PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 87(101) (2010), 1–7 DOI: 10.2298/PIM1001001D

THE KERZMAN–STEIN OPERATOR FOR THE ELLIPSE

Milutin R. Dostanić

Communicated by Stevan Pilipović

Abstract. We give, in the case of ellipse, a simple connection between the spectrum of the Kerzman–Stein operator and the eccentricity of the ellipse.

1. Introduction

Let Ω ⊂ ℂ be a bounded domain with twice differentiable boundary and τ = τ(ξ) be the unit tangent vector at a ξ ∈ ∂Ω, oriented positively with respect to Ω. If |dξ| is the arclength measure on ∂Ω, then the Kerzman–Stein operator is defined (on L^{2}(∂Ω)) by

Af(z) = ∫_{∂Ω} A(z, ξ) f(ξ) |dξ|, where A(z, ξ) = \frac{1}{2πi} ( \frac{τ(ξ)}{ξ − z} − \frac{\overline{τ(z)}}{\overline{ξ} − \overline{z}} ) .

The space L^{2}(∂Ω) is defined using the hermitian inner product

(f, g) = ∫_{∂Ω} f(ξ) \overline{g(ξ)}|dξ|.

The kernel A(·, ·) is bounded at the diagonal—the apparent singularities cancel each other. Having in mind that A is bounded on ∂Ω, it follows that A is a Hilbert– Schmidt operator. The operator A can be represented in the form A = C − C^{∗}, where C : L^{2}(∂Ω) → L^{2}(∂Ω) is a (singular) Cauchy operator defined by

Cf(z) = \frac{1}{2πi} ∫_{∂Ω} \frac{f(ξ)}{ξ − z} dξ.

(Here, integral is interpreted in the sense of Principal Value.)

2010 Mathematics Subject Classification: Primary 47G10.

Key words and phrases: Kerzman–Stein operator, eigenvalues, singular values.

Partially supported by MNZZS Grant ON144010.

From this, it follows that A^{∗} = −A and the spectrum of the operator A is discrete and imaginary, except for an accumulation point at zero. The operator A first appeared in the paper of Kerzman and Stein [3] where they gave a new method for computing the Riemann map in one dimension. In a later article [4], Kerzman formulated some problems concerning the operator A, including the following.

Problem 1.1 (Kerzman 1979). Relate the spectrum of the Kerzman–Stein operator to the geometry of the domain.

The first paper giving some results concerning to above problem is [1]. Bolt therein proved the following theorem.

Theorem 1.1. The Kerzman–Stein operator for an ellipse has eigenvalues ± iλ_{n} where each ± iλ_{n} has multiplicity two. If ellipse is parametrized by t ↦ e^{it} + ρe^{−it} (0 < ρ < 1) and λ_{1} ⩾ λ_{2} ⩾ ⋯⩾ 0, then λ_{n} = β_{n}ρ^{2n−1} (1 + o(1)), ρ → 0+, where 0 < β_{n} ⩽ 1 and it can be computed explicitly.

In this paper (in the case of ellipse) we give a connection between the spectrum of the operator A and the eccentricity of ellipse. This connection is expressed through the Hilbert–Schmidt norm of the operator A using a double integral which depends on the eccentricity of an ellipse. This integral is hard for computation and we express it only in the form of a series.

2. Main result

Let Δ = {z : |z| > 1} and Ω = {(x, y) : x^{2}/a^{2} + y^{2}/b^{2} < 1} .

Theorem 2.1. a) If λ_{n}(A) denotes the eigenvalues of the operator A for ellipse ∂Ω, then

∑_{n} λ_{n}^{2}(A) = −\frac{1}{2π^{2}} ∫_{∂Δ} ∫_{∂Δ} \frac{|1 − ωz^{2}| |1 − ωξ^{2}| − |1 − ωzξ|^{2}}{|ξ − z|^{2}|1 − ωzξ|^{2}} |dz| |dξ|,

where ω = \frac{1 − \sqrt{1 − e^{2}}}{1 + \sqrt{1 − e^{2}}} and e denote the eccentricity of the ellipse ∂Ω.

b) For ellipses with small (enough) eccentricity, the eccentricity is uniquely determined by the Hilbert–Schmidt norm of operator A (or by the spectrum of the operator A).

