12. Proof of the division theorem

We shall give the proof of Theorem 2.1 for analytic
functions. In doing so we shall consider analytic functions
not of real but of complex variables. Passing to the complex domain, as is often the case, greatly simplifies the argument, but
does not thereby restrict the results obtained: representations for
real-analytic functions are obtained from the corresponding representations for complex-analytic ones (which we shall henceforth call holomorphic) by restriction to the real axis.
Let F(z, y) : C × Cⁿ → C be a holomorphic function having at the
origin multiplicity k < ∞, and let G(z, y) : C × Cⁿ → C be an
arbitrary holomorphic function. Let us also define the functions
k—1
P_k(z, a) = z^k + Σ aᵢzⁱ, a = (a₀,…,a_{k-1}) ∈ Cᵏ. (12.1)
i=0
LEMMA 12.1. For any functions G(z, y) and P_k(z, a) there exists
a neighborhood of zero in the space C × Cⁿ × Cᵏ, in which the
representation
G(z, y) = q(z, y, a) P_k(z, a) + r(z, y, a), (12.2)
k—1
r(z, y, a) = Σ rⱼ(y, a) zʲ, (12.3)
j=0
holds, where q(z, y, a) : C × Cⁿ × Cᵏ → C and rⱼ(y, a) : Cⁿ × Cᵏ → C are some
holomorphic functions.
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PROOF. It is clear that
k—1
P_k(ζ, a) — P_k(z, a) = (ζ—z) Σ bᵢ(ζ,a) zⁱ,
where bᵢ(ζ, a) : C × Cᵏ → C are holomorphic functions. Hence
P_k(ζ, a) — P_k(z, a)
—————————————
ζ — z
and, consequently,
G(z, y) G(ζ, y) P_k(ζ, a) — P_k(z, a)
——— = ——— — ———————————————— ×
ζ—z
G(ζ, y) (P_k(ζ, a) — P_k(z, a))
——— + ————————————————————— (12.4)
ζ—z
In the complex plane of the variable ζ, choose a closed contour Γ, going once around a fixed point z. For definiteness
we may take Γ to be the circle of radius ρ centered at z, i.e.
consisting of the points ζ = z + ρe^{it}, where i is the imaginary unit, t a real parameter. Using the integral formula of Cauchy, well known from complex analysis, and taking (12.4) into account, we obtain
G(z, y) = (1/2πi) ∮_Γ [G(ζ,y)/(ζ—z)] dζ = P_k(z, a) ×
(1/2πi) ∮_Γ [G(ζ,y)/((ζ—z) P_k(ζ,a))] dζ + r(z, y, a) = P_k(z, a) q(z, y, a) + r(z, y, a),
where r(z, y, a) has the form given in (12.3), and the functions
q(z, y, a) = (1/2πi) ∮_Γ [G(ζ,y)/((ζ—z) P_k(ζ,a))] dζ and rⱼ(y, a) = (1/2πi) ∮_Γ [G(ζ,y) ζ^{k—1—j} /P_k(ζ,a)] dζ.
1 A reader unfamiliar with this theorem may find it in any decent
textbook on the theory of functions of a complex variable.
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To complete the proof it remains to show that all the functions in (12.5) are holomorphic in some neighborhood of the origin
of the space C × Cⁿ × Cᵏ. It is clear that this condition will be
satisfied if all the integrals in (12.5) are taken over a contour Γ that
contains no zeros of the function P_k(ζ, a).
Note that at a = 0 the polynomial P_k(ζ, a) equals ζᵏ and, consequently, has the single root ζ = 0 of multiplicity k. Since the
roots of the polynomial P_k(ζ, a) depend continuously on a, for any ε > 0
there exists a δ > 0 such that all the roots of P_k(ζ, a), for |a| < δ, lie
within the ε-neighborhood of the point ζ = 0. Choose ε so that
the inequality ρ > 2ε holds, and the corresponding value of δ. Then,
for all (z, a) for which |z| ≤ ε and |a| < δ, for the points
ζ = z + ρe^{it} the inequality |ζ| > ρ — ε > ε holds, showing that
all the roots of the polynomial P_k(ζ, a) lie strictly inside the circle Γ.
Thus we have proved representation (12.3) for all (z, a) for
which |z| ≤ ε and |a| < δ. |
By the definition of multiplicity, (2.2) holds for F(z, y), and, consequently, (2.1) holds for the function F(z, 0). This yields the representation
F(z, 0) = zᵏ F₀(z), F₀(0) ≠ 0, (12.6)
with some holomorphic function F₀ (see also Problem 1.2).
THEOREM 12.1. For any functions F(z, y) and G(z, y) possessing the properties indicated above, there exists a neighborhood of zero
in the space C × Cⁿ, in which the representation
k—1
G(z, y) = q(z, y) F(z, y) + r(z, y), r(z, y) = Σ rⱼ(y) zʲ, j=0
holds, where q(z, y): C × Cⁿ → C and rⱼ(y): Cⁿ → C are holomorphic functions.
