ON ASYMPTOTIC SOLUTIONS OF ANALYTIC EQUATIONS

JACEK STASICA

imuj preprint 2003/01

Abstract. Sufficient conditions for the existence of analytic solution in complex and real cases are given.

1. Introduction.

For fixed holomorphic function F (z, t) defined in a neighborhood of (a, b) ∈ ℂ_{z,t}^{n+1} we show the existence of a number N ∈ ℕ such that the existence of a continuous quasi-solution g of order at least N (i.e. f(z, g(z)) = O|z−a|^{N}) implies the existence of holomorphic solution (i.e. f(z, h(z)) ≡ 0 with some holomorphic function h defined in a neighborhood of a, and h(a) = b). We are also able to estimate the number N effectively.

The famous Artin theorem says that the existence of a formal solution of the system of analytic equations implies the existence of an analytic one. Our theorem implies Artin theorem in the case of one equation. Moreover we give some sufficient conditions for existence of Nash solution in real case. As an application we obtain a straightforward proof of the theorem saying that the set of smooth points of a semi-algebraic set is semi-algebraic as well. This theorem about semialgebraicity plays an important role in semialgebraic geometry. In [Ł1] S. Łojasiewicz proved that for locally semialgebraic sets the notion of a Nash smooth point coincides with analytic one. Moreover if Γ ⊂ ℝ^{n} is an analytic submanifold, then for the germ Γ_{a} of Γ at a ∈ Γ we have the equivalence:

Γ_{a} is a Nash ⇔ Γ_{a} is semialgebraic.

Hence the semialgebraicity of smooth points one can obtain also following Łojasiewicz’s method for the analogous theorem for semianalytic sets.

1991 Mathematics Subject Classification. 14P10, 14P20, 32A10, 32A60.

Key words and phrases. Analytic function, holomorphic function, asymptotic solution, semialgebraic set, smooth point, topographic submanifold.

This paper is partially supported by the KBN grant number 159/P03/2001/21.

For the convenience of the reader we recall some basic definitions.

Consider a monic polynomial P in t ∈ ℂ whose coefficients a_{1}, …, a_{n} are holomorphic functions in an open subset Ω of ℂ^{n}, i.e. the function

P (z, t) = t^{n} + a_{1}(z)t^{n−1} + ⋯ + a_{n}(z).

The function

D(P)(z) = ∏_{i<j}(t_{i}(z) − t_{j}(z))^{2} = (−1)^{\binom{k}{2}} ∏_{j=1}^{k} \frac{∂P}{∂t} (z, t_{j}(z)),

where t_{1}(z), …, t_{k}(z) is the complete sequence of roots of the polynomial t → P (z, t) is called the discriminant of the polynomial P.

We say that a holomorphic function f(z, t) in a neighborhood of zero in ℂ^{n} × ℂ is t-regular if f(0, t) ≢ 0 in a neighborhood of zero in ℂ.

Let P (z, t) be a distinguished polynomial for which DP (z) ≡ 0 in some neighborhood 0 ∈ ℂ^{n}. In such a case there exists a distinguished polynomial redP (z, t) for which D(redP)(z) ≢ 0 and {P = 0} = {redP = 0} in some neighborhood of 0.

We say that a subset of ℝ^{n} is semi-algebraic if it is described by polynomials on ℝ^{n}. Thus, the class of semi-algebraic subsets of ℝ^{n} is the algebra of subset of ℝ^{n} which is generated by the family of sets {P > 0}, where P is a polynomial. Equivalently, E ⊂ ℝ^{n} is semi-algebraic if there are polynomials P_{i} and P_{ij}, i = 1, …, p, j = 1, …, q, such that

E = ⋃_{i=1}^{p}{x | P_{i}(x) = 0, P_{ij}(x) > 0, j = 1, …, q}.

Let G be an open subset of ℝ^{n}. We say that an analytic function f : G ⟶ ℝ is a Nash function at the point a ∈ G, if and only if W (x, f(x)) = 0 in a neighborhood of a with a polynomial W ≢ 0 in ℝ_{x,t}^{n+1}. A Nash function on G is an analytic function on G, which is Nash at each point of G.

