SOME ISOPERIMETRIC INEQUALITIES AND THEIR APPLICATION TO PROBLEMS ON POLYNOMIALS

V. Bentkus and A. Dubickas

Abstract. We prove some sharp inequalities for the product of the Lebesque integral over some power of a positive absolutely continuous and non–decreasing function (or some linear combination of its distinct powers) and the Lebesque integral over the square of its derivative. These inequlities are releted to some problems for polynomials with small Mahler measure. As an application we give a lower bound for the logarithmic height of a non–cyclotomic algebraic number in terms of its degree.

Key words: isoperimetric inequalities, Mahler’s measure, linear form in one logarithm.

2000 Mathematics Subject Classification: 11R04, 11R09, 26A46, 26D10.

1. Introduction.

Let L be the space of all absolutely continuous and non–decreasing functions G in the interval [0, 1] satisfying G(0) = 0 , G(1) = 1 such that the derivative G′ (which, by the Lebesque theorem [16], exists almost everywhere) is in L^{2}[0, 1] . For a real number λ set

x_{λ}(G) = ∫_{0}^{1}G(t)^{λ}dt∫_{0}^{1}G′(t)^{2}dt.

We begin with the following statement.

Theorem 1. Given a real number λ > −2 , and G ∈ L, we have x_{λ}(G) ⩾ 4/(λ + 2)^{2} for λ in the range −2 < λ < 2 , and x_{λ}(G) > 4/(λ + 2)^{2} for λ ⩾ 2 . Furthemore, for every λ > −2 the respective inequality is sharp.

Corollary 1. If G ∈ L then x_{1}(G) ⩾ 4/9 , and this inequality is sharp.

Corollary 2. If G ∈ L then x_{2}(G) > 1/4 , and this inequality is sharp.

Our purpose here is to investigate which values can be taken by the functional

F_{p}(G) = (∫_{0}^{1}(x_{1}(G) − τx_{2}(G))^{p/(p−1)}dτ)^{(p−1)/p},

where p > 1 . In two extreme cases p → 1 and p → ∞, set, taking the limits, F_{1}(G) := x_{1}(G) and F_{∞}(G) := x_{1}(G) − (1/2)x_{2}(G) .

In the next section we describe a number theoretic problem on small Mahler measure in which the functionals of this type arise. In Section 3 we formulate three isoperimetric problems, and remind two more problems on polynomials in which they arise. In particular, Corollary 1 solves the first isoperimetric problem. In Section 4 we describe the values taken by the linear functional in the first two moments, namely ax_{1}(G) + bx_{2}(G) . This contains the answer to the second question. In Section 5 a solution to the third problem is given. Some general facts and observations about the behavior of the functional F(x_{1}(G), x_{2}(G), …, x_{m}(G)) are given in Section 6. In Sections 7, 8, and 9 we prove Theorems 1, 2, and 3 stated in Sections 1, 4, and 5, respectively. Some computational results are given in the final Section 10. To simplify the notation we often (especially in the proofs) write I(f) for the integral ∫_{0}^{1}f(t)dt.

2. A number theoretic problem.

Given a fixed constant σ > 1 , Lehmer [17] asked whether there is a monic polynomial with integer coefficients such that the modulus of the product of its zeroes outside the unit circle is strictly greater than 1, but smaller than or equal to σ. The modulus of the above product (of roots outside the unit circle) is called the Mahler measure of the monic polynomial, named after Mahler who investigated this quantity in [19] and later. It is believed since, that there is no such polynomial if σ is sufficiently close to 1. The smallest known value for σ, namely σ_{0} = 1.17628082 …, for which such polynomial exists was found by Lehmer himself. Here σ_{0} is the only positive root of the polynomial z^{10} + z^{9} − z^{7} − z^{6} − z^{5} − z^{4} − z^{3} + z + 1 which is greater than 1.

In 1857, Kronecker [15] showed that if the Mahler measure of a monic polynomial P (z) with integer coefficients is equal to 1 then the polynomial is a product of cyclotomic polynomials and a power of z. Note that the Mahler measure of the monic polynomial with integer coefficients is never smaller than 1, since the product of all its non–zero roots equals 1 in absolute value. In modern terms, therefore, Lehmer’s conjecture is usually reformulated as follows: for every irreducible non–cyclotomic polynomial P , of degree d ⩾ 2 , we have M(P) ⩾ σ_{1} . Here M(P) stands for the Mahler measure of P , and σ_{1} > 1 is either σ_{0} or may be a smaller constant. Also, M(P) is a good example of “height”, namely certain measure of complexity of a polynomial. See, for instance, a book of Everest and Ward [14] on introduction to the Mahler measure and Chapter 3 of Walschmidt’s book [29] on introduction to the theory of heights.

The maximum of the absolute values of roots of P is another example of the height. This is called the house of P . Suppose we are given an integer d ⩾ 2 and a positive constant δ. Schinzel and Zassenhaus [26] asked whether there is an irreducible monic polynomial, of degree d, with integer coefficients such that its house is greater than 1, but smaller than or equal to 1 + δ/d.

Although these two problems are not yet solved, there are some lower bounds on the minimum taken over the Mahler measures (and over the houses, respectively) of all non–cyclotomic monic integer polynomials of, say, degree d in terms of d. See, for instance, Blanksby [2], Blanksby and Montgomery [3], Stewart [28], Dobrowolski [8]. The best unconditional (depending on d only) both asymptotic (when d is sufficiently large) and uniform (for all d) bounds were obtained using the method of Dobrowolski.

