Proceedings of the Steklov Institute of Mathematics, Vol. 218, 1997, pp. 53–68. Translated from Trudy Matematicheskogo Instituta imeni V. A. Steklova, Vol. 218, 1997, pp. 58–73. Original Russian Text Copyright \begin{matrix}c \\ ◯ \end{matrix} 1997 by Bernik, Borbat. English Translation Copyright \begin{matrix}c \\ ◯ \end{matrix} by ���� �����/Interperiodica Publishing (Russia).

Simultaneous Approximation of Zero by Values of Integral Polynomials

V. I. Bernik and V. N. Borbat

Received February, 1997

The direction of studies formulated in the title was indicated by the works of Sprindzhuk [6, 7] in which he posed the basic problem of this direction, solved particular cases of the problem, and pointed out some applications. The conjectures by V. G. Sprindzhuk were proved in 1980 [3]. Later various generalizations and applications were obtained [4, 9], but the method itself does not allow one in principle to pass from a power function on the right-hand side of the inequality to a logarithmic one; this does not allow one in principle to obtain many refined characteristics of classical sets appearing in the theory of transcendental numbers.

In this paper, we obtain a two-dimensional analog of the theorem in [3]; this analog can be considered as a proof of a two-dimensional generalization of the conjecture by Baker [8], which was proved in [1].

Suppose that P (x) = a_{n}x^{n} + a_{n−1}x^{n−1} + ⋯ + a_{1}x + a_{0} is a polynomial with integer coefficients, H = H(P) = max_{0≤i≤n} |a_{i}| is the height of P (x), the function Ψ(x) monotonically decreases for x > 0, and ∑_{H=1}^{∞} Ψ(H) < ∞.

Theorem. The system of inequalities

\{ \begin{matrix}|P (ω_{1})| < H^{−w_{1}}Ψ^{v_{1}}(H) \\ |P (ω_{2})| < H^{−w_{2}}Ψ^{v_{2}}(H), \end{matrix}  (1)

where w_{1} +w_{2} = n−2, v_{1} +v_{2} = 1, has only a finite number of solutions in polynomials P (x) ∈ ℤ[x] for almost all (w_{1}, w_{2}) ∈ ℝ^{2}.

We introduce some notation and give several lemmas, which are necessary for further arguments. Denote by c(n) positive functions depending only on n. We shall realize operations over c(n) by the formal rules

c(n) + c(n) = c(n), c(n)c(n) = c(n),

the sense of these rules consists in the fact that the sum and the product are again a certain function depending on n. Since in the theorem one speaks on the finiteness or infiniteness of the number of solutions to the system of inequalities (1), we shall assume that H > H_{0}, where H_{0} is a sufficiently large positive integer. Further, we shall assume that ω_{1} and ω_{2} both are transcendental numbers since the measure of (ω_{1}, ω_{2}) ∈ ℝ^{2} for which at least one of the numbers ω_{1} and ω_{2} is algebraic is equal to zero.

Lemma 1. Let P_{1}(x), …, P_{l}(x) ∈ ℝ[x] be polynomials and let P (x) = P_{1}(x)P_{2}(x) ⋯ P_{l}(x). Then

c_{1}(n)H(P_{1})H(P_{2}) ⋯ H(P_{l}) < H(P) < c_{2}(n)H(P_{1})H(P_{2}) ⋯ H(P_{l}).

Lemma 1 is proved, for example, in [7].

Lemma 2. Let G ⊂ ℝ^{2} be some bounded domain and let B ⊂ G be a measurable set in the plane, μB > c_{1}(n)μG, where μB and μG are the Lebesgue measures of the sets B and G in ℝ^{2}. Let further the inequality

|P (ω_{1})P (ω_{2})| < H^{−w}

with deg P (x) ≤ n hold for (ω_{1}, ω_{2}) ∈ B. Then the inequality

|P (ω_{1})P (ω_{2})| < c_{2}(n)H^{−w}

holds for all (ω_{1}, ω_{2}) ∈ G.

Lemma 2 is an analog of Lemma 10 in [2] and is proved with the help of the Lagrange interpolation formula.

Denote by P_{n}(H) the class of irreducible polynomials with the condition a_{n}(P) = H. Suppose that P_{n} = ⋃_{H=1}^{∞} P_{n}(H). Let P (x) ∈ P_{n}(H) and let κ_{1}, κ_{2}, …, κ_{n} be roots of this polynomial. We shall assume that the roots κ_{1}, κ_{2}, …, κ_{n} are ordered in such a way that Re κ_{1} ≤Re κ_{2} ≤⋯≤Re κ_{n}. In the case of the equality Re κ_{l} = Re κ_{t}, we shall previously write the root whose modulus of the imaginary part is less than that of the other ones, and in the case where the moduli of the imaginary parts are equal, we put previously the root whose imaginary part is positive. Choose any two roots κ_{11} and κ_{21}. We order all other roots with respect to each of these roots in the following way:

|κ_{11} − κ_{12}| ≤ |κ_{11} − κ_{13}| ≤ ⋯ ≤ |κ_{11} − κ_{1n}|, |κ_{21} − κ_{22}| ≤ |κ_{21} − κ_{23}| ≤ ⋯ ≤ |κ_{21} − κ_{2n}|.

Introduce the following notation:

|κ_{11} − κ_{1i}| = H^{−μ_{i}}, i = 2, …, n, |κ_{21} − κ_{2s}| = H^{−Θ_{s}}, s = 2, …, n.

Fix ε. Set ε_{1} = εd^{−1}, where d = d(n) is a sufficiently large value. Set T = [ε_{1}^{−1}]. Define integers l_{i} and s_{i} by the inequalities

\frac{l_{i} − 1}{T} ≤ μ_{i} < \frac{l_{i}}{T} , i = 2, …, n, \frac{s_{i} − 1}{T} ≤ Θ_{i} < \frac{s_{i}}{T} , i = 2, …, n.

We shall assign an integral vector \overline{s}_{1,2} = \overline{s} = (l_{2}, …, l_{n}, s_{2}, …, s_{n}) to a fixed pair of roots (κ_{11}, κ_{21}) of the polynomial P (x). All polynomials P (x) ∈ P_{n}(H) having one and the same vector \overline{s} are joined into the class P_{n}(H, \overline{s}). Let

p_{i} = \frac{l_{i+1} + ⋯ + l_{n}}{T} , q_{i} = \frac{s_{i+1} + ⋯ + s_{n}}{T} , i = 1, …, n − 1,

S(κ_{11}) = {ω_{1} ∈ ℝ : min_{1≤l≤n} |ω_{1} − κ_{l}| = |ω_{1} − κ_{11}|},

S(κ_{21}) = {ω_{2} ∈ ℝ : min_{1≤s≤n} |ω_{2} − κ_{s}| = |ω_{2} − κ_{21}|}.

