Some peculiarities of approximation methods for singular integral equations with conjugation

V. D. Didenko, B. Silbermann and S. Roch^{1}

Approximation methods for singular integral operators with continuous coefficients and conjugation on curves with corners are investigated with respect to their stability. Particular emphasis is devoted to index constraints for the local stability conditions. It turns out that, if an associated local operator is Fredholm, then the absolute value of its Fredholm index is necessarily bounded by 2, and this maximal value is attained in some instances (whereas it is known that in case of pure singular integral equations Fredholmness of the associated local operators already implies vanishing index).

Proposed short running title:

Singular integral equations with conjugation

AMS Subject Classification:

Primary: 65R20, Secondary: 47G10

Keywords:

Singular integral equations with conjugation, quadrature methods, local index constraints

1 The work of the first author was partially supported by the German Academic Exchange Service (DAAD) and by the German Research Foundation (Deutsche Forschungsgemeinschaft).

1 Introduction

Many problems in mathematical physics can be described by singular integral equations which contain both the unknown function and its complex conjugate, see, e.g., [10,11,16,17,15,23]. The presence of the complex conjugation leads to numerous new effects which have no counterparts for equations without conjugation. Concerning the applicability of certain approximation methods, some of these effects have been observed in [8,9] for the equation

(Au)(t) ≡ a(t)u(t) + \frac{b(t)}{πi} ∫_{Γ} \frac{u(τ)dτ}{τ − t} + c(t)\overline{u(t)} + \frac{d(t)}{πi} ∫_{Γ} \frac{\overline{u(τ)}dτ}{τ − t}

− \frac{e(t)}{πi}\overline{∫_{Γ} \frac{u(τ)dτ}{τ − t}} − \frac{r(t)}{πi}\overline{∫_{Γ} \frac{\overline{u(τ)}dτ}{τ − t}}  (1)

+ ∫_{Γ} k_{1}(t, τ)u(τ)dτ + ∫_{Γ} k_{2}(t, τ)\overline{u(τ)}dτ = f(t), t ∈ Γ,

where a, b, c, d, e, f, g, k_{1}, k_{2} are given continuous functions, Γ is an open or closed piecewise smooth curve in the complex plane ℂ, and where the operator A is considered as acting on a weighted L^{2}-space. The double layer potential equation

a(t)u(t) + \frac{b(t)}{π} ∫_{Γ} u(τ)\frac{d}{dn_{τ}} log |t − τ|dΓ_{τ} + (T u)(t) = f(t), t ∈ Γ, (2)

where T is a compact operator and n_{τ} refers to the inner normal to Γ at the point τ is a prominent example of equations of the form (1) (set c = d = e = 0 and b = r in (1)). In the present paper we consider another special case of (1) where we suppose the coefficients e and r to be identically equal to zero on Γ:

(Au)(t) ≡ a(t)u(t) + \frac{b(t)}{πi} ∫_{Γ} \frac{u(τ)dτ}{τ − t} + c(t)\overline{u(t)} + \frac{d(t)}{πi} ∫_{Γ} \frac{\overline{u(τ)}dτ}{τ − t}  (3)

+ ∫_{Γ} k_{1}(t, τ)u(τ)dτ + ∫_{Γ} k_{2}(t, τ)\overline{u(τ)}dτ = f(t), t ∈ Γ.

We start with describing a class of approximation methods for equation (3) and with establishing the corresponding stability criteria which proceeds in complete analogy to [9,14]. For the stability condition, we have to associate with every concrete approximation method for (3) (which will be specified later on), and with every point τ of Γ, a certain ”model” operator A_{τ} which reflects the local behaviour of the sequence of the approximation operators at the point τ. Then the stability criterion says that the approximation method is stable if and only if the operator A as well as all operators A_{τ} are invertible.

The invertibility of the local operators A_{τ} (as well as that of the operator A itself) can be checked effectively only in a few rare situations. However, which will be done in the present paper, the Fredholm properties of these operators, and the behaviour of their Fredholm indices, can be characterized completely. Analogous problems were studied for the first time in [9] where the basic space is L^{2}(Γ) without weight. Particularly, we observed in [9] that the indices κ(A_{τ}) of the operators A_{τ} can be different from zero if τ is a corner point of Γ, but that in any case the inequality

|κ(A_{τ})| ≤ 1 (4)

holds. This fact has no counterpart for singular integral equations without conjugation where it is known that the Fredholmness of the operators A_{τ} is independent of the angle at τ, and that it automatically implies the vanishing of the index of A_{τ}, cf. [19,20]. Moreover, the latter result remains valid for singular integral operators without conjugation acting on L^{2}-spaces with power weight.

An essential change of the Fredholm properties can be observerd for the local operators A_{τ} which correspond to equation (3) on weighted spaces. In this case, we will show that the indices of A_{τ} will not only depend on the values of the angles, but also on the weight function at the point τ ∈ Γ. More precisely, if the angle ω_{τ} of Γ at τ belongs to one of the intervals (0, π/2] or [3π/2, 2π), then the index of the corresponding operator A_{τ} is subject to estimate (4) for all possible weight functions. However, if ω_{τ} ∈ (π/2, π) ∪ (π, 3π/2), then

|κ(A_{τ})| ≤ 2, (5)

and the equality sign in (5) holds for some combinations of coefficients, weights and angles.

We proceed with describing necessary and sufficient conditions which guarantee that the weight function can be specified in such a way that the index of every operator A_{τ} becomes equal to zero in the corresponding weighted space. The case ind A_{τ} = 0 for all τ is of particular interest because it allows to apply certain cutting off techniques which regularize the approximation systems in such a way that the sequence of the regularized systems becomes stable. The applicability of cutting off techniques is discussed in the concluding Section 5.

The case where the indices of some of the local operators are non-zero is obviously unacceptable for practically relevant approximation methods. On the other hand, the estimate (5) shows that the Fredholm properties of the local operators cannot be arbitrarily bad. So there is some hope that index cancellation techniques could help to manage also these situations. But this is beyond the scope of this paper.

We will illustrate the mentioned pecularities only for one special, quadrature formula based, approximation method for solving (3). But clearly, analogous effects appear, and can be studied in a similar manner, for other discretizations of (3) as well. Spline collocation and qualocation methods can serve as further examples (compare [9]). In this sense, we would like to consider this paper as a case study which can help to understand some peculiarities of approximation methods for singular integral operators with conjugation.

Throughout this paper, let L_{add}(X) stand for the set of all additive (i.e. A(x_{1}+ x_{2}) = Ax_{1} + Ax_{2} for x_{1}, x_{2} ∈ X) and continuous operators acting on some Banach space X.

2 A quadrature method and its stability

Let Γ be a simple closed curve, and let γ : ℝ → Γ be a 1-periodic parametrization of Γ. We suppose that there are points u_{1} < u_{2} < … < u_{n_{0}} = u_{1} + 1 such that γ is twice continuously differentiable on each of the intervals (u_{j}, u_{j+1}), j = 1, 2, …, n_{0}−1, and that the derivatives γ′ and γ′′ possess finite one-sided limits γ′(u_{j} ± 0) and γ′′(u_{j} ± 0) such that |γ′(u_{j} − 0)| = |γ′(u_{j} + 0)|, but, possibly, arg γ′(u_{j}−0) ≠ arg γ′(u_{j}+0) at every point u_{j}, j = 1, …, n_{0}−1. By ω = ω(t) we denote the function

ω(t) := ∏_{j=1}^{n_{0}−1} |t − γ(u_{j})|^{−ρ_{j}}, t ∈ Γ,

with ρ_{j} ∈ (−1/2, 1/2) for all j. Under these conditions, the operator A of (3) acts boundedly on the Lebesgue space L_{ω}^{2}(Γ) of all complex-valued measurable functions x with

||x||_{L_{ω}^{2}} := (∫_{Γ} |x(t)|^{2}ω^{2}(t)dt)^{1/2} < ∞

which is considered as a Banach space over the field ℝ.

