An integral that counts the zeros of a function

Lemma 1.1

Let f : ℝ → ℝ be a 2π-periodic C²function with only simple zeros, i.e. points x withf(x) = 0 ≠ f′(x). Then, the number n of zeros of f in [0,2π)equals twice the winding number of the curve γ: [0, 2π) → ℝ², x ↦ (f′(x), f(x)) with respect to the origin. Hence

Definition 2.1

A zero x₀ ∈ (a, b) of a function f ∈ C⁰(a, b) ∩ C¹((a, b)∖{x₀}) will be called admissible provided

If f extends continuously to a (or b) and f(a) = 0 (or f(b) = 0), we will say that f has an admissible zero in a (or b) if

Remarks

1. An admissible zero is necessarily an isolated zero. In fact, if the zero x₀ is an accumulation point of zeros of f then, by Rolle’s Theorem, it is also an accumulation point of zeros of f′ and the limits in Definition 2.1 cannot be plus or minus infinity.

2. The condition on the limits given in (2.1) is in fact equivalent to

   limx→x0ddxln⁡|f(x)|=∞.(2.2)

   Indeed, if 2.2) holds true, it follows that x₀ is an isolated zero of f, hence f does not change its sign on (x₀,x₀+ε) and on (x₀-ε,x₀) for ε > 0 small enough. Moreover 0 < |f(x)| < |f′(x)| on a punctured neighborhood of x₀. Hence, f′ cannot change sign and the claim follows by distinction of cases. The condition (2.2) is slightly more compact than (2.1), however, (2.1) is easier to handle in the calculations below.

3. A simple zero x₀ ∈ (a, b) of f ∈ C¹(a, b), i.e.f(x₀) = 0 and f′(x₀) ≠ 0 is admissible. It suffices to consider x₀ = 0:

   limx↘0f′(x)f(x)=limx↘0f′(0)+o(1)f(0)+xf′(0)+o(x)=limx↘01x⋅f′(0)+o(1)f′(0)+o(1)=∞.

   The limit x ↗ 0 is analogous.

4. If f(x₀) = f′(x₀) = 0 and f′ is monotone on (x₀,x₀+ϵ) and on (x₀−ϵ,x₀) for some ϵ > 0, then x₀ is an admissible zero: Indeed, for x₀ < x < x₀+ϵ and f′ non-decreasing (if f′ is non-increasing consider -f) on (x₀, x₀+ϵ), we have f(x) = ∫x0xf′(t) dt ≤ (x−x₀)f′(x) and thus f′(x)f(x)≥1x−x0→∞ for x ↘ x₀. The argument for the limit x ↗ x₀ is analogous.

5. If f ∈ C^k(a, b) and x₀ ∈ (a, b) is a zero of multiplicity k > 1, i.e.f^(ℓ)(x₀) = 0 for all ℓ = 0,…, k−1 and f^(k)(x₀) ≠ 0, then x₀ is admissible. This follows easily by an iterated application of L–Hôpital’s rule. Hence the zeros of real-analytic functions and a fortiori zeros of polynomials are admissible.

6. If f(x) = |x−x₀|^αg(x) for a C¹-function g with g(x₀) ≠ 0 and 0 < α ∈ ℝ, then x₀ is an admissible zero of f.

7. Every f ∈ C¹([a, b]) can be extended to f̃ ∈ C¹(I), where I ⊃ [a, b] is an open interval and the limits

   limx↗af′(x)f(x) and limx↘bf′(x)f(x)(2.3)

   can be defined via f̃, provided f(a),f(b) ≠ 0. If f has an admissible zero in a (or b), f can be extended antisymmetrically with respect to a (or b) to an extension f̃ for which a (or b) is an admissible zero. We will henceforth use this particular extension when computing limits like in (2.3).

Example 2.2

1. The function f₁ ∈ C⁰(ℝ) ∩ C^∞(ℝ∖{0}), x ↦ |x| has an admissible zero in x = 0 (see Remark 6 above).

2. The C^∞-function

   f2(x):=exp−1x2,x≠00,x=0,

   has an admissible zero of infinite multiplicity at x = 0 (see Remark 4 above).