In order to formulate these results we shall use the standard notation. Let d ⩾ 2 be a positive integer. Since the set of all monic irreducible polynomials of degree d is isomorphic to the set of all algebraic integers of degree d, from now on, as usual, we denote by M(α) the Mahler measure of the minimal polynomial of α. Set also \overline{|α|} = max_{1⩽k⩽d} |α_{k}| for the house of α. Here α_{1}, …, α_{d} (with α one of these) are the conjugates of α, namely the roots of its minimal polynomial P of degree d. It is now known that for every positive ε there is a d(ε) such that if α is a non–cyclotomic algebraic number of degree d, d ⩾ d(ε) , then

log M(α) > (\frac{9}{4} − ε)(\frac{log log d}{log d})^{3}, log \overline{|α|} > (\frac{64}{π^{2}} − ε)\frac{1}{d}(\frac{log log d}{log d})^{3}

(see [18] and [9], respectively). Earlier, Dobrowolski [8] proved the first inequality with the constant 1 instead of 9/4. Cantor and Straus [7] and independently Rauch [23] improved the constant to 2. The result of Smyth [27] reduces both problems to the reciprocal algebraic integers, namely those of even degree which have the set of conjugates invariant over the action of the map z → z^{−1} . The result of Louboutin, therefore, implies the second inequality with the constant 9/2.

The proofs of these results involve either Siegel’s lemma (see e.g. Bombieri and Vaaler [4] for a modern version of Siegel’s lemma) or certain confluent determinants. It was shown in [12] that these constants 1, 2, 9/4, 9/2, 64/π^{2} can all be obtained via the values of certain functional F . The latter can be defined on the set of the absolutely continuous functions on [0, 1] monotonically decreasing in this interval from 1 to 0.

Let p be a positive number. Set

M_{p}(α) = (\frac{1}{d} ∑_{k=1}^{d} | log |α_{k}||^{p})^{1/p}.

This quantity M_{p}(α) is a height on the multiplicative group of algebraic integers. We can call it the logarithmic height of α. The logarithmic height at p = 1 is exactly twice the Weil height for reciprocal integers. Indeed, if α is reciprocal then

M_{1}(α) = \frac{2 log M(α)}{d},

and

M_{∞}(α) := lim_{p→∞} M_{p}(α) = log \overline{|α|}.

Also, for any reciprocal α, the function p → M_{p}(α) is non–decreasing (in p). The inequality of the form inf dM_{p}(α) ⩾ f_{p}, therefore, lies between the conjecture of Lehmer and the conjecture of Schinzel and Zassenhaus. Here, p > 1 , f_{p} is a positive constant which depends only on p and the infimum is taken over every non–cyclotomic reciprocal algebraic integer of degree d.

A contruction of the confluent interpolation determinant which allows us to estimate this infimum from below (in terms of d) for any p ⩾ 1 is given in [12]. The results of Section 2 in [12] are given in terms of non–increasing functions. The map t → 1 − t allows us to reformulate these results in terms of the non–decreasing functions.

3. Three isoperimetric problems.

In this section, all of the infimums are taken over every function in L.

Problem 1. Find ζ_{1} = inf x_{1}(G) .

Problem 2. Find ζ_{∞} = inf{x_{1}(G) − (1/2)x_{2}(G)}.

Problem 3. Given p > 1 , find

ζ_{p} = inf (∫_{0}^{1}(x_{1}(G) − τx_{2}(G))^{p/(p−1)}dτ)^{(p−1)/p}.

Note that Corollary 1 settles Problem 1: ζ_{1} = 4/9 . Our main result in [12] is the inequality

dM_{p}(α) > (\frac{2}{F_{p}(G)} − ε)(\frac{log log d}{log d})^{3}

(see Theorem 2 in [12]). In particular, for every positive ε there is a d(ε) such that for every non–cyclotomic algebraic integer α of degree d ⩾ d(ε)

log M(α) > (\frac{1}{x_{1}(G)} − ε)(\frac{log log d}{log d})^{3},

log \overline{|α|} > (\frac{2}{F_{∞}(G)} − ε)\frac{1}{d}(\frac{log log d}{log d})^{3}.

Here, G ∈ L is in our disposition. Generally speaking, we can replace F_{p}(G) , x_{1}(G) , and F_{∞}(G) by ζ_{p} , ζ_{1} , and ζ_{∞} , respectively.

As for the inequality for the Mahler measure, the choice of G(t) = t gives the constant 1/x_{1}(G) = 1 obtained by Cantor and Straus [7]. The choice G(t) = t^{2} gives the constant 1/x_{1}(G) = 9/4 obtained by Louboutin [18]. In case of the house, the choices of G(t) = t, G(t) = t^{2} , and G(t) = 1 − cos(πt/2) give the constants 6, 45/7, and 64/π^{2} , respectively.

Problem 2 has an application to estimate of the linear form in one logarithm log α. See the papers of Mignotte [20], Mignotte and Waldschmidt [21], Bugeaud, Mignotte and Normandin [6], Amoroso [1], and also [10], [11]. With the methods used in [10] one can prove that for α ≠ 1 we have the inequality

| log α| > exp \{ − (\sqrt{2F_{∞}(G)} + o(1))d\sqrt{h(α) log(1/h(α))}\},

where o(1) tends to zero as the Weil height of α, namely h(α) = (1/d) log M(α) , tends to zero. The choice of G(t) = t gives the constant \sqrt{2/3} (as in [6]). A more subtle choice G(t) = 1 − cos(πt/2) gives the constant π/4 .

