Singularities of the Szegö kernel for certain weakly pseudoconvex domains in ℂ^{2}

Friedrich Haslinger* Journal of Functional Analysis 129 (1995), 406-427.

Abstract. We study the Szegö kernel for a class of weakly pseudoconvex domains in ℂ^{2}. Starting from an integral representation of the Szegö kernel on the boundary of the domain we derive a complete asymptotic expansion of the main part of the integral kernel, which gives, near boundary points of finite type, a full description of the singularities up to a bounded function. Finally we point out how one can extend the results to more general domains.

*Partially supported by a FWF–grant P8624–PHY of the Austrian Ministry of Sciences.

1980 Mathematics Subject Classification (1991 Revision). Primary 32F35; Secondary 32F40, 32M15.

Key words and phrases. Szegö kernel, pseudoconvex domains.

1. Introduction.

In this paper we consider the Szegö kernel of certain unbounded domains of ℂ^{2}. Let p : ℂ ⟶ ℝ_{+} denote a C^{1}–function and define Ω_{p} ⊆ ℂ^{2} by

Ω_{p} = {(z_{1}, z_{2}) ∈ ℂ^{2} : ℑ(z_{2}) > p(z_{1})}.

Weakly pseudoconvex domains of this kind were investigated by Nagel, Rosay, Stein and Wainger [NRSW1,NRSW2] , where estimates for the Szegö and the Bergman kernel of the domain were made in terms of the nonisotropic pseudometric defined in [N]. For the case where p(z) = |z|^{k} , k ∈ ℕ, Greiner and Stein [GS] found an explicit expression for the Szegö kernel of Ω_{p}, in which one can recognize the form of the pseudometric used for the nonisotropic estimates (see [D,K]). If p is a subharmonic function, which depends only on the real or only on the imaginary part of z, then one can find analogous expressions and estimates in [N].

Let H^{2}(∂Ω_{p}) denote the space of all functions f ∈ L^{2}(∂Ω_{p}), which are holomorphic in Ω_{p} and such that

sup_{y>0} ∫_{ℂ} ∫_{ℝ} |f(z, t + ip(z) + iy)|^{2} dλ(z)dt < ∞,

where dλ denotes the Lebesgue measure on ℂ. We identify ∂Ω_{p} with ℂ × ℝ, and denote by S((z, t), (w, s)), z, w ∈ ℂ , s, t ∈ ℝ, the Szegö kernel of H^{2}(∂Ω_{p}).

We will consider the case where the function p is of the form

p(z) = \frac{|x|^{α}}{α} ,

for α > 1 and z = x + iy.

In [HAS] we studied the the connection between the Szegö kernel of these unbounded domains in ℂ^{2} and the Bergman kernel of weighted spaces of entire functions of one complex variable.

The starting point is the following integral representation of the Szegö kernel on the boundary ∂Ω_{p} :

S((z, t), (w, s)) = \frac{1}{4π^{2}} ∫_{0}^{∞} ∫_{ℝ} \frac{τ exp(τ(η(z + \overline{w}) − p(z) − p(w) − i(s − t)))}{∫_{ℝ} exp(2τ(rη − p(r))) dr} dη dτ,

which can be found in a sligthly different form in Nagel [N] ( see also [HAS]). This formula for the Szegö kernel is closely connected with the CR structure of ∂Ω_{p}. (see Nagel [N], Hanges [HAN])

In the first part of this paper we consider the integral

F_{τ} (η) = ∫_{ℝ} exp(2τ(rη − p(r))) dr ,

where τ > 0 , η ∈ ℂ and p(r) = |r|^{α}/α.

In [CG] the above expression for the Szegö kernel is used to show that the operator \overline{∂}_{b} fails to be analytic hypoelliptic, in a modified sense, on the hypersurface \{(z_{1}, z_{2}) : ℑ(z_{2}) = [ℜ(z_{1})]^{m}}, for any even integer m ≥ 4.

We also point out that integrals of similar type recently appeared in the study of certain sums of squares of vector fields which fail to be analytic hypoelliptic (see Christ [C]) and also in the study of singular solutions for sums of squares of vector fields (see Hanges and Himonas [HANHI]).

We show that F_{τ} is an entire function of order α′ and of type 2τ/α′, where

1/α + 1/α′ = 1.

From this observation we develop a full asymptotic expansion of the integral

∫_{ℝ} \frac{τ exp[τη(z + \overline{w})]}{∫_{ℝ} exp[2τ(rη − p(r))] dr} dη

as τ tends to infinity.

The main point is a certain duality connected with the function p. The socalled Young–conjugate p^{∗} of p is defined by

p^{∗}(s) = sup{rs − p(r) : r ≥ 0} ,

and we have p^{∗∗} = p. For our case

p^{∗}(s) = \frac{s^{α′}}{α′} .

We use only methods from asymptotic analysis, the saddle point method and the method of stationary phase. N. Hange [HAN] used the theory of Fourier integral operators developed in [BMS] and [MS] to determine the asymptotic expansion for the Szegö kernel for bounded strictly pseudoconvex domains whose boundary (near 0) is given by the equation

ℑw = φ(z, \overline{z}),

here φ is a real valued, real analytic function of (z, \overline{z}) with φ(0, 0) = 0.