To improve the convergence rate we consider discretizations based on graded meshes. For, given α ≥ 1, choose a 1-periodic function g : ℝ → ℝ which satisfies the following conditions:

1) g(j/n_{0}) = u_{j} for all j = 1, 2, …, n_{0} − 1,

2) g(s) = g(j/n_{0})+|s−j/n_{0}|^{α}sign(s−j/n_{0}) for all s in some neighbourhood U_{s_{j}} of the point s_{j} = j/n_{0}, with U_{s_{j}} ∩ U_{s_{k}} = ∅ if j ≠ k, and

3) the function g is strictly monotonically increasing on [0, 1], and its restriction onto the set [s_{j}, s_{j+1}] \ (U_{s_{j}} ∩ U_{s_{j+1}}) is a twice differentiable function for every j.

Let n = ln_{0} with l ∈ ℕ, choose numbers δ, ε ∈ (0, 1) with δ ≠ ε, define γ̃(s) = γ(g(s)), and for each k = 0, 1, …, n − 1, put

t_{k}^{(n)} := γ̃ (\frac{k + δ}{n} ) , τ_{k}^{(n)} := γ̃ (\frac{k + ε}{n}) , Δt_{j}^{(n)} = \frac{1}{n}γ̃ (\frac{j + δ}{n} ) . (6)

Applying a shifted trapezoidal rule to the singular integrals in (3) and then collocating at the differently shifted points τ_{k}^{(n)} leads to the following system of algebraic equations

[a(τ_{k}^{(n)}) − b(τ_{k}^{(n)})i cot(π(ε − δ))]ξ_{k}^{(n)} + \frac{b(τ_{k}^{(n)})}{πi} ∑_{j=0}^{n−1} \frac{Δt_{j}^{(n)}}{t_{j}^{(n)} − τ_{k}^{(n)}} ξ_{j}^{(n)}

+ [c(τ_{k}^{(n)}) − d(τ_{k}^{(n)})i cot(π(ε − δ))]\overline{ξ_{k}^{(n)}} + \frac{d(τ_{k}^{(n)})}{πi} ∑_{j=0}^{n−1} \frac{Δt_{j}^{(n)}}{t_{j}^{(n)} − τ_{k}^{(n)}} \overline{ξ_{j}^{(n)}}

+ ∑_{j=0}^{n−1} k_{1}(t_{k}^{(n)}, t_{j}^{(n)})Δt_{j}^{(n)}ξ_{j}^{(n)} + ∑_{j=0}^{n−1} k_{2}(t_{k}^{(n)}, t_{j}^{(n)})Δt_{j}^{(n)}\overline{ξ_{j}^{(n)}}

= f(τ_{k}^{(n)}),  k = 0, 1, ⋯, n − 1,  (7)

where the ξ_{k}^{(n)} are thought of as approximations of the values of the solution u of (3) at the points t_{k}^{(n)} (for details we refer to [9,14,20]). If χ_{j}^{(n)}(t) denotes the characteristic function of the subarc [γ̃(j/n), γ̃((j + 1)/n)) of Γ, then the function

x_{n}(t) = ∑_{j=0}^{n−1} ξ_{j}^{(n)}χ_{j}^{(n)}(t), t ∈ Γ. (8)

can be viewed as an approximate solution of equation (3).

The stability of the approximation method (7)-(8) can be studied by means of localizing techniques as in [9,14,20]. For, we associate with every point τ ∈ Γ a model equation A^{τ}x^{τ} = f^{τ}, which locally represents the equation (3), as well as a model approximation method for the model equation

A_{n}^{τ}x_{n}^{τ} = f_{n}^{τ}, (9)

which locally represents (7). This association can be done as follows: Given τ ∈ Γ, let s_{τ} refer to that point of [0, 1) with γ̃(s_{τ}) = τ, denote by ω_{τ} the number

ω_{τ} := arg(−γ̃′(s_{τ} − 0)/γ̃′(s_{τ} + 0)), ω_{τ} ∈ [0, 2π),

and write Γ_{τ} for the angle Γ_{τ} := e^{iω_{τ}} [0, +∞)∪[0, +∞) with the ray e^{iω_{τ}} [0, +∞) being directed to the origin, while [0, +∞) is directed away from it. Then define L^{2}(Γ_{τ}, ρ_{τ}) as the real Banach space of all complex-valued Lebesgue measurable functions on Γ_{τ} such that

∫_{Γ_{τ}} |x(t)|^{2}|t|^{−2ρ_{τ}} (t)dt < +∞,

where ρ_{τ} = ρ_{j} if τ = γ(u_{j}) (j = 1, 2, …, n_{0} − 1) and ρ_{τ} = 0 at all other points τ. Further introduce the operators

(S_{Γ_{τ}} x)(t) := \frac{1}{πi} ∫_{Γ_{τ}} \frac{x(u)du}{u − t}, (Mx)(t) := \overline{x(t)}, t ∈ Γ_{τ},

as well as

A^{τ} := a(τ)I + b(τ)S_{Γ_{τ}} + c(τ)M + d(τ)S_{Γ_{τ}} M, (10)

which act boundedly on L^{2}(Γ_{τ}, ρ_{τ}). Now, the model equation has the form

A^{τ}x^{τ} = f^{τ}, x^{τ} ∈ L^{2}(Γ_{τ}, ρ_{τ}), (11)

with a function f^{τ} belonging to the Banach space R^{2}(Γ_{τ}, ρ_{τ}) of all functions which are Riemann integrable on every finite subcurve of Γ_{τ} and have a finite Riemann norm

∥f∥_{R^{2}(Γ_{τ} ,ρ_{τ})} := ∥f∥_{L^{2}(Γ_{τ} ,ρ_{τ})} + (∑_{k=0}^{∞} sup_{t∈[k,k+1]} |f(t)|^{2}|t|^{−2ρ_{τ}} )^{1/2}

+ (∑_{k=0}^{∞} sup_{t∈e^{iωτ} [k,k+1]} |f(t)|^{2}|t|^{−2ρ_{τ}} )^{1/2} .

For the approximate solution of equation (11), introduce points t̃_{k}^{(n)} and τ̃_{k}^{(n)}, k ∈ ℤ, by

t̃_{k}^{(n)} := \{ \begin{matrix}(\frac{k+δ}{n} )α & if & k ≥ 0 \\ − (\frac{k+δ}{n} )^{α} e^{iω_{τ}} & if & k < 0 \end{matrix} , τ̃_{k}^{(n)} := \{ \begin{matrix}(\frac{k+ε}{n})α & if & k ≥ 0 \\ − (\frac{k+ε}{n})^{α} e^{iω_{τ}} & if & k < 0 \end{matrix} ,

and determine approximate values ξ_{k}^{(n)} of the exact solution x^{τ}(t̃_{k}^{(n)}), k ∈ ℤ, of (11) by solving the infinite model system

[a(τ) − b(τ)i cot(π(ε − δ))]ξ_{k}^{(n)} + \frac{b(τ)}{πi} ∑_{j=−∞}^{+∞} \frac{Δt̃_{j}^{(n)}}{t̃_{j}^{(n)} − τ̃_{k}^{(n)}} ξ_{j}^{(n)}

+ [c(τ) − d(τ)i cot(π(ε − δ))]\overline{ξ_{k}^{(n)}} + \frac{d(τ)}{πi} ∑_{j=−∞}^{+∞} \frac{Δt̃_{j}^{(n)}}{t̃_{j}^{(n)} − τ̃_{k}^{(n)}} \overline{ξ_{j}^{(n)}}

= f(τ̃_{k}^{(n)}) for k ∈ ℤ  (12)

where

Δt̃_{j}^{(n)} = \{ \begin{matrix}\frac{α}{n} · (\frac{j + δ}{n} )^{α−1} & if & j ≥ 0 \\ −\frac{α}{n} · (\frac{j + δ}{n} )^{α−1}e^{iω_{τ}} & if & j < 0 \end{matrix} .

