KRONECKER’S THEOREM, MAHLER’S MEASURE, AND LEHMER’S PROBLEM

VSEVOLOD F. LEV

Abstract. Following are introductory notes, presenting basic notions and results on the subjects, listed in the title.

1. Reciprocal polynomials

Definition . The reciprocal of a polynomial f ∈ ℂ[x] is the polynomial f^{∗}(x) = x^{n}f(x^{−1}), where n = deg f. A polynomial f is called reciprocal if f^{∗} = f; alternatively, this means that the coefficients a_{i} of f satisfy a_{n−i} = a_{i} for i = 0, …, n.

Remark . Sometimes, f is called reciprocal if f^{∗}(x) = af(x), where a is a constant. It is easy to show, however, that in this case a = ±1 and moreover, if f^{∗}(x) = −f(x) (that is if a = −1), then f(x) = (x − 1)f_{0}(x), where f_{0}(x) is reciprocal in our sense: f_{0}^{∗}(x) = f_{0}(x).

Observe, that if f ∈ ℝ[x] then |f^{∗}(x)| = |f(x)| for any x on the unit circle: this is because in this case |f^{∗}(x)| = |f(1/x)| = |f(x̄)| = |\overline{f(x)}| = |f(x)|.

Since the definition implies f(−1) = (−1)^{n}f(−1), it follows that f(−1) = 0 for any reciprocal polynomial f of odd degree, whence (x + 1) | f(x). Moreover, by the following lemma any such polynomial is of the form f(x) = (x + 1)f_{0}(x), where f_{0} is a reciprocal polynomial of even degree.

Lemma 1. If g is a reciprocal divisor of a reciprocal polynomial f (that is, g | f and both f and g are reciprocal), then f/g is reciprocal also.

Proof. Let n = deg f and m = deg g, so that n − m = deg f/g. Then

x^{n−m}f(x^{−1})/g(x^{−1}) = (x^{n}f(x^{−1}))/(x^{m}g(x^{−1})) = f(x)/g(x).
□

Similarly, it is also true that a product of reciprocal polynomials is reciprocal; however, the product can be reciprocal without each factor to be reciprocal, as in (x − α)(x − α^{−1}) = x^{2} − (α + α^{−1})x + 1.

Corollary 1. Roots of a reciprocal polynomial f(x) ∈ ℂ[x] of even degree deg f = 2m come in pairs:

f(x) = ∏_{i=1}^{m}(x − α_{i})(x − α_{i}^{−1}).

Proof. We use induction on m. Let α be a root of f. We notice that 0 = f(α) = α^{2m}f(1/α), whence 1/α is a root of f also. If α ≠ ±1, then f is divisible by (x−α)(x− 1/α), which is reciprocal, and the quotient is reciprocal by the above lemma, so we can use the induction hypothesis. Now if α = ±1, then from f′(x) = 2mx^{2m−1}f(x^{−1}) − x^{2m−2}f′(x^{−1}) it follows that α is a multiple root and f is divisible by (x − α)^{2} which is reciprocal, so we can apply the induction hypothesis to f/(x − α)^{2}. □

On the other hand, if f(x) ∈ ℚ[x] is an irreducible polynomial with rational coefficients and f(α) = f(1/α) = 0 for some α ∈ ℂ, then f is reciprocal (whence all roots of f come in pairs), with the only exception f(x) = a(x − 1): indeed, in this case both f and f^{∗} are constant multiples of the minimal polynomial of α, thus cf(x) = f^{∗}(x) = x^{deg f}f(x^{−1}) and substituting x = 1 we see that either c = 1, f(x) = f^{∗}(x), or f(1) = 0.

Corollary 2. A non-reciprocal irreducible polynomial f ∈ ℚ[x] cannot have roots on the unit circle, except for f(x) = a(x − 1).

Proof. If |α| = 1 and α is a root of f, then so is \overline{α} = 1/α. □

Theorem 1. For any reciprocal polynomial f(x) ∈ ℂ[x] of even degree deg f = 2m there is a unique polynomial g(x) ∈ ℂ[x] such that

f(x) = x^{m}g (x + \frac{1}{x}) .

Moreover,

i) deg g = m; ii) the coefficients of g lie in the ℤ-module, spanned by the coefficients of f, and vice versa; in particular, if M is one of ℝ, ℚ, or ℤ, then f[x] ∈ M[x] iff g ∈ M[x];

iii) the leading coefficient of g equals to the leading coefficient of f; iv) if f(x) has rational coefficients and is irreducible over ℚ, then so is g.