3. An example of an isolated zero which is not admissible is given by the C^∞-function

   f3(x):=f2(x)(sin⁡(1x3)+2),

   which vanishes (together with all derivatives) in 0 but the corresponding limits (2.1) do not exist.

Definition 2.3

A function f : [a, b] → ℝ belongs to 𝓐^k([a, b]), k ∈ ℕ, if the following holds:

1. f ∈ C⁰([a, b]).

2. f has only admissible (and therefore finitely many) zeros x₁ < … < x_n and f|_(xi,xi+1) (i = 1,…,n−1), f|_(a,x1) and f|_(xn,b) are of class C^k+1.

3. There exists a partition a = y₁ < y₂ < … < y_m = b such that f|_(yi,yi+1) is of class C^k+2 for all i = 1,…,m−1.

If f ∈ 𝓐⁰([a, b]), f will be called admissible.

Remarks

1. Observe that 𝓐^k+1([a, b]) ⊂ 𝓐^k([a, b]) for all k ∈ ℕ by construction.

2. Every analytic function is in 𝓐^∞([a, b]).

3. f:[−1,1] → ℝ, x ↦ |x| is in 𝓐^∞([a, b]).

4. If f is admissible, then x ↦ (f′(x),f(x)) is not necessarily a continuous curve.

As a building block of the intended results we need the following: For σ ∈ [−∞,∞], let

where h:ℝ → ℝ is any piecewise continuous function such that the improper integral ∫−∞∞h(x)dx=1. Then we have the following theorem (recall (2.3) in order to make sense of the limits that appear).

Theorem 2.4

Let f ∈ 𝓐⁰([a, b]). The number of zeros n(f) of f in [a, b] is given by

and the number of zeros n̊(f) of f in (a, b) by

Proof

Consider first the case, where f(a),f(b) ≠ 0. Then the zeros of f are given by a < x₁ < x₂ < … < x_n(f) < b. The integrand of

is a priori undefined whenever f vanishes or whenever f″ is undefined. We decompose the integral and compute the resulting improper integrals using unilateral limits. Since f is admissible, we have

for all j = 1,…, n(f)−1, where H_x: = H(f′(x)/f(x)). Integrating over a neighborhood of a point y where f″ is undefined does not introduce further boundary terms since lim_{x ↘ y} H_x − lim_{x↗ y}H_x = 0. Hence

and therefore

The computation above suggests that n(f) > 1 but one can check that formula (2.6) holds true for n(f) = 1 and n(f) = 0 as well.

If f has zeros in a and b and therefore x₁ = a, x_n(f) = b, computation (2.5) gives

According to (2.3), limx↘bHx−limx↗aHx=1 H_x = 1 and (2.7) becomes

and hence (2.8) counts the zeros of f in [a, b] since it reduces to (2.6) if f(a),f(b) ≠ 0 and one can check that the remaining cases f(a) = 0≠ f(b) and f(a) ≠ 0 = f(b) are also covered. Let now

Since

we conclude that n̊(f) counts the zeros of f in (a, b).□

Remarks

1. Putting g(x) : = f′(x)/f(x), the integrand in Theorem 2.4 reads −(h ∘ g)(x) g′(x).With respect to the signed Borel-Lebesgue-Stieltjes-Measure dg(x): = g′(x) dx (see [9]), the integral can be written more compactly as

   −∫abh(g)dg.

2. If h(x): = 1/(π(1+x²)), i.e. h equals the Cauchy Density and f is an admissible 2π-periodic function, then the number n of zeros of f in [0,2π) equals

   n= 1π[∫02πf′(x)2−f(x)f″(x)f(x)2+f′(x)2dx+limx↗2πarctanf′(x)f(x)−limx↗0arctanf′(x)f(x)]== 1π∫02πf′(x)2−f(x)f″(x)f(x)2+f′(x)2dx,(2.9)

   since the integral-free terms cancel out in this case. In this way we obtain Lemma 1.1 as a corollary of Theorem 2.4. Observe that a 2π-periodic C² function with an odd number of zeros on [0,2π) gives rise to a curve x ↦ (f′(x),f(x)) having a half-integer valued winding number. This idea, further developed, leads to a generalized version of the Residue Theorem (see [3]).