Remark 2.1. The Bolt result relates to the asymptotic behavior of the eigenvalues of the Kerzman–Stein operator for the family of ellipses (not only one ellipse) “tending" to unit circle.

Proof. a) For our proof it is suitable to put a = α + β, b = α − β, α > β > 0. Then ω = β/α, and the function φ(ξ) = αξ + β/ξ realizes a conformal mapping of Δ onto ℂ ∖ \overline{Ω} . Let A_{0} : L^{2}(∂Δ) → L^{2}(∂Δ) and V : L^{2}(∂Ω) → L^{2}(∂Δ) be linear operators defined by

A_{0}f(z) = ∫_{∂Δ} K_{0}(z, ξ)f(ξ) |dξ|, V f(z) = f(φ(z))\sqrt{φ′(z)}.

Here K_{0}(z, ξ) = \sqrt{φ′(z)}·\overline{\sqrt{φ′(ξ)}}·A (φ(z), φ(ξ)) . Let us denote the eigenvalues of the operator A_{0} by λ_{n}(A_{0}). The operator V is an isometry and we have λ_{n}(A) = λ_{n}(A_{0}) and so, ∑ _{n} λ_{n}^{2}(A) = ∑ _{n} λ_{n}^{2}(A_{0}). Having on mind that the eigenvalues of the operator A (A_{0}) is imaginary, we have

(2.1)  ∑_{n} λ_{n}^{2}(A) = ∑_{n} λ_{n}^{2}(A_{0}) = − ∑_{n} s_{n}^{2}(A_{0}) = − ∫_{∂Δ} ∫_{∂Δ} |K_{0}(z, ξ)|^{2}|dz| |dξ|

(with s_{n}(T), we denote the singular values of the operator T; for more information about singular values see [2]). If ξ ∈ ∂Δ we have

τ(φ(ξ)) = \frac{φ′(ξ)}{|φ′(ξ)|} \frac{dξ}{|dξ|} = iξ \frac{φ′(ξ)}{|φ′(ξ)|}

and then we can transform K_{0}(·, ·) to the following form

(2.2)  K_{0}(z, ξ) = \frac{ξ}{2π} \frac{\sqrt{φ′(ξ)}\sqrt{φ′(z)}}{φ(ξ) − φ(z)} + \frac{\overline{z}}{2π} \overline{( \frac{\sqrt{φ′(ξ)}\sqrt{φ′(z)}}{φ(ξ) − φ(z)} )} .

Having on mind that for z, ξ ∈ ∂Δ, we have

\frac{\sqrt{φ′(ξ)}\sqrt{φ′(z)}}{φ(ξ) − φ(z)} = \frac{\sqrt{α − β/z^{2}} \sqrt{α − β/ξ^{2}}}{αξ + β/ξ − αz − β/z} = \frac{\sqrt{1 − ω\overline{z}^{2}} \sqrt{1 − ω\overline{ξ}^{2}}}{(ξ − z)(1 − ω\overline{z} \overline{ξ})}

and from (2.2) we conclude

K_{0}(z, ξ) = \frac{ξ}{2π} \frac{\sqrt{1 − ω\overline{z}^{2}} \sqrt{1 − ω\overline{ξ}^{2}}}{(ξ − z)(1 − ω\overline{z} \overline{ξ})} + \frac{\overline{z}}{2π} \frac{\sqrt{1 − ωz^{2}} \sqrt{1 − ωξ^{2}}}{(\overline{ξ} − \overline{z})(1 − ωzξ)}

i.e.,

(2.3) K_{0}(z, ξ) = \frac{ξ}{2π(ξ − z)} [ \frac{(1 − ω\overline{z}^{2})^{\frac{1}{2}} (1 − ω\overline{ξ}^{2})^{\frac{1}{2}}}{1 − ω\overline{z} \overline{ξ}} − \frac{(1 − ωz^{2})^{\frac{1}{2}} (1 − ωξ^{2})^{\frac{1}{2}}}{1 − ωzξ} ]