Problem 2 also has an application to the problem of finding an integer polynomial with large number of cyclotomic factors. This problem was first studied by Schinzel [25], see also Pinner and Vaaler [22]. The constant 3\sqrt{2}/π, which was obtained in [13], generally speaking, can be replaced by 3/(4\sqrt{ζ_{∞}}) .

One of the conclusions of Corollaries 1 and 3 is the following: unless there is a new idea, one cannot improve the best known asymptotic constants 9/4 (that of [18]), 64/π^{2} (that of [9]), π/4 (that of [10]), and 3\sqrt{2}/π (that of [13]), by just replacing the function G by a “better” one. Ideologically similar to this paper is that of Brisebarre [5], where a similar conclusion (although based just on extensive computation near the point of extremum) concerning the irrationality measures of log 2 and π/\sqrt{3} was obtained.

4. Linear form in two moments.

Given three real numbers a, b, and c, we are interested whether there is a function G ∈ L such that

ax_{1}(G) + bx_{2}(G) = c.

We say that a triple (a, b, c) ∈ ℝ^{3} is admissible if there is a function G ∈ L which is a solution to the above equation. To describe the admissible triples, note that if (a, b, c) is (is not) admissible then the triple (ψa, ψb, ψc) is (is not) admissible for every real number ψ ≠ 0 . For this reason, it is convenient to describe the admissible triples via points in 2− dimensional projective space ℙ^{2} . We can, therefore, assume that b = 0 or b = ±1 and describe the points (a : 0 : c) ∈ ℙ^{2} , (a : 1 : c) ∈ ℙ^{2} , (a : −1 : c) ∈ ℙ^{2} with non–negative a so that (ψa, ψb, ψc) ∈ ℝ^{3} , where b ∈ {0, −1, 1} and ψ ∈ ℝ, are the admissible triples.

Theorem 2. Let a ⩾ 0 and let ψ ≠ 0 . The triple (ψa, 0, ψc) ∈ ℝ^{3} is admissible if and only if one of the following is true

(i) a = 0, c = 0; (ii) a > 0, c ⩾ 4a/9. The triple (ψa, ψ, ψc) ∈ ℝ^{3} is admissible if and only if one of the following is true (iii) a = 0, c > 1/4; (iv) a > 0, c ⩾ (1/16)((a + 2)\sqrt{a + 1} − a^{2} log(\sqrt{1/a} + \sqrt{1 + 1/a}))^{2}. Finally, the triple (ψa, −ψ, ψc) ∈ ℝ^{3} is admissible if and only if one of the following is true

(v) a = 0, c < −1/4; (vi) 0 < a < 1 , and c is any real number; (vii) a ⩾ 1, c ⩾ (1/64)(a^{2} arccos(1 − 2/a) − 2(a − 2)\sqrt{a − 1})^{2}.

It is easy to see, by (ii), that the smallest c for which the triple (1, 0, c) is admissible equals 4/9 (this gives another proof of Corollary 1). Also, Corollary 2 is contained in (iii). On applying (vii) with ψ = 1/2, a = 2 , we deduce that ζ_{∞} is equal to π^{2}/32 which gives a solution to the second problem.

Corollary 3. If G ∈ L then

ζ_{∞} = min ∫_{0}^{1}(G(t) − (1/2)G(t)^{2})dt∫_{0}^{1}G′(t)^{2}dt = π^{2}/32.

It is worth pointing out that, by (vi), the linear form 0.99x_{1}(G) − x_{2}(G) takes any real value (as G ∈ L), whereas, by (vii), the values of the form x_{1}(G) − x_{2}(G) all lie in the interval [π^{2}/64, ∞) .

5. Computation of ζ_{p} .

Recall that

F_{p}(G) = (∫_{0}^{1}(x_{1}(G) − τx_{2}(G))^{p/(p−1)}dτ)^{(p−1)/p},

and

ζ_{p} = inf F_{p}(G),

where p > 1 and the infimum is taken over every function G ∈ L.

For every ω in the interval [0, π/2] we define the set of functions Ω = {G_{ω}(t)} as follows. Denote

G_{ω}(t) = \frac{1 − cos(ωt)}{1 − cos(ω)} = (\frac{sin(ωt/2)}{sin(ω/2)})^{2}

for every ω in the interval (0, π/2] , and G_{0}(t) = t^{2} . (Note that, for every fixed real number t, G_{ω}(t) tends to t^{2} as ω → 0 .) Clearly, Ω is a subset of L. The proof of the next theorem is the most difficult in this paper. Apparently, the standard methods cannot be applied in the proof. We show that, for every fixed p > 1 , the infimum ζ_{p} of the functional F_{p}(G) , taken over every every function G ∈ L, is actually attained at some function G_{ω} = G_{ω}(t) .

Theorem 3. For every fixed p > 1 , we have

ζ_{p} = min F_{p}(G_{ω}),

where the minimum is taken over every ω in the interval [0, π/2] .

It reduces the problem of finding ζ_{p} to computation of minimum of the function F_{p}(G_{ω}) in one variable in the interval. We give some numerical values of the computation in Section 10. From Corollary 1 and Corollary 3 it follows that Theorem 3 also holds for p = 1 (then ω = 0 ) and for p = ∞ (then, respectively, ω = π/2 .) As p increases from 1 to ∞, the respective ω increases from 0 to π/2 .