If χ̃_{k}^{(n)} stands for the characteristic function of the interval [(k/n)^{α}, ((k + 1)/n)^{α}] in case k ≥ 0 and of the interval [e^{iω_{τ}} (k/n)^{α}, e^{iω_{τ}} ((k + 1)/n)^{α}] in case k < 0, then the approximate solution of (11) on the whole curve Γ_{τ} is given by

x_{n}^{τ}(t) = ∑_{k∈ℤ} ξ_{k}^{(n)}χ̃_{k}^{(n)}(t), t ∈ Γ_{τ}. (13)

One can think of the infinite model system (12) - (13) as an operator equation

A_{n}^{τ}x_{n}^{τ} = f_{n}^{τ}, (14)

with x_{n}^{τ}, f_{n}^{τ} ∈ L^{2}(Γ_{τ}, ρ_{τ}) and A_{n}^{τ} ∈ L_{add}(L^{2}(Γ_{τ}, ρ_{τ})). Note that the boundedness of A_{n}^{τ} follows from the corresponding results of [14], Chapters 2 and 3, or of [19]. But it is more convenient for our purposes to translate this system into an equation on accordingly chosen little l^{2}-spaces.

Let l̃_{ν}^{2}, ν ∈ ℝ, be the Banach space of all two-sided sequences {ξ_{k}}_{k∈ℤ} of complex numbers such that

||{ξ_{k}}_{k∈ℤ}||l̃_{ν}^{2} := (∑_{k∈ℤ} |ξ_{k}|^{2}(|k| + 1)^{2ν})^{1/2} < ∞.

Set r_{τ} := (α − 1)/2 − ρ_{τ}α. It is well known (see, e.g., [14,20]) that there is a constant c > 0 such that

\frac{1}{c}∥ ∑_{k∈ℤ} ξ_{k}χ̃_{k}^{(n)}∥_{L^{2}(Γ_{τ} ,ρ_{τ})} ≤ n^{−r_{τ} −1/2}∥\{ξ_{k}\}_{k∈ℤ}∥l̃_{rτ}^{2} ≤ c∥ ∑_{k∈ℤ} ξ_{k}χ̃_{k}^{(n)}∥_{L^{2}(Γ_{τ} ,ρ_{τ})} (15)

for every sequence {ξ_{k}} ∈ l̃_{r_{τ}}^{2} . The de Boor inequalities (15) allow to identify the system (12) with an operator equation

Ã_{n}^{τ}ξ_{n} = η_{n} (16)

where ξ_{n} = {ξ_{k}^{(n)}}_{k∈ℤ} and η_{n} = {f(τ̃_{k}^{(n)})}_{k∈ℤ} belong to l̃_{r_{τ}}^{2} and Ã_{n}^{τ} ∈ L_{add}(l̃_{r_{τ}}^{2} ). Moreover, again due to (15), the sequence {A_{n}^{τ}}_{n∈ℕ} in (14) is stable if and only if the sequence {Ã_{n}^{τ}}_{n∈ℕ} in (16) is so. Now the advantage of considering the model problem on l̃_{r_{τ}}^{2} becomes obvious: Namely, the operators Ã_{n}^{τ}, n ∈ ℕ turn out to be independent of n, thus, the sequence {Ã_{n}^{τ}}_{n∈ℕ} is actually constant, and so this sequence is stable if and only if one of its elements, say the operator Ã_{1}^{τ}, is invertible.

Applying the local principle analogously to [9,14] one arrives at the following result.

Theorem 1 Let a, b, c, d ∈ C(Γ), and let k_{1}, k_{2} ∈ C(Γ × Γ). The quadrature method (7) – (8) is stable if and only if the operator A ∈ L_{add}(L^{2}(Γ_{τ}, ρ_{τ})) as well as all operators Ã_{1}^{τ} ∈ L_{add}(l̃_{r_{τ}}^{2} ), τ ∈ Γ, are invertible.

The difference between the situation examined in [9] and the previous theorem is that what we study here is weighted spaces (i.e. ρ ≠ 0) and non-uniformly graded meshes (i.e. α ≠ 0). These modifications do not influence seriously the proof of Theorem 1, but they essentially change the Fredholm properties of the Ã_{1}^{τ} as we will see now.

3 Local operators and their indices

As mentioned above, both the Fredholm properties and the index of the local operators Ã_{1}^{τ} which correspond to pure singular integral equations without conjugation are independent both of the angles ω_{τ} and of the weights ρ_{τ} (see [19,20]). On the other hand, as we observed in [9] (where the weight function was supposed to be identically 1), the index of the operators Ã_{1}^{τ} which correspond to singular integral equations with conjugation can differ from zero; however, it can take the values -1, 0 and 1 only. Now we are going to examine how the change of the weight function influences the index.

We start with representing the local operators Ã_{1}^{τ} as a 2 × 2 operator matrix. To this end we identify the Hilbert space l̃_{r_{τ}}^{2} with the direct sum l_{r_{τ}}^{2} ⊕ l_{r_{τ}}^{2} of the corresponding Hilbert spaces of one-sided sequences. Then every operator on l̃_{r_{τ}}^{2} corresponds to a 2 × 2 matrix whose entries are operators on l_{r_{τ}}^{2} .

Further, given ν, let T_{ν} stand for the smallest closed subalgebra of L(l_{ν}^{2}) which contains all Toeplitz operators generated by piecewise constant functions, and let T_{ν}^{2×2} be the subalgebra of L(l̃_{ν}^{2}) ≅ L(l_{ν}^{2} ⊕ l_{ν}^{2}) ≅ L^{2×2}(l_{ν}^{2}) which consists of all 2 × 2 matrices with entries in T_{ν}. (Recall that the Toeplitz operator T (a) with generating function a ∈ L^{∞} is defined via its matrix representation with respect to the standard basis of l_{ν}^{2} by T (a) = (a_{j−k})_{j,k=0}^{∞} with a_{k} referring to the kth Fourier coefficient of a. If a is a piecewise constant function, then T (a) is a bounded operator on l_{ν}^{2} if −1/2 < ν < 1/2.)

Lemma 2 Every operator Ã_{1}^{τ} as in (16) can be written as

Ã_{1}^{τ} = Ã^{1,τ} + Ã^{2,τ} \overline{M}, (17)

where Ã^{1,τ}, Ã^{2,τ} ∈ T_{r_{τ}}^{2×2}, and where \overline{M} is the operator

\overline{M}\{\{ξ_{k}\}_{k=0}^{∞}, \{η_{k}\}_{k=0}^{∞}\} = \{\{\overline{ξ_{k}}\}_{k=0}^{∞}, \{\overline{η_{k}}\}_{k=0}^{∞}.\}

Proof. Fix a point τ ∈ Γ and let for brevity

a = a(τ), b = b(τ), c = c(τ), d = d(τ), ρ = ρ_{τ}, ω = ω_{τ}.