(because z, ξ ∈ ∂Δ). From (2.1) and (2.3) we obtain

(2.4)  ∑_{n} λ_{n}^{2}(A) = −\frac{1}{4π^{2}} ∫_{∂Δ} ∫_{∂Δ} \frac{1}{|ξ − z|^{2}} |\overline{D} − D|^{2} |dz| · |dξ|,

where

D = \frac{\sqrt{1 − ωz^{2}} \sqrt{1 − ωξ^{2}}}{(1 − ωzξ)} .

Now, we prove that

(2.5)  ∫_{∂Δ} ∫_{∂Δ} \frac{1 − D^{2}}{|ξ − z|^{2}} |dz| |dξ| = 0.

4 DOSTANIĆ

Indeed, we have

∫_{∂Δ} ∫_{∂Δ} \frac{1 − A^{2}}{|ξ − z|^{2}} |dz| |dξ|= ∫_{∂Δ} ∫_{∂Δ} \frac{ω(z − ξ)^{2}}{|ξ − z|^{2}(1 − ωzξ)^{2}} |dz| |dξ|

= ∫_{∂Δ} ∫_{∂Δ} \frac{ω(z − ξ)^{2}}{(ξ − z)(\overline{ξ} − \overline{z})(1 − ωzξ)^{2}} \frac{dz}{iz} \frac{dξ}{iξ}

= ω ∫_{∂Δ} ∫_{∂Δ} \frac{dzdξ}{(1 − ωzξ)^{2}} = 0.

From (2.4) and (2.5) we get

∑_{n} λ_{n}^{2}(A) = −\frac{1}{4π^{2}} ∫_{∂Δ} ∫_{∂Δ} \frac{2(|D|^{2} − 1)}{|ξ − z|^{2}} |dz| |dξ|

i.e.

(2.6)  ∑_{n} λ_{n}^{2}(A) = −\frac{1}{2π^{2}} ∫_{∂Δ} ∫_{∂Δ} \frac{|1 − ωz^{2}| |1 − ωξ^{2}| − |1 − ωzξ|^{2}}{|ξ − z|^{2}|1 − ωzξ|^{2}} |dz| |dξ|.

b) Let

G(ω) = −\frac{1}{2π^{2}} ∫_{∂Δ} ∫_{∂Δ} \frac{|1 − ωz^{2}| |1 − ωξ^{2}| − |1 − ωzξ|^{2}}{|ξ − z|^{2}|1 − ωzξ|^{2}} |dz| |dξ|.

Now, we prove that the function G is monotone on some interval [0, ω_{0}], ω_{0} ∈ (0, 1). From that, the statement b) of Theorem 2 follows.

Let |z| = 1, 0 ⩽ t < 1. Then we have

|1 − tz^{2}| = (1 − tz^{2})^{\frac{1}{2}} (1 − t \overline{z}^{2})^{\frac{1}{2}} = ∑_{n=0}^{∞} A_{n}(z) t^{n}

where

A_{n}(z) = ∑_{k=0}^{n} \binom{1/2}{k} (−z^{2})^{k} \binom{1/2}{n − k} (−\overline{z}^{2})^{n−k} = (−1)^{n}z^{−2n} ∑_{k=0}^{∞} \binom{1/2}{k} \binom{1/2}{n − k} z^{4k}.

So, if |z| = |ξ| = 1, then we obtain

(2.7)  |1 − tz^{2}| |1 − t ξ^{2}| = ∑_{n=0}^{∞} B_{n}(z, ξ)t^{n}

where B_{n}(z, ξ) = ∑ _{k=0}^{n} A_{k}(z)A_{n−k} (ξ). If we put z = ξ in (2.7), we get

B_{n}(z, z) = ∑_{k=0}^{n} A_{k}(z)A_{n−k} (z) = 0 for n ⩾ 3.