6. General observations.

Suppose that the function F(x_{1}, x_{2}, …, x_{m}) in m variables is continuous in (0, ∞)^{m} . Which values are taken by the functional F(G) = F(x_{1}(G), …, x_{m}(G)) , where G ∈ L ? We call the functional continuous if it is a continuous function in x_{1}, …, x_{m} . Let V (F) denotes the set of the values of F(G) with G ∈ L.

Observation 1. For every continuous functional F the set V (F) is connected.

Proof. If F(G_{1}) = c and F(G_{2}) = d, where say c < d, then F(μG_{1} + (1 − μ)G_{2}) takes all values in the interval [c, d] , as μ runs over the intverval [0, 1] , because this function is continuous in μ. Also, μG_{1} + (1 − μ)G_{2} remains in the same class of functions L as do G_{1} and G_{2} , which completes the proof.

We say that the function F(x_{1}, x_{2}, …, x_{m}) in m variables is positively homogenius if for every non–zero number y we have

F(x_{1}y, x_{2}y, …, x_{m}y) = y^{r}F(x_{1}, x_{2}, …, x_{m})

with a positive number r. We call the functional positively homogenius if it is a positively homogenius function in x_{1}, …, x_{m} .

Observation 2. Suppose F is a positively homogenius continuous functional, and c ∈ V (F) , where c ≠ 0 . Then [c, ∞) ∈ V (F) if c > 0 (or (−∞, c] ∈ V (F) if c < 0 , respectively).

Proof. There is no loss of generality to assume that c > 0 , as the proof in other case follows by considering −F(G) . Let G be the function at which F takes the value c. Take two points t_{1} < t_{2} in the interval (0, 1) so that G(t_{1}) < G(t_{2}) . Suppose ε is a small positive constant and consider the function G_{ε} defined as follows: G_{ε}(t) = G(t) outside the interval [t_{1}, t_{2}] ; G_{ε}(t) increases linearly on the interval [t_{1}, t_{1} + ε] from G(t_{1}) to G(t_{2}) ; G_{ε}(t) is equal to G(t_{2}) on [t_{1} + ε, t_{2}] . Note that I(G_{ε}) is greater than I(G) and I(G′_{ε}) tends to ∞ as ε tends to zero. It is also clear that if t_{2} is sufficiently close to t_{1} then F(G_{ε}) > 0 . By a standard epsilon delta argument, one can easily show that F(G_{ε}) → ∞ as ε → 0 . On the other hand, on applying Observation 1, we see the set V (F) is connected, which proves the observation.

We clearly do need some condition on F except for continuity in Observation 2 as, for instance, the example of F(x_{1}, x_{2}, …, x_{m}) = sin(x_{1} + … + x_{m}) shows.

Observation 3. Let c be a real number in the interval (0, 1) . For every function G(t) ∈ L let G_{c}(t) be the function defined to be 0 on [0, c] , and G((t − c)/(1 − c)) on [c, 1] . Then G_{c}(t) ∈ L, and x_{λ}(G) = x_{λ}(G_{c}) for every λ.

Proof. Note that I(G_{c}^{λ}) = (1 − c)I(G^{λ}) and I(G_{c}^{′2}) = I(G^{′2})/(1 − c) . Hence
x_{λ}(G_{c}) = I(G_{c}^{λ})I(G_{c}^{′2}) = (1 − c)I(G^{λ})I(G^{′2})/(1 − c) = x_{λ}(G),

and the result follows (as the first claim is trivial).

The main consequence of this result is the following one. Suppose the function G(t) ∈ L is equal to zero in the interval [0, c] , and is positive in (c, 1] . On applying the inverse transformation to that of Observation 3, we can replace G by another function from L which is positive in (0, 1] , and which gives the same value of the functional x_{λ} at it (and so the same value of every functional F(x_{1}, x_{2}, …, x_{m}) ) as is its value at G. We can therefore assume that G is always positive in (0, 1] .

7. Proof of Theorem 1.

Lemma 1. (Causchy–Swarz [24]) For any two functions f and g in the space L^{2}[0, 1] the inequality I(|fg|)^{2} ⩽ I(f^{2})I(g^{2}) holds with equality if and only if there is a constant c such that f(t) = cg(t) almost everywhere in [0, 1] .

On applying Lemma 1 to x_{λ} we deduce that

x_{λ}(G) = I(G^{λ})I(G^{′2}) ⩾ I(G′G^{λ/2})^{2}.

Since

I(G′G^{λ/2}) = ∫_{0}^{1}G^{λ/2}dG = \frac{2}{λ + 2}

for λ > −2 (and equals ∞ for λ ⩽ −2 ) we obtain the inequality as required. The equality holds if almost everywhere in the interval [0, 1] we have

G′(t) = cG^{λ/2}(t)

with a constant c. Suppose that G′ exists in (0, 1) . After solving the above diferential equation and substituting G(0) = 0 , G(1) = 1 , we obtain the function G(t) = t^{2/(2−λ)} if −2 < λ < 2 . These are clearly not the only functions for which the equality holds (see Observation 3).