2. Asymptotic solutions.

Lemma 1. Let K_{0}, K be open discs in ℂ, \overline{K_{0}} ⊂ K and a ∈ K_{0}. Let P (z, t) be a monic polynomial with holomorphic coefficients in K. Let us assume that D(P)(z) ≠ 0 in K \ {a}. Then each holomorphic in K \ K_{0} function h for which P (z, h(z)) ≡ 0 extends holomorphically on K.

Proof. Each of the open sets B_{ν} = K ∩ Π_{ν}, where Π_{1}, Π_{2} and Π_{3}, Π_{4} are open halfplanes cut of from ℂ by two different lines passing by a is homeomorphic with an open disc, hence each of the sets W_{B_{ν}} = W ∩(B_{ν} ×ℂ), where W = {P (z, t) = 0}, is a finite sum of graphs of holomorphic functions on B_{ν}. Therefore the restrictions of function h to the open connected sets B_{0ν} = (K \ \overline{K_{0}}) ∩ Π_{ν} has holomorphic extensions h_{ν} on B_{ν}, which are compatible. Thus ⋃ h_{ν} is the extension of function h, which from the Riemann Theorem extends holomorphically by a. □

Lemma 2. Let P (z, t) be a monic polynomial of degree k with holomorphic coefficients for which D(redP)(z) ≠ 0 in an open set G ⊂ ℂ_{z}^{n}. Let us define

τ(z) := \frac{1}{2} min_{i≠j} |t_{i}(z) − t_{j}(z)|

and

η(z, t) := min_{i} |t − t_{i}(z)|,

where t_{1}(z), …, t_{l}(z) is the sequence of all different roots of the polynomial t ↦ P (z, t). Then if there exists a continuous function g for which η(z, g(z)) < τ(z) in G, then there exists a holomorphic function h such that P (z, h(z)) ≡ 0 and |h(z) − g(z)| ≤ |P (z, g(z))|^{\frac{1}{k}} in G.

Proof. It follows from the assumption that for each a ∈ G there exists the uniquely determined root h(a) of the polynomial t ↦ P (a, t) such that |h(a) − g(a)| < τ(a). Hence |h(a) − g(a)| = η(a, g(a)) ≤ |P (a, g(a))|^{\frac{1}{k}}. Since g is a continuous function, following implicit function theorem, the root h(z) must coincide in some neighborhood of a with a holomorphic root t(z) of the polynomial P. □

Theorem 1. Let F (z, t) be a holomorphic function in a neighborhood of (a, b) ∈ ℂ_{z,t}^{n+1}, t-regular in (a, b) and let L > 0. Then there exists N ∈ ℕ such that, if F (z, g(z)) = O(|z − a|^{N}) for z → a with some continuous in a neighborhood of a function g, g(a) = b, then there exists a holomorphic, in a neighborhood of a, function h, h(a) = b such that F (z, h(z)) ≡ 0. Moreover, if in addition, the function g is holomorphic we have h(z) − g(z) = O(|z − a|^{L}) for z → a.

Proof. We can assume that a = b = 0, and by the Weierstrass Preparation Theorem that F (z, t) = t^{k} + a_{1}(z)t^{k−1} + ⋯ + a_{k}(z) is a distinguished polynomial of degree k > 0 with bounded holomorphic coefficients a_{j} in some Ω × Δ open neighborhood of 0 ∈ ℂ^{n} where Ω ⊂ ℂ_{u}^{n−1} and Δ ⊂ ℂ_{s}^{1}, z = (u, s). We can also assume that polynomial redF has bounded holomorphic coefficients in Ω × Δ and that {F = 0} = {redF = 0}. Moreover, by the Weierstrass Preparation Theorem we can assume that D(redF) = HD_{0} in Ω × Δ, where D_{0}(u, s) = s^{r} + b_{1}(u)s^{r−1} + ⋯+b_{r}(u) is a distinguished polynomial with holomorphic, bounded in Ω coefficients b_{j} and inf |H| > 0 in Ω × Δ. Finally we can assume that redD_{0} has holomorphic bounded coefficients in Ω. Let us take τ and η from Lemma 2 for the polynomial redF in the set D(redF) ≠ 0. Of course in this set we have |D(redF)(z)| ≤ Mτ(z) with some constant M > 1 and |F (z, t)| ≥ η(z, t)^{k}. Hence, due to Lemma 2 for each open set G ⊂ {D(redF) ≠ 0} the inequality