If |z| = 1, by differentiation of both sides of the equality

(1 − tz^{2})^{\frac{1}{2}} (1 − tz^{−2} )^{\frac{1}{2}} = ∑_{n=0}^{∞} A_{n}(z)t^{n}

with respect to z and multiplying the resulting equality with (1 − tz^{2})^{\frac{1}{2}} (1 − tz^{−2} )^{\frac{1}{2}} we get

∑_{n=0}^{∞} t^{n}( ∑_{k=0}^{n} A′_{k}(z)A_{n−k} (z)) = t (z^{−3} − z)

i.e., ∑ _{k=0}^{n} A′_{k}(z)A_{n−k} (z) = 0 for n ⩾ 2. So, we obtained (for |z| = 1)

(2.8)  ∑_{k=0}^{n} A′_{k}(z)A_{n−k} (z) = 0,  ∑_{k=0}^{n} A_{k}(z)A_{n−k} (z) = 0 for n ⩾ 3.

From (2.7) we get

(2.9) |1 − ωz^{2}| |1 − ωξ^{2}| − |1 − ωzξ|^{2}

= (B_{1}(z, ξ) + \overline{z}\overline{ξ} + zξ) ω + (B_{2}(z, ξ) − 1)ω^{2} + ∑_{n=3}^{∞} B_{n}(z, ξ)ω^{n}, (0 < ω < 1)

By direct calculation we obtain

(2.10)  \begin{matrix}B_{1}(z, ξ) + \overline{z}\overline{ξ} + zξ = −\frac{1}{2}(z − ξ)^{2} − \frac{1}{2}(\overline{z} − \overline{ξ})^{2}, \\ B_{2}(z, ξ) − 1 = −\frac{1}{8}(z − ξ)^{2}(z + ξ)^{2}(1 − \overline{z}^{2} \overline{ξ}^{2})^{2}, \end{matrix}

where z, ξ ∈ ∂Δ. From (2.8) it follows that B_{n}(z, z) = 0, \frac{∂B_{n}}{∂ξ} |_{ξ=z} = 0 for n ⩾ 3 and

(2.11)  B_{n}(z, ξ) = \frac{1}{2!} \frac{∂^{2}B_{n}}{∂ξ^{2}} |_{ξ=z} (ξ − z)^{2} + ⋯

From (2.9), (2.10), and (2.11), it follows that the integral on the right-hand side in (2.6) is not singular and so we have

G′(ω) = −\frac{1}{2π^{2}} ∫_{∂Δ} ∫_{∂Δ} \frac{1}{|ξ − z|^{2}} \frac{d}{dω} ( \frac{|1 − ωz^{2}| |1 − ωξ^{2}|}{|1 − ωzξ|^{2}} ) |dz| |dξ|.

Since

\frac{d}{dω} ( \frac{|1 − ωz^{2}| |1 − ωξ^{2}|}{|1 − ωzξ|^{2}} )

= \frac{|1 − ωz^{2}| |1 − ωξ^{2}|}{|1 − ωzξ|^{2}} Re[ \frac{d}{dω} (ln(1 − ωz^{2}) + ln(1 − ωξ^{2}) − 2 ln(1 − ωzξ)) ]

= −\frac{|1 − ωz^{2}| |1 − ωξ^{2}|}{|1 − ωzξ|^{2}} Re[ ∑_{n=1}^{∞} ω^{n−1} (ξ^{n} − z^{n})^{2}],

6 DOSTANIĆ

we get

G′(ω) = ∑_{n=1}^{∞} ω^{n−1} Re( \frac{1}{2π^{2}} ∫_{∂Δ} ∫_{∂Δ} \frac{|1 − ωz^{2}| |1 − ωξ^{2}|}{|ξ − z|^{2}|1 − ωzξ|^{2}} (ξ^{n} − z^{n})^{2}|dz| |dξ|) .