Proof. Write f(x) = a_{m}x^{2m} + ⋯ + a_{1}x^{m+1} + a_{0}x^{m} + a_{1}x^{m−1} + ⋯ + a_{m}. Then

f(x) = x^{m} (a_{m} (x^{m} + \frac{1}{x^{m}} ) + ⋯ + a_{1} (x + \frac{1}{x}) + a_{0})

= x^{m} (a_{m} (x + \frac{1}{x})^{m} + ⋯) ,

and the existence of g follows by induction on m. The uniqueness is easy: if g_{1}(x + x^{−1}) = g_{2}(x + x^{−1}), then letting t = x + x^{−1} we get g_{1}(t) = g_{2}(t) for all t. Plainly, if g(x) ∈ M[x], where M is a ℤ-submodule of ℂ, then f(x) ∈ M[x], and the converse follows from the above computation. If g ∈ ℚ[x] is reducible, then clearly so is f. □

Theorem 1 gives another proof of the fact that the roots of reciprocal polynomials of even degree come in pairs (α, 1/α): indeed, if α_{i} are the roots of g, then the roots of f are those of the quadratic polynomials x^{2}−α_{i}x+1, and the latter satisfy x_{i1}x_{i2} = 1. Since the roots of polynomials with real coefficients come in conjugate pairs also, we have the following lemma.

Lemma 2. Roots of a reciprocal polynomial of even degree with real coefficients come in quadruples (α, \overline{α}, α^{−1}, \overline{α}^{−1}). The quadruple becomes a pair if α lies on the real line or on the unit circle.

When do the exceptional cases occur?

Lemma 3. The roots of the polynomial x^{2} − αx + 1 are real iff α ∈ ℝ and |α| ≥ 2. The roots lie on the unit circle iff α ∈ ℝ and |α| ≤ 2.

Proof. If α ∈ ℝ, |α| ≥ 2 then the roots are real; conversely, if the roots are real, then so is α (which equals their sum), and we must have |α| ≥ 2. Next, if α ∈ ℝ, |α| ≤ 2, then the roots are conjugate (as their sum is real) and their product equals 1, hence the roots are on the unit circle; on the other hand, if the roots are on the unit circle, they are conjugate (their product equals 1), hence α (which is their sum) is real, and we must have |α| ≤ 2. □

Example 1. The sequence of cyclotomic polynomials f(x) = Φ_{m}(x) (for m ≥ 3, to ensure that deg f = φ(m) is even) determines the sequence of polynomials g_{m}(x) with the first elements

g_{3}(x) = x + 1, g_{4}(x) = x, g_{5}(x) = x^{2} + x − 1,

g_{6}(x) = x − 1, g_{7}(x) = x^{3} + x^{2} − 2x − 1, g_{8}(x) = x^{2} − x, g_{9}(x) = x^{3} − 3x + 1, g_{10}(x) = x^{2} − x − 1.

4 VSEVOLOD F. LEV

As

Φ_{m}(x) = ∏_{\begin{matrix}1≤r<m \\ (r,m)=1 \end{matrix}}(x − ζ_{m}^{r}) = ∏_{\begin{matrix}1≤r<m/2 \\ (r,m)=1 \end{matrix}} (x^{2} − 2x cos 2π \frac{r}{m} + 1)

(where ζ_{m} = e(1/m)), we have

g_{m}(x) = ∏_{\begin{matrix}1≤r<m/2 \\ (r,m)=1 \end{matrix}} (x − 2 cos 2π \frac{r}{m}) .

2. Kronecker’s theorem

Theorem 2 (Kronecker). An algebraic integer, all of whose conjugates are inside or on the unit circle, is a root of unity. Equivalently, if f(x) = (x−α_{1}) … (x−α_{n}) ∈ ℤ[x] is irreducible and |α_{i}| ≤ 1 (i = 1, …, n), then f is a cyclotomic polynomial.