Straightforward computation shows that the operator Ã_{1}^{τ} can be written in the form (17) with

Ã^{1,τ} = (A_{k,r}^{1,τ})_{k,r=1}^{2} , Ã^{2,τ} = (A_{k,r}^{2,τ})_{k,r=1}^{2} ,

where

A_{1,1}^{1,τ} = (a − ib cot(π(ε − δ)))I + \frac{b}{πi} (\frac{α(j + δ)^{α−1}}{(j + δ)^{α} − (k + ε)^{α}} )_{k,j=0}^{∞} ,

A_{2,1}^{1,τ} = \frac{b}{πi} (\frac{α(j + δ)^{α−1}}{(j + δ)^{α} − (k + 1 − ε)^{α}} e^{iω})_{k,j=0}^{∞} ,

A_{1,2}^{1,τ} = \frac{b}{πi} (\frac{α(j + 1 − δ)^{α−1}e^{iω}}{(j + 1 − δ)^{α}e^{iω} − (k + ε)^{α}} )_{k,j=0}^{∞} ,

A_{2,2}^{1,τ} = (a − ib cot(π(δ − ε)))I − \frac{b}{πi} (\frac{α(j + 1 − δ)^{α−1}}{(j + 1 − δ)^{α} − (k + 1 − ε)^{α}} )_{k,j=0}^{∞} .

Replacing a and b by c and d, respectively, one gets analogous representations for the operators A_{k,r}^{2,τ}, k, r = 1, 2.

Let further Λ^{z}, z ∈ ℂ, refer to the operator of multiplication by the diagonal matrix Λ^{z} := ((j + 1)^{z}δ_{j,k})_{k,j=0}^{+∞}. Evidently, Λ^{z} is a linear isometry from l_{z}^{2} onto l^{2} (where, clearly, l^{2} = l_{0}^{2}). Therefore, Ã_{1}^{τ} is a continuous additive operator on l_{r_{τ}}^{2} ⊕ l_{r_{τ}}^{2} if and only if the operator

Â_{1}^{τ} := ( \begin{matrix}Λ^{r_{τ}} & 0 \\ 0 & Λ^{r_{τ}} \end{matrix} ) Ã_{1}^{τ} ( \begin{matrix}Λ^{−r_{τ}} & 0 \\ 0 & Λ^{−r_{τ}} \end{matrix} )

belongs to L_{add}^{2×2}(l^{2}). Since Λ^{z}\overline{M} = \overline{M}Λ^{z} for every real z, the operator Â_{1}^{τ} can be rewritten in the form

Â_{1}^{τ} = Â^{1,τ} + Â^{2,τ} \overline{M}, (18)

with Â^{i,τ} = Λ^{(α−1)/2−ρα}Ã^{i,τ}Λ^{(−α+1)/2+ρα} for i = 1, 2. To complete the proof one has to check that the latter operators belong to the algebra T_{0}^{2×2}. This has actually be done in [20], pp. 46-49, and [14], Sections 2.4 and 2.11.

We further need the following result; a proof is in [9].

Lemma 3 (a) If Â ∈ T_{0}, then \overline{M}Â\overline{M} is in T_{0} again.

(b) The operator Ã_{1}^{τ} in (17) is Fredholm on the real space l̃^{2} if and only if the operator

A_{1,τ} := ( \begin{matrix}Â^{1,τ} & Â^{2,τ} \\ \overline{M}Â^{2,τ} \overline{M} & \overline{M}Â^{1,τ} \overline{M} \end{matrix} ) (19)

is Fredholm on the corresponding complex space. In that case,

ind_{ℂ} A_{1,τ} = ind_{ℝ} Ã_{1}^{τ}.

In what follows we will fix the point τ ∈ Γ. Thus, we drop the subscript τ in A_{1,τ}, but we will indicate the dependence of that operator from α, ρ and ω by writing A_{1}^{α,ρ,ω} in place of A_{1,τ}. The Fredholmness of the operator A_{1}^{α,ρ,ω} can be studied via the Gohberg/Krupnik symbol calculus for the algebra T_{0}^{2×2}. For that goal, we let f^{ν} stand for the function

f^{ν}(e^{i2πs}) = 2e^{iπν(s−1)} \frac{sin(−πνs)}{sin(−πν)} − 1 for 0 ≤ s < 1,

and we set f^{∗}(t) := \overline{f(\overline{t})}, β := β(μ) = sin(πθ), γ := γ(μ) = −i cot(πθ), and θ := θ(μ) = \frac{1}{2} − ρ + \frac{i}{2πα} log \frac{μ}{1 − μ}, μ ∈ [0, 1]. Then the symbol Sym (A_{1}^{α,ρ,ω}) of the operator A_{1}^{α,ρ,ω}, which is a 4 × 4-matrix-valued function defined on T × [0, 1], is given as follows (see [1,9,13,14,20]):

If t ≠ 1 and μ ∈ [0, 1], then (Sym A_{1}^{α,ρ,ω})(t, μ) is equal to

( \begin{matrix}a + bf^{(ε−δ)}(t) & 0 & c + df^{(ε−δ)}(t) & 0 \\ 0 & a − bf^{(δ−ε)}(t) & 0 & c − df^{(δ−ε)}(t) \\ c̄ + d̄f^{(ε−δ)∗}(t) & 0 & ā + b̄f^{(ε−δ)∗}(t) & 0 \\ 0 & c̄ − d̄f^{(δ−ε)∗}(t) & 0 & ā − b̄f^{(δ−ε)∗}(t) \end{matrix} ) . (20)

If t = 1 and μ ∈ [0, 1], then (Sym A_{1}^{α,ρ,ω})(1, μ) is given by

( \begin{matrix}a + bγ & i\frac{b}{β} e^{i(ω−π)θ} & c + dγ & i\frac{d}{β} e^{i(ω−π)θ} \\ −i\frac{b}{β} e^{i(π−ω)θ} & a − bγ & −i\frac{d}{β} e^{i(π−ω)θ} & c − dγ \\ \overline{c} − \overline{d}γ & −i\frac{\overline{d}}{β} e^{i(π−ω)θ} & \overline{a} − \overline{b}γ & −i\frac{\overline{b}}{β} e^{i(π−ω)θ} \\ i\frac{\overline{d}}{β} e^{i(ω−π)θ} & \overline{c} + \overline{d}γ & i\frac{\overline{b}}{β} e^{i(ω−π)θ} & \overline{a} + \overline{b}γ \end{matrix} ) . (21)

We prepare the investigation of the Fredholmness of the local operator A_{1}^{α,ρ,ω} by mentioning two important properties of that operator.

Lemma 4 If t ≠ 1, then det{Sym A_{1}^{α,ρ,ω}(t, μ)} is real and non-negative for arbitrary μ ∈ [0, 1] and a, b, c, d ∈ ℂ.

A proof of this assertion is in our paper [9]. Here we offer another proof which is essentially simpler than the former one. Put ν = ε − δ and multiply the matrix (20) both from the left and from the right hand side by the matrix

J := ( \begin{matrix}1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{matrix} ) .

This transformation does not influence the determinant, and the new matrix has the form

( \begin{matrix}a + bf^{ν}(t) & c + df^{ν}(t) & 0 & 0 \\ \overline{c} + \overline{d}f^{ν∗}(t) & \overline{a} + \overline{b}f^{ν∗}(t) & 0 & 0 \\ 0 & 0 & a − bf^{−ν}(t) & c − df^{−ν}(t) \\ 0 & 0 & \overline{c} − \overline{d}f^{−ν∗}(t) & \overline{a} − \overline{b}f^{−ν∗}(t) \end{matrix} ) .

Its determinant is just

[(a + bf^{ν}(t))(\overline{a} + \overline{b}f^{ν∗}(t)) − (c + df^{ν}(t))(\overline{c} + \overline{d}f^{ν∗}(t))]
· [(a − bf^{−ν}(t))(\overline{a} − \overline{b}f^{−ν∗}(t)) − (c − df^{−ν}(t))(\overline{c} − \overline{d}f^{−ν∗}(t))],  (22)

and taking into account the identities

f^{ν∗}(t) = −f^{ν}(t), \overline{f^{ν}(t)} = f^{−ν}(t),

we can write (22) in the form

[(a + bf^{ν}(t))(\overline{a} − \overline{b}f^{ν}(t)) − (c + df^{ν}(t))(\overline{c} − \overline{d}f^{ν}(t))]
· [\overline{(a + bf^{ν}(t))}(a − b\overline{f^{ν}(t)}) − \overline{(c + df^{ν}(t))}(c + d\overline{f^{ν}(t)})],

which is obviously real and non-negative.