(i)  |F (z, g(z))| < M^{−k}|D(redF)(z)|^{k}

with some continuous function g on G gives us the existence of a holomorphic root h of the polynomial F in G, such that

(ii)  |h(z) − g(z)| ≤ |F (z, g(z))|^{\frac{1}{k}} in G.

Put |u| = σ. The roots of the polynomial s ↦ D_{0}(u, s), are O(σ^{\frac{1}{r}}). Hence, taking M sufficiently big, there exists δ > 0 such that for σ > 0 sufficiently small on the set

G_{σ} := B_{σ} × (K′_{σ} \ \overline{K_{σ}}),

where B_{σ} = \{|u| < 2δ\}, K_{σ} = \{|s| < Mσ^{\frac{1}{r}} \}, K′_{σ} = \{|s| < 2Mσ^{\frac{1}{r}} \}, we have the inequality |D_{0}(u, s)| ≥ δσ and G_{σ} ⊂ Ω × Δ. Let D(redD_{0})(u) = ∑_{|s|≥q} c_{s}u^{s}. Hence there exists ε ∈ (0, 1) such that for sufficiently small σ > 0 we have

max_{|u|=σ} |D(redD_{0})(u)| ≥ 3εσ^{q}.

It follows that for some u_{0}, |u_{0}| = σ, we have |D(redD_{0})(u)| ≥ 3εσ^{q}. But when σ > 0 is sufficiently small, the roots s_{i}(u_{0}) of the polynomial s ↦ D_{0}(u_{0}, s) fulfils |s_{i}(u_{0})| ≤ \frac{1}{2} therefore |s_{i}(u_{0}) − s_{j}(u_{0})| ≥ 3εσ^{q} for s_{i}(u_{0}) ≠ s_{j}(u_{0}). Observe that for

K_{σ_{i}} = {|s − s_{i}(u_{0})| < εσ^{k}}

we have \overline{K_{σ_{i}}} ∩\overline{K_{σ_{j}}} = ∅ (for different roots) and we have in the complement of ⋃_{i} \overline{K_{σ_{i}}} the inequality |D_{0}(u, s)| ≥ ε^{r}σ^{rq}. Finally diminishing δ > 0, when σ is sufficiently small, we have

|D(redF)(z)| ≥ δσ^{rq}

in

G = G_{σ} ∪ (U_{σ} × (K_{σ} \ ⋃_{i} \overline{K_{σ_{i}}})) ⊂ Ω × Δ

for some open neighborhood U_{σ} ⊂ B_{σ} of the point u_{0}. Taking N > krq, the condition F (z, g(z)) = O(|z|^{N}) implies the inequality (i) for σ sufficiently small. Then there exists a holomorphic root h of the polynomial F in G. According to the Lemma 1, if u ∈ U_{σ}, the function h(u, ·) extends holomorphically on K_{σ′}. Hence h extends holomorphically on U_{σ} × K_{σ′} (using Cauchy formula for h(u, ·) on {|s| = \frac{3}{2}Mσ^{\frac{1}{2}} \}). Thus, from Hartogs Lemma, h extends holomorphically on B_{σ} × K_{σ′}. In case g is holomorphic let us take moreover N > kLr. According to (ii) we have |h(z) − g(z)| ≤ M̃|z|^{Lr}, with some M̃, in the set G_{σ} when σ is small enough. Hence in G_{σ} we have the inequality |h(z) − g(z)| ≤ M′σ^{L} with some constant M′ and due to maximum principle this inequality holds also in B_{σ} × K_{σ′}. So h − g = O(|z|^{L}). □

Remark. From the above proof it is easy to see that if F (z, t) is a polynomial, monic in t, then there exists N ∈ ℕ the same for all points z ∈ ℂ^{n}.

