z^3-\dots|\leqslant 2c_1+2c_3|z|^2+\dots,$$ set $c_1=a+3c_3|z|^2+8c_4|z|^3+\dots$ for some non-negative $a$ and use triangle inequality many times, we reduce the inequality to a family of inequalities $$|(k^2-1)|z|^k+(k+1)z^k|+|(k^2-1)|z|^k-(k-1)z^k|\leqslant (2(k^2-1)+2)|z|^k$$ (this inequality goes with coefficient $c_{k+1}$), but this is a partial case of lemma with $\alpha=1/(k-1),\beta=1/(k+1)$.

• That's amazing, thank you for the proof! I understand now your idea of pulling the terms from the triangle inequality: I couldn't guess to push the condition $\phi_1(|z|) \leq 1$ in terms of coefficient $c_1 \geq \ldots$, now it is clear. As for the lemma, I also couldn't come with the idea of how to bound the sum of two absolute values. It is still non-obvious to me, in geometrical sense (I tried to re-prove it using inversion but also didn't succeed). Would be nice if you explain how did you guess the lemma :) – Sergey Dovgal Jun 14 '16 at 13:48
• Partial case of the lemma arises when $f$ has only two non-zero coefficients, say $c_1 z+c_5 z^5$. So it arises naturally (of course, I did not know a priori what are necessary conditions for $\alpha,\beta$, it is a luck that they are exactly what we are given. But probably not just a luck.) – Fedor Petrov Jun 14 '16 at 17:27