Lemma 5 For any α_{1}, α_{2} ∈ ℝ, α_{1} ≥ 1, α_{2} ≥ 1, the operators A_{1}^{α_{1},ρ,ω} and A_{1}^{α_{2},ρ,ω} are simultaneously Fredholm or not. In case they are Fredholm,

ind A_{1}^{α_{1},ρ,ω} = ind A_{1}^{α_{2},ρ,ω}. (23)

Proof. A closer look at the symbol of the operator A_{1}^{α,ρ,ω} reveals that only the matrix (21) depends on the grading parameter α via the function θ. But the image of the function θ = θ(μ), μ ∈ [0, 1], coincides with 1/2 − ρ + iℝ which is independent on α.

So we can suppose here and hereafter that α = 1, and we will write A_{1}^{ρ,ω} in place of A_{1}^{1,ρ,ω}. Recall from [9] that the index of A_{1}^{0,ω} can take the values -1, 0 or 1 only. Our next goal is to verify that this assertion remains true for ρ ∈ (0, 1/2) as well. This proof is essentially different from that given in [9] for case ρ = 0.

Lemma 6 The determinant of Sym A_{1}^{ρ,ω}(1, μ) can be represented as

det{Sym A_{1}^{ρ,ω}(1, μ)} = U + VΦ^{2}(μ),

with certain real numbers U, V, and with Φ denoting the function

Φ(μ) = Φ_{ρ,ω}(μ) = \frac{sin((ω − π)θ)}{sin(πθ)}, μ ∈ [0, 1],

where θ = θ(μ) is as above.

Proof. Expanding the determinant according to the second order minors of the first two rows gives

det{Sym A_{1}^{ρ,ω}(1, μ)} =

= [(a + bγ)(a − bγ) − \frac{b^{2}}{β^{2}} ][(\overline{a} + \overline{b}γ)(\overline{a} − \overline{b}γ) − \frac{\overline{b^{2}}}{β^{2}} ]

+ [(c + dγ)(c − dγ) − \frac{d^{2}}{β^{2}} ][(\overline{c} + \overline{d}γ)(\overline{c} − \overline{d}γ) − \frac{\overline{d^{2}}}{β^{2}} ]

+ [(a + bγ)(c − dγ) − \frac{bd}{β^{2}} ][\frac{\overline{bd}}{β^{2}} − (\overline{a} − \overline{b}γ)(\overline{c} + \overline{d}γ)]

+ [(\overline{a} + \overline{b}γ)(\overline{c} − \overline{d}γ) − \frac{\overline{bd}}{β^{2}} ][\frac{bd}{β^{2}} − (a − bγ)(c + dγ)]

+ \frac{1}{β^{2}} (ad − bc)(\overline{ad − bc})e^{i2(π−ω)θ}

+ \frac{1}{β^{2}} (ad − bc)(\overline{ad − bc})e^{i2(ω−π)θ}.  (24)

The first line on the right hand side of (24) is equal to

[a^{2} − b^{2}γ^{2} − \frac{b^{2}}{β^{2}} ][\overline{a}^{2} − \overline{b}^{2}γ^{2} − \frac{\overline{b^{2}}}{β^{2}} ] =

= [a^{2} − b^{2}(− cot^{2}(πθ) + \frac{1}{sin^{2}(πθ)})][\overline{a}^{2} − \overline{b}^{2}(− cot^{2}(πθ) + \frac{1}{sin^{2}(πθ)})]

= |a^{2} − b^{2}|^{2};

analogously, the second one coincides with |c^{2} − d^{2}|^{2}. For the third line, abbreviate ac − bd and cb − ad by N and P, respectively, and note that

(a + bγ)(c − dγ) − \frac{bd}{β^{2}} = N + P γ, \frac{\overline{bd}}{β^{2}} − (\overline{a} − \overline{b}γ)(\overline{c} + \overline{d}γ) = −\overline{N} + \overline{P} γ,

which finally gives −|N|^{2}−2i Im(P \overline{N})γ+|P |^{2}γ^{2} for the third and, analogously, −|N|^{2} + 2i Im(P \overline{N})γ + |P |^{2}γ^{2} for the forth line. Putting all pieces together we find

det{Sym A_{1}^{ρ,ω}(1, μ)} =

= |a^{2} − b^{2}|^{2} + |c^{2} − d^{2}|^{2} − 2|N|^{2} + 2|P |^{2}γ^{2} + 2|P |\frac{cos(2(ω − π)θ)}{sin^{2}(πθ)}

= |a^{2} − b^{2}|^{2} + |c^{2} − d^{2}|^{2} − 2|N|^{2} + 2|P |^{2} (1 − 2 (\frac{sin((ω − π)θ)}{sin(πθ)})^{2})

= U + V Φ^{2}(μ)

with

V = −4|P |^{2}, U = |a^{2} − b^{2}|^{2} + |c^{2} − d^{2}|^{2} + 2(|P |^{2} − |N|^{2}) (25)

as desired.

Corollary 7 If cb − ad = 0 and if the operator A_{1}^{ρ,ω} is Fredholm, then its index is zero.

The condition cb − ad = 0 is in particular true for singular integral equations without conjugation (where c = 0, d = 0). The corresponding assertion for the index of A_{1}^{ρ,ω} is well-known, see, e.g., [19]. Thus, Lemma 6 provides us with another proof of this result.

Corollary 8 If ω = π and if the operator A_{1}^{ρ,π} is Fredholm, then its index is zero.

It is obvious from Lemma 6 that the winding number of det Sym A_{1}^{ρ,ω}(1, μ) and, thus, the index of the operator A_{1}^{ρ,ω} depend essentially on the properties of the function

Φ(μ) = \frac{sin((ω − π)θ)}{sin(πθ)}, θ = \frac{1}{2} − ρ + \frac{i}{2π} log \frac{μ}{1 − μ}.

Of particular interest for index computation is the location of the zeros of the real and imaginary part of the function Φ in the open interval (0, 1) which will be our next concern. The results of those (somewhat tedious) considerations are summarized in Lemmas 12 (imaginary part) and 18 (real part) below.

Since θ = α + iβ with α = 1/2 − ρ and β = (1/2π) log(μ/(1 − μ)), one has

Φ(μ) = \frac{sin((ω − π)θ)\overline{sin(πθ)}}{sin(πθ)\overline{sin(πθ)}} = \frac{sin((ω − π)θ) sin(π\overline{θ})}{| sin(πθ)|^{2}}
= \frac{cos((ω − 2πα) + iωβ) − cos(ωα + i(ω − 2π)β)}{2| sin(πθ)|^{2}}
= \frac{cos((ω − 2π)α) cosh(ωβ) − cos(ωα) cosh((ω − 2π)β)}{2| sin(πθ)|^{2}}

− i \frac{sin((ω − 2π)β) sinh(ωβ) − sin(ωα) sinh((ω − 2π)β)}{2| sin(πθ)|^{2}};

thus, the real and imaginary parts of the function Φ have the form

Re Φ(μ) = \frac{cos((ω − 2π)α) cosh(ωβ) − cos(ωα) cosh((ω − 2π)β)}{2| sin(πθ)|^{2}}, (26)

Im Φ(μ) = \frac{sin(ωα) sinh((ω − 2π)β) − sin((ω − 2π)β) sinh(ωβ)}{2| sin(πθ)|^{2}}. (27)