We will show now that there is no function G ∈ L for which the above differential equation holds almost everywhere if λ ⩾ 2 . Suppose that there is one. By Observation 3, then there is another one, say G, which is positive on (0, 1] , and for which the functional takes the same value 4/(λ+2)^{2} . The latter satisfies the differential equation with some (may be different) c. This new c is positive, since G(t) > 0 on (0, 1] . Consider first the case λ = 2 . In the interval (0, 1] we define the function h(t) = log G(t) . Then h′(t) = c almost everywhere in [0, 1] . Consequently,

− log G(ε) = log G(1) − log G(ε) = ∫_{ε}^{1} h′(t)dt = c(1 − ε).

The right–hand side here is bounded above by c, whereas the left–hand side tends to ∞ as ε → 0 , a contradiction. In the case λ > 2 , one can argue similarly, by taking the function h(t) = G(t)^{1−λ/2} .

All said above implies that in the range −2 < λ < 2 the inequality is sharp and becomes an equality with, say, the function G(t) = t^{2/(2−λ)} . This also shows that the inequality in the range λ ⩾ 2 is strict. We need, therefore, to construct some sequence of functions in L to show that the strict inequality is sharp. In the case λ = 2 it suffices to take G_{n}(t) = t^{n} . Then x_{2}(G_{n}) = n^{2}/(4n^{2} − 1) tends to 1/4 as n → ∞.

Suppose that λ > 2 . We define G_{n}(t) to be n^{(λ−4)/(λ−2)}t in the interval [0, 1/n] and G_{n}(t) = (n+1−nt)^{2/(2−λ)} in the interval [1/n, 1] . Then, by an easy computation,

I(G_{n}^{λ}) = \frac{λ − 2}{(λ + 2)n} − \frac{λ − 2}{(λ + 2)n^{2λ/(λ−2)}} + \frac{1}{(λ + 1)n^{(3λ−2)/(λ−2)}} ,

I(G_{n}^{′2}) = \frac{4n}{λ^{2} − 4} + n^{(λ−6)/(λ−2)} − \frac{4}{(λ^{2} − 4)n^{4/(λ−2)}} .

So that the product of these, namely x_{λ}(G_{n}) , tends to 4/(λ + 2)^{2} if n → ∞, as required. This completes the proof.

8. Proof of Theorem 2.

We begin with case b = 0 . If a = 0 then, clearly, c = 0 . Let a > 0 . From Theorem 1 and from Observation 2, it follows that x_{1}(G) takes all values in the interval [4/9, ∞) . This proves that if b = 0 then only the alternatives for (a, 0, c) to be an admissible triple are (i) and (ii).

Suppose now that b = 1 . If a = 0 then, from Theorem 1 and from Observation 1 again, we obtain that (0, 1, c) is an admissible triple if c > 1/4 . This gives the alternative (iii). Let a > 0 . On applying Lemma 1, we deduce that

ax_{1}(G) + x_{2}(G) = I(aG + G^{2})I(G^{′2}) ⩾ I(G′\sqrt{aG + G^{2}})^{2}.

By a standard computation of the integral,

I(G′\sqrt{aG + G^{2}}) = ∫_{0}^{1}\sqrt{aG + G^{2}}dG = \frac{(a + 2)\sqrt{a + 1} − a^{2} log (\sqrt{1/a} + \sqrt{1 + 1/a})}{4}.

It suffices to take the square of the last expression to obtain the lower bound for c as stated in (iv).

Let θ be a positive solution of the equation cosh θ = 1 + 2/a. Then with the function G(t) = (a/2)(cosh(θt) − 1) the inequality becomes an equality. Once again we remark that this choice is not unique. One can apply the inverse transformation to that of Observation 3 to obtain many such functions. On applying Observation 2, we complete the proof in the case b = 1 .

Finally, let b = −1 . If a = 0 then we get the alternative (v) analogously to (iii). We postpone the less trivial case a < 1 and suppose now that a ⩾ 1 . On applying Lemma 1, we obtain

ax_{1}(G) − x_{2}(G) = I(aG − G^{2})I(G^{′2}) ⩾ I(G′\sqrt{aG − G^{2}})^{2}.

Once again, by a standard computation of the integral,

I(G′\sqrt{aG − G^{2}}) = ∫_{0}^{1}\sqrt{aG − G^{2}}dG = \frac{−2(a − 2)\sqrt{a − 1} + a^{2} arccos(1 − 2/a)}{8},

and by taking a square, we obtain the lower bound as in (vii). With the function, for instance, G(t) = (a/2)(1 − cos(θt)) , where θ = arccos(1 − 2/a) , the equality is true throughout. Thus, we have the alternative (vii), by Observation 2.

We are left with the case 0 < a < 1 . It suffices to show that ax_{1}(G) − x_{2}(G) takes a positive and a negative value. Then, by Observations 1 and 2, this expression takes every real value. Let us consider the functions G_{n}(t) = nt on the interval [0, 1/n] and G_{n}(t) = 1 on [1/n, 1] . Then, as I(G_{n}) = 1 − 1/(2n) , I(G_{n}^{2}) = 1 − 2/(3n) , and I(G_{n}^{′2}) = n, we obtain that

ax_{1}(G_{n}) − x_{2}(G_{n}) = (a − 1)n − \frac{a}{2} + \frac{2}{3} < 0

for n sufficiently large.

It suffices to prove that this expression also takes a positive value. Let us fix a positive number δ to be chosen later. Set

G_{n}(t) = t exp{−n(1 − t)^{δ}}.