From the asymptotic expansion we can determine the behavior of the Szegö kernel in a neighborhood of a boundary point (z_{0}, s_{0}) of finite type (here we suppose α > 4/3) :

S = h_{1}/ψ^{2} + h_{0}/ψ − h_{−1} log ψ + O(1),

where

ψ((z, t), (w, s)) = p(z) + p(w) − 2 (\frac{z + \overline{w}}{2} )^{α} /α + i(s − t) ,

if ℜz_{0} > 0 and

ψ((z, t), (w, s)) = p(z) + p(w) − 2 (−\frac{z + \overline{w}}{2} )^{α} /α + i(s − t) ,

if ℜz_{0} < 0 and where the functions h_{1}, h_{0}, h_{−1} can be computed by methods from asymptotic analysis.

Results of this type were obtained by Ch. Fefferman [F] for the case of bounded strictly pseudoconvex domains, see also Boutet de Monvel and Sjöstrand [BMS] and N. Hange [HAN].

We conclude the paper by pointing out how one can extend the results for more general weight functions p.

2. An important parameterintegral.

We consider the integral

F_{τ} (η) = ∫_{ℝ} exp(2τ(rη − p(r))) dr ,

where τ > 0 , η ∈ ℂ and p(r) = |r|^{α}/α.

Proposition 1. F_{τ} is an entire function of order α′ and of type 2τ/α′, where

1/α + 1/α′ = 1.

Proof. We can write F_{τ} in the form

F_{τ} (η) = ∫_{ℝ} ∑_{k=0}^{∞} \frac{(2τrη)^{k}}{k!} exp(−2τp(r)) dr

= ∑_{k=0}^{∞} \frac{(2τη)^{k}}{k!} ∫_{ℝ} r^{k} exp(−2τp(r)) dr.

The integral

∫_{ℝ} r^{k} exp(−2τp(r)) dr

is zero for k odd and equals to

2 ∫_{0}^{∞} r^{k} exp(−2τp(r)) dr

for k even.

By the integral representation of the Γ− function we obtain

2 ∫_{0}^{∞} r^{k} exp(−2τp(r)) dr = \frac{2}{α}Γ (\frac{k + 1}{α}) (\frac{α}{2τ} )^{\frac{k+1}{α}} .

Now set a_{k} = 0, if k is odd and

a_{k} = \frac{4(2τ)^{k}}{αk!} Γ (\frac{k + 1}{α}) (\frac{α}{2τ} )^{\frac{k+1}{α}} ,

otherwise. Then an easy computation shows that

lim sup_{k→∞} \frac{log k}{−\frac{1}{k} log |a_{k}|} = α′,

which proves that F_{τ} has order α′ (see [M]). In addition we have

\frac{1}{eα′} lim sup_{k→∞} k|a_{k}|^{α′/k} = \frac{2τ}{α′}

and therefore F_{τ} has type 2τ/α′ (see [M]). □

(See also [CG]) This result is somehow the key for the following asymptotic analysis of the Szegö kernel.

3. The asymptotic expansion of the main term.

As already mentioned in the introduction we use the following integral representation of the Szegö kernel on ∂Ω_{p} × ∂Ω_{p} :

S((z, t), (w, s)) = \frac{1}{4π^{2}} ∫_{0}^{∞} ∫_{ℝ} \frac{τ exp(τ(η(z + \overline{w}) − p(z) − p(w) − i(s − t)))}{∫_{ℝ} exp(2τ(rη − p(r))) dr} dη dτ,

where

p(z) = \frac{|x|^{α}}{α} ,

for α > 1 and z = x + iy.

We want to determine the asymptotic behavior of the integral

∫_{ℝ} \frac{τ exp[τη(z + \overline{w})]}{∫_{ℝ} exp[2τ(rη − p(r))] dr} dη

as τ tends to infinity.

First we consider the integral

∫_{ℝ} exp[2τ(rη − p(r))] dr,

which equals to the sum

∫_{0}^{∞} exp[2τ(rη − p(r))] dr + ∫_{0}^{∞} exp[2τ(−rη − p(r))] dr.

For the first summand we apply the saddlepoint method (see [B] and [O]).

Lemma 1. Let η > 0, then

∫_{0}^{∞} exp[2τ(rη − p(r))] dr = τ^{−1/2} exp(2τp^{∗}(η)) [\sqrt{π}φ_{1}(η) + O(τ^{−1})] ,

as τ → ∞, where p^{∗}(η) = η^{α′}/α′ , 1/α + 1/α′ = 1 and

φ_{1}(η) = [(α − 1)η^{(α−2)/(α−1)}]^{−1/2} .

Proof. Set P_{η}(r) = rη − p(r), then P′_{η}(r) = η − p′(r) and P′′_{η}(r) = −p′′(r).

The equation P′_{η}(r) = 0 has the unique solution t_{η} = η^{1/(α−1)} and we have

P_{η}(t_{η}) = p^{∗}(η).

Now, by Young’s inequality

−P_{η}(r) + p^{∗}(η) = −rη + p(r) + p^{∗}(η) > 0,

for all r > 0 except for r = t_{η}, where the expression equals to zero.

This means that all the assumptions of Theorem 7.1, pg.127 in [O] are satisfied and therefore we have

∫_{0}^{∞} exp[2τ(rη − p(r))] dr = 2 exp(2τp^{∗}(η)) [Γ(1/2) τ^{−1/2}(4p′′(t_{η}))^{−1/2} + O(τ^{−3/2})] ,

which immediately gives the desired result. □

Lemma 2. Let η > 0, then for n ∈ ℕ we have

∫_{0}^{∞} exp[−2τ(rη + p(r))] dr =

∑_{k=0}^{n−1} \frac{(−1)^{k}}{k!} Γ(αk + 1)α^{−k}η^{−αk−1}(2τ)^{(1−α)k−1} + O(τ^{(1−α)n−1}),

as τ → ∞.