Let

K_{n}(ω) = \frac{1}{2π^{2}} ∫_{∂Δ} ∫_{∂Δ} \frac{|1 − ωz^{2}| |1 − ωξ^{2}|}{|ξ − z|^{2}|1 − ωzξ|^{2}} (ξ^{n} − z^{n})^{2}|dz| |dξ|.

It is easy to see that K_{n} is real and so we have

(2.12)  G′(ω) = ∑_{n=1}^{∞} ω^{n−1} K_{n}(ω).

Using the binomial expansion we obtain (for |z| = 1)

(2.13)  |1 − ωz^{2}| = ∑_{k∈ℤ} c_{k}(ω)z^{2k},

where

c_{n}(ω) = ∑_{\begin{matrix}k,l ⩾ 0 \\ k−l = n \end{matrix}} \binom{1/2}{k} \binom{1/2}{l} (−ω)^{k+l} .

It is clear that c_{−n} (ω) = c_{n}(ω). If n ⩾ 0, we can easily conclude that

(2.14)  c_{n}(ω) = (−1)^{n}ω^{n} ∑_{ν=0}^{∞} \binom{1/2}{ν} \binom{1/2}{ν + n} ω^{2ν}.

Since |\binom{1/2}{n}| ⩽ L_{1}n^{−3/2} , (L_{1} does not depend on n) from (2.14) we obtain

(2.15)  |c_{n}(ω)| ⩽ \frac{L_{2}}{(1 + |n|)^{3/2}} ω^{|n|} , (0 < ω < 1)

where the constant L_{2} does not depend on n and ω. Using (2.13) we calculate K_{n}. Namely

(2.16)  K_{1}(ω) = −\frac{2}{1 − ω^{2}} ∑_{n∈ℤ} c_{n}^{2}(ω) · ω^{|2n+1|} ,

(2.17) K_{n}(ω) = −\frac{2}{1 − ω^{2}}

× [n ∑_{m∈ℤ} c_{m}^{2}(ω) ω^{|n+2m|} + 2 ∑_{r=1}^{n−1} (n − r) ∑_{m∈ℤ} c_{m}(ω) c_{m+r} (ω) ω^{|n+2m+r|} ]

in the case n ⩾ 2. From (2.14) and (2.16) it follows that

(2.18)  K_{1}(ω) = −\frac{2ω}{1 − ω^{2}} (1 + o(1)), ω → 0 + .

If n ⩾ 2, from (2.15) and (2.17) we obtain the estimate

(2.19)  |K_{n}(ω)| ⩽ L_{3} n ω^{n} \frac{1}{1 − ω^{2}}

where the constant L_{3} does not depend on n and ω ∈ (0, 1). From (2.12), (2.18) and (2.19) we conclude that there exists ω_{0} ∈ (0, 1) such that G′(ω) < 0 for ω ∈ (0, ω_{0}) i.e., the function G is monotone on [0, ω_{0}]. Theorem 2 is proved. □

References

[1] M. Bolt, Spectrum of the Kerzman–Stein operator for ellipse, Integral Equations Oper. Theory 57 (2007), 167–184.

[2] I. C. Gohberg, M. G. Krein, Theory of linear nonselfadjoint operators, Trans. Math. Monographs 18, Am. Math. Soc., Providence, R.I., 1969.

[3] N. Kerzman and E. M. Stein, The Cauchy kernel, the Szegö kernel and the Riemann mapping function, Math. Ann. 236(1) (1978), 85–93.

[4] N. Kerzman, Singular integrals in complex analysis; in: Harmonic analysis in Euclidean spaces, Part 2, Williamstown/Massachusetts 1978, Proc. Symp. Pure Math. 35, pp. 3–41 (1979).

Matematički fakultet (Received 14 09 2009) Beogradski univerzitet (Revised 04 02 2010) Begrad Serbia

domi@matf.bg.ac.rs