Theorem 2. Let Q(x, t) be a complex polynomial on ℝ_{x,t}^{n+1}, monic on t. Then there exists m > 0 such that if f, g are complex, continuous roots: Q(x, f(x)) = Q(x, g(x)) = 0 in a neighborhood of a ∈ ℝ^{n}, then the following implication holds:

f(x) − g(x) = O(|x − a|^{m}) ⇒ f = g

in a neighborhood of a.

Proof. We can assume that a = 0 and the discriminant D(Q)(z) ≠ 0 (replacing Q by redQ). Obviously, when |x| ≤ 1 we have |t′′(x) − t′(x)| ≥ M|D(Q)(x)| with some constant M for different roots t′′(x), t′(x) of the polynomial t ↦ Q(x, t). Let us take an open ball B = {|x| < R}, R < 1, in which f and g are defined. The number of connected components of the set B ∩ {D(Q) ≠ 0} is finite and each connected component is semi-algebraic. Let S be one of them. Assume, that 0 ∈ \overline{S}. The function

d : r ↦ sup_{S(r)} |D(r)|,

where S(r) = S ∩ {|x| = r}, is positive semi-algebraic. Hence d(r) ≥ r^{m−1} in (0, δ) with common m > 0, δ > 0 for all such components S. The assumption f ≢ g in each neighborhood of zero implies f ≠ g on one of this components S, because the set {f = g} ∩ S is open-closed in S. But |f(x) − g(x)| ≥ M|D(Q)(x)| on S which implies sup_{S(r)} |f(x) − g(x)| ≥ Mr^{m−1} for sufficiently small r. The last inequality contradicts our assumption that f(x) − g(x) = O(|x|^{m}). □

Theorem 3. Let Q be a real polynomial on ℝ_{x,t}^{n+1}, monic on t and let be L > 0. Then there exists N ∈ ℕ such that the following implication holds: if Q(x, ψ(x)) = O(|x − a|^{N}) for x → a, with some real analytic function ψ defined in a neighborhood of a ∈ ℝ^{n}, then there exists an analytic function φ defined in a neighborhood of a ∈ ℝ^{n} such that Q(x, φ(x)) ≡ 0 and φ − ψ = O(|x − a|^{L}) for x → a.

Proof. Replacing Q and ψ by their complexification, and taking m from Theorem 2 we get (due to Theorem 1 with L > m) a holomorphic function φ defined in a neighborhood of a such that Q(z, φ(z)) ≡ 0 and φ − ψ = O(|z − a|^{m}). But then also Q(z, \overline{φ}(z)) ≡ 0 holds. Moreover \overline{φ} − ψ = O(|z − a|^{m}) which implies \overline{φ} − φ = O(|z − a|^{m}). According to Theorem 2 we get \overline{φ} = φ hence the function φ is real. □

3. Semialgebraicity of smooth points.

Lemma 3. Let E be a semialgebraic subset of ℝ^{n}. If int E = ∅, then some nonzero polynomial on ℝ^{n} vanishes on E.

Proof. We have E = ⋃_{i=1}^{q} B_{i} with B_{i} = {P_{i} = 0} ∩ {P_{ij} > 0} with some polynomials P_{i} and P_{ij}. Now we can assume that each B_{i} is nonempty. Then P_{i} ≠ 0, since otherwise B_{i} should be open. Hence P_{1} · . . . · P_{q} is a required polynomial. □

Lemma 4. Every semialgebraic set E ⊂ ℝ^{n} is contained in an algebraic one V ⊂ ℝ^{n} of the same dimension.