Proof. From Taylor’s Theorem with the exact remainder we get

e^{−2τp(r)} = ∑_{k=0}^{n−1} \frac{(−1)^{k}}{k!} (2τp(r))^{k} + R_{n},

where

|R_{n}| ≤ \frac{(2τp(r))^{n}}{n!} .

Next we multiply with e^{−2τrη} and integrate with respect to r

∫_{0}^{∞} exp[−2τ(rη + p(r))] dr = ∑_{k=0}^{n−1} \frac{(−1)^{k}}{k!} ∫_{0}^{∞}(2τp(r))^{k}e^{−2τrη} dr + ∫_{0}^{∞} R_{n}e^{−2τrη} dr.

By the integral representation of the Γ–function we have

∫_{0}^{∞}(2τp(r))^{k}e^{−2τrη} dr = α^{−k}(2τη)^{−(αk+1)}Γ(αk + 1),

which proves the lemma. □

Finally we remark that for the case η = 0 we get again by the integral representation of the Γ–function

∫_{0}^{∞} exp[−2τp(r)] dr = \frac{1}{α}Γ(1/α) (\frac{α}{2τ} )^{1/α} .

Combining the last results we get now

Lemma 3. For each η ∈ ℝ \ {0}

(1) \{∫_{ℝ} exp[2τ(rη − p(r))] dr\}^{−1} = τ^{1/2} exp(−2τp^{∗}(η)) [(\sqrt{π}φ_{1}(η))^{−1} + O(τ^{−1})] ,

as τ → ∞, where p^{∗}(η) = |η|^{α′}/α′ , 1/α + 1/α′ = 1 and

φ_{1}(η) = [(α − 1)|η|^{(α−2)/(α−1)}]^{−1/2} .

Proof. For η > 0 we obtain from Lemma 1 and Lemma 2 that

∫_{ℝ} exp[2τ(rη − p(r))] dr = τ^{−1/2} exp(2τp^{∗}(η)) [\sqrt{π}φ_{1}(η) + O(τ^{−1})] , and for η < 0 by interchanging the roles of Lemma 1 and Lemma 2

∫_{ℝ} exp[2τ(rη − p(r))] dr = τ^{−1/2} exp(2τp^{∗}(η)) [\sqrt{π}φ_{1}(η) + O(τ^{−1})] . Now by [HE] pg.369 \{∫_{ℝ} exp[2τ(rη − p(r))] dr\}^{−1} = τ^{1/2} exp(−2τp^{∗}(η)) [(\sqrt{π}φ_{1}(η))^{−1} + O(τ^{−1})] . □ The next step is to determine the asymptotic behavior of the integral

∫_{ℝ} exp(τη(z + \overline{w}) − 2τp^{∗}(η))φ_{2}(η) dη,

as τ → ∞, where φ_{2} = φ_{1}^{−1}.

We set 2ζ = z + \overline{w} and write the above integral in the form ∫_{0}^{∞} exp(2τηζ − 2τp^{∗}(η))φ_{2}(η) dη + ∫_{0}^{∞} exp(−2τηζ − 2τp^{∗}(η))φ_{2}(η) dη.

Lemma 4. Let ζ = r_{0}e^{iθ_{0}}, r_{0} ≠ 0 and suppose that α > 4/3 and that

−\frac{π}{2} (1 + \frac{1}{α′} ) < θ_{0} < \frac{π}{2} (1 + \frac{1}{α′} ) .

Then

∫_{0}^{∞} exp(−2τηζ − 2τp^{∗}(η))φ_{2}(η) dη =

c_{α} ∑_{k=0}^{n−1} \frac{(−1)^{k}}{k!} Γ(α′k + β + 1)α^{,−k}ζ^{−α′k−β−1}(2τ)^{k(1−α′)−β−1} + O(τ^{n(1−α′)−β−1}),

as τ → ∞, where c_{α} = (α − 1)^{1/2} and β = \frac{α−2}{2(α−1)} = −\frac{α′}{2} + 1.

Proof. First observe that the integral

∫_{0}^{∞} exp(−2τηζ − 2τp^{∗}(η))φ_{2}(η) dη

can be replaced by

∫_{L} exp(−2τηζ − 2τη^{α′}/α′)φ_{2}(η) dη,

where L is the infinite line starting from the origin and making an angle \frac{π}{2α′} − ϵ, for a small ϵ > 0, with the positive real axis.

This follows from Cauchy’s theorem and the fact that for η = Re^{iθ} we have | exp(−2τηζ − 2τp^{∗}(η))| ≤

exp (−2τRr_{0} cos(θ + θ_{0}) − 2τR^{α′}/α′ cos(α′θ))

and that cos(α′θ) > 0 for 0 ≤ θ < \frac{π}{2α′} .

Now again from Taylor’s Theorem with the exact remainder we get

exp (−2τ \frac{η^{α′}}{α′} ) = ∑_{k=0}^{n−1} \frac{(−1)^{k}}{k!} (\frac{2τ}{α′} )^{k} η^{α′k} + R_{n},

where

|R_{n}| ≤ (\frac{2τ}{α′} )^{n} |η|^{α′n}.

