Bound for the modulus of a complex number obtained from approximation of products

Question

I am currently working with a paper of Denef where he claims that

If $a_1, …, a_n$ are complex numbers and there exists a constant $c ≥ 1$ such that $$\prod_{i = 1}^n |a_i + j| ≤ c$$ for all integers $0 ≤ j ≤ n - 1$, then $|a_i| < 2^n c$.

He gives the hint to consider two cases:

1. $∃ j_0 : ∀ i : |a_i + j_0| ≥ \frac{1}{2}$ and
2. $∀ j : ∃ i_j : |a_{i_j} + j| < \frac{1}{2}$.

In the first case I deduced that $$ c ≥ \prod_{i = 1}^n |a_i + j_0| ≥ 2^{-n}, $$ which can be transformed to $$2^n c ≥ \prod_{i = 1}^n (\underbrace{|a_i + j_0|2}_{≥ 1}) ≥ 2 |a_i + j_0|.$$

But I don't know how to proceed from here. A second approach was to consider the product as the factorisation of a polynomial in $j$. Then one can obtain a Vandermonde-style linear system for the coefficients of this polynomial and derive an upper bound for their modulus of the form $λc$. But $λ$ depends on the $a_i$.

Answer 1

The hint number 2 seems to me uncorrect, if the claim holds for any set of complex numbers $a_1, \ldots, a_n$. Say $n>5$ and one chooses j=5, then why should be always an $a_{i_j}$ that makes $|a_{i_j} + j| < 1/2$ correct? By the way, further manipulating your last inequality in the case of all $a_i$ satisfies $|a_i+j| \ge 1/2$ and $|a_i+j| \ge |a_i|$, you get: \begin{equation} 2^n c \ge 2 |a_i + j_0| \ge 2 |a_i| \gt |a_i|. \end{equation} But this is valid only in this case. In general, the product for a particular $j$ ($n>1$, $n=1$ is triv.) could be separated for three factors. One, which contains the product of such $a_i$, for which $|a_i+j| \ge 1/2$ and $|a_i+j|>|a_i|$, another which contains $a_i$s with $|a_i+j|\ge 1/2$, but $|a_i+j|<|a_i|$ and the last in which $|a_i+j|<1/2$ and $|a_i+j|<|a_i|$.

Answer 2

I present a nearly complete solution to my question.

Note that if $0 ≤ j_1, j_2 ≤ n - 1$ have the property that $|a_i + j_k| < \frac{1}{2}$ then $$|j_1 - j_2| ≤ |a_i + j_1| + |a_i + j_2| < \frac{1}{2} + \frac{1}{2} = 1$$ and thus $j_1 = j_2$.

I will argue by induction on $n$ and notice that the claim is trivial for $n = 1$. Thus, we may assume that $n > 1$.

Let us assume that the second case is true, i.e. we have that $$∀ j \;∃ i_j : |a_{i_j} + j| < \frac{1}{2}.$$ Since we allow only $n$ values for both $j$ and $i_j$ and by the observation two different values of $j$ must have two different witnesses $i_j$, we find that there must exist a bijective function $f: \{0, …, n- 1\} → \{1, …, n\}$ with the property that $$|a_{f(j)} + j| < \frac{1}{2}$$ and for all indices $j' ≠ j$ we must have $$|a_{f(j)} + j'| ≥ \frac{1}{2}.$$

Set $i_0 := f(n - 1)$ then we know for all $0 ≤ j ≤ n - 2$ that $|a_{i_0} + j| ≥ 1/2$ and therefore that $$c ≥ \prod_{i = 1}^n |a_i + j| ≥ \frac{1}{2} \prod_{\substack{i = 1\\i ≠ i_0}}^n |a_i + j|$$

Thus, $a_1, …, a_{i_0 - 1}, a_{i_0 + 1}, …, a_n$ satisfy the assumption of the induction hypothesis (w.r.t. $c' := 2c$) and hence $$|a_i| < 2^{n - 1} 2 c.$$

To reach a contradiction assume that $|a_{i_0}| ≥ 2^n c$ then $$\frac{1}{2} > |a_{i_0} + {n - 1}| ≥ |a_{i_0}| - |n - 1| ≥ 2^n c - n + 1 ≥ 2^n - n + 1 ≥ 2.$$ Thus, concluding the second case.

As for the first case, we assume that $$∃ j_0 \; ∀ i : |a_i + j_0| ≥ \frac{1}{2}.$$ Then, as I have already mentioned we have for all $i ≤ k ≤ n$ that $$2^nc ≥ \prod_{i= 1}^n \underbrace{2|a_i + j_0|}_{≥ 1} ≥ 2|a_k + j_0|.$$ Now if we can choose $j_0 = 0$, we are done. Otherwise, there exists an index $i_0$ such that $|a_{i_0}| < 1/2$. But then $|a_{i_0} + j| > 1/2$ for all $1 ≤ j ≤ n - 1$. Now we have that $$c ≥ \prod_{i = 1}^n |a_i + j| ≥ \frac{1}{2} \prod_{\substack{i = 1\\i ≠ i_0}}^n |a_i + j|$$ is satisfied for all $1 ≤ j ≤ n - 1$. Set $b_i := a_i + 1$ for all $i ≠ i_0$. Then we can use the induction hypothesis on the $b_i$ to obtain that $$|a_i + 1| = |b_i| < 2^n c.$$ Hence, $|a_i| < 2^n c + 1$ which is a bit weaker than the claimed result.

Does anyone see how to get rid of the trailing $+1$?