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS  Vol. 218  1997

Lemma 3. The inequality

|P (ω_{1})P (ω_{2})| < H^{−n+2−δ}  (2)

has only a finite number of solutions in reducible integral polynomials of degree not exceeding n for any δ > 0 and almost all (ω_{1}, ω_{2}) ∈ ℝ^{2}.

Proof. We shall consider ω_{1} ∈ S(κ_{11}) and ω_{2} ∈ S(κ_{21}). Denote by S(n) the set of (ω_{1}, ω_{2}) for which inequality (2) holds infinitely many times. Let M(H) be the set of polynomials P (x) satisfying the condition H(P) = H. Set M_{t} = ⋃_{2^{t}≤H<2^{t+1}} M(H), M = ⋃ M_{t}, t > c(n) ln H_{0}(δ). If inequality (2) for P (x) ∈ M_{t} holds for all (ω_{1}, ω_{2}) in a certain domain S(P), then by Lemma 1, there exists a polynomial P_{1}(x) which divides P (x) and which satisfies the inequality

|P_{1}(ω_{1})P_{1}(ω_{2})| < c(n)H(P_{1})^{−n+2−δ},  deg P_{1} ≤ n − 1,  (3)

on the set S_{1}(P), μS_{1}(P) > c(n)μS(P). By Lemma 2, inequality (3) holds for all (ω_{1}, ω_{2}) ∈ S(P), maybe with another value of c(n). Consider three cases appeared.

1. If H(P_{1}) > 2^{tε/(2(n+1)n^{2})}, then using the metric theorem in [3], we obtain that the set S = ⋃_{p∈M} S(P) can be covered by a system of domains whose total measure does not exceed an arbitrary given positive number.

2. If H(P_{1}) ≤ 2^{tε/(2(n+1)n^{2})} and

|P_{1}(ω_{1})P_{1}(ω_{2})| < 2^{−tε/n}  (4)

on the set B_{1}(P), μB_{1}(P) > c(n)μS(P), then again applying Lemma 2, we obtain that

|P_{1}(ω_{1})P_{1}(ω_{2})| < c(n)2^{−tε/n}

for (ω_{1}, ω_{2}) ∈ S(P). Further, we have μS(P) < c(n)2^{−tε/n}. Since the number of polynomials P_{1}(x) with the condition H(P_{1}) ≤ 2^{tε/(2(n+1)n^{2})} does not exceed c(n)2^{tε/(2n^{2})}, we have

∑_{P (x),H(P_{1})≤2^{tε/(2(n+1)n^{2})}} 1 < c(n)2^{−tε/(2n^{2})}.

Since the series ∑_{t} 2^{−tλ} converges for λ > 0, we use the Borel–Cantelli lemma in order to complete the proof.

3. Assume that inequality (4) does not hold on the set B_{1}(P) with the condition μB_{1}(P) > c(n)μS(P). Then the inequality |P_{1}(ω_{1})P_{1}(ω_{2})| > 2^{−tε/n} holds on the set B_{2}(P) = S(P) \ B_{1}(P), μB_{2}(P) > c(n)μS(P). Set P_{2}(x) = P (x)/P_{1}(x). Using Lemmas 1 and 2, we obtain from inequality (2) that

|P_{2}(ω_{1})P_{2}(ω_{2})| < c(n)H^{−n+2−δ−ε/n}, c(n)2^{t(1−ε)/(2(n+1)n^{2})} < H(P_{2}) < 2^{t+1}.

Again, we apply the metric theorem in [3].

Lemma 4. We can consider only polynomials P (x) ∈ P_{n} in the system of inequalities (1).

The passing to polynomials whose leading coefficient is equal to the height is realized in the same way as in [3]. The passing to irreducible polynomials is realized by application of Lemma 3.

Lemma 5. The number of the classes P_{n}(H, \overline{s}) is finite and depends only on n and ε.

Lemma 5 is proved in [3].

Lemma 6. Let P (x) ∈ P_{n}(H) and let ω ∈ S(κ_{11}). Then

|ω − κ_{11}| ≤ 2^{n} \frac{|P (ω)|}{|P′(κ_{11})|},  (5)
|ω − κ_{11}| ≤ min_{2≤j≤n}(2^{n} \frac{|P (ω)|}{|P′(κ_{11})|}|κ_{11} − κ_{12}| ⋯ |κ_{11} − κ_{1j}|)^{1/j}.  (6)

Lemma 6 is proved in [3].

Lemma 7. Let P (x) ∈ P_{n}(H, \overline{s}). Then

|P^{(l)}(κ_{11})| < c(n)H^{1−p_{l}+(n−l)ε_{1}}, l = 1, …, n − 1.

Lemma 7 is proved in [3].

Lemma 8. Let δ > 0 be a certain real number and let n ≥ 2 be a positive integer. Let P (x) and Q(x) be integral relatively prime polynomials of degree at most n and let max(H(P), H(Q)) ≤ H. If for all (ω_{1}, ω_{2}) in a certain rectangle K = I_{1} × I_{2}, |I_{1}| = H^{−η_{1}}, η_{1} > 0, |I_{2}| = H^{−η_{2}}, η_{2} > 0, the inequalities

max(|P (ω_{1}), Q(ω_{1})|) < H^{−τ_{1}}, max(|P (ω_{2}), Q(ω_{2})|) < H^{−τ_{2}}

hold, then

τ_{1} + τ_{2} + 2 + 2(max(τ_{1} + 1 − η_{1}, 0) + max(τ_{2} + 1 − η_{2}, 0)) ≤ 2n + δ.

Lemma 8 is proved in [5].

Lemma 9. Let B(δ, w_{1}, w_{2}, v_{1}, v_{2}) be the set of real vectors \overline{ω} = (ω_{1}, ω_{2}) for which the system of inequalities (1) has an infinite number of solutions in polynomials P (x) ∈ P_{n} with the condition |κ_{i}−κ_{j}| > δ for any i and j and an arbitrary but fixed δ > 0. Then we have μB(δ, w_{1}, w_{2}, v_{1}, v_{2}) = 0 for any δ > 0.

The proof of Lemma 9 is slightly different from that of Lemma 10 in [3].

Lemma 10. Let B_{2}(w_{1}, w_{2}, v_{1}, v_{2}) be the set of \overline{ω} for which the system of inequalities (1) has an infinite number of solutions in polynomials P (x) ∈ P_{2}. Then μB(δ, w_{1}, w_{2}, v_{1}, v_{2}) = 0.