Next we multiply with e^{−2τηζ} and integrate with respect to η

∫_{L} exp(−2τηζ − 2τη^{α′}/α′)φ_{2}(η) dη =

∑_{k=0}^{n−1} \frac{(−1)^{k}}{k!} (\frac{2τ}{α′} )^{k} ∫_{L} φ_{2}(η)e^{−2τηζ}η^{α′k} dη + ∫_{L} φ_{2}(η)e^{−2τηζ}R_{n} dη.

The last integrals over L are convergent as long as ℜ(ηζ) > 0 on L. This holds if

−\frac{π}{2} < θ + θ_{0} < \frac{π}{2} ,

which, by our assumption on L means if

−\frac{π}{2} (1 + \frac{1}{α′} ) < θ_{0} < \frac{π}{2} (1 − \frac{1}{α′} ) .

Since the integrals over L take conjugate values for conjugate values of ζ, the angle may be extended to

−\frac{π}{2} (1 + \frac{1}{α′} ) < θ_{0} < \frac{π}{2} (1 + \frac{1}{α′} ) .

By the integral representation of the Γ–function (see [HE], pg.33) we obtain

∫_{L} φ_{2}(η)e^{−2τηζ}η^{α′k} dη = c_{α}Γ(α′k + β + 1)(2τζ)^{−α′k−β−1},

if α′k + β + 1 > 0 for each k = 0, 1, 2, .., which is true since α > 4/3. This proves the lemma. □

Remark. An investigtion of the remainder term shows that the last asymptotic expansion holds uniformly in ζ , for ζ belonging to a sufficiently small neighborhood of a fixed ζ_{0}, which does not leave the angle

−\frac{π}{2} (1 + \frac{1}{α′} ) < θ_{0} < \frac{π}{2} (1 + \frac{1}{α′} ) .

Lemma 5. Let ζ = r_{0}e^{iθ_{0}}, r_{0} ≠ 0 and suppose that

−\frac{π}{2α} < θ_{0} < \frac{π}{2α}.

Then

∫_{0}^{∞} exp(2τηζ − 2τp^{∗}(η))φ_{2}(η) dη = exp(2τζ^{α}/α) ∑_{k=0}^{∞} Γ(k + 1/2)a_{2k}τ^{−(k+1/2)},

as τ → ∞, where the coefficients a_{2k} depend on ζ,

a_{0} = (α − 1)ζ^{α−2}.

We always take the principle branches of the functions of the form ζ^{γ}, γ ∈ ℝ.

Proof. We use again the saddlepoint method . First we replace the integral

∫_{0}^{∞} exp(2τηζ − 2τη^{α′}/α′)φ_{2}(η) dη

by

∫_{L} exp(2τηζ − 2τη^{α′}/α′)φ_{2}(η) dη,

where L is the infinite line starting from the origin and making an angle \frac{θ_{0}}{α′−1} with the positive real axis. This follows again from Cauchy’s Theorem and the estimate

| exp(2τηζ − 2τη^{α′}/α′)| ≤ exp(2τRr_{0} cos(θ_{0}α) − 2τR^{α′}/α′ cos(θ_{0}α)),

where η = R exp (\frac{iθ_{0}}{α′−1}) lies on L. Our assumption implies that cos(θ_{0}α) > 0. The saddle point of the function in the exponential term of the integral is

η_{0} = ζ^{1/(α′−1)},

which lies on L. Now we calculate the real part of

2τ(η_{0}ζ − η_{0}^{α′}/α′ − ηζ + η^{α′}/α′)

and obtain for η = R exp (\frac{iθ_{0}}{α′−1}) on L

2τ cos(θ_{0}α)(r_{0}^{α}/α − r_{0}R + R^{α′}/α′) > 0,

by Young’s inequality , except for η = η_{0}.

Then all assumptions of Theorem 7.1 in [O] are satisfied. For the computation of the coefficient a_{0} one has to notice that

\frac{1}{2} (\frac{α − 2}{(α − 1)(α′ − 1)} − \frac{α′ − 2}{α′ − 1}) = α − 2,

and that

(\frac{α − 1}{α′ − 1})^{1/2} = α − 1.

This gives the desired result. □

Remark.A careful analysis of the proof of Theorem 6.1 ([O], pg.125) shows that by application of the inverse theorem for holomorphic functions also in this case the asymptotic expansion holds uniformly in ζ, for ζ belonging to a sufficiently small neighborhood of a fixed ζ_{0}, which does not leave the angle

−\frac{π}{2α} < θ_{0} < \frac{π}{2α}.

Now we are left to consider the asymptotic behavior of the integral

∫_{0}^{∞} exp(2τηζ − 2τη^{α′}/α′)φ_{2}(η) dη

on the critical lines θ_{0} = ±\frac{π}{2α}.

The critical lines θ_{0} = ±\frac{π}{2} (1 + \frac{1}{α′} ) for the integral

∫_{0}^{∞} exp(−2τηζ − 2τη^{α′}/α′)φ_{2}(η) dη

correspond to the first case after the transformation ζ ↦ −ζ.

In this case we use the method of stationary phase.

Lemma 6. Suppose that 1 < α < 4 and that

θ_{0} = ±\frac{π}{2α}.

Then for ζ = r_{0} exp(iπ/(2α)) ≠ 0

∫_{0}^{∞} exp(2τηζ − 2τη^{α′}/α′)φ_{2}(η) dη ≍ \sqrt{π}(α − 1)ζ^{α−2}τ^{−1/2} exp(2τζ^{α}/α),

as τ → ∞.

