����������� ������ 51 (1999), 91–96

UDK 517.984.5

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RELATIONSHIP BETWEEN THE OPERATORS OF CAUCHY AND LOGARITHMIC POTENTIAL TYPE

Milutin R. Dostanić

Abstract. Asymptotic behaviour of singular values, together with the estimate of the remainder term, of Cauchy operator and the operator of logarithmic potential type, as well as the corresponding trace formula are determined.

1. Introduction and notation

Let Ω ⊂ ℂ be bounded domain. Denote by L^{2} (Ω) the space of complex valued

functions in Ω for which the norm ∥f∥_{2} = (∫_{Ω} |f (ξ)|^{2} dA (ξ))^{1/2} is finite. Here

dA (ξ) denotes the area measure in Ω. The operators L and C: L^{2} (Ω) → L^{2} (Ω) defined by

Lf (z) = −\frac{1}{2π} ∫_{Ω} ln |z − ξ| f (ξ) dA (ξ)

and

Cf (z) = −\frac{1}{π} ∫_{Ω} \frac{f (ξ)}{ξ − z} dA (ξ)

will be called the operator of logarithmic potential type and Cauchy operator respectively. It is well known that C and L are compact operators on L^{2} (Ω) and L > 0 iff Ω is inside the unit disc; see [1]. The spectral properties of the operators L and C are studied in many papers (see [1], [2], [5], [7]).

In what follows we denote by ∫_{Ω} M (z, ξ) · dA (ξ) the integral operator acting on L^{2} (Ω) whose kernel is M (z, ξ).

For compact operator T we denote by s_{n} (T) the n-th eigenvalue of the operator |T | = (T^{∗}T)^{1/2}, i.e. s_{n} (T) = λ_{n} (|T |). By c_{p} we denote the set of compact operators T such that

|T |_{p} = (∑_{k≥1} s_{k}^{p} (T))^{1/p}.

In particular c_{1} is the set of nuclear operators.

AMS Subject Classification: 47 B 10, 47 G 10, 45 C 05

Keywords and phrases: Cauchy integral operator, operator of logarithmic potential type, singular values, trace formula

92 M. Dostanić

2. Results The operators L and \frac{C^{∗}C}{4} have similar spectral asymptotics. Let λ_{n} (L) and

λ_{n} (\frac{C^{∗}C}{4} ) be eigenvalues of the operators L and \frac{C^{∗}C}{4} arranged in the decreasing order, according to their multiplicity.

Theorem 1. If Ω ⊂ ℂ is a bounded simply connected domain with analytic boundary, then the following asymptotic formulae hold: a)

λ_{n} (L) = \frac{|Ω|}{4πn} + O (\frac{1}{n^{4/3}} )

b)

λ_{n} (\frac{C^{∗}C}{4} ) = \frac{|Ω|}{4πn} + O (\frac{1}{n^{4/3}} )

Proof. For the eigenvalues λ_{n} of the problem

−Δu = λu, u|_{∂Ω} = 0 (∗)

the following Weyl formula holds

\frac{1}{λ_{n}} = \frac{|Ω|}{4πn} − \frac{|∂Ω|}{\sqrt{4π |Ω|}} \frac{1}{n^{3/2}} + o (\frac{1}{n^{3/2}} ) .

Here |∂Ω| denotes the length of the |∂Ω|. By the inversion of the boundary problem

−Δu = λu, u|_{∂Ω} = 0

we get

Sf (z) = ∫_{Ω} G (z, ξ) f (ξ) dA (ξ)

where G (·, ·) is the Green function of the problem (∗).

It is clear that s_{n} (S) = \frac{1}{λ_{n}}. In terms of conformal mapping h : Ω → D (D is the unit disc) we have

G (z, ξ) = −\frac{1}{2π} ln |\frac{h (z) − h (ξ)}{1 − \overline{h (ξ)}h (z)}| .

(Observe that the function h can be analytically continued to some neighborhood of \overline{Ω}. The same holds for φ (z) = h^{−1} (z). Therefore, φ′ (z) is a bounded function on \overline{D.})

Then we get

S = −\frac{1}{2π} ∫_{Ω} ln |\frac{h (z) − h (ξ)}{z − ξ}| · dA (ξ) + \frac{1}{2π} ∫_{Ω} ln |1 − \overline{h (ξ)}h (z)| · dA (ξ) + L

= S_{1} + S_{2} + L.

From the Birman-Solomjak theorem on the growth of singular values of operators with an analytic kernel, [3], we obtain

s_{n} (S_{1}) = O (e^{−c_{0}\sqrt{n}}) (c_{0} does not depend on n).  (1)

Since s_{n} (∫_{D} ln |1 − \overline{ξ}z| · dA) = O(1/n^{2}) and

s_{n} (S_{2}) = s_{n}(\frac{1}{2π} ∫_{D} φ′ (z) \overline{φ′ (ξ)} ln |1 − \overline{ξ} z| · dA)

we have s_{n} (S_{2}) = O(1/n^{2}) and from (1) and properties of singular values of the sum of two operators we get

s_{n} (S_{1} + S_{2}) = O (\frac{1}{n^{2}} ) .