Lemma 10 is proved analogously to Lemma 11 in [3]. As in [3], it is proved that any set of \overline{ω} for which the system of inequalities (1) holds infinitely many times in polynomials P (x) ∈ P_{n} with condition |κ_{11}(P) − κ_{21}(P)| ≤ δ for any δ > 0 has zero measure. Therefore, in what follows, we can assume that |κ_{11}(P) − κ_{21}(P)| > δ for a certain arbitrary but fixed δ > 0.

Denote by L(w_{1}, w_{2}, v_{1}, v_{2}) the set of (ω_{1}, ω_{2}) for which the system of inequalities (1) has an infinite number of solutions in polynomials P (x) ∈ ℤ[x].

We divide the proof of the theorem into nine cases.

Proposition 1. If

l_{2}T^{−1} + p_{1} ≤ w_{1} + v_{1} + 1,  (7)
s_{2}T^{−1} + q_{1} ≤ w_{2} + v_{2} + 1,  (8)
n − 1 + 4nε_{1} < l_{2}T^{−1} + p_{1} + s_{2}T^{−1} + q_{1},

then μL(w_{1}, w_{2}, v_{1}, v_{2}) = 0.

Proof. We use the inequality Ψ(H) < H^{−1} which can easily be obtained. Then system (1) takes the form

\{ \begin{matrix}|P (ω_{1})| < H^{−w_{1}−v_{1}} \\ |P (ω_{2})| < H^{−w_{2}−v_{2}}. \end{matrix}  (9)

Define P_{t}(\overline{s}) = ⋃_{2^{t}≤H<2^{t+1}} P_{n}(H, \overline{s}). Since a_{n} = H and the inequality |κ_{i}| ≤ 1 + H/|a_{n}| is known, we have |κ_{i}| ≤ 2. Then it follows from inequality (2) for j = n that all (ω_{1}, ω_{2}) ∈ ℝ^{2} for which the system of inequalities (9) holds are in the interior of the square [−3; 3] × [−3; 3]. Divide this square into equal rectangles K with sides H^{−η_{1}} and H^{−η_{2}}, η_{1} = w_{1} + v_{1} + 1 − p_{1} − ε_{1}/2, η_{2} = w_{2} + v_{2} + 1 − q_{1} − ε_{1}/2. We shall say that the polynomial P (x) belongs to the rectangle K if there exists (ω_{1}, ω_{2}) ∈ K such that |P (ω_{1})| < H^{−w_{1}−v_{1}}, |P (ω_{2})| < H^{−w_{2}−v_{2}}. Assume that at most one polynomial P (x) ∈ P_{t}(\overline{s}) belongs to each rectangle K. Then we obtain from relation (5) that the measure of the set of ω_{1} ∈ S(κ_{11}) for which the first inequality of system (9) holds does not exceed c(n)2^{−t(w_{1}+v_{1}+1−p_{1})}, and the measure of the set of ω_{2} ∈ S(κ_{21}) for which the second inequality of system (9) holds does not exceed c(n)2^{−t(w_{2}+v_{2}+1−q_{1})}. Therefore, the measure of the set of (ω_{1}, ω_{2}), ω_{1} ∈ S(κ_{11}), ω_{2} ∈ S(κ_{21}) for which the system of inequalities (9) holds does not exceed c(n)2^{−t(n+1−p_{1}−q_{1})}. The number of polynomials does not exceed the number of rectangles K. Hence the measure of the set of (ω_{1}, ω_{2}) for which the system of inequalities (9) holds for at least one polynomial P (x) ∈ P_{t}(\overline{s}) is estimated from above by

c(n)2^{−t(−n−1+p_{1}+q_{1}+w_{1}+v_{1}+1−p_{1}−ε_{1}/2+w_{2}+v_{2}+1−q_{1}−ε_{1}/2)} < c(n)2^{−tε_{1}}.

Since the series ∑_{t=1}^{∞} 2^{−tε_{1}} converges, the proof of the theorem in this case follows from the Borel– Cantelli lemma.

Assume now that there exist rectangles K such that they contain two or more polynomials P (x) ∈ P_{t}(\overline{s}). Let P (x), Q(x) ∈ K; then there exist points (ω_{11}, ω_{21}) and (ω_{12}, ω_{22}) belonging to K such that

max(|P (ω_{11})|, |Q(ω_{12})|) < H^{−w_{1}−v_{1}},  max(|P (ω_{21}), |Q(ω_{22})|) < H^{−w_{2}−v_{2}}.  (10)

Suppose that κ_{11}(P), κ_{21}(P), κ_{11}(Q), and κ_{21}(Q) are the nearest to ω_{11}, ω_{21}, ω_{12}, and ω_{22} roots of the polynomials P (x) and Q(x), respectively. We obtain from relations (6) and (10) that

max(|ω_{11} − κ_{11}(P)|, |ω_{12} − κ_{11}(Q)|) < c(n)H^{−w_{1}−v_{1}−1+p_{1}},
max(|ω_{21} − κ_{21}(P)|, |ω_{22} − κ_{21}(Q)|) < c(n)H^{−w_{2}−v_{2}−1+q_{1}}.

Therefore,

|κ_{11}(P) − κ_{11}(Q)| ≤ |κ_{11}(P) − ω_{11}| + |ω_{11} − ω_{12}| + |ω_{12} − κ_{11}(Q)|
< c(n)(H^{−w_{1}−v_{1}−1+p_{1}} + H^{−η_{1}}) < c(n)H^{−η_{1}},  (11)
|κ_{21}(P) − κ_{21}(Q)| ≤ |κ_{21}(P) − ω_{21}| + |ω_{21} − ω_{22}| + |ω_{22} − κ_{21}(Q)|
< c(n)(H^{−w_{2}−v_{2}−1+q_{1}} + H^{−η_{2}}) < c(n)H^{−η_{2}}.  (12)

Let us estimate the difference |κ_{11}(P)−κ_{1i}(Q)|, i = 2, …, n, taking into account that inequality (7) implies l_{2}T^{−1} − ε_{1} < η_{1}. We have

\begin{aligned}|κ_{11}(P) − κ_{1i}(Q)| & ≤ |κ_{11}(P) − κ_{11}(Q)| + |κ_{11}(Q) − κ_{1i}(Q)| \\ & < c(n)(H^{−η_{1}} + H^{−l_{i}T^{−1}+ε_{1}}) < c(n)H^{−l_{i}T^{−1}+ε_{1}}, \end{aligned}

∏_{i=1}^{n} |κ_{11}(P) − κ_{1i}(Q)| < c(n)H^{−η_{1}−p_{1}+(n−1)ε_{1}}.  (13)

Analogously, taking into account that inequality (8) implies s_{2}T^{−1} − ε_{1} < η_{2}, we obtain

|κ_{21}(P) − κ_{2i}(Q)| < c(n)H^{−s_{i}T^{−1}+ε_{1}},  i = 2, …, n.  (14)