Proposition 2.5

Let h:ℝ → ℝ be continuous and h(x) ∼Cx2for|x|→∞.Then, the integrand in Theorem 2.4

is continuous if f ∈ Cⁿ([a, b]), n ≥ 2, has only zeros of multiplicity ≤ n.

Proof

It suffices to show that I is continuous in 0 if x = 0 is a zero of f of multiplicity n. Then, by Taylor expansion, we have

where r_i are continuous functions with lim_{x → 0}r_i(x) = 0. Using these expressions in I, we get

for continuous functions s_i with lim_{x → 0}s_i(x) = n. Thus

for x → 0.

Definition 2.6

The multiplicity μ_f(x₀) of a zero x₀ of f ∈ 𝓐⁰ is defined to be

Example 2.7

1. A function f ∈ Cⁿ, n ≥ 2 with 0 = f(x₀) = f′(x₀) = … = f⁽ⁿ⁻¹⁾(x₀) ≠ f⁽ⁿ⁾(x₀) has a zero of multiplicity n in x₀: the Definition 2.6 is compatible with the usual notion of multiplicity.

2. The function f(x) = |x|^r, r > 0 has a zero of multiplicity r in x = 0.

3. The function

   f(x)=1ln⁡|x|,x≠0    0,x=0

   has a zero of multiplicity 0 in x = 0.

4. The function f₂ in Example 2.2.2 has a zero in x = 0 with μ_f2(0) = ∞.

Proposition 2.8

Let h:ℝ → ℝ be a piecewise continuous function such that h(x) = O(1/x²) for |x| → ∞ and let f ∈ 𝓐⁰([a, b]) ∩ W^2,1(a, b) have only zeros of positive multiplicity in the sense of Definition 2.6. Furthermore we assume that for each zero x₀we have a neighborhood U such that either f″(x)≡ 0 on U∖{x₀} or

where z₁,z₂,… denote the countably many zeros of f″ in U. Then

Proof

Choose neighborhoods U₁,…, U_n of the n zeros of f, which do not (with the possible exception of the respective zero itself) contain singular points of f″ or zeros of f′ and let

Since |f| ≥ η for some η > 0 on the complement U^c and W^2,1(a, b)↪ C¹([a, b]) we can estimate

Consider now wlog the neighborhood U_i of the zero x_i = 0 and assume U_i = (−ε,ε) for some ε > 0. We need to show that I|_(−ε,ε) ∈ L¹. Since h(x) = O(1/x²) for |x| → ∞, there exists a constant C > 0 such that

Note that ff″/f′² ∈ L¹(−ε,ε) if and only if N ∈ BV(−ε,ε), where N(x) = x−f(x)/f′(x) denotes the Newton-Operator of f and BV(−ε,ε) denotes the space of functions g:(−ε,ε) → ℝ of bounded variation. It follows from the admissibility of the zero that N:(−ε,ε)∖{0} → ℝ can be continuously extended to N(0) = 0 and it holds that

for x ≠ 0. Let μ > 0 denote the multiplicity of the zero according to Definition 2.6. It holds that

According to the mean value theorem we have N(x)/x = N′(ξ) for some ξ between 0 and x and deduce that N ∈ C¹(−ε,ε). The Taylor expansion of N around x = 0 is given by

In any case there exists a constant K > 0 such that

We will now show that N ∈ BV([0,ε)), the argument on (−ε,0] being similar. We start by noticing that N is absolutely continuous on [δ,ε) for every 0 < δ < ε since x, f(x) and f′(x) are absolutely continuous and f′(x) ≠ 0 on [δ,ε). In particular, N ∈ BV([δ,ε)) for every 0 < δ < ε.