Proof. We carry out the argument for θ_{0} = \frac{π}{2α}. We set ζ = ω^{α′−1}, then the above integral has the form

∫_{0}^{∞} exp(2τηω^{α′−1} − 2τη^{α′}/α′)φ_{2}(η) dη

and after the substitution η = ωu

ω ∫_{L} exp(2τω^{α′}u − 2τω^{α′}u^{α′}/α′)φ_{2}(ωu) du,

here L is the infinite line starting from the origin and making an angle −π/(2α′) with the positive real axis.

With the help of Cauchy’s Theorem we show that the last integral equals to

ω ∫_{0}^{∞} exp(2τω^{α′}u − 2τω^{α′}u^{α′}/α′)φ_{2}(ωu) du.

For this purpose we have to show that the expression

R ∫_{− \frac{π}{2α′}}^{0} exp [ℜ (2τr_{0}^{α}R exp(i(π/2 + t)) − 2τr_{0}^{α} \frac{R^{α′}}{α′} exp(i(π/2 + α′t)))] ×

|φ_{2}(ωRe^{it})| dt

tends to zero as R tends to infinity.

If we compute the function φ_{2}, we see that the last integral equals to

C ∫0^{\frac{π}{2}} R^{−(α′−4)/2} exp[2τr_{0}^{α}(R sin(\frac{t}{α′}) − \frac{R^{α′}}{α′} sin t)] dt,

where C is independent of R. Since

\frac{2t}{π} ≤ sin t ≤ t , for 0 ≤ t ≤ \frac{π}{2} ,

we can now estimate by

∫0^{\frac{π}{2}} R^{−(α′−4)/2} exp[2τr_{0}^{α}(R sin(\frac{t}{α′}) − \frac{R^{α′}}{α′} sin t)] dt ≤

R^{−(α′−4)/2} [−\frac{πα′}{2τr_{0}^{α}(2R^{α′} − πR)} exp(τr_{0}^{α}(2R^{α′} − πR)/α′) + \frac{πα′}{2τr_{0}^{α}(2R^{α′} − πR)}] .

The last expression tends to zero as R tends to infinity, if

−(α′ − 4)/2 − α′ < 0,

which is equivalent to α < 4.

Hence we can continue to consider the integral

ω ∫_{0}^{∞} exp(2τω^{α′}u − 2τω^{α′}u^{α′}/α′)φ_{2}(ωu) du =

ω exp(2τ(α′ − 1)ω^{α′}/α′) ∫_{0}^{∞} exp [−\frac{2τω^{α′}}{α′} (u^{α′} − α′u + α′ − 1)] φ_{2}(ωu) du.

Since

ω^{α′} = ir_{0}^{α},

we can write the last integral in the form

∫_{0}^{∞} exp [−i\frac{2τr_{0}^{α}}{α′} (u^{α′} − α′u + α′ − 1)] φ_{2}(ωu) du,

and finally use the method of stationary phase (see [O]). We first split the integral into

∫_{0}^{∞} = ∫_{0}^{1} + ∫_{1}^{∞},

then an easy computation shows that for both integrals all assumptions of Theorem 13.1 in [O] are satisfied for the critical point u = 1. The contribution of both integrals to the asymptotic expansion equals to

\frac{\sqrt{π}}{2} ζ^{−1}(α − 1)τ^{−1/2},

which immediately gives the desired result. □

Finally we have to consider the case when ζ = 0.

Lemma 7. Suppose that α > 4/3. Then

∫_{ℝ} exp(−2τ|η|^{α′}/α′)φ_{2}(η) dη = (α − 1)^{1/2} (\frac{2}{α′} )^{\frac{α′−4}{2α′} +1} τ^{\frac{α′−4}{2α′}} Γ (\frac{3α − 4}{2α}) .

Proof. This follows from the integral representation of the Γ–function. □

Finally we have to put the things together. We want to determine the asymptotic behavior of the integral

I(τ, z, \overline{w}) = ∫_{ℝ} exp(τη(z + \overline{w}) − 2τ|η|^{α′}/α′)φ_{2}(η) dη =

∫_{0}^{∞} exp(τη(z + \overline{w}) − 2τη^{α′}/α′)φ_{2}(η) dη + ∫_{0}^{∞} exp(−τη(z + \overline{w}) − 2τη^{α′}/α′)φ_{2}(η) dη =

I_{1}(τ, z, \overline{w}) + I_{2}(τ, z, \overline{w}),

as τ → ∞.

Proposition 2. Let z + \overline{w} ≠ 0 and arg(z + \overline{w}) = θ_{0}.