PROOF. Let us apply representation (12.3) to the functions F(z, y) and G(z, y) with the same P_k(z, a), k = n + 1. We obtain
F(z, y) = q_F(z, y, a) P_{k+n}(z, a) + f(z, y, a),
k+n
f(z, y, a) = Σ fⱼ(y, a), j=0
(12.8)
G(z, y) = q_G(z, y, a) P_{k+n}(z, a) + g(z, y, a),
k+n
g(z, y, a) = Σ gⱼ(y, a), j=0
(12.9)
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where q_F, q_G : C × Cⁿ × C^{k+n} → C and fⱼ, gⱼ : Cⁿ × C^{k+n} → C are holomorphic
functions. From (12.1) and (12.6) it follows that all fⱼ(0) = 0 and g_k(0) ≠ 0.
In a neighborhood of the origin of the space Cⁿ × C^{k+n} consider the system of equations
f₀(y, a) = 0, …, f_{k+n}(y, a) = 0 (12.10)
with respect to the unknowns a = (a₀,…,a_{k+n}). By the implicit
function theorem, for the (unique) solvability of system (12.10) with respect to a it suffices to show that the Jacobian matrix
(∂fᵢ/∂aⱼ)(0, 0) (12.11)
is nondegenerate at the origin. To this end, substitute y = 0 into (12.8),
whereby, taking (12.6) into account, we obtain the identity
q_F(z, 0, a) P_{k+n}(z, a) + f(z, 0, a) = zᵏ F₀(z).
Differentiating this last identity with respect to aⱼ, j = 0,…,k+n, and then setting a = 0, we obtain an identity in z:
∂q_F/∂aⱼ (z,0,0) zᵏ + q_F(z,0,0) zʲ + Σ ∂fᵢ/∂aⱼ(0,0) zⁱ = 0. (12.12)
Fix an index j ∈ {0,…,k+n} and consider the coefficients
of the monomials zⁱ for all i < j. It is clear that for any i < j the monomial
zⁱ occurs only in the last sum in (12.12), while for i = j the monomial
from the last sum has, in addition, the monomial q_F(0,0,0)... From (12.12) we thus find:
∂fᵢ/∂aⱼ(0,0) = 0 for i < j.
This means that the matrix (12.11) is triangular, and all the elements on its main diagonal equal —q_F(0,0,0) ≠ 0. Consequently, the matrix (12.11) is nondegenerate, and by the implicit function theorem there exists, in a neighborhood of the origin, a holomorphic mapping
α: Cⁿ → C^{k+n}, such that α(0) = 0 and fⱼ(y, α(y)) = 0 for all j ∈ {0,…,k+n}.
The identities (12.8), (12.9) hold for all (z, y), and therefore
one may substitute into them a = α(y). Then from (12.8), taking q_F(0,0,0) ≠ 0 into account,
we obtain:
F(z, y)
—————————— = P_{k+n}(z, α(y)) (12.13)
q_F(z,y,α(y))
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Using (12.13), from (12.9) we obtain
G(z, y) = q_G(z, y, α(y)) P_{k+n}(z, α(y)) + g(z, y, α(y)) =
q_G(z, y, α(y))
——————————— F(z, y) + g(z, y, α(y)) = q(z, y) F(z, y) + r(z, y),
q_F(z, y, α(y))
where
q_G(z, y, α(y))
q(z, y) = ———————————, rⱼ(y) = gⱼ(y, α(y)).
q_F(z, y, α(y))
The theorem is proved. |
PROBLEM 12.1. Prove that the functions q(z, y) and r(z, y) in representation (12.7) are unique, i.e. that for each pair F, G they are determined uniquely.
PROOF OF THEOREM 2.1. The division theorem 2.1 is obtained
as a simple corollary of Theorem 12.1. Indeed, let us write representation (12.7) for the function G(z, y) = z^{k+1}:
k
z^{k+1} = q(z, y) F(z, y) + r(z, y), r(z, y) = Σ rⱼ(y) zʲ. (12.14) j=0
Equating the coefficients of the monomials z^{k+1} on the left- and right-hand sides of equality (12.14), and taking (12.6) into account, we obtain q(0,0) ≠ 0. Let us move the term r(z, y) from the right- to the left-hand side of equality (12.14),
make the change of coefficients rⱼ(y) = —a_j(y), and finally multiply
both sides of the resulting equality by h(z, y) = 1/q(z, y). As a result we
obtain the required representation (2.3). |
REMARK 12.1. Lemma 12.1 and Theorem 12.1 can be carried over
to the class of smooth real functions. This requires
additional effort and special techniques, similar to those presented
by us in Section 5.3. The reader can find the details in [9, 15].