We have from relations (12) and (14) that

∏_{i=1}^{n} |κ_{21}(P) − κ_{2i}(Q)| < c(n)H^{−η_{2}−q_{1}+(n−1)ε_{1}}.  (15)

Similarly, we estimate

∏_{i=1}^{n} |κ_{12}(P) − κ_{1i}(Q)| < c(n) ∏_{i=1}^{n}(|κ_{12}(P) − κ_{11}(P)| + |κ_{11}(P) − κ_{11}(Q)| + |κ_{11}(Q) − κ_{1i}(Q)|)
< c(n)H^{−l_{2}T^{−1}−p_{1}+nε_{1}},  (16)
∏_{i=1}^{n} |κ_{22}(P) − κ_{2i}(Q)| < c(n)H^{−s_{2}T^{−1}−q_{1}+nε_{1}}.  (17)

Since the polynomials P (x) and Q(x) are from P_{t}(\overline{s}), they have no common roots and the modulus of their resultant is |R(P, Q)| ≥ 1. We have from relations (13), (15)–(17) that

1 ≤ |R(P, Q)| < c(n)2^{2tn} ∏_{1≤i,j≤n} |κ_{i}(P) − κ_{j}(Q)|
< c(n)2^{t(2n−η_{1}−η_{2}−2p_{1}−2q_{1}−l_{2}T^{−1}−s_{2}T^{−1}+(4n−2)ε_{1})} < c(n)2^{−tε_{1}}.

The obtained inequality for large t leads one to a contradiction, and this indicates that the rectangles K to which at most one polynomial belongs do not exist.

Proposition 2. If inequalities (7) and (8) hold and if

3 − \frac{ε}{2} ≤ l_{2}T^{−1} + p_{1} + s_{2}T^{−1} + q_{1} ≤ n − 1 + 4nε_{1},  (18)

then

μL(w_{1}, w_{2}, v_{1}, v_{2}) = 0.

Proof. Again, we pass from the system of inequalities (1) to the system of inequalities (9). We set

k = n + 1 − l_{2}T^{−1} − p_{1} − s_{2}T^{−1} − q_{1}.  (19)

Assume that

{k} > ε.  (20)

Then we obtain from relations (18)–(20) that n − [k] ≥ 3. Taking into account relations (8), (9), and (18), we can assume that at least one of the following two inequalities holds:

l_{2}T^{−1} + p_{1} ≤ w_{1} + v_{1} + 1 − 2(n + 1)ε_{1},  (21)
s_{2}T^{−1} + q_{1} ≤ w_{2} + v_{2} + 1 − 2(n + 1)ε_{1}.  (22)

Divide the square [−3, 3]×[−3, 3] into equal rectangles with sides H^{−σ_{1}} and H^{−σ_{2}}. If inequality (21) holds, and this is not the case for inequality (22), then we set σ_{1} = l_{2}T^{−1} + 2(n + 1)ε_{1}, σ_{2} = s_{2}T^{−1}; otherwise we set σ_{1} = l_{2}T^{−1}, σ_{2} = s_{2}T^{−1} + 2(n + 1)ε_{1}. If both inequalities (21) and (22) hold, then we set σ_{1} = l_{2}T^{−1} + (n + 1)ε_{1}, σ_{2} = s_{2}T^{−1} + (n + 1)ε_{1}. For example, suppose that both inequalities (21) and (22) hold. Other cases are considered analogously. Denote by N(K) the number of polynomials belonging to K. If N(K) < c(n)H^{ν}, ν = k − 1 − 0, 1ε, then the total measure of the set of (ω_{1}, ω_{2}) ∈ [−3, 3] × [−3, 3] for which the system of inequalities (9) holds for at least one P (x) ∈ P_{n}(H, \overline{s}) is estimated from above by

c(n)H^{−w_{1}−v_{1}−1+p_{1}−w_{2}−v_{2}−1+q_{1}+l_{2}T^{−1}+s_{2}T^{−1}+(2n+2)ε_{1}+k−1−0.1ε} < c(n)H^{−1−0.1ε+(2n+2)ε_{1}}
< c(n)H^{−1−ε_{1}}.

Since the series ∑_{H=1}^{∞} H^{−1−ε_{1}} converges, in order to complete the proof of Proposition 2 in this case, it is sufficient to apply the Borel–Cantelli lemma.

It remains to assume that there exist rectangles K such that N(K) > c(n)H^{[k]−1}H^{{k}−0.1ε}. We fix one of such rectangles K and enumerate P_{1}(x), …, P_{m}(x) belonging to K. Two such polynomials

\begin{aligned}P_{i}(x) & = Hx^{n} + a_{n−1}^{(i)}x^{n−1} + ⋯ + a_{1}^{(i)}x + a_{0}^{(i)}, \\ P_{j}(x) & = Hx^{n} + a_{n−1}^{(j)}x^{n−1} + ⋯ + a_{1}^{(j)}x + a_{0}^{(j)}, \end{aligned} 1 ≤ i < j ≤ m,

are united in one class if

a_{n−1}^{(i)} = a_{n−1}^{(j)}, …, a_{n−[k]+1}^{(i)} = a_{n−[k]+1}^{(j)}.

Let us apply the Dirichlet box principle. Since the number of different classes does not exceed c(n)H^{[k]−1}, among c(n)H^{ν} polynomials, there are at least c(n)H^{{k}−0.1ε} polynomials belonging to one and the same class. We enumerate these polynomials P_{0}(x), …, P_{l}(x), l = H^{{k}−0.1ε} and form new polynomials

t_{1}(x) = P_{1}(x) − P_{0}(x), …, t_{l}(x) = P_{l}(x) − P_{0}(x).

All polynomials t_{i}(x) are different from each other, have the degree at most n − [k] and height not more than 2H. Expand any polynomial P (x) belonging to K into the Taylor series in the neighborhood of the root κ_{11}(P):

P (ω_{1}) = P′(κ_{11})(ω_{1} − κ_{11}) + \frac{1}{2!}P′′(κ_{11})(ω_{1} − κ_{11})^{2} + ⋯ + \frac{1}{n!}P^{(n)}(κ_{11})(ω_{1} − κ_{11})^{n}.  (23)

Since P (x) ∈ K, there exists a point (ω_{01}, ω_{02}) ∈ K such that |P (ω_{01})| < H^{−w_{1}−v_{1}} and |P (ω_{02})| < H^{−w_{2}−v_{2}}. We obtain from inequality (5)

|ω_{01} − κ_{11}| < c(n)H^{−w_{1}−v_{1}−1+p_{1}}.