We will now distinguish two cases: If f″≡ 0 on (0,ε), then N ≡ 0 and we are done. In the remaining case we first consider the case when the set of zeros of f″ in (0,ε) is empty: Then N is monotone on [0,ε) and hence N ∈ BV([0,ε)). Otherwise the zeros of f″ in [0,ε) are given by z₁ > z₂ > … and we may set δ: = z₁. According to (2.11) and since the zeros of f″ are precisely the zeros of N′ we can estimate the total variation of N on (z_k+1,z_k) by

The total variation of N on [0,ε) is bounded by

where the series converges by assumption and the integral is finite since N ∈ BV([δ,ε)). We conclude that N ∈ BV([0,ε)), which finishes the proof.□

Remark

The key estimate (2.11) in the proof above follows from the admissibility and the positive multiplicity of the zeros. We will however formulate a variant of Proposition 2.8 below (Proposition 2.10), which covers admissible functions that have zeros of ill-defined multiplicity for which (2.11) still holds true: Take e.g. the C¹ function f:x ↦ x³ ≤ ft(sin(1/x) + 2) + x which has an admissible zero in x = 0, but for which μ_f(0) does not exist, however, (2.11) holds true since f(x)/(xf′(x)) is bounded near 0 – in fact

Example 2.7.3. shows an admissible function for which (2.11) does not hold true. In the mentioned example, the first derivative is unbounded. But even functions with higher regularity may behave in such a pathological way near an admissible zero, that (2.11) does not hold true, as the following example shows:

Example 2.9

Let

Then f(x) = ∫0xk(t) dt is of class C³ and has an admissible zero in x = 0 but f(x)/(xf′(x)) is unbounded near 0.

Proposition 2.10

Let h:ℝ → ℝ be a piecewise continuous function such that h(x) = O(1/x²) for |x| → ∞ and let f ∈ 𝓐⁰([a, b])∩ W^2,1(a, b) be such that that for every zero x₀of f there exists a relatively open neighborhood U ⊂ [a, b] such that

on U ∖{x₀} and such that either f″≡ 0 on U∖{x₀}, or

where z₁,z₂,… denote the countably many zeros of f″ in U∖{x₀}. Then

Proof

Choose neighborhoods U₁,…, U_n of the n zeros of f, which do not (with the possible exception of the respective zero itself) contain singular points of f″ or zeros of f′ such that (2.12) holds on each punctured neighborhood. As in the proof of Proposition 2.8 we obtain ∥I∥_L¹(U^c) < ∞, where U = U₁∪ … ∪ U_n and the estimate (2.10). Let wlog 0 be a zero of f and let (−ε,ε) be its respective neighborhood for some ε > 0. As in the proof of Proposition 2.8, we are done if we show that N ∈ BV([0,ε)). The condition 0 < |f(x)/(xf′(x))| < K͠ on (−ε,ε)∖{0} implies that

from which we conclude that N extends continuously to [0,ε) (where N(0) = 0) and

This is just estimate (2.11) with K = K͠ + 1. The rest of the proof is exactly the same as the one of Proposition 2.8.□

Lemma 3.1

Let all the zeros of f ∈ 𝓐⁰([a, b])∩ C²([a, b]) have well-defined multiplicities. Then there exists c ∈ ℝ such that g₁has no poles.

Proof

If x₀ is a zero of f, we have that g₁(x) → μ_f(x₀) as x → x₀. In other words g₁ extends continuously to the zeros of f. Hence there are open neighborhoods of the zeros of f, where g₁ has no poles. On the complement of these neighborhoods, there exists a number δ > 0 such that |f(x)| ≥ δ. Hence f′(x)²+cf(x)² ≥ f′(x)²+cδ². If we choose c > δ⁻²∥ff″∥_{C⁰([a, b])}, then g₁ has no poles. In particular, if f is analytic, this choice of c ensures that g₁ is analytic as well.□