The common denominator of (26) and (27) can be written as

| sin(πθ)|^{2} = sin(πθ) sin(π\overline{θ}) = \frac{cos(π(θ − \overline{θ})) − cos(π(θ + \overline{θ}))}{2}

= \frac{cosh(2πβ) − cos(2πα)}{2} = \frac{1}{4} [(\frac{μ}{1 − μ} + \frac{1 − μ}{μ}) + 2 cos(2πρ)]

= \frac{(2μ − 1)^{2} + 4μ(1 − μ) cos^{2}(πρ)}{4μ(1 − μ)},

whence via straightforward calculation follows

Re Φ(μ) = \frac{μ(1 − μ)}{R(μ)} [((\frac{μ}{1 − μ})^{δ} + (\frac{1 − μ}{μ})^{δ}) cos(2π(δ − 1)α)

− ((\frac{μ}{1 − μ})^{δ−1} + (\frac{1 − μ}{μ})^{δ−1}) cos(2πδα)] ,  (28)

Im Φ(μ) = \frac{μ(1 − μ)}{R(μ)} [((\frac{μ}{1 − μ})^{δ−1} − (\frac{1 − μ}{μ})^{δ−1}) sin(2πδα)

− ((\frac{μ}{1 − μ})^{δ} − (\frac{1 − μ}{μ})^{δ}) sin(2π(δ − 1)α)]  (29)

with

R(μ) := (2μ − 1)^{2} + 4μ(1 − μ) cos^{2}(πρ), −1/2 < ρ < 1/2,

and δ := ω/(2π), δ ∈ (0, 1).

Let us start with considering the imaginary part of Φ. For brevity, introduce a new variable x by x = (1−μ)/μ and set y = 2πα, i.e. x runs through (1, +∞), and y lies in (0, 2π). Because the function μ ↦ μ(1 − μ)/R(μ) has no zeros in the interval (0, 1), the zeros of Im Φ are completely determined by those of the function

ψ_{δ}(x) := (x^{1−δ} − x^{δ−1}) sin(δy) + (x^{−δ} − x^{δ}) sin((1 − δ)y). (30)

The following property of the function ψ_{δ} is obvious:

Lemma 9 Let δ be in (1/2, 1) and y = π/δ. Then the function ψ_{δ} has no zeros in (1, ∞).

Thus, when investigating the function ψ_{δ} in case δ > 1/2, we can restrict ourselves to the case when y ≠ π/δ.

Write the function ψ_{δ} as

ψ_{δ}(x) = sin(δy)(x^{−δ} − x^{δ})(ϕ_{δ}(x) − A_{δ}(y)) (31)

with

A_{δ}(y) := \frac{sin((1 − δ)y)}{sin(δy)} and ϕ_{δ}(x) := \frac{x^{1−δ} − x^{δ−1}}{x^{−δ} − x^{δ}} . (32)

Further, given a function f, denote by I_{f}(a, b) the image of the interval (a, b) under that function.

Lemma 10 Let ϕ_{δ} be the function defined in (32). Then

I_{ϕ_{δ}}(1, ∞) = \{ \begin{matrix}((1 − δ)/δ, ∞) & if & δ ∈ (0, 1/2) \\ (0, (1 − δ)/δ) & if & δ ∈ (1/2, 1) \end{matrix} . (33)

Proof. The function ϕ_{δ} is continuous on (1, ∞) with one-sided limits

lim_{x→1+0} ϕ_{δ}(x) = (1 − δ)/δ,  lim_{x→+∞} ϕ_{δ}(x) = \{ \begin{matrix}+∞ & if & δ ∈ (0, 1/2) \\ 0 & if & δ ∈ (1/2, 1) \end{matrix} ,

and the elementary inequality

\frac{x^{α} − x^{−α}}{α} < \frac{x^{β} − x^{−β}}{β} , (34)

which is valid for x > 1 and 0 < α < β < 1, yields

ϕ_{δ}(x) < \frac{1 − δ}{δ} if x ∈ (1, ∞) and δ ∈ (0, 1/2),

and

\frac{1 − δ}{δ} < ϕ_{δ}(x) if x ∈ (1, ∞) and δ ∈ (1/2, 1)

(put α = δ, β = 1 − δ and α = 1 − δ, β = δ in (34), respectively). A similar discussion of the function A_{δ} gives Lemma 11 Let A_{δ} be defined by (32) and let y ∈ (0, 2π).

(a) If δ ∈ (0, 1/2), then A_{δ}(y) < (1 − δ)/δ for all y ∈ (0, 2π).

(b) If δ ∈ (1/2, 1), then A_{δ}(y) > (1 − δ)/δ for all y ∈ (0, π/δ), and A_{δ}(y) < 0 for all y ∈ (π/δ, 2π). Proof. For 0 < α < β < π one has

\frac{sin β}{β} < \frac{sin α}{α}. (35)

If δ ∈ (0, 1/2), and if we set α = δy and β = (1 − δ)y in (35), then this inequality yields for all y ∈ (0, π/(1 − δ))

A_{δ}(y) = \frac{sin((1 − δ)y)}{sin(δy)} < \frac{1 − δ}{δ} ,

which shows that assertion (a) is correct in this case, whereas in case y ∈ (π/(1 − δ), 2π) one has sin((1 − δ)y) ≤ 0 which verifies (a) for those y, too.

If we analogously put α = (1 − δ)y and β = δy with y ∈ (0, π/δ) in (35), then we get the first estimate in (b), and the second one is evident.

Lemma 12 Let δ ∈ (0, 1/2) ∪ (1/2, 1), and ρ ∈ (−1/2, 1/2). Then the function μ ↦ Im Φ(μ) has no zero in the intervals (0, 1/2) and (1/2, 1).

Proof. The assertion concerning the interval (0, 1/2) follows from Lemmas 9, 10 and 11, and for (1/2, 1) one can make use of the evident identity

Im Φ(μ) = −Im Φ(1 − μ), μ ∈ (0, 1/2).

Now we turn over to the real part of the function Φ which is given by (28). As above, we introduce a new variable x := (1 − μ)/μ, set y := 2πα, and consider the function

ψ̂_{δ}(x) = (x^{δ} + x^{−δ}) cos((δ − 1)y) − (x^{δ−1} + x^{1−δ}) cos(δy), x ∈ (1, +∞).

For δ ∈ (1/4, 1/2) ∪ (1/2, 3/4), there is only one number y ∈ (0, 2π) such that cos(δy) vanishes, whereas in case δ ∈ (3/4, 1) there are exactly two different zeros y_{1}, y_{2} ∈ (0, 2π) of the equation cos(δy) = 0. But in any case, if cos(δy) = 0, then the sign of the function ψ̂_{δ} is constant on the interval (1, ∞). Therefore, in what follows we suppose that y ∈ (0, 2π) as well as δ ∈ (0, 1/2) ∪ (1/2, 1) are chosen in such a way that cos(δy) ≠ 0. Under this assumption, the function ψ̂_{δ} can be written as

ψ̂_{δ}(x) = (x^{δ} + x^{−δ}) cos(δy)(Â_{δ} − ϕ̂_{δ}(x)) (36)

where

ϕ̂_{δ}(x) = \frac{x^{1−δ} + x^{δ−1}}{x^{δ} + x^{−δ}} and Â_{δ} = Â_{δ}(y) = \frac{cos((δ − 1)y)}{cos(δy)}. (37)

Lemma 13 Let ϕ̂_{δ}(x) be the function defined in (37). Then

Iϕ̂_{δ}(1, ∞) = \{ \begin{matrix}(1, ∞) & if & δ ∈ (0, 1/2) \\ (0, 1) & if & δ ∈ (1/2, 1). \end{matrix}

The proof is the same as that of Lemma 10 with the only difference that the inequality

x^{β} + x^{−β} < x^{α} + x^{−α} for all 0 < β < α < 1 and x > 1

should be used instead of (34).