If (ω_{1}, ω_{2}) ∈ K, then |ω_{1} − ω_{01}| < H^{−σ_{1}}. Hence we have

|ω_{1} − κ_{11}| < c(n)H^{−l_{2}T^{−1}−(n+1)ε_{1}}.  (24)

Using Lemma 7 and inequality (24), we obtain

\begin{aligned}|P′(κ_{11})(ω_{1} − κ_{11})| & < c(n)H^{1−p_{1}−l_{2}T^{−1}−2ε_{1}}, \\ |P^{(i)}(κ_{11})(ω_{1} − κ_{11})^{i}| & < c(n)H^{1−p_{i}+(n−i)ε_{1}−il_{2}T^{−1}−i(n+1)ε_{1}} \\ & < c(n)H^{1−p_{1}−l_{2}T^{−1}−2ε_{1}},  i = 2, …, n − 1, \\ |P^{(n)}(κ_{11})(ω_{1} − κ_{11})^{n}| & < c(n)H^{1−nl_{2}T^{−1}−n(n+1)ε_{1}} < c(n)H^{1−p_{1}−l_{2}T^{−1}−2ε_{1}}. \end{aligned}  (25)

It follows from relations (23) and (25) that

|P (ω_{1})| < c(n)H^{1−p_{1}−l_{2}T^{−1}−2ε_{1}}.  (26)

Expanding the polynomial P (x), which belongs to K, into the Taylor series in a neighborhood of the root κ_{21}(P), we obtain in a similar way that

|P (ω_{2})| < c(n)H^{1−q_{1}−s_{2}T^{−1}−2ε_{1}}.  (27)

Since inequalities (26) and (27) hold for any P (x) ∈ P_{n}(H, \overline{s}) belonging to K, for any polynomial t_{i}(x), 1 ≤ i ≤ l, the following inequalities hold:

|t_{i}(ω_{1})| < c(n)H^{1−p_{1}−l_{2}T^{−1}−2ε_{1}},  |t_{i}(ω_{2})| < c(n)H^{1−q_{1}−s_{2}T^{−1}−2ε_{1}}.  (28)

Consider the following three possibilities that appear:

(a) All t_{i}(x) = a_{i}t(x), a_{i} ∈ ℤ. Then there is a polynomial R(x) among all polynomials t_{i}(x) whose height does not exceed c(n)H^{1−{k}+0.1ε}. We obtain from relations (28) that

|R(ω_{1})| < c(n)H(R)^{(1−p_{1}−l_{2}T^{−1}−2ε_{1})/(1−{k}+0.1ε)},
|R(ω_{2})| < c(n)H(R)^{(1−q_{1}−s_{2}T^{−1}−2ε_{1})/(1−{k}+0.1ε)}.

Since, under condition (20), we have

\frac{p_{1} + l_{2}T^{−1} + q_{1} + s_{2}T^{−1} − 2 + 4ε_{1}}{1 − \{k\} + 0.1ε} > n − [k] − 1,

in this case, in order to complete the proof, it is sufficient to apply the metric theorem in [3]. (b) There are reducible polynomials among t_{i}(x). We apply here Lemma 3 since

p_{1} + l_{2}T^{−1} + q_{1} + s_{2}T^{−1} − 2 + 4ε_{1} > n − [k] − 2.

(c) There are at least two polynomials among the polynomials t_{i}(x) which have no common roots. Then we apply Lemma 8. Here, under condition (20), we have

τ_{1} = p_{1} + l_{2}T^{−1} − 1 + 2ε_{1}, τ_{2} = q_{1} + s_{2}T^{−1} − 1 + 2ε_{1},
η_{1} = l_{2}T^{−1} + (n + 1)ε_{1}, η_{2} = s_{2}T^{−1} + (n + 1)ε_{1}.

We obtain

p_{1} + q_{1} − (4n − 8)ε_{1} ≤ −2 + 2{k} + l_{2}T^{−1} + s_{2}T^{−1} + δ.  (29)

Since p_{1} ≥ l_{2}T^{−1} and q_{1} ≥ s_{2}T^{−1}, for δ < 2 − 2{k} − (4n − 8)ε_{1}, inequality (29) leads one to a contradiction. The case 0 ≤ {k} ≤ ε requires some modifications, which are connected with the choice of the parameters. It is clear that under the validity of relations (7), (8), and (18), at least one of the inequalities

l_{2}T^{−1} + p_{1} ≤ w_{1} + v_{1} + 0.2,  s_{2}T^{−1} + q_{1} ≤ w_{2} + v_{2} + 0.2.  (30)

holds. For example, suppose that the first of inequalities (30) holds, then we set δ_{1} = l_{2}T^{−1} + 0.8, δ_{2} = s_{2}T^{−1}, ν = k − 1.8 − ε_{1} and realize similar arguments.

Proposition 3. If relations (7) and (8) hold and if

ε < l_{2}T^{−1} + p_{1} + s_{2}T^{−1} + q_{1} < 3 − \frac{ε}{2},  (31)

then μL(w_{1}, w_{2}, v_{1}, v_{2}) = 0.

Proof. Let us unite the polynomials P (x) = Hx^{n} + a_{n−1}x^{n−1} + ⋯ + a_{1}x + a_{0} from the set P_{n}(H, \overline{s}) into one class P_{n}(H, \overline{s}, \overline{a}) if they have one and the same vector \overline{a} = (a_{n−1}, …, a_{3}). Let σ_{1}(P) be the set of real numbers ω_{1} satisfying the inequality

|ω_{1} − κ_{11}| < 2^{n}H^{−w_{1}}Ψ^{v_{1}}(H)|P′(κ_{11})|^{−1},

σ_{2}(P) be the set of real numbers ω_{2} satisfying the inequality

|ω_{2} − κ_{21}| < 2^{n}H^{−w_{2}}Ψ^{v_{2}}(H)|P′(κ_{21})|^{−1},

τ_{1}(P) be the set of real numbers ω_{1} satisfying the inequality

|ω_{1} − κ_{11}| < 2^{n}H^{−α_{1}}|P′(κ_{11})|^{−1},

τ_{2}(P) be the set of real numbers ω_{2} satisfying the inequality

|ω_{2} − κ_{21}| < 2^{n}H^{−α_{2}}|P′(κ_{21})|^{−1},

where α_{1} and α_{2} are chosen in the following way:

l_{2}T^{−1}+p_{1} < 1+α_{1}− \frac{ε_{1}}{5} , s_{2}T^{−1}+q_{1} < 1+α_{2}− \frac{ε_{1}}{5} , α_{1}+α_{2} = 1, α_{1} ≤ w_{1}+v_{1}, α_{2} ≤ w_{2}+v_{2}.