Theorem 3.2

Let f:[a, b] → ℝ be an analytic function and choose c in the definition of g₁such that g₁is analytic. If h(x) = O(1/x²) for |x| → ∞, then I_g1, I_g2 ∈ L^∞(a, b) and if f has n_ℓ zeros of multiplicity ℓ in [a, b] and n̊_ℓ zeros of mutliplicity ℓ in (a, b), then

Proof

We first prove the L^∞-bounds: It suffices to show that I_g1 and I_g2 are bounded near the zeros of f. Let x₀ be a zero of multiplicity k and write (locally) f(x) = (x−x₀)^kj(x), where j is analytic and j(x₀)≠0. Since

we find the limits

If

is bounded near x₀, the claim follows. Since ≤ |h≤ (f′(x)}/{f(x))|≤ C ⋅f(x)²}/{f′(x)² and

we obtain I_g1 ∈ L^∞(a, b). For I_g2, observe that

and therefore

Proceeding as for g₁ we find

and hence I_g2 ∈ L^∞(a, b). The computation of the integrals is done as in the proof of Theorem 2.4.□

Remark

If f ∈ 𝓐¹([a, b])∩ C²([a, b]) only has zeros of well-defined multiplicities and if the set of zeros of f in (a, b) is given by N̊ and the set of zeros of f in [a, b] by N, then

Lemma 3.3

Let 𝓝 be the set of sequences with natural entries of which only finitely many are non-zero. Then the map 𝓕:𝓝 → ℝ defined by 𝓕(k₁,…) = ∑ℓ=1∞kℓexp1ℓis injective.

Proof

The difference 𝓕(k₁,… )−𝓕( k1′,… ) is equal to the finite sum

If this sum vanishes, k_ℓ = kℓ′ for all ℓ by the von Lindemann-Weierstrass theorem (see [7, §3]).□

Corollary 3.4

Let f:[a, b] → ℝ be analytic. If f has n_{ℓ zeros of multiplicity ℓ in} [a, b] and n̊_ℓ zeros of mutliplicity ℓ in (a, b), then

Example 3.5

Let f(x) = cos(2 x) + x² Sin (2x) −12ex+x−24. Using Theorem 2.4 and 3.2 on [0,2π] we obtain

and we conclude that f has two zeros in (0,2π) and a double zero on the boundary of [0,2π].

Example 3.6

Let f(x) = x⁷−2x⁶+x⁵−x³+2x²−x have n_ℓ zeros of multiplicity ℓ on ℝ. By Theorem 2.4 and 3.2 on ℝ (observe that the boundary terms of the integrals cancel out in this case) we find that

Hence (n₁,…) either equals (1,2,0,…) or (2,0,1,…). In particular n_ℓ = 0, for ℓ ≥ 4. Using again Theorem 3.2 we get

Since 1⋅ e + 2 ⋅ e ≈ 6.0157 and 2⋅ e + 1⋅ e3 ≈ 6.8322 we conclude that f has two simple zeros and one of multiplicity 3.

Theorem 4.1

Let f satisfy the assumptions in Theorem 2.4. If

then one can replace the integral in Theorem 2.4 by the finite sum T_N(I) and round the result to the closest integer to get the values n(f) and n̊(f),respectively.

Example 4.2

Let f:ℝ → ℝ, x ↦ J₀(x), be the zeroth Bessel function of the first kind. If h is the Cauchy density, one can verify that ∥I″∥L∞<1π. We want to compute the number of zeros of J₀ on [0,2π] by Theorem 4.1. It suffices to employ the trapezoidal rule with only

equidistant intervals. We find

and thus

and hence, J₀ has two zeros on [0,2π].

If we compute the number of zeros of J₀ on [0,100π], we have to choose

(Actually, a finer analysis shows that a much smaller N suffices). In this case, we get

and

hence we conclude that J₀ has n = 100 zeros on [0,100π], in accordance with the well known distribution of zeros of J₀. Surprisingly, the routine CountRoots of Mathematica ^™ is giving up on this simple problem after giving it some thought.