Proof. For k = 1, 2, … consider the polynomials f_{k}(x) = (x − α_{1}^{k}) … (x − α_{n}^{k}) ∈ ℤ[x]. As |α_{i}| ≤ 1, the coefficients of all these polynomials are uniformly bounded, and therefore f_{k} = f_{l} for two distinct positive integers k and l. Thus, there is a permutation σ ∈ S_{n} such that α_{i}^{k} = α_{σi}^{l}, i = 1, …, n. Fix an index j and let r be the order of this index: σ^{r}j = j. Then

α_{j}^{k} = α_{σj}^{l}, α_{j}^{k^{2}} = (α_{σj}^{k})^{l} = α_{σ^{2}j}^{l^{2}} , …, α_{j}^{k^{r}} = α_{σ^{r}j}^{l^{r}} = α_{j}^{l^{r}},

and it follows that α_{j} is a root of unity. □

Algebraic integers in Kronecker’s theorem cannot be replaced by arbitrary algebraic numbers: if α_{1}, …, α_{n} are conjugates, then α_{1}/N, …, α_{n}/N are also conjugates and all lie in the unit circle for any sufficiently large rational integer N. Equivalently, the condition that f is monic cannot be dropped: consider f(x) = 2x^{2} − x + 2.

It can happen that all, but two roots of an irreducible polynomial f(x) ∈ ℤ[x] are on the unit circle.

Example 2. Let p ∈ ℤ[x] be a polynomial of degree n − 1 = deg p ≥ 2 with all the roots in the open interval (−2, 2). For a sufficiently large integer A, define g(x) = x^{n} + Ap(x). It is easy to verify that g has at least n − 1 roots in (−2, 2): indeed, if 2 < u_{1} < … < u_{n} ⊆ (−2, 2) are chosen so that each interval I_{i} := (u_{i}, u_{i+1}) contains precisely one root of p, then p(u_{i})p(u_{i+1}) < 0 (i = 1, …, n − 1) whence g(u_{i})g(u_{i+1}) < 0 (i = 1, …, n − 1) (as g(u_{i}) and p(u_{i}) are of the same sign for A large enough) and therefore each I_{i} contains a root of g. If we set f(x) = x^{n}g(x + 1/x), then by Theorem 1 and Lemma 3 the polynomial f has integer coefficients, is monic, reciprocal, and each of the n − 1 roots of g on (−2, 2) gives rise to a couple of

KRONECKER’S THEOREM, MAHLER’S MEASURE,  AND LEHMER’S PROBLEM  5

conjugate roots of f on the unit circle. Moreover, if f is reducible over ℚ then one of its irreducible factors has all of its roots on the unit circle, hence f is divisible by a cyclotomic polynomial of order m < 2n and

f(ζ_{m}) = 0 (ζ_{m} = e^{2πi/m}),
g(ζ_{m} + ζ_{m}^{−1}) = g(2 cos 2π/m) = 0,
Ap(2 cos 2π/m) = −2^{n} cos^{n} 2π/m.

This can hold, however, for a finite number of A only; thus, for A large f is irreducible and moreover, the two roots of f not on the unit circle are real, as so is the nth root of g.

On the other hand, as we saw above a non-reciprocal irreducible polynomial f(x) ∈ ℚ[x] cannot have roots on the unit circle (except the polynomials f(x) = a(x − 1)).

Suppose that g(x) ∈ ℤ[x] is monic, irreducible and all its roots are real and less then 2 in absolute value. Then f(x) = x^{deg g}g(x + 1/x) is also monic and irreducible, and all its roots are on the unit circle. By Kronecker’s theorem, f is then a cyclotomic polynomial. It follows that there are no polynomials g with the above listed properties, other than those considered in Example 1.

Lehmer’s problem is to determine whether the roots of a monic (irreducible) noncyclotomic polynomial f(x) = (x − α_{1}) … (x − α_{n}) are bounded away from the unit circle in the sense that

∏_{|α_{i}|>1} |α_{i}| > 1 + ε

with a positive absolute constant ε. It is conveniently reformulated in terms of Mahler’s measure.

3. Mahler’s measure

Definition. The Mahler’s measure of a polynomial f(x) ∈ ℂ[x] is defined by

m(f) = e^{∫_{0}^{1} ln |f(e^{2πit})| dt}.

Thus, m(f) measures the height of f on the unit circle. Equivalently, if f(x) = a_{n}(x − α_{1}) … (x − α_{n}), then

m(f) = |a_{n}| ∏_{i=1}^{n} max{1, |α_{i}|}.

The equivalence follows from

∫_{0}^{1} ln |f(e^{2πit})| dt = ℜ ∫_{0}^{1} ln f(e^{2πit}) dt = ℜ\frac{1}{2πi} ∫_{|z|=1} \frac{ln f(z)}{z} dz

6 VSEVOLOD F. LEV

and Jensen’s formula. Thus, m(f) also measures how far are the roots of f from the unit circle.