Clearly, I(G_{n}^{′2}) > 0 . Also, on replacing n(1 − t)^{δ} by t, we obtain that

I(G_{n}) = ∫_{0}^{1}t exp\{−n(1 − t)^{δ}\}dt = \frac{1}{δn^{1/δ}} ∫_{0}^{n} t^{1/δ−1}(1 − (t/n)^{1/δ}) exp\{−t\}dt.

It follows that

I(G_{n}) = \frac{Γ(1/δ)(1 + ε_{n})}{δn^{1/δ}},

where ε_{n} → 0 as n → ∞. Here Γ is the gamma function.

Also,

I(G_{n}^{2}) = ∫_{0}^{1}t^{2} exp\{−2n(1−t)^{δ}\}dt = \frac{1}{δ(2n)^{1/δ}} ∫_{0}^{2n} t^{1/δ−1}(1−(t/(2n))^{1/δ})^{2} exp\{−t\}dt.

Consequently,

I(G_{n}^{2}) = \frac{Γ(1/δ)(1 + μ_{n})}{δ(2n)^{1/δ}},

where μ_{n} → 0 as n → ∞. It is now convenient to take δ so that

δ < \frac{log 2}{log(1/4a)} < \frac{1}{2}.

Then

\frac{I(G_{n})}{I(G_{n}^{2})} > 2^{1/δ−1} > \frac{2}{a}

for n sufficiently large. It follows that ax_{1}(G_{n}) − x_{2}(G_{n}) > x_{2}(G_{n}) > 0 , and the proof of Theorem 2 is completed.

9. Proof of Theorem 3.

On the set Ω , the quantity I(G^{2})/I(G) can only have values in the interval

[\frac{3}{5}, \frac{3}{2} − \frac{1}{π − 2}].

Indeed, the value at the left endpoint 3/5 is attained at the function G_{0}(t) whereas the value at the right endpoint

ν = \frac{3}{2} − \frac{1}{π − 2} = 0.624030803 …

is attained at G_{π/2}(t) . The above claim now follows by continuity and from the direct (but cumbersome) proof that the function I(G_{ω}^{2})/I(G_{ω}) is increasing in ω. We will give the proof of this fact later in Lemma 4.

Set

H_{p}(ϱ) = (∫_{0}^{1}(1 − ϱτ)^{p/(p−1)}dτ)^{(p−1)/p} = (\frac{(p − 1)(1 − (1 − ϱ)^{(2p−1)/(p−1)})}{(2p − 1)ϱ})^{(p−1)/p}.

Lemma 2. For every fixed p > 1 the function (1 − ϱ/2)/H_{p}(ϱ) is decreasing (in ϱ) in the interval [0, 1] .

Proof. Note that

\frac{(p − 1)2^{p/(p−1)}}{2p − 1} (\frac{1 − ϱ/2}{H_{p}(ϱ)})^{p/(p−1)} = \frac{(1 + t)^{σ}(1 − t)}{1 − t^{σ+1}},

where σ = p/(p − 1) and 1 − ϱ = t. We need, therefore, to show that the function on the right–hand side is increasing in t. Its derivative is equal to

(1 + t)^{σ−1}(1 − t^{σ+1})^{2}((σ − 1)(1 − t^{σ+1}) − (σ + 1)(t − t^{σ})).

It is non–negative if

1 − t^{σ+1} − \frac{σ + 1}{σ − 1}(t − t^{σ}) ⩾ 0.

The latter function equals 1 at t = 0 and equals 0 at t = 1 . Its derivative

−(σ + 1)t^{σ} − \frac{σ + 1}{σ − 1}(1 − σt^{σ−1})

is non–positive on [0, 1] if (σ − 1)t^{σ} − σt^{σ−1} + 1 is non–negative. This is the case, because the derivative of the latter σ(σ − 1)(t^{σ} − t^{σ−1}) is non–positive and so it is decreasing from 1 to 0 on the interval [0, 1] . This completes the proof of Lemma 2.

From now on, we consider the functional

F_{p}(G) = (∫_{0}^{1}(x_{1}(G) − τx_{2}(G))^{p/(p−1)}dτ)^{(p−1)/p} = x_{1}(G)H_{p}(I(G^{2})/I(G)).

The functional satisfies the conditions of the observations, provided that p > 1 . Also, for every fixed number ϱ let F_{p}(G, ϱ) = x_{1}(G)H_{p}(ϱ) . Then ζ_{p} = inf F_{p}(G, ϱ) , where the infimum is taken over every number ϱ in (0, 1) and over every function G ∈ L such that ϱ = I(G^{2})/I(G) . We will first show that for every ϱ either inside or outside the interval [0.6, ν] the values of F_{p}(G, ϱ) are greater than or equal to some value of the functional with ϱ in this interval and with the function G in Ω .

Lemma 3. Suppose ϱ is a fixed real number in the interval [3/5, ν] . Then there is an ω ∈ [0, π/2] such that

inf x_{1}(G) = x_{1}(G_{ω}),

where the infimum is taken over every function G ∈ L such that ϱ = I(G^{2})/I(G) .

Proof. Fix ϱ in the above interval and take an ω so that ϱ = I(G_{ω}^{2})/I(G_{ω}) . Such ω exists, by to be proved later Lemma 4. From the proof of Theorem 2 (vii) on substituting a = 2/(1 − cos ω) we now have that

x_{1}(G)(1 − \frac{ϱ(1 − cos ω)}{2}) ⩾ \frac{(2ω − sin(2ω))^{2}}{32(1 − cos ω)^{3}}

with equality if G = G_{ω} . This shows that for every fixed ϱ the infimum inf x_{1}(G) is attained at a function G_{ω} ∈ Ω , as claimed.