Clearly, σ_{1}(P) ⊂ τ_{1}(P) and σ_{2}(P) ⊂ τ_{2}(P). We obtain from relation (5) that all ω_{1} ∈ S(κ_{11}) and ω_{2} ∈ S(κ_{11}) satisfying the inequalities

|P (ω_{1})| < H^{−w_{2}}Ψ^{v_{2}}(H),  |P (ω_{2})| < H^{−w_{2}}Ψ^{v_{2}}(H),  |P (ω_{1})| < H^{−α_{1}},  |P (ω_{2})| < H^{−α_{2}}

belong to the sets σ_{1}(P), σ_{2}(P), τ_{1}(P), and τ_{2}(P), respectively. Let (ω_{1}, ω_{2}) ∈ τ_{1}(P)×τ_{2}(P). Then we have

\begin{aligned}|P′(κ_{11})(ω_{1} − κ_{11})| & < 2^{n}H^{−α_{1}}, \\ |\frac{1}{i!}P^{(i)}(κ_{11})(ω_{1} − κ_{11})^{i}| & < c(n)H^{1−p_{i}+(n−i)ε_{1}−iα_{1}−i+ip_{1}} \\ & < c(n)H^{−α_{1}},  i = 2, …, n − 1, \\ |\frac{1}{n!}P^{(n)}(κ_{11})(ω_{1} − κ_{11})^{n}| & < c(n)H^{1−nα_{1}−n+np_{1}} < c(n)H^{−α_{1}}. \end{aligned}  (32)

Using expansion (23) of the polynomial P (x), for ω_{1} ∈ τ_{1}(P), we obtain from inequalities (32)

|P (ω_{1})| < H^{−α_{1}}.

Analogously, for ω_{2} ∈ τ_{2}(P), we obtain

|P (ω_{2})| < H^{−α_{2}}.

The domain Δ(P) = τ_{1}(P) × τ_{2}(P) is called nonessential if there is a polynomial Q(x) in the class P (H, \overline{s}, \overline{a}) such that

μ(Δ(P) ∩ Δ(Q)) ≥ \frac{1}{2}μΔ(P).

Otherwise, the domain Δ(P) is called essential.

If the domain Δ(P) is essential, then each point (ω_{1}, ω_{2}) ∈ [−3, 3] × [−3, 3] belongs to not more than four essential domains, and therefore,

∑_{P (x)∈P_{n}(H,\overline{s},\overline{a})} μΔ(P) < const .

We obtain from the inequality μ(σ_{1}(P) × σ_{2}(P)) < μΔH^{−n+3}Ψ(H) that

∑_{P (x)∈P_{n}(H,\overline{s},\overline{a})} μ(σ_{1}(P) × σ_{2}(P)) < c(n)H^{−n+3}Ψ(H).

Since the series ∑_{H=1}^{∞} Ψ(H) converges, by the Borel–Cantelli lemma, the set of (ω_{1}, ω_{2}) which occur in an infinite number of essential domains Δ(P) has zero measure.

If the domain Δ(P) is nonessential, then

|τ_{1}(P) ∩ τ_{1}(Q)| ≥ \frac{1}{2}|τ_{1}(P)|, |τ_{2}(P) ∩ τ_{2}(Q)| ≥ \frac{1}{2}|τ_{2}(P)|,

and on the intervals J_{1} = τ_{1}(P)∩τ_{1}(Q) and J_{2} = τ_{2}(P)∩τ_{2}(Q), the polynomial R(x) = P (x)−Q(x) satisfies the conditions

|R(ω_{1})| < c(n)H^{−α_{1}},  |R(ω_{2})| < H^{−α_{2}},  deg R(x) ≤ 2.  (33)

If |κ_{1}(R)−κ_{2}(R)| > δ, where δ is an arbitrary, but fixed number, then the height of the polynomial R(x) satisfies the inequalities

H(R) < c(n)|P′(κ_{11})|,  H(R) < c(n)|P′(κ_{21})|.  (34)

We obtain from relations (31), (33), and (34) that for (ω_{1}, ω_{2}) ∈ J_{1} × J_{2},

|R(ω_{1})R(ω_{2})| < c(n)H^{−1/(1−ε)} < c(n)H^{−1−ε}.

Further, using Lemma 2 and the metric theorem in [3], we conclude that the set of (ω_{1}, ω_{2}) which belong to an infinite number of nonessential domains Δ(P) has zero measure. In the case |κ_{1}(R) − κ_{2}(R)| < δ, we argue similarly as we did when proving Lemma 10.

Proposition 4. If relations (7) and (8) hold and if

l_{2}T^{−1} + p_{1} + s_{2}T^{−1} + q_{1} < ε,

then μL(w_{1}, w_{2}, v_{1}, v_{2}) = 0.

Proof. We unite polynomials P (x) = Hx^{n} + a_{n−1}x^{n−1} + ⋯ + a_{1}x + a_{0} from the set P_{n}(H, \overline{s}) into one class P_{n}(H, \overline{s}, \overline{β}) if they have one and the same vector \overline{β} = (a_{n−1}, …, a_{2}). Define σ_{1}(P), σ_{2}(P) as in Proposition 3, while τ_{1}(P) and τ_{2}(P) are defined as the sets of real numbers ω_{1} and ω_{2} satisfying the inequalities

|ω_{1} − κ_{11}| < 2^{−n−1}(n + 1)^{−1}|P′(κ_{11})|^{−1},

and

|ω_{2} − κ_{21}| < 2^{−n−1}(n + 1)^{−1}|P′(κ_{21})|^{−1}

respectively. Using the Taylor series expansion (23) of the polynomial P (x) and the inequality |P^{(i)}(ω)| < i! 3^{n}(n + 1)H for ω_{1} ∈ [−3, 3], we obtain

|P′(κ_{11})(ω_{1} − κ_{11})| < 2^{−n−1}(n + 1)^{−1},
\frac{1}{i!}|P^{(i)}(κ_{11})(ω_{1} − κ_{11})^{i}| < 3^{n}(n + 1)H × 2^{−i(n−1)}(n + 1)^{−i} × H^{−i+εi}
< (n + 1)^{−1}2^{−n−1}, i = 2, …, n,
|P (ω_{1})| < 2^{−n−1}.

Similarly, for ω_{2} ∈ [−3, 3], we obtain

|P (ω_{2})| < 2^{−n−1}.

If the domain Δ(P) is essential, then we obtain

∑_{P (x)∈P_{n}(H,\overline{s})} μ(σ_{1}(P) × σ_{2}(P)) < c(n) ∑_{\overline{β}} ∑_{P (x)∈P_{n}(H,\overline{s},\overline{β})} μΔ(P)H^{−n+2}Ψ(H) < c(n)Ψ(H)

since

μ(σ_{1}(P), σ_{2}(P)) ≤ c(n)μΔ(P)H^{−n+2}Ψ(H).