Both definitions readily imply multiplicativity of Mahler’s measure: m(fg) = m(f)m(g). From the second definition it is obvious that m(f) ≥ |a_{n}|, and m(f) ≥ |a_{0}| follows from

|a_{n}| ∏_{|α_{i}|>1} |α_{i}| ≥ |a_{n}| ∏_{i=1}^{n} |α_{i}| = |a_{0}|.

Moreover, if f(x) = a_{n}x^{n} + ⋯ + a_{0}, then

|a_{n−i}| = |a_{n}| | ∑_{k_{1}<...<k_{i}} α_{k_{1}} … α_{k_{i}} | ≤ \binom{n}{i}m(f),  (1)

whence

m(f) ≥ \binom{n}{i}^{−1}|a_{i}|; i = 0, …, n.  (2)

The symmetry in i is not occasional: if f^{∗}(x) = x^{deg f}f(x^{−1}) is the reciprocal of f, then any of the two definitions above readily shows that

m(f^{∗}) = m(f).

Write L(f) := |a_{n}| + ⋯ + |a_{0}| (the length of f). Summing up (1) over i = 0, …, n, we get

m(f) ≥ 2^{−n}L(f).

(Though this bound is attained for f(x) = (x ± 1)^{n}, it is difficult to believe that it cannot be improved.) In the opposite direction, it is obvious from the first definition of Mahler’s measure that

m(f) ≤ L(f);

indeed, |f(x)| ≤ L(f) for any x on the unit circle. The latter estimate can be somewhat improved as follows. First, notice that

∫_{0}^{1} |f(e^{2πit})|^{2} dt = |a_{n}|^{2} + ⋯ + |a_{0}|^{2} ≥ |a_{0}|^{2} = |f(0)|^{2}.

Next, enumerate the roots of f so that α_{1}, …, α_{s} are precisely those roots outside the unit circle, and consider the auxiliary polynomial g(x) = a_{n}(x − α_{1}) … (x − α_{s})(\overline{α}_{s+1}x − 1) … (\overline{α}_{n}x − 1). As |x − α| = |1 − \overline{x}α| = |\overline{α}x − 1| for any x on the unit circle, we have

|a_{n}|^{2} + ⋯ + |a_{0}|^{2} = ∫_{0}^{1} |f(e^{2πit})|^{2} dt

= ∫_{0}^{1} |g(e^{2πit})|^{2} dt ≥ |g(0)|^{2} = |a_{n}|^{2}|α_{1}|^{2} … |α_{s}|^{2} = (m(f))^{2};

KRONECKER’S THEOREM, MAHLER’S MEASURE,  AND LEHMER’S PROBLEM  7

therefore,

m(f) ≤ \sqrt{|a_{n}|^{2} + ⋯ + |a_{0}|^{2}}.

The following lemma gives an interpretation of Mahler’s measure as a resultant (or rather as a limit of resultants), provided the polynomial has no roots on the unit circle.

Lemma 4. Suppose that f(x) ∈ ℂ[x] has no roots on the unit circle. Then

m(f) = lim_{N→∞} |R[f(x), x^{N} − 1]|^{1/N},

where R[f, g] is the resultant of the polynomials f and g.

Proof.

ln m(f) = ∫_{0}^{1} ln |f(e^{2πiθ})| dθ

= lim_{N→∞} \frac{1}{N} ∑_{j=0}^{N−1} ln |f(e^{2πij/N})|

= lim_{N→∞} ln | ∏_{j=0}^{N−1} f(e^{2πij/N})|^{1/N}

= ln lim_{N→∞} |R[f(x), x^{N} − 1]|^{1/N}.

□

Corollary 3. Under the conditions of the lemma,

m(f) = lim_{p→∞} |R[f, Φ_{p}]|^{1/p},

where p runs over prime numbers, and Φ_{p}(x) = x^{p−1} + ⋯ + 1 is the cyclotomic polynomial of order p.