Fix a number ϱ, 0 < ϱ < 1 , and consider all functions G ∈ L so that ϱ = I(G^{2})/I(G) . For every such a ϱ the function H_{p}(ϱ) is increasing in p. Also, from integral representation of H_{p}(ϱ) it easily seen that for every fixed p the function H_{p}(ϱ) is decreasing in ϱ. If ϱ < 3/5 then for every G ∈ L, by Corollary 1, we have

F_{p}(G, ϱ) = x_{1}(G)H_{p}(ϱ) ⩾ \frac{4H_{p}(ϱ)}{9} > \frac{4H_{p}(3/5)}{9} = F_{p}(G_{0}, 3/5).

On the other hand, if ϱ > ν then for every function G ∈ L

F_{p}(G, ϱ) = x_{1}(G)H_{p}(ϱ) > x_{1}(G_{π/2})H_{p}(ν) = F_{p}(G_{π/2}, ν).

Indeed, if the opposite inequality is true, namely if there is a function G ∈ L such that

x_{1}(G)H_{p}(ϱ) ⩽ x_{1}(G_{π/2})H_{p}(ν),

then, by Lemma 2, we would have that

x_{1}(G)(1 − ϱ/2) < x_{1}(G_{π/2})(1 − ν/2).

Here, the right–hand side is equal to x_{1}(G)−(1/2)x_{2}(G) and the left hand side is equal to x_{1}(G_{π/2}) − (1/2)x_{2}(G_{π/2}) = ζ_{∞} , a contradiction to Corollary 3.

We conclude that if ϱ is outside the interval [0.6, ν] , then the infimum of F_{p}(G) is not attained. For every ϱ in the interval [0.6, ν] there is an ω in the interval [0, π/2] such that ϱ = I(G^{2})/I(G) with G ∈ Ω , and, vice versa, for every ω in [0, π/2] there is a ϱ = ϱ_{ω} for which ϱ_{ω} = I(G^{2})/I(G) . By Lemma 3, for every ϱ in the interval [0.6, ν] the infimum of the functional F_{p}(G, ϱ) is actually the minimum and is attained at a certain function G from Ω . Supose now that the infimum of F_{p}(G) , where G ∈ L, is attained as the limit of some sequence of functions {G_{n}}. We already know that there is no generality to assume that I(G_{n}^{2})/I(G_{n}) is bounded below by 3/5 and bounded above by ν . Hence there is a subsequence (by abuse of notation, we denote it by {G_{n}} again) such that

I(G_{n}^{2})/I(G_{n}) → ϱ^{∗},

as n → ∞ with ϱ^{∗} ∈ [3/5, ν] . As we are interested in the infimum, there is no loss of generality again, to consider the functions such that, e.g., x_{1}(G_{n}) ⩽ 10 . There is a number x_{1}^{∗} , therefore, so that on a certain subsequence {G_{n}} (here, we do not change the notation once again) x_{1}(G_{n}) → x_{1}^{∗} as n → ∞. Let ω^{∗} be such that ϱ^{∗} = I(G_{ω^{∗}}^{2})/I(G_{ω^{∗}}) . By Lemma 3, we have that

x_{1}(G_{ω^{∗}}) ⩽ x_{1}^{∗}.

On the other hand, by the definition of x_{1}^{∗} and ϱ^{∗} , we have the inequality x_{1}^{∗}H_{p}(ϱ^{∗}) ⩽ x_{1}(G_{ω^{∗}})H_{p}(ϱ^{∗}) . This is only possible if last two inequalities are both equalities, thus

ζ_{p} = x_{1}^{∗}H_{p}(ϱ^{∗}) = x_{1}(G_{ω^{∗}})H_{p}(ϱ^{∗}) = min F_{p}(G_{ω}).

We now conclude the proof of Theorem 3, by establishing the following statement.

Lemma 4. The function

I(G_{ω}^{2})/I(G_{ω})

is increasing in ω in the interval [0, π/2] .

Proof of the lemma. By a standard computation of integrals, we find that

\frac{I(G_{ω}^{2})}{I(G_{ω})} = \frac{6ω − 8 sin ω + sin(2ω)}{4(1 − cos ω)(ω − sin ω)}.

To check that this formula is correct at ω = 0 , one may use the Taylor expansions of the involved trigonometric functions. The right–hand side here is equal to

\frac{3}{2(1 − cos ω)} − \frac{sin ω}{2(ω − sin ω)}.

The derivative of this expression is equal to

\frac{−3 sin ω}{2(1 − cos ω)^{2}} + \frac{sin ω − ω cos ω}{2(ω − sin ω)^{2}} .

It is positive if

(sin ω − ω cos ω)(1 − cos ω)^{2} − 3 sin ω(ω − sin ω)^{2} > 0

for ω ∈ (0, π/2] . On dividing by the positive function 2 sin(ω/2) , we obtain the function

2(sin ω − ω cos ω)( sin(ω/2))^{3} − 3(ω − sin ω)^{2} cos(ω/2).