Then we apply the Borel–Cantelli lemma. If the domain Δ(P) is nonessential, then we have

|R(ω_{1})| < 2^{−n},  |R(ω_{2})| < 2^{−n},  deg R(x) ≤ 1  (35)

for the polynomial R(x) = P (x)−Q(x). Hence, the distance from (ω_{1}, ω_{2}) for which inequalities (35) hold and the line ω_{1} = ω_{2} does not exceed a given positive number. In order to complete the proof, we apply Lemma 2.

Proposition 5. If

l_{2}T^{−1} + p_{1} > w_{1} + v_{1} + 1,  (36)
s_{2}T^{−1} + q_{1} ≤ w_{2} + v_{2} − 1 + 2nε_{1},  (37)
s_{2}T^{−1} + q_{1} > 2 − w_{1} − v_{1} − (n − 1)ε_{1},

then μL(w_{1}, w_{2}, v_{1}, v_{2}) = 0.

Proof. We pass from the system of inequalities (1) to the system of inequalities (9). We set

k = w_{2} + v_{2} + 1 − s_{2}T^{−1} − q_{1}.

If {k} > ε, then we set σ_{1} = w_{1} + v_{1} + 1 − p_{1}, σ_{2} = s_{2}T^{−1}, and ν = k − 1 − 0.1ε; then we proceed as in the proof of Proposition 2. If 0 ≤ {k} ≤ ε, then we set σ_{1} = w_{1} + v_{1} + 1 − p_{1}, σ_{2} = s_{2}T^{−1} + 0.8, ν = k − 1.8 − ε_{1} and proceed just as in the proof of Proposition 2.

Proposition 6. If relations (36) and (37) hold and if

ε < s_{2}T^{−1} + q_{1} < 2 − w_{1} − v_{1} − (n − 1)ε_{1},

then μL(w_{1}, w_{2}, v_{1}, v_{2}) = 0.

Proof. We unite the polynomials P (x) ∈ P_{n}(H, \overline{s}) into one class P_{n}(H, \overline{s}, \overline{a}) if they have equal coefficients of the powers x^{n−1}, …, x^{3}. Define the domains σ_{1}(P) and σ_{2}(P) as in Proposition 3. Let τ_{1}(P) = σ_{1}(P) and let τ_{2}(P) be the set of real ω_{2} satisfying the inequality

|ω_{2} − κ_{21}| < 2^{n}H^{−1−w_{1}}Ψ^{v_{1}}(H)|P′(κ_{21})|^{−1}.

Further, we proceed as in the proof of Proposition 3.

Proposition 7. If relations (36) and (37) hold and if

s_{2}T^{−1} + q_{1} < ε,

then μL(w_{1}, w_{2}, v_{1}, v_{2}) = 0.

Proof. We unite the polynomials P (x) from the set P_{n}(H, \overline{s}) into one class P_{n}(H, \overline{s}, \overline{β}) if they have one and the same vector \overline{β} = (a_{n−1}, …, a_{2}). Define σ_{1}(P), σ_{2}(P), and τ_{1}(P) as in Proposition 6. Let τ_{2}(P) be the set of real ω_{2} satisfying the inequality

|ω_{2} − κ_{21}| < 2^{−n−1}(n + 1)^{−1}H^{−w_{1}}Ψ^{v_{1}}(H)|P′(κ_{21})|^{−1}.

Then we proceed as in the proof of Proposition 4.

Proposition 8. If inequality (36) holds and if

s_{2}T^{−1} + q_{1} > w_{2} + v_{2} + 1,  (38)

then μL(w_{1}, w_{2}, v_{1}, v_{2}) = 0.

Proof. We pass from the system of inequalities (1) to the system of inequalities (9). We can obtain from inequalities (36) and (38) that

l_{2}T^{−1} > \frac{w_{1} + v_{1} + 1 − p_{2} − 1/T}{2} ,  (39)
s_{2}T^{−1} > \frac{w_{2} + v_{2} + 1 − q_{2} − 1/T}{2} .  (40)

Let us show that there exist two integers m and r (2 ≤ m ≤ n − 1, 2 ≤ r ≤ n − 1) for which the following system of inequalities holds:

\frac{l_{m}}{T} \frac{w_{1} + v_{1} + 1 − p_{m} − 1/T}{m} ≥ \frac{l_{m+1}}{T} ,  (41)
\frac{s_{r}}{T} \frac{w_{2} + v_{2} + 1 − q_{r} − 1/T}{r} ≥ \frac{s_{r+1}}{T} .  (42)

The left-hand side of inequality (41) follows from inequality (39) for m = 2. If

\frac{w_{1} + v_{1} + 1 − p_{2} − 1/T}{2} ≥ \frac{l_{3}}{T} ,

then inequality (41) holds for m = 2. If

\frac{l_{3}}{T} > \frac{w_{1} + v_{1} + 1 − p_{2} − 1/T}{2} ,

then, taking into account the fact that p_{2} = l_{3}T^{−1} + p_{3}, we arrive at the inequality

3l_{3}T^{−1} > w_{1} + v_{1} + 1 − p_{3} − \frac{1}{T}

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS  Vol. 218  1997

66 BERNIK, BORBAT

and therefore,

\frac{l_{3}}{T} > \frac{w_{1} + v_{1} + 1 − p_{3} − 1/T}{3} .

If now

\frac{w_{1} + v_{1} + 1 − p_{3} − 1/T}{3} ≥ \frac{l_{4}}{T} ,

then we obtain that inequality (41) holds for m = 3. Otherwise, we pass to the inequality

\frac{l_{4}}{T} > \frac{w_{1} + v_{1} + 1 − p_{4} − 1/T}{4}

and so on. Clearly, we arrive at m for which relation (41) holds, since |κ_{11} −κ_{21}| > δ and therefore, for a certain m ≤ n − 1 we have l_{s}T^{−1} < 2/T, p_{s−1} < 2(s − 2)/T for s ≥ m. Similarly, we can prove the existence of r for which inequality (42) holds. Assume that

|κ_{11}(P) − κ_{11}(Q)| < H^{−(w_{1}+v_{1}+1−p_{m}−1/T)/m},  (43)
|κ_{21}(P) − κ_{21}(Q)| < H^{−(w_{2}+v_{2}+1−q_{r}−1/T)/r},  (44)

where P (x), Q(x) ∈ P_{t}(\overline{s}). Then it follows from the inequalities

\frac{l_{2}}{T} ≥ \frac{l_{3}}{T} ≥ ⋯ ≥ \frac{l_{m}}{T} > \frac{w_{1} + v_{1} + 1 − p_{m} − 1/T}{m}

and from inequality (43) that

|κ_{1i}(P) − κ_{1j}(Q)| ≤ |κ_{1i}(P) − κ_{11}(P)| + |κ_{1j}(Q) − κ_{11}(Q)|
≤ |κ_{1i}(P) − κ_{11}(P)| + |κ_{1j}(Q) − κ_{11}(Q)| + |κ_{11}(P) − κ_{11}(Q)|
≤ 3H^{−(w_{1}+v_{1}+1−p_{m}−1/T)/m}  (45)

for any i, j, 2 ≤ i ≤ m, 2 ≤ j ≤ m. Therefore we obtain from relation (45) that

∏_{1≤i,j≤m} |κ_{1i}(P) − κ_{1j}(Q)| < c(n)H^{−m(w_{1}+v_{1}+1−p_{m}−1/T)}.  (46)