Now, we need the following simple generalization of the Ky-Fan theorem (it can be deduced in the same way as the Ky-Fan theorem [9], see also [8]).

Lemma 1. If A and B are compact operators such that

s_{n} (A) = \frac{a}{n^{α}} + O (n^{−β}) α + 1 > β > α > 0
s_{n} (B) = O (n^{−β_{1}}) β_{1} ≥ \frac{β}{α + 1 − β} ,

then

s_{n} (A + B) = \frac{a}{n^{α}} + O (n^{−β})

Applying Lemma 1 to the operators S and − (S_{1} + S_{2}) we get the statement of Theorem 1 a).

For the proof of Theorem 1 b) we observe that the operator R = \frac{C^{∗}C}{4} − L has the following kernel

X (z, ξ) = \frac{1}{4π^{2}i} ∫_{∂Ω} \frac{ln |t − z|}{t − ξ} dt

and hence

s_{n} (R) = O (\frac{1}{n^{2}} ) .  (2)

(Indeed, the kernel of the operator \frac{C^{∗}C}{4} is equal to the function

K(z, ξ) = \frac{1}{4π^{2}} ∫_{Ω} \frac{dA (t)}{(ξ − t) (\overline{z} − \overline{t})},

so applying the Cauchy-Green formula we obtain

K(z, ξ) = −\frac{1}{2π} ln |z − ξ| + X (z, ξ)

i.e. the kernel of the operator R is equal to X (z, ξ).)

Let us show (2). It is enough to prove that for the singular values of the operator R: L^{2} (D) → L^{2} (D) with the kernel

∫_{∂D} \frac{ln |φ (t) − φ (z)|}{φ (t) − φ (ξ)} · φ′ (t) dt

there holds s_{n} (R_{1}) = O (1/n^{2}). Observe that

∫_{∂D} \frac{ln |φ (t) − φ (z)|}{φ (t) − φ (ξ)} · φ′ (t) dt

= ∫_{∂D} \frac{(t − ξ)φ′ (t)}{φ (t) − φ (ξ)} · \frac{ln |t − z|}{t − ξ} dt + ∫_{∂D} ln |\frac{φ (t) − φ (z)}{t − z}| · \frac{φ′ (t) − φ′ (ξ)}{φ (t) − φ (ξ)} dt

+ φ′ (ξ) ∫_{∂D} ln |\frac{φ (t) − φ (z)}{t − z}| · (\frac{1}{φ (t) − φ (ξ)} − \frac{1}{t − ξ} ) dt

+ φ′ (ξ) ∫_{∂D} \frac{ln |\frac{φ(t)−φ(z)}{t−z} |}{t − ξ} dt = K_{1} + K_{2} + K_{3} + K_{4}.

The functions K_{2}, K_{3}, K_{4} are real analytic on a neighborhood of \overline{D} × \overline{D} and thus the corresponding integral operators have exponential decrease of singular values. (Theorem of Birman-Solomjak). The function (t, ξ) ⟼ (t − ξ) \frac{φ′(t)}{φ(t)−φ(ξ)} is analytic on a neighborhood of \overline{D} × \overline{D} and the rate of growth of the singular values of the operator with the kernel K_{1} is equal to the rate of growth of the singular values of the operator with the kernel

∫_{∂D} \frac{ln |t − z|}{t − ξ} dt (= πi ln (1 − ξ \overline{z})) .

Since s_{n} (∫_{D} ln (1 − ξ \overline{z}) · dA (ξ)) = O(1/n^{2}) we obtain (2).

It follows that s_{n} (R − S_{1} − S_{2}) = O (\frac{1}{n^{2}} ) and applying Lemma 1 to the operators S and R − S_{1} − S_{2} we obtain the statement of Theorem 1 b).

From Theorem 1 it follows that the function

Ψ (z) = ∑_{n=1}^{∞} (λ_{n}^{z} (\frac{C^{∗}C}{4} ) − λ_{n}^{z} (L))

(λ^{z} = e^{z ln λ} ; ln λ = ln |λ| + i arg λ, 0 ≤ arg λ < 2π) is analytic for Re z > 2/3.

Theorem 2. If Ω ⊂ ℂ is a bounded simply connected domain with analytic boundary, then the following trace formula

∑_{n=1}^{∞} (λ_{n} (\frac{C^{∗}C}{4} ) − λ_{n} (L)) = −\frac{1}{8π^{2}} ∫_{∂Ω} ∫_{∂Ω} ln^{2} |z − ξ| dzd\overline{ξ}

holds.