(i) If α > 4/3 and

−\frac{π}{2α} < θ_{0} < \frac{π}{2α} ,

then

I(τ, z, \overline{w}) = \sqrt{π}(α − 1) (\frac{z + \overline{w}}{2} )^{α−2} τ^{−1/2} exp [2τ (\frac{z + \overline{w}}{2} )^{α} /α] (1 + O(τ^{−1}));

(ii) if α > 4/3 and

\frac{π}{2} (1 + \frac{1}{α′} ) < θ_{0} < \frac{π}{2} (3 − \frac{1}{α′} ) ,

then

I(τ, z, \overline{w}) = \sqrt{π}(α − 1) (−\frac{z + \overline{w}}{2} )^{α−2} τ^{−1/2} exp [2τ (−\frac{z + \overline{w}}{2} )^{α} /α] (1 + O(τ^{−1}));

(iii) if α > 4/3 and

\frac{π}{2α} < θ_{0} < \frac{π}{2} (1 + \frac{1}{α′} ) ,

or if

\frac{π}{2} (3 − \frac{1}{α′} ) < θ_{0} < \frac{π}{2} (4 − \frac{1}{α}) ,

then I(τ, z, \overline{w}) =

(α − 1)^{1/2}Γ(−α′/2 + 2)τ^{α′/2−2} [(z + \overline{w})^{α′/2−2} + (−z − \overline{w})^{α′/2−2}] + O(τ^{−α′/2−1});

(iv) if 1 < α < 4 and θ_{0} = ±\frac{π}{2α} then

I(τ, z, \overline{w}) = \sqrt{π}(α − 1) (\frac{z + \overline{w}}{2} )^{α−2} τ^{−1/2} exp [2τ (\frac{z + \overline{w}}{2} )^{α} /α] (1 + O(τ^{−1}));

(v) if 1 < α < 4 and θ_{0} = ±\frac{π}{2} (1 + \frac{1}{α′}) then

I(τ, z, \overline{w}) = \sqrt{π}(α − 1) (−\frac{z + \overline{w}}{2} )^{α−2} τ^{−1/2} exp [2τ (−\frac{z + \overline{w}}{2} )^{α} /α] (1 + O(τ^{−1}));

for each case as τ → ∞.

Proof. (i) We apply Lemma 5 to I_{1}(τ, z, \overline{w}), and observe that the asymptotic expansion of I_{2}(τ, z, \overline{w}) (Lemma 4) is negligible in angle (i) in comparison with the asymptotic expansion of the first summand.

(ii) After writing −z − \overline{w} = z′ + \overline{w′}, we have

−\frac{π}{2α} ≤ arg(z′ + \overline{w′}) ≤ \frac{π}{2α},

and we can apply Lemma 5 to I_{1}(τ, z′, \overline{w′}). The asymptotic expansion of I_{2}(τ, z′, \overline{w′}) is again negligible in comparison with the asymptotic expansion of the first summand.

(iii) For the first case I_{2}(τ, z, \overline{w}) remains in the angle of Lemma 4. For I_{1}(τ, z, \overline{w}) we write −z − \overline{w} = z′ + \overline{w′}, then

I_{1}(τ, z, \overline{w}) = I_{2}(τ, z′, \overline{w′}),

and

−\frac{π}{2} (1 + \frac{1}{α′} ) < arg(z′ + \overline{w′}) < −\frac{π}{2α},

which means that we can apply Lemma 4 for I_{2}(τ, z′, \overline{w′}). The second case is analogous to the first.

(iv) and (v) Apply Lemma 6.

Observe that

\frac{π}{2} (4 − \frac{1}{α}) ≡ −\frac{π}{2α} mod (2π),

which means that we covered the whole plane. Further we have

\frac{π}{2} (1 + \frac{1}{α′} ) = \frac{π}{2} (2 − \frac{1}{α}) and \frac{π}{2} (3 − \frac{1}{α′} ) = \frac{π}{2} (2 + \frac{1}{α}) . □

4. The main results on the Szegö kernel.

Now we proceed to state the main results

Proposition 3. Let S((z, t), (w, s)) denote the Szegö kernel of H^{2}(∂Ω_{p}) .

Then

S((z, t), (w, s)) = \frac{1}{4π^{2}} ∫_{0}^{∞} exp(−τ(p(z) + p(w) + i(s − t)))×

∫_{ℝ} \frac{τ exp[τη(z + \overline{w})]}{∫_{ℝ} exp[2τ(rη − p(r))] dr} dη dτ

and

G(z, \overline{w}, τ) = ∫_{ℝ} \frac{τ exp[τη(z + \overline{w})]}{∫_{ℝ} exp[2τ(rη − p(r))] dr} dη

has an asymptotic expansion of the following form: first suppose that z + \overline{w} ≠ 0 and set θ_{0} = arg(z + \overline{w}),

(i) if α > 4/3 and

−\frac{π}{2α} < θ_{0} < \frac{π}{2α} ,

then

G(z, \overline{w}, τ) ∼ ∑_{k=−1}^{∞} g_{−k}(z, \overline{w}, τ)τ^{−k}, as τ → ∞,

with

g_{1}(z, \overline{w}, τ) = (α − 1) (\frac{z + \overline{w}}{2} )^{α−2} exp [2τ (\frac{z + \overline{w}}{2} )^{α} /α] ,

and

g_{−k}(z, \overline{w}, τ) = h_{−k}(z, \overline{w}) exp [2τ (\frac{z + \overline{w}}{2} )^{α} /α] ,

for k = 0, 1, ..., where h_{−k} are analytic functions of z+\overline{w} with singularities at z+\overline{w} = 0; (ii) if α > 4/3 and

\frac{π}{2} (1 + \frac{1}{α′} ) < θ_{0} < \frac{π}{2} (3 − \frac{1}{α′} ) ,

then

G(z, \overline{w}, τ) ∼ ∑_{k=−1}^{∞} g_{−k}(z, \overline{w}, τ)τ^{−k}, as τ → ∞,

with

g_{1}(z, \overline{w}, τ) = (α − 1) (−\frac{z + \overline{w}}{2} )^{α−2} exp [2τ (−\frac{z + \overline{w}}{2} )^{α} /α] ,

and

g_{−k}(z, \overline{w}, τ) = h_{−k}(z, \overline{w}) exp [2τ (−\frac{z + \overline{w}}{2} )^{α} /α] ,

for k = 0, 1, ..., where h_{−k} are analytic functions of −(z + \overline{w}) with singularities at z + \overline{w} = 0;