Proof. Let π_{α} : ℝ^{n} ⟶ L_{α} be the natural projections onto the spaces L_{α} = ℝ_{x_{α_{1}},...,x_{α_{k+1}}}^{k+1}, where α = (α_{1}, …, α_{k+1}), 1 ≤ α_{1} < ⋯ < α_{k+1} ≤ n, and k = dim E. Then π_{α}(E) are semialgebraic of dimension at most k. Hence π_{α}(E) ⊂ {P_{α} = 0} for a nonzero polynomial P_{α}. Therefore E ⊂ V = ⋂_{α}{P_{α} ∘ π_{α} = 0}. Moreover dim V = k, since otherwise V would contain a semialgebraic leaf Γ ^{1} of dimension k + 1 and so there would exists an α such that int π_{α}(Γ) ≠ ∅ that would imply P_{α} ≡ 0. □

We say that a point a ∈ E is a smooth point of dimension k of the set E if it has a neighborhood in E which is an analytic submanifold of dimension k. By definition the dimension of E is equal to the maximum of the dimension of its smooth points.

Note that every polynomial on ℝ^{n} of degree r > 0 is monic of degree r with respect to each of its variables in some coordinate system.

1We call semialgebraic leaf any analytic submanifold which is a semialgebraic set.

Lemma 5. Let E ⊂ ℝ_{x}^{n} be a semialgebraic set of dimension ≤ k < n. Then in some linear coordinate system E is contained in a Weierstrass set

{P_{k+1}(u, x_{k+1}) = ⋯ = P_{n}(u, x_{n}) = 0},  (∗)

where u = (x_{1}, …, x_{k}), P_{j} is a monic polynomial on ℝ_{u,x_{j}}^{k+1}, j = k + 1, …, n. (Such a coordinate system we call a regular system for E).

Proof. By the previous lemma it suffices to give the proof for an algebraic set V ⊃ E of dimension k. Let P_{n} ⊃ ⋯ ⊃ P_{k+1} ⊃ P_{k} denote rings of polynomials on ℝ_{u,x_{k+1},...,x_{n}}^{n}, …, ℝ_{u,x_{k+1}}^{k+1}, ℝ_{u}^{k} (after suitable identification). Let us denote by R_{j} the ring of restrictions of polynomials from P_{j} to V. Let I = {P ∈ P_{n} | P_{|V} = o}. Changing in ℝ^{n}, …, ℝ^{k+1} successively coordinate system we find successively in I ∩ P_{j} monic polynomials with respect to x_{j}. It means in particular that w_{j} = x_{j}_{|V} is integral over R_{j−1}. Evidently R_{j} = R_{j−1}[w_{j}], it follows that w_{n}, …, w_{k+1} are integral over R_{k}, which means that there exist monic polynomials P_{j} ∈ P_{k}[x_{j}], j = n, …, k + 1, P_{j}_{|V} = 0. □

We say that the submanifold Γ ⊂ ℝ^{n} is topographic if it is the graph of an analytic mapping of an open subset of ℝ^{k} into the space ℝ^{n−k}.

Theorem 4. Let E ⊂ ℝ_{u,v}^{k+1} be a semialgebraic set contained in the Weierstrass set (∗). Then the set

Λ = {x ∈ E | U ∩ E is k-topographic submanifold for some nghb U of x}

is semialgebraic.

Proof. For a ∈ ℝ_{u}^{k}, b ∈ ℝ_{v}^{l} and δ, ε > 0 let us denote U_{abδε} = B(a, δ) × B(b, ε), where B(a, δ) = {|u − a| < δ}, B(b, ε) = {|v − b| < ε}. Let (a, b) ∈ E. The set E_{abδε} = E ∩ U_{abδε} is the graph of some continuous function B(a, δ) ⟶ ℝ_{v}^{l} if and only if \overline{E} ∩ U_{abδε} ⊂ E and for u ∈ B(a, δ) we have (u, v) ∈ \overline{E} ∩ U_{abδε} for exactly one v. Thus the set

F = {(a, b, δ, ε) | E_{abδε} is the graph of some continuous function on B(a, δ)}

is semialgebraic.