It is equal to zero at zero, so it suffices to show that its derivative, namely

sin(ω/2)(\frac{9}{2} + \frac{3}{2}ω^{2} − 8ω sin ω − \frac{5}{4}ω sin(2ω) − \frac{9}{2} cos(2ω)),

is positive on (0, π/2] .

We consider the Taylor series of the function

ϕ(ω) = \frac{9}{2} + \frac{3}{2}ω^{2} − 8ω sin ω − \frac{5}{4}ω sin(2ω) − \frac{9}{2} cos(2ω) := ∑_{k=0}^{∞} a_{k}ω^{k}

at ω = 0 , and show that the sign of ϕ in the interval (0, 3) is the same as that of the first non–zero coefficient. Using the expansions sin z = ∑_{k=1}^{∞}(−1)^{k−1}z^{2k−1}/(2k − 1)! , and cos z = ∑_{k=1}^{∞}(−1)^{k−1}z^{2k−2}/(2k − 2)! , we see that a_{k} with k odd are all equal to zero. Also, a_{0} and a_{2} are both equal to zero. For the remaining coefficients we have the formulae

a_{2k} = (−1)^{k−1}(\frac{9}{k} 4^{k−1} − 8 − \frac{5}{2}4^{k−1})/(2k − 1)!.

Here, k ⩾ 2 . One can now easily compute more coefficients like a_{4} = 0 , a_{6} = 0 , a_{8} = 1/210 , a_{10} = −13/25200 , a_{12} = 43/1663200 . The sign of a_{2k} is (−1)^{k} .

The function ϕ can, therefore, be written as follows

ϕ(ω) = ∑_{k=0}^{∞} a_{k}ω^{k} = \frac{ω^{8}}{210} − \frac{13ω^{10}}{25200} + \frac{43ω^{12}}{1663200} + ∑_{k=7}^{∞} a_{2k}ω^{2k}.

Note that

4^{k−1} < (5/2)4^{k−1} + 8 − (9/k)4^{k−1} < 4^{k}

if k ⩾ 6 . Since 0 < ω < 3 , we deduce the following inequality for the absolute values in the range k ⩾ 6 :

|a_{2k}ω^{2k}| = (−1)^{k}a_{2k}ω^{2k} > \frac{4^{k−1}ω^{2k}}{(2k − 1)!} > \frac{4^{k+1}ω^{2k+2}}{(2k + 1)!} > |a_{2k+2}ω^{2k+2}|,

because 12 × 13 = 156 > 144 = 16 × 9 . This inequality is also true for k = 4 , as 1/210 > (9 × 13)/25200 , and for k = 5 , as 13/25200 > (9 × 43)/1663200 . The function ϕ(ω) is, therefore, positive in the interval (0, 3) . The latter contains the interval (0, π/2) , thus the proof of Lemma 4 is completed.

10. Computations.

Here, we give the table of computations of the minimum of the functional F_{p} , by Theorem 3. The values presented in the table are those of 2/ζ_{p} , which is more convenient for the main application: the numerical value in estimate of M_{p}(α) from below (see Section 2). All 10 given decimal digits for the value of ζ_{p} are true (as well as are those for the extremal value of the respective ω = ω_{p} ). The obtained function is rapidly increasing from 4.5 to 64/π^{2} = 6.4845557 . . .. Recall that the extremal function can be obtained by substituting ω into the formula G_{ω}(t) = (1 − cos(ωt))/(1 − cos ω).

p :=  ω :=  2/ζ_{p} :=
1.00  0.0000000000  4.5000000000
1.0001  0.0258179014  4.5039161853
1.01  0.2562351948  4.6871099991
1.05  0.5563990344  5.0888506114
1.10  0.7602978811  5.3866751418
1.20  1.0038538044  5.7246332387
1.30  1.1433469232  5.9086733046
1.40  1.2304592813  6.0225832793
1.50  1.2890206221  6.0994671085
1.60  1.3307724922  6.1546640810
1.70  1.3619264506  6.1961422045
1.80  1.3860135013  6.2284183470
1.90  1.4051699438  6.2542332844
2.00  1.4207572438  6.2753424528
3.00  1.4936805684  6.3755271804
4.00  1.5189484290  6.4108669595
5.00  1.5317510186  6.4289105418
6.00  1.5394848289  6.4398572661
7.00  1.5446620536  6.4472055338
8.00  1.5483703914  6.4524790707

9.00  1.5511572347  6.4564477950
10.0  1.5533280681  6.4595426331
20.0  1.5624996028  6.4726509701
50.0  1.5675745301  6.4799275275
99.9  1.5691993484  6.4822607841

From the table we see that, for instance,

(\frac{1}{d} ∑_{k=1}^{d} | log |α_{k}||^{p})^{1/p} > \frac{6}{d}(\frac{log log d}{log d})^{3}

if p ⩾ 1.4 .

Acknowledgements. This paper was initiated when the second named author was visiting and using the facilities of the Department of Mathematics at Bielefeld University. His research was partially supported by the Lithuanian State Science and Studies foundation. The research of the first named author was supported by the Max Planck Institut for Mathematics, Bonn. We also thank to Denis Vasilyev who performed the computations presented in Section 10.

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V. Bentkus Institute of Mathematics and Informatics Akademijos 4 2600 Vilnius, Lithuania e-mail: bentkus@takas.lt

A. Dubickas Department of Mathematics and Informatics Vilnius University Naugarduko 24 2600 Vilnius, Lithuania e-mail: arturas.dubickas@maf.vu.lt