(iii) if 1 < α < 4 and θ_{0} = ±\frac{π}{2α} then

G(z, \overline{w}, τ) ∼ ∑_{k=−1}^{∞} g_{−k}(z, \overline{w}, τ)τ^{−k}, as τ → ∞,

with

g_{1}(z, \overline{w}, τ) = (α − 1) (\frac{z + \overline{w}}{2} )^{α−2} exp [2τ (\frac{z + \overline{w}}{2} )^{α} /α] ,

and

g_{−k}(z, \overline{w}, τ) = h_{−k}(z, \overline{w}) exp [2τ (\frac{z + \overline{w}}{2} )^{α} /α] ,

for k = 0, 1, ..., where h_{−k} are analytic functions of z+\overline{w} with singularities at z+\overline{w} = 0; (iv) if 1 < α < 4 and θ_{0} = ±\frac{π}{2} (1 + \frac{1}{α′}) then

G(z, \overline{w}, τ) ∼ ∑_{k=−1}^{∞} g_{−k}(z, \overline{w}, τ)τ^{−k}, as τ → ∞,

with

g_{1}(z, \overline{w}, τ) = (α − 1) (−\frac{z + \overline{w}}{2} )^{α−2} exp [2τ (−\frac{z + \overline{w}}{2} )^{α} /α] ,

and

g_{−k}(z, \overline{w}, τ) = h_{−k}(z, \overline{w}) exp [2τ (−\frac{z + \overline{w}}{2} )^{α} /α] ,

for k = 0, 1, ..., where h_{−k} are analytic functions of −(z + \overline{w}) with singularities at z + \overline{w} = 0;

(v) if α > 4/3 and

\frac{π}{2α} < θ_{0} < \frac{π}{2} (1 + \frac{1}{α′} ) ,

or

\frac{π}{2} (3 − \frac{1}{α′} ) < θ_{0} < \frac{π}{2} (4 − \frac{1}{α}) ,

then

G(z, \overline{w}, τ) ∼ ∑_{k=0}^{∞} g_{k}(z, \overline{w})τ^{γ(α,k)}, as τ → ∞,

where γ(α, k) is a function of α and the summation index k , decreasing in k to −∞, with γ(α, 0) = (α′ − 3)/2 , γ(α, 1) = −(α′ + 1)/2 and where

g_{0}(z, \overline{w}) = (π(α − 1))^{1/2}Γ(−α′/2 + 2) [(z + \overline{w})^{α′/2−2} + (−z − \overline{w})^{α′/2−2}] ,

and g_{k}(z, \overline{w}) are analytic functions of z + \overline{w} with singularities at z + \overline{w} = 0, for k = 1, 2....

Finally if z + \overline{w} = 0 and 1 < α < 4 then

G(z, \overline{w}, τ) ∼ ∑_{k=0}^{∞} a_{k}τ^{δ(α,k)}, as τ → ∞,

where δ(α, k) is a function of α and the summation index k , decreasing in k to −∞, with δ(α, 0) = 2(α′ − 1)/α′ and where

a_{0} = (\frac{α − 1}{π})^{1/2} (\frac{2}{α′} )^{\frac{α′−4}{2α′} +1} Γ (\frac{3α − 4}{2α}) ,

and all further coefficients a_{k} depend only on α.

Proof. Use Proposition 2 and Lemma 1, 4, 5, 6 and 7.

Remark. The functions h_{−k} , g_{k} and γ(α, k) appearing in Proposition 3 can be computed explicitly using methods from asymptotic analysis (see [HE] and [O] ).

Now we investigate the singularities of the Szegö kernel of H^{2}(∂Ω_{p}) on the diagonal of ∂Ω_{p} × ∂Ω_{p}.

In the following we interpret the expression

S((z, t), (w, s)) = \frac{1}{2π} ∫_{0}^{∞} exp(−τ(p(z) + p(w) + i(s − t)))G(z, \overline{w}, τ) dτ

as an oscillatory integral defining a distribution on ∂Ω_{p} × ∂Ω_{p}, especially, for the asymptotic expansion, we interpret the appearing oscillatory integrals as ”finite part” integrals using the classical formulas

pf ∫_{0}^{∞} e^{−τa}τ^{k} dτ = \frac{(−1)^{−k}}{(−k − 1)!}a^{−k−1} (log a + γ − ∑_{j=1}^{−k−1} \frac{1}{j}) ,

where k = −1, −2, ... , a ∈ ℂ, a ≠ 0, ℜa ≥ 0 and γ denotes Euler’s constant (see [Z] ).

Proposition 4. Suppose that α > 4/3 and let S((z, t), (w, s)) denote the Szegö kernel of H^{2}(∂Ω_{p}) and let z_{0} ∈ ℂ with ℜz_{0} > 0, s_{0} ∈ ℝ. Then

S = h_{1}/ψ^{2} + h_{0}/ψ − h_{−1} log ψ + O(1)

in a sufficiently small neighborhood of ((z_{0}, s_{0}), (z_{0}, s_{0})), where

ψ((z, t), (w, s)) = p(z) + p(w) − 2 (\frac{z + \overline{w}}{2} )^{α} /α + i(s − t)

and the functions h_{1}, h_{0}, h_{−1} are defined in Proposition 3 (i) (here we have to multiply them with \frac{1}{4π^{2}} ).