Let us take m from Theorem 2, the same, for all polynomials P_{j}, and N > m from Theorem 3 also the same for all polynomials P_{j}, and L = m. For c = {c_{α}}_{|α|≤N}, where c_{α} ∈ ℝ^{l} and α ∈ ℕ^{k}, we define the polynomial mapping P_{c} : u ↦ ∑_{|α|} c_{α}u^{α}. In the sequel we write (c_{0}, c′) = c. For a ∈ ℝ_{u}^{k}, c, C, N, δ > 0 we define the subset of ℝ_{u,v}^{k+l}:

W_{acCNδ} = {u ∈ B(a, δ), |v − P_{c}(u − a)| ≤ C|u − a|^{N}}.

It is enough to show that

Λ = {(a, b) ∈ ℝ_{u,v}^{k+l} | (a, b, δ, ε) ∈ F and E_{abδε} ⊂ W_{a(b,c′)CNδ} with some δ, ε, c′, C}

(∗∗)

because the latest set is semi-algebraic. Let (a, b) ∈ Γ. Obviously for some sufficiently small δ, ε > 0 we have (a, b, δ, ε) ∈ F and the set E_{abδε} is the graph of some Nash function φ; taking c′ such that x ↦ P_{(b,c′)}(x − a) is N-th Taylor polynomial for φ we have E_{abδε} = φ ⊂ W_{a(b,c′)CNδ} with some C when δ is sufficiently small.

Now, let us assume, that the condition (∗∗) holds. Then E_{abδε} is the graph of some continuous function ψ : B(a, δ) ⟶ ℝ_{v}^{l}. Hence Q(u, ψ(u)) = 0 on B(a, δ), where Q = (P_{k+1}, …, P_{n}). The graph of ψ is contained in the set W_{acCNδ}. Thus ψ(u)−P_{c}(u−a) = O(|u−a|^{N}). It follows that Q(u, P_{c}(u−a)) = O(|u−a|^{N}). Due to the Theorem 3 there exists an analytic function φ defined in a neighborhood of a such that Q(u, φ(u)) ≡ 0 and φ(u)−P_{c}(u−a) = O(|u−a|^{m}). Hence φ−ψ = O(|u−a|^{m}) and according to the Theorem 2 we have φ = ψ in a neighborhood of a. □

Theorem 5. Let E ⊂ ℝ^{n} be a semi-algebraic set. Then the set of its smooth points of dimension k E^{(k)}is semi-algebraic as well.

Proof. It is enough to prove the theorem for k = dim E, because we can proceed by induction.

For each φ : ℝ^{n} ⟶ ℝ^{n} isomorphism we define

λ(φ) = φ^{−1}(e_{k+1}) ∧ ⋯ ∧ φ^{−1}(e_{n}) ∈ Λ^{n−k}ℝ^{n},

where e_{1}, …, e_{n} is a canonical basis of ℝ^{n}. We say that a sequence of linear isomorphisms φ_{1}, …, φ_{r} of ℝ^{n} is complete if λ(φ_{1}), …, λ(φ_{r}) generate Λ^{n−k}ℝ^{n}. One can easily prove that:

(1) there exists complete sequence,

(2) any sequence that is sufficiently closed (in the natural topology) to a complete one is complete,

(3) if φ_{1}, …, φ_{r} is a complete sequence and the set E is smooth of dimension k at a then φ_{ν}(E) is k-topographic at φ_{ν}(a) for some ν.

We find a complete sequence φ_{1}, …, φ_{r} such that each φ_{i} is regular system for E. According to the Theorem 5 the set R_{ν} of points in which φ_{ν}(E) is k-topographic is a semi-algebraic set. Hence

E^{(k)} = ⋃ φ_{ν}^{−1}(R_{ν})

is semi-algebraic as finite sum of semi-algebraic sets. □

References

[A] M. Artin On the solutions of analytic equations. Invent. Math. 5, 277-291 (1968) [Ł1] S. Łojasiewicz Ensembles semi-analitiques. IHES, Bures-sur-Yvette (1965) [Ł2] S. Łojasiewicz Introduction to complex analytic geometry. Birkhäuser (1991)

(Jacek Stasica) Instytut Matematyki, Uniwersytet Jagielloński, ul. Reymonta 4, 30-059 Kraków, Poland

  E-mail address: jacek.stasica@im.uj.edu.pl