Proof. R[f(x), x^{p} − 1] = R[f(x), Φ_{p}(x)]R[f(x), x − 1]. □

Let f(x) = a(x − α_{1}) … (x − α_{n}). As

|R[f, x^{N} − 1]| = |a|^{N}|1 − α_{1}^{N}| … |1 − α_{n}^{N}|

(the resultant is the product of x^{N} − 1 over the roots of f times the Nth power of the leading coefficient of f), Lemma 4 gives yet another proof of the equivalence of two definitions of Mahler’s measure, provided the polynomial in question has no roots of modulus 1. In fact, it is easily seen that if p is any polynomial such that p(0) ≠ 0, and if f has no roots on the unit circle, then

lim_{N→∞} |R[f(x), p(x^{N})]|^{1/N} = (m(f))^{deg p}.

8 VSEVOLOD F. LEV

4. Lehmer’s problem

In terms of Mahler’s measure, the question of Lehmer is:

Does there exist an absolute constant ε > 0 such that m(f) > 1 + ε for any polynomial f(x) ∈ ℤ[x], not all of whose roots are roots of unity?

Clearly, it is suffices to consider monic irreducible polynomials f only. (Recall that m(fg) = m(f)m(g) and m(f) ≥ |a_{n}|, where a_{n} is the leading coefficient of f.)

It is easy to see that for any n ≥ 2 there exists ε_{n} > 0 such that m(f) > 1 + ε_{n} for any monic irreducible f of degree deg f = n; indeed, by (2) there are only finitely many polynomials f of degree n such that m(f) < 2, say. (Alternatively, one can observe that there are only finitely many algebraic integers of degree n which lie with all their conjugates in a bounded domain of the complex plane, as the coefficients of the minimal polynomials of all these algebraic integers are bounded; therefore, we see again that there are only finitely many f of degree n such that m(f) ≤ 2.) Some explicit estimates for ε_{n} are mentioned below. The point of Lehmer’s problem is, therefore, to establish estimates which are uniform in the degree n.

A simple way to estimate ε_{n} effectively (which, however, leads to extremely poor estimates) is to examine carefully the proof of Kronecker’s theorem. Given a polynomial f(x) = (x − α_{1}) … (x − α_{n}) ∈ ℤ[x], we define f_{k}(x) := (x − α_{1}^{k}) … (x − α_{n}^{k}) = x^{n} + a_{n−1}^{(k)}x^{n−1} + ⋯ + a_{1}^{(k)}x + a_{0}^{(k)}. Then by (2) we have |a_{i}^{(k)}| ≤ \binom{n}{i}m^{k}(f), whence for k = 1, …, K the integer n-tuple (a_{n−1}^{(k)}, …, a_{0}^{(k)}) lies in the region {(x_{1}, …, x_{n}) ∈ ℝ^{n}: |x_{i}| ≤ \binom{n}{i}m^{K}(f)\}. Since the total number of integer points in this region is

∏_{i=0}^{n−1} (2\binom{n}{i}m^{K}(f) + 1) ≤ 3^{n}m^{Kn}(f) ∏_{i=0}^{n−1} \binom{n}{i}

≤ 3^{n}m^{Kn}(f) (\frac{1}{n} ∑_{i=0}^{n−1} \binom{n}{i})^{n}

< 2^{n^{2}}m^{Kn}(f)

(we assumed here n ≥ 3 for simplicity), we have 2^{n^{2}}m^{Kn}(f) > K for any integer K ≥ 2, unless some of the polynomials f_{k} coincide in which case f is cyclotomic, as in the proof of Kronecker’s theorem. Choosing K = 2^{2n^{2}} we get

2^{n^{2}}m^{n2^{2n^{2}}}(f) > 2^{2n^{2}},
n2^{2n^{2}} ln m(f) > n^{2} ln 2,
m(f) − 1 > ln m(f) > 2^{−2n^{2}},

KRONECKER’S THEOREM, MAHLER’S MEASURE,  AND LEHMER’S PROBLEM  9

whence

m(f) > 1 + 4^{−n^{2}}.

An improvement was obtained by Schinzel and Zassenhaus (A refinement of two theorems of Kronecker, Michigan Math. J. 12 (1965), 81–85) who proved that if f(x) = (x − α_{1}) … (x − α_{n}) ∈ ℤ[x] is monic and has r real roots, then

max_{1≤i≤n} |α_{i}| ≥ 1 + 2^{−n−4+r};

plainly, this yields m(f) ≥ 1 + 2^{−n−4+r}. The idea behind the proof is as follows. Let α_{μ} be the root of f of the maximum absolute value. If α_{μ} is real, consider the non-zero integer

∏_{i=1}^{n}(α_{i}^{2} − 1);  (3)

provided that the roots of f are “very close” to the unit circle, the factors corresponding to real α_{i} are “very small”, while the rest are “not too large”, which yields a contradiction. If α_{μ} = re^{2πiθ} is not real then it can happen that none of the factors (α_{i}^{2} − 1) are small, and instead of (3) we consider the product

∏_{i=1}^{n}(α_{i}^{q} − 1),

where q is chosen using Dirichlet’s approximation theorem so that qθ is “almost integer”, while q is relatively small.

