Lemma 1 -- coefficient-sum geometric dichotomy

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Statement

Let $0<\alpha<\pi/2$, let $b_1,\ldots,b_{n-1}\in\mathbb C$, and put

$$A_0=1, \qquad A_k=1+b_1+\cdots+b_k.$$

For every $1\leq k\leq n-1$, at least one of the following inequalities holds:

$$|A_{k-1}-kb_k|\geq \sin\alpha\,|A_{k-1}|, \tag{1}$$

or

$$|A_k|\geq |A_{k-1}|+\cos\alpha\,|b_k|. \tag{2}$$

Moreover, if (2) holds for every $1\leq k\leq s$, where $s\leq n-1$, then

$$|A_s|>\cos\alpha \left(1+|b_1|+\cdots+|b_s|\right). \tag{3}$$

Proof of the dichotomy

Fix $k$ and abbreviate $u=A_{k-1}$ and $v=b_k$. If $u=0$, then (1) is automatic and (2) follows from $1\geq\cos\alpha$. If $v=0$, then (1) and (2) are both automatic. We may therefore suppose that $u,v\neq0$.

Let $\theta\in[0,\pi]$ be the smaller angle between the two vectors $u$ and $v$ in the complex plane. Suppose first that $\theta\leq\alpha$. Direct expansion gives

$$\begin{aligned} |u+v|^2- \left(|u|+\cos\alpha\,|v|\right)^2 ={}&\sin^2\alpha\,|v|^2\\ &+2|u||v|(\cos\theta-\cos\alpha)\geq0. \end{aligned}$$

Taking nonnegative square roots yields (2).

Now suppose that $\theta>\alpha$. If $\theta<\pi/2$, resolve $u-kv$ parallel and perpendicular to the line through $v$:

$$|u-kv|^2 =\left(k|v|-|u|\cos\theta\right)^2 +|u|^2\sin^2\theta >|u|^2\sin^2\alpha.$$

If instead $\theta\geq\pi/2$, then $\operatorname{Re}(u\overline v)\leq0$, and hence

$$|u-kv|^2 =|u|^2+k^2|v|^2-2k\operatorname{Re}(u\overline v) \geq |u|^2 >|u|^2\sin^2\alpha.$$

Thus (1) holds whenever $\theta>\alpha$, completing the dichotomy.

Iteration

If (2) holds for $k=1,\ldots,s$, repeated application starting from $|A_0|=1$ gives

$$|A_s|\geq1+\cos\alpha \left(|b_1|+\cdots+|b_s|\right).$$

Because $0<\cos\alpha<1$, the right-hand side is strictly greater than

$$\cos\alpha \left(1+|b_1|+\cdots+|b_s|\right),$$

which proves (3).

Source scope

This is Lemma 1 on printed pp. 210--211, physical PDF pp. 224--225, of the published volume scan. The page's (1) and (2) are the print's (3) and (4), and its (3) is the print's unlabeled closing assertion. In the print the $b_k$ are the coefficients of the polynomial with roots $z_2,\ldots,z_n$ fixed in the proof of Theorem 1; the argument uses no property of them, so the lemma is recorded here for arbitrary complex $b_1,\ldots,b_{n-1}$. The print also notes (p. 211) that the geometric step can be replaced by its Lemma 2 applied with $A=k$ and $z=b_k/(1+b_1+\cdots+b_{k-1})$. The source summarizes the first part as an elementary geometric consideration; the squared-distance calculations above make both angular regions and all degenerate cases explicit.

Used by. [[analysis/biro_1994_problem_turan_concerning_sums_powers_complex/theorem_1|Theorem 1]].

Bears on. Problem 519.