For i ≤ m and j > m it follows from relation (41) that

|κ_{1i}(P)−κ_{1j}(Q)| ≤ |κ_{1i}(P)−κ_{11}(P)|+|κ_{11}(P)−κ_{11}(Q)|+|κ_{11}(Q)−κ_{1j}(Q)| < c(n)H^{−l_{j}/T} . (47)

Also, we obtain the following inequality for i > m, j ≤ m, which is analogous to inequality (47):

|κ_{1i}(P) − κ_{1j}(Q)| < c(n)H^{−l_{j}/T} .  (48)

It follows from relations (47) and (48) that

∏_{max(i,j)>m} |κ_{1i}(P) − κ_{1j}(Q)| < c(n)H^{−2mp_{m}}.  (49)

If now we use the inequalities

\frac{s_{2}}{T} ≥ \frac{s_{3}}{T} ≥ ⋯ ≥ \frac{s_{r}}{T} > \frac{w_{2} + v_{2} + 1 − q_{r} − 1/T}{r}

and inequalities (42) and (44), then we obtain the following results which are analogous to inequalities (46) and (49):

∏_{1≤i,j≤m} |κ_{2i}(P) − κ_{2j}(Q)| < c(n)H^{−r(w_{2}+v_{2}+1−q_{r}−1/T)}, ∏_{max(i,j)>m} |κ_{2i}(P) − κ_{2j}(Q)| < c(n)H^{−2rq_{r}}.

(50)

The polynomials P (x) and Q(x) belonging to P_{t}(\overline{s}) have no common roots, and thus, |R(P, Q)| ≥ 1. We have from inequalities (46), (49), and (50) that

1 ≤ |R(P, Q)| ≤ c(n)2^{t(2n−m(w_{1}+v_{1}+1−p_{m}−1/T)−2mp_{m})}2^{t(−r(w_{2}+v_{2}+1−q_{r}−1/T)−2rq_{r})}
< c(n)2^{t(2n−m(w_{1}+v_{1}+1−1/T))}2^{t(−r(w_{2}+v_{2}+1−1/T))} < c(n)2^{−t(2−(m+r)/T)}.  (51)

For large t, inequality (51) is contradictory. Therefore there are no roots κ_{11}(Q) and κ_{21}(Q) of any other polynomial Q(x) ∈ P_{t}(\overline{s}) in the rectangle

|ω_{1} − κ_{11}(P)| < H^{−(w_{1}+v_{1}+1−p_{m}−1/T)/m}, |ω_{2} − κ_{21}(P)| < H^{−(w_{2}+v_{2}+1−q_{r}−1/T)/r}.

This means that the number of polynomials P (x) ∈ P_{t}(\overline{s}) does not exceed

c(n)2^{t((w_{1}+v_{1}+1−p_{m}−1/T)/m+(w_{2}+v_{2}+1−q_{r}−1/T)/r)}.  (52)

It follows from relation (5) that the measure of the set of (ω_{1}, ω_{2}), ω_{1} ∈ S(κ_{11}), ω_{2} ∈ S(κ_{21}) for which the system of inequalities (9) holds does not exceed

c(n)2^{−t((w_{1}+v_{1}+1−p_{m})/m+(w_{2}+v_{2}+1−q_{r})/r)}.  (53)

We conclude from expressions (52) and (53) that the measure of the set of (ω_{1}, ω_{2}) for which the system of inequalities (9) holds for at least one polynomial P (x) ∈ P_{t}(\overline{s}) does not exceed

c(n)2^{−t(1/(mT)+1/(rT))} < c(n)2^{−2t/(T n)}.  (54)

Since the series ∑_{t=1}^{∞} 2^{−tλ} converges for any λ > 0, Proposition 8 follows from inequality (54) and the Borel–Cantelli lemma.

Proposition 9. If inequalities (36) hold and if

w_{2} + v_{2} − 1 + 2nε_{1} < s_{2}T^{−1} + q_{1} ≤ w_{2} + v_{2} + 1,

then μL(w_{1}, w_{2}, v_{1}, v_{2}) = 0.

Proof. Again, we pass to the system of inequalities (9). Divide the square [−3, 3] × [−3, 3] into equal rectangles K with sides H^{−η_{1}}, H^{−η_{2}}, η_{1} = (w_{1} +v_{1} +1−p_{m} −1/T)/m, η_{2} = w_{2} +v_{2} +1−q_{1} − ε_{1}/2, where m is chosen in such a way that inequality (41) holds. Suppose that not more than one polynomial P (x) ∈ P_{t}(\overline{s}) belongs to each rectangle K. Then the measure of the set of (ω_{1}, ω_{2}), ω_{1} ∈ S(κ_{11}), ω_{2} ∈ S(κ_{21}) for which the system of inequalities (9) holds for at least one polynomial P (x) ∈ P_{t}(\overline{s}) is estimated from above by

c(n)2^{t(−(w_{1}+v_{1}+1−p_{m})/m−w_{2}−v_{2}−1+q_{1}+(w_{1}+v_{1}+1−p_{m}−1/T)/m+w_{2}+v_{1}−q_{1}−ε_{1}/2)} < c(n)2^{−t(1/(mT)+ε_{1}/2)}.

Since the series ∑_{t=1}^{∞} 2^{−t(1/(mT)+ε_{1}/2)} converges, we apply the Borel–Cantelli lemma in order to complete the proof. Assume now that there exist rectangles K such that they contain two or more polynomials P (x) ∈ P_{t}(\overline{s}). Let P (x) and Q(x) belong to K. Since P (x), Q(x) ∈ P_{t}(\overline{s}), they have no common roots. Then, using inequalities (15), (17), (46), and (49), we obtain

1 ≤ |R(P, Q)| ≤ c(n)2^{t(2n−m(w_{1}+v_{1}+1−p_{m}−1/T)−2mp_{m})}2^{t(−η_{2}−2q_{1}−s_{2}T^{−1}+(2n−1)ε_{1})}
< c(n)2^{t(2n−2w_{1}−2v_{1}−2−2/T −w_{2}−s_{2}T^{−1}−q_{1}+(2n−0.5)ε_{1}−v_{2})} < c(n)2^{−2t/T} .

This inequality is contradictory for large t. Consequently, there are no rectangles K to which more than one polynomial belong.

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Translated by S. Vakhromeev