Let us now examine the function Â_{δ} given by (37). The behaviour of the derivative

Â′_{δ}(y) = \frac{sin((2δ − 1)y) + (2δ − 1) sin y}{2 cos^{2}(δy)} (38)

of that function depends essentially on the parameter δ. Therefore, the cases δ ∈ (0, 1/4], δ ∈ (1/4, 1/2), δ ∈ (1/2, 3/4) and δ ∈ [3/4, 1) will be considered separately in the subsequent four lemmata.

Lemma 14 If δ ∈ (0, 1/4], then IÂ_{δ}(0, 2π) ∩ Iϕ̂_{δ}(1, ∞) = ∅.

Proof. To start with, suppose δ ∈ (0, 1/4). Let η̂_{δ} denote the numerator of (38). Its derivative

η̂′_{δ}(y) = 2(2δ − 1) cos(δy) cos((1 − δ)y) (39)

has the two zeros

y_{1} := \frac{π}{2(1 − δ)} ∈ (\frac{π}{2} , \frac{2π}{3} ) , y_{2} := \frac{3π}{2(1 − δ)} ∈ (\frac{3π}{2} , 2π) ,

in (0, 2π), and η̂′_{δ}(y) is negative for y ∈ (0, y_{1}) ∪ (y_{2}, 2π) and positive for y ∈ (y_{1}, y_{2}). Particularly, η̂_{δ} is a monotonically decreasing function on both intervals (0, y_{1}) and (y_{2}, 2π). Since η̂_{δ}(+0) = 0 and η̂_{δ}(2π −0) = sin(4πδ) > 0, one further has

Â′_{δ}(y) < 0 for all y ∈ (0, y_{1}) and Â′_{δ}(y) > 0 for all y ∈ (y_{2}, 2π).

These inequalities imply that the function Â_{δ} is monotonically decreasing on (0, y_{1}) and monotonically increasing on (y_{2}, 2π). Since Â_{δ}(+0) = Â_{δ}(2π−0) = 1 we thus obtain Â_{δ}(y) < 1 for all y ∈ (0, y_{1}) ∪ (y_{2}, 2π). Moreover, it is easy to see that Â_{δ}(y) ≤ 0 for all y ∈ [y_{1}, y_{2}]. Hence, IÂ_{δ}(0, 2π) ⊆ (−∞, 1), which proves together with Lemma 13 the claim in case δ ∈ (0, 1/4). If, finally, δ = 1/4, then y_{2} = 2π, and Â_{δ}(y) is negative everywhere on the interval (y_{1}, 2π).

Lemma 15 Let δ ∈ (1/4, 1/2) and set y_{1} := π/(2δ) ∈ (π, 2π) and y_{2} := π/(2(1 − δ)) ∈ (2π/3, π). Then

(a) IÂ_{δ}(0, y_{1}) ∩ Iϕ̂_{δ}(1, ∞) = ∅.

(b) IÂ_{δ}(y_{1}, 2π) ⊆ Iϕ̂_{δ}(1, ∞). Moreover, for every y ∈ (y_{1}, 2π), there is only one x ∈ (1, ∞) such that ϕ̂_{δ}(x) = Â_{δ}(y).

Proof. The points y_{1} and y_{2} are zeros of the derivative η̂′_{δ} in (39), and η̂′_{δ}(y) < 0 for all y ∈ (0, y_{2}) ∪ (y_{1}, 2π) but η̂′_{δ}(y) > 0 if y ∈ (y_{2}, y_{1}). Hence, the function η̂_{δ} is monotonically decreasing on (0, y_{2}) ∪ (y_{1}, 2π) and monotonically increasing on (y_{2}, y_{1}). Since η̂_{δ}(+0) = 0 and η̂_{δ}(y_{1}) = 2δ sin(π/2δ) < 0, the derivative Â′_{δ} is negative on (0, y_{1}) ∪ (y_{1}, 2π). Therefore, the function Â_{δ} is monotonically decreasing on the intervals (0, y_{1}) and (y_{1}, 2π). Moreover, Â_{δ}(+0) = 1. Combining these observations with Lemma 13, we obtain assertion (a).

Now consider the function Â_{δ} on the interval (y_{1}, 2π). It is continuous there, and has the one-sided limits Â_{δ}(y_{1} + 0) = ∞ and Â_{δ}(2π − 0) = 1. Taking into account Lemma 13 once again, we find that Â_{δ}(y) ∈ Iϕ̂_{δ}(1, ∞) for every y ∈ (y_{1}, 2π), and the uniqueness of an element x such that ϕ̂_{δ}(x) = Â_{δ}(y) follows from the strict monotonicity of the function ϕ̂_{δ}.

Analogous discussions lead to the following two lemmas.

Lemma 16 Let δ ∈ (1/2, 3/4), and set y_{1} := π/(2δ) ∈ (2π/3, π) and y_{2} := π/(2(1 − δ)) ∈ (π, 2π). Then

(a) IÂ_{δ}((0, y_{1}) ∪ (y_{1}, y_{2})) ∩ Iϕ̂_{δ}(1, ∞) = ∅.

(b) IÂ_{δ}(y_{2}, 2π) ⊆ Iϕ̂_{δ}(1, ∞). Moreover, for every y ∈ (y_{2}, 2π), there is only one x ∈ (1, ∞) such that ϕ̂_{δ}(x) = Â_{δ}(y) .

Lemma 17 Let δ ∈ (3/4, 1) and set y_{1} := π/(2δ) and y_{2} := 3π/(2δ). Then

IÂ_{δ}((0, y_{1}) ∪ (y_{1}, y_{2}) ∪ (y_{2}, 2π)) ∩ Iϕ̂_{δ}(1, ∞) = ∅.

Summarizing the preceding lemmas and taking into account the identity

Re Φ(1 − μ) = Re Φ(μ), for μ ∈ (0, 1/2)

we arrive at the following characterization of the zeros of the real part of the function Φ.

Lemma 18 Let Φ be the function defined in Lemma 6.

(a) If δ ∈ (0, \frac{1}{4}]∪[\frac{3}{4}, 1) and ρ ∈ (−\frac{1}{2}, \frac{1}{2}), then the real part Re Φ of the function Φ has no zeros in (0, \frac{1}{2}) ∪ (\frac{1}{2}, 1).

(b) If δ ∈ (\frac{1}{4}, \frac{1}{2}) and ρ ∈ [\frac{1}{2} − \frac{1}{4δ}, \frac{1}{2}), then Re Φ has no zeros on (0, \frac{1}{2}) ∪ (\frac{1}{2}, 1), whereas in case ρ ∈ (−\frac{1}{2}, \frac{1}{2} − \frac{1}{4δ}) the function Re Φ has exactly one zero in (0, \frac{1}{2}) and one in (\frac{1}{2}, 1).

(c) If δ ∈ (\frac{1}{2}, \frac{3}{4}) and ρ ∈ [\frac{1}{2} − \frac{1}{4(1−δ)}, \frac{1}{2}), then Re Φ has no zeros on (0, \frac{1}{2}) ∪ (\frac{1}{2}, 1), whereas in case ρ ∈ (−\frac{1}{2}, \frac{1}{2} − \frac{1}{4(1−δ)}), the function Re Φ has exactly one zero in (0, \frac{1}{2}) and one in (\frac{1}{2}, 1).

Corollary 19 If δ ∈ (0, 1/2) ∪ (1/2, 1) and ρ ≥ 0, then the function Re Φ has no zeros on (0, 1/2) ∪ (1/2, 1).

Here is the main result of this section.