Thus, for a generic f the result of Schinzel and Zassenhaus shows that m(f) > 1 + 2^{−n−4}, while for those f with “almost all” roots real, it settles the Lehmer’s problem completely. (If all roots of f are real, and f is not cyclotomic, then as remarked above some of the roots of f are outside [−2, 2], whence m(f) > 2.) In fact, it follows from a result of Bilu (Uniform distribution of algebraic numbers close to the unit circle, Vestzi AN BSSR) that if m(f) is small, then the arguments of α_{i} are uniformly distributed in [0, 2π].

In 1979 E. Dobrowolsky (but better see D.C. Cantor and E.G. Straus, On a conjecture of D.C. Lehmer, Acta Arithmetica XLII 1982, 97–100) got a much better estimate, specifically,

m(f) > 1 + c (\frac{ln ln n}{ln n})^{3} .

Dobrowolski’s constant c was later improved by Cantor/Straus and Dubickas.

In 1971 Smyths showed that the answer to Lehmer’s question is positive at least in the “non-reciprocal case”. Indeed, Smyth proved that if f(x) ∈ ℤ[x] is a nonreciprocal polynomial, then m(f) ≥ m(x^{3} − x − 1) ≈ 1.325. We present a simplified version of Smyth’s argument which gives a somewhat weaker estimate m(f) ≥ \sqrt{5/4}; it is attributed by G. Everest (Notices of the AMS 20(3) (1998)) to Stewart.

We can assume that f(x) is irreducible and monic and moreover,|f(0)| = 1; consequently, f(x) has no roots on the unit circle, f(x) = ∏(x−α) and f^{∗}(x) = ∏(1−αx), where α runs over the roots of f(x). Consider

F (x) := \frac{f(x)}{f^{∗}(x)} = ∏_{|α|<1} \frac{x − α}{1 − αx} / ∏_{|α|>1} \frac{1 − αx}{x − α} = \frac{u(x)}{v(x)}.

Plainly, both u(x) and v(x) are analytic in the unit circle and for |x| ≤ 1 we can write u(x) = u_{0} + u_{1}x + u_{2}x^{2} + ⋯ and v(x) = v_{0} + v_{1}x + v_{2}x^{2} + · · ·.

Since f(x) is real, its roots come in conjugate pairs whence u(x) = ∏_{|α|<1}(x − α)/(1 − ᾱx). If |x| = 1 then |x − α| = |1 − αx̄| = |1 − ᾱx|, and this shows that |u(x)| = 1 in this case. Similarly, |v(x)| = 1 for |x| = 1 and therefore

|u_{0}|^{2} + |u_{1}|^{2} + |u_{2}|^{2} + ⋯ = |v_{0}|^{2} + |v_{1}|^{2} + |v_{2}|^{2} + ⋯ = 1.  (4)

Observe, that |u_{0}| = |v_{0}| = 1/m(f); to complete the proof we want to show that |u_{i}| and |v_{i}| are “not too small” for i ≥ 1. To this end we notice that F (x) has Taylor expansion with integer coefficients, hence we can write F (x) = F_{0} + F_{k}x^{k} + F_{k+1}x^{k+1} + ⋯ where |F_{0}| = 1, k ≥ 1 is an integer, and |F_{k}| ≥ 1. Now from u(x) = v(x)F (x) comparing the coefficients of x^{k} we get u_{k} = v_{0}F_{k} + v_{k}F_{0} so that |u_{k} − F_{0}v_{k}| ≥ |v_{0}| = 1/m(f), implying max{|u_{k}|, |v_{k}|} ≥ 1/(2m(f)). Substitution to (4) gives (1/m(f))^{2} + (1/(2m(f)))^{2} ≤ 1 whence m(f) ≥ \sqrt{5/4}, as claimed.

Department of Mathematics, University of Haifa at Oranim, Tivon 36006, Israel.

  E-mail address: seva@math.haifa.ac.il