If ℜz_{0} < 0, then

S = h_{1}/ψ̃^{2} + h_{0}/ψ̃ − h_{−1} log ψ̃ + O(1)

in a sufficiently small neighborhood of ((z_{0}, s_{0}), (z_{0}, s_{0})), where

ψ̃((z, t), (w, s)) = p(z) + p(w) − 2 (−\frac{z + \overline{w}}{2} )^{α} /α + i(s − t)

and the functions h_{1}, h_{0}, h_{−1} are defined in Proposition 3 (ii) (here we have to multiply them with \frac{1}{4π^{2}} ).

Proof. We use Proposition 3 and set

χ(z, \overline{w}) = (\frac{z + \overline{w}}{2} )^{α} /α.

Now it follows from Proposition 3 (i) and the remark following Lemma 5 that there exists a neighborhood U of (z_{0}, z_{0}) such that

|G(z, \overline{w}, τ) − exp(2τχ(z, \overline{w})) (h_{1}(z, \overline{w})τ + h_{0}(z, \overline{w}) + h_{−1}(z, \overline{w})τ^{−1})| ≤ C/τ^{2},

for each (z, w) ∈ U and for each τ ≥ N_{U}, where C > 0 is a constant depending on U. Note that since z_{0} + \overline{z_{0}} = 2ℜz_{0} > 0 it is possible to find a neighborhood U of (z_{0}, z_{0}) such that for each (z, w) ∈ U

−\frac{π}{2α} < arg(z + \overline{w}) < \frac{π}{2α} .

Now we claim that the neighborhood U can be chosen such that

ℜ(p(z) + p(w) − χ(z, \overline{w})) ≥ 0,

for each (z, w) ∈ U. For this purpose we write z = x + iy and w = u + iv , then it is sufficient to show that

ℜ((x + iy + u − iv)^{α}) ≤ 2^{α}(x^{α} + u^{α}).

Choose U so small that for each (z, w) ∈ U,

1 + tan^{2} θ ≤ 2^{2/α},

where θ = arg(z + \overline{w}).

Then for each (z, w) ∈ U we have

ℜ((x + iy + u − iv)^{α}) ≤ ((x + u)^{2} + (y − v)^{2})^{α/2} ≤ (x + u)^{α}(1 + tan^{2} θ)^{α/2}

≤ 2(x + u)^{α} ≤ 2^{α}(x^{α} + u^{α}),

in the last step we used the inequality

(x + u)^{α} ≤ 2^{α−1}(x^{α} + u^{α}),

which follows by convexity of the function t ↦ t^{α}.

This means that

ℜ(ψ((z, t), (w, s))) = ℜ(p(z) + p(w) − χ(z, \overline{w})) ≥ 0,

for (z, w) ∈ U hence we can apply the above formulas for oscillatory integrals and since

|G(z, \overline{w}, τ) − exp(2τχ(z, \overline{w})) (h_{1}(z, \overline{w})τ + h_{0}(z, \overline{w}) + h_{−1}(z, \overline{w})τ^{−1})| ≤ C/τ^{2},

for (z, w) ∈ U, we see that the contribution in the asymptotic expansion to the Szegö kernel from the remaining parts is a bounded function and obtain therefore the desired conclusion.

The second part of the Proposition follows by Proposition 3 (ii) and by analogous arguments as used in the first part.(Compare [BMS] and [HAN] .) □

Remarks.

1.)It is possible to extend the above results to more general weight functions p depending only on the real part of z. The main point is the involutivity of the Legendre transform p^{∗} of the function p (we called the Legendre transform p^{∗} of p the Young– conjugate); in more general terms one has to find a point r(s) such that \frac{d}{dr} (p(r)−sr) = 0 for r = r(s) and then one can define the Legendre transform p^{∗} by

p^{∗}(s) = sr(s) − p(r(s)).

Under suitable assumptions the Legendre transform is involutive, i.e. p^{∗∗} = p, which for instance is the case if p is a convex function with appropriate growth behavior.

The convergence of the corresponding integrals of the asymptotic expansions can be expressed in terms of the function which corresponds to the function φ_{2} appearing in Lemma 5. One has also to suppose that p is real analytic in a neighborhood of the point z_{0} with ℜz_{0} ≠ 0 in order to extend Lemma 5. and Proposition 4.

2.) For p(z) = x^{2}/2 everything can be computed explicitly:

∫_{ℝ} exp(2τ(rη − r^{2}/2)) dr = (π/τ)^{1/2} exp(τη^{2}),

and

S((z, t), (w, s)) = \frac{1}{4π^{2}} (\frac{1}{4}(z + \overline{w})^{2} − \frac{1}{8}(z + \overline{z})^{2} − \frac{1}{8}(w + \overline{w})^{2} − i(s − t))^{−2}

(see [HAS] ).

In this case Ω_{p} is strictly pseudoconvex and the term of exponent −1 as well as the logarithmic term in Proposition 4. disappear.

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Institut für Mathematik , Universität Wien, Strudlhofgasse 4, A-1090 Wien, Austria E-mail: HAS@PAP.UNIVIE.AC.AT