In the proof of this theorem we need the following

Relationship between the operators of Cauchy and logarithmic potential type 95

Lemma 2. [6] If C and D are positive compact operators on some complex Hilbert space H such that |C − D|^{Re z} ∈ c_{1} (0 < Re z < 1), then C^{z} − D^{z} ∈ c_{1} and

|C^{z} − D^{z}|_{1} ≤ \frac{|sin πz|}{sin π (Re z)} ||C − D|^{Re z}|_{1} .

Proof of Theorem 2. From (2) it follows that for every k ∈ N holds

s_{n} ((\frac{C^{∗}C}{4} )^{k} − L^{k}) = O (\frac{1}{n^{k+1}} )

and by Lemma 2, the function

z ⟼ tr ((\frac{C^{∗}C}{4} )^{z} − L^{z}) = tr [((\frac{C^{∗}C}{4} )^{k})^{z/k} − (L^{k})^{z/k}]

is analytic for \frac{1}{k+1} < Re (\frac{z}{k} ) < 1 i.e. \frac{k}{k+1} < Re z < k. So, the function

z ⟼ tr ((\frac{C^{∗}C}{4} )^{z} − L^{z})

is analytic for Re z > 1/2.

If Re z > 1, then both operators (\frac{C^{∗}C}{4} )^{z} and L^{z} are nuclear and hence

tr ((\frac{C^{∗}C}{4} )^{z} − L^{z}) = tr (\frac{C^{∗}C}{4} )^{z} − tr L^{z} = Ψ (z) .

Hence

Ψ (z) ≡ tr ((\frac{C^{∗}C}{4} )^{z} − L^{z})

for Re z > \frac{2}{3}. Particularly, for z = 1 we obtain

∑_{n=1}^{∞} (λ_{n} (\frac{C^{∗}C}{4} ) − λ_{n} (L)) = tr (\frac{C^{∗}C}{4} − L) = tr R.

Having in mind that R ∈ c_{1,} by [4], Theorem 3.1 it follows

tr R = \frac{1}{4π^{2}i} ∫_{Ω} dA (z) ∫_{∂Ω} \frac{ln |z − t|}{z − t} dt = \frac{1}{4π^{2}i} ∫_{∂Ω} dt ∫_{Ω} \frac{ln |z − t|}{z − t}dA (z)

Let Ω_{ε} = Ω\ {z : |z − t| < ε} , t ∈ ∂Ω. Since

∫_{Ω} \frac{ln |z − t|}{z − t} dA (z) = − lim_{ε→0} ∫_{Ω_{ε}} \frac{ln |z − t|}{z − t}dA (z) = − lim_{ε→0} ∫_{Ω_{ε}} \frac{∂}{∂z} (ln^{2} |z − t|) dA (z)

according to Green’s formula, [10], we get

∫_{Ω} \frac{ln |z − t|}{z − t} dA (z) = − lim_{ε→0} (−\frac{1}{2i} ∫_{∂Ω_{ε}} ln^{2} |z − t| d\overline{z}) = \frac{1}{2i} ∫_{∂Ω} ln^{2} |z − t| d\overline{z}

So,

tr R = \frac{1}{4π^{2}i} ∫_{∂Ω} dt \frac{1}{2i} ∫_{∂Ω} ln^{2} |z − t| d\overline{z} = −\frac{1}{8π^{2}} ∫_{∂Ω} ∫_{∂Ω} ln^{2} |z − t| dt d\overline{z}.

Remark. In the case Ω = D (D is the unit disc) it was proved in [1] that the spectra of the operators \frac{C^{∗}C}{4} and L are equal and we may expect that

∑_{n=1}^{∞} (λ_{n} (\frac{C^{∗}C}{4} ) − λ_{n} (L)) = 0.

But direct calculation shows that

∫_{∂D} ∫_{∂D} ln^{2} |z − ξ| dz d\overline{ξ} = 2π^{2} .

The explanation lies in the fact that the operators \frac{C^{∗}C}{4} and L have the same spectra (in the case Ω = D ) but their multiplicities differ.

Corollary. Assume the boundary of a simply connected bounded domain G_{ε} is given by z (θ) = (1 + εφ (θ)) e^{iθ}. Then applying Theorem 2 to the operators \frac{C^{∗}C}{4} and L (acting on L^{2} (G_{ε}) ) we have

∑_{n=1}^{∞} (λ_{n} (\frac{C^{∗}C}{4} ) − λ_{n} (L)) = −\frac{1}{8π^{2}} ∫_{∂D} ∫_{∂D} ln^{2} |z − t| dz d\overline{t}+ε·A+O (ε^{2}) , ε → 0

where A depends only on φ and φ′.

So the operators \frac{C^{∗}C}{4} and L on domains which are close to the unit disc, either have different spectra or if they have the same spectrum the multiplicities of eigenvalues are not the same.

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(received 01.02.1999.)

Faculty of Mathematics, Studentski trg 16, P.O.Box 550, 11000 Beograd, Yugoslavia E-mail: domi@matf.bg.ac.yu