Theorem 20 Let ρ ∈ (−1/2, 1/2), ω ∈ (0, 2π), and let the operator A_{1}^{ρ,ω} be Fredholm. Then the index κ := κ(A_{1}^{ρ,π}) of A_{1}^{ρ,ω} satisfies the estimates summarized in the following table. This table has to be read as follows: If the angle ω and the weight ρ belong to the mentioned intervals, then κ is subject to the corresponding estimate in the last column.

Angle  Weight  Index
0 < ω ≤ \frac{π}{2}  −\frac{1}{2} < ρ < \frac{1}{2}  |κ| ≤ 1
\frac{π}{2} < ω < π −\frac{1}{2} < ρ < \frac{1}{2} − \frac{π}{2ω}  |κ| ≤ 2
\frac{π}{2} < ω < π \frac{1}{2} − \frac{π}{2ω} ≤ ρ < \frac{1}{2}  |κ| ≤ 1
ω = π  −\frac{1}{2} < ρ < \frac{1}{2}  |κ| = 0
π < ω < \frac{3π}{2} −\frac{1}{2} < ρ < \frac{1}{2} − \frac{π}{2(2π−ω)} |κ| ≤ 2
π < ω < \frac{3π}{2}  \frac{1}{2} − \frac{π}{2(2π−ω)} ≤ ρ < \frac{1}{2} |κ| ≤ 1
\frac{3π}{2} ≤ ω < 2π −\frac{1}{2} < ρ < \frac{1}{2}  |κ| ≤ 1

Proof. Lemma 6 implies that the imaginary part of the determinant of the symbol of A_{1}^{ρ,ω} has the form

Im det{Sym A_{1}^{ρ,ω}(1, μ)} = −8|N|^{2}Im Φ(μ)Re Φ(μ). (40)

According to Lemmas 12 and 18, the function Im Φ · Re Φ can possess either three (μ = 0, μ = 1/2, μ = 1) or five zeros on [0, 1]. Since det{Sym A_{1}^{ρ,ω}(t, μ)} is real for t ≠ 1 and μ ∈ [0, 1] (cf. Lemma 4), the curve

{det{Sym A_{1}^{ρ,ω}(1, μ)}, 0 ≤ μ ≤ 1} ∪ {det{Sym A_{1}^{ρ,ω}(t, 0)}, t ≠ 1}

can perform no more than one (in case of three zeros) or two (in case of five zeros) rotations arround the origin.

Corollary 21 Let ω ∈ (0, 2π), ρ ∈ [0, 1/2), and let the operator A_{1}^{ρ,ω} be Fredholm. Then |κ(A_{1}^{ρ,ω})| ≤ 1.

Figure 1 shows the image of the interval [0, 1] under the function Φ^{2} in the case ω = 5π/4, ρ = −0.49. Observe that at the first glance, this figure does not give any hint that there might be a second coil of the curve. However, after Theorem 20 was proved we returned to this example and, as the result

\begin{aligned}0 & 250 & 500 & 0 & −250 & 250 \\ & & Figure 1 \end{aligned}  \begin{aligned}0 & −5 & −10 & 0 & −2 & 2 \\ & & Figure 2 \end{aligned}

of a more careful consideration, we observed that, in a neighbourhood of the origin, the curve actually behaves as in Figure 2.

4 On the vanishing of the index of the local operators

Here we are going to discuss criteria for the vanishing of the index κ(A_{1}^{ρ,ω}), which is a necessary condition for the applicability of the modified quadrature methods considered in Section 5. The following result shows indeed that, in case the estimate |κ(A_{1}^{ρ,ω})| ≤ 1 can be guaranteed, it is possible to force that κ(A_{1}^{ρ,ω}) vanishes by appropriately choosing the weight function ρ. Recall that U and V are given by (25).

Theorem 22 Let ω ∈ (0, π) ∪ (π, 2π), assume the operator A_{1}^{0,ω} to be Fredholm, and let one of the following conditions be satisfied:

(i) ω ∈ (0, π] ∪ [3π/2, 2π) and ρ ∈ (−1/2, 1/2),
(ii) ω ∈ (π/2, π) and ρ ∈ [−1/2 − π/(2ω), 1/2), or
(iii) ω ∈ (π, 3/2π) and ρ ∈ [−1/2 − π/(2(2π − ω)), 1/2).

Then the following assertions are true:

(a) The operator A_{1}^{ρ,ω} is Fredholm if and only if

t_{2} = t_{2}(ρ) := t_{1} − 4|cb − ad|^{2} \frac{sin^{2}((ω − π)(1/2 − ρ))}{cos^{2}(πρ)} ≠ 0, (41)

with t_{1} := U = |a^{2} − b^{2}|^{2} + |c^{2} − d^{2}|^{2} + 2(|cb − ad|^{2} − |ac − bd|^{2}).

(b) If the operator A_{1}^{ρ,ω} is Fredholm, then its index vanishes if and only if t_{1} · t_{2} > 0.

Proof. Let A_{1}^{0,ω} be a Fredholm operator. Then t ↦ det{Sym A_{1}^{0,ω}(t, 0)}, t ≠ 1, is a real-valued function without zero. Since

det{Sym A_{1}^{ρ,ω}(t, 0)}|_{t≠1} = det{Sym A_{1}^{0,ω}(t, 0)}|_{t≠1},

the operator A_{1}^{ρ,ω} is Fredholm if and only if the origin does not lie on the arc

Γ̃ := {det{Sym A_{1}^{ρ,ω}(1, μ)}, μ ∈ [0, 1]} = {U + V Φ^{2}(μ), μ ∈ [0, 1]}.

Each of the conditions (i) – (iii) implies that Γ̃ possesses only two points in common with the real axis which correspond to the parameters μ = 0 and μ = 1/2. We denote these points by t̃_{1} and t̃_{2}. Since

Φ^{2}(μ) = (Re Φ(μ))^{2} − (Im Φ(μ))^{2} + 2iRe Φ(μ)Im Φ(μ)

with Re Φ and Im Φ given by (28) and (29), respectively, and since

Re Φ(0) = Im Φ(0) = Im Φ(1/2) = 0,

Re Φ(1/2) = \frac{sin((ω − π)(1/2 − ρ))}{cos(πρ)}.

we conclude that t̃_{1} = t_{1} and t̃_{2} = t_{2}. However, t_{1} cannot be zero since A_{1}^{0,ω} is a Fredholm operator. This finishes the proof of assertion (a), and that of (b) is obvious now.

Thus, under each of the hypotheses (i) – (iii) of Theorem 22, the operator A_{1}^{ρ,ω} is Fredholm, and if t_{1} · t_{2} < 0, then necessarily κ(A_{1}^{ρ,ω}) ≠ 0. Obviously, since the location of the point t_{2} on the real line depends on the weight ρ, one might try in this case to look for some new weight ρ̃ such that κ(A_{1}^{ρ̃,ω}) = 0. The following corollary establishes conditions when such a choice is possible.

Corollary 23 Let one of conditions (i) – (iii) of the previous theorem be satisfied, and suppose that t_{1} · t_{2} < 0. Then one can choose a weight ρ ∈ (−1/2, 1/2) such that κ(A_{1}^{ρ,ω}) = 0 if and only if one of the following conditions is fulfilled:

(a) ω ∈ (0, π/2] ∪ [3π/2, 2π) and

0 ∈ (−∞, U − 4|cb − ad|^{2}(1 − π/ω)^{2}).

(b) ω ∈ (π/2, π) and

0 ∈ (U − 4|cb − ad|^{2} cot^{2}(π^{2}/2ω), U − 4|cb − ad|^{2}(1 − π/ω)^{2}).

(c) ω ∈ (π, 3π/2) and

0 ∈ (U − 4|cb − ad|^{2} cot^{2}(π^{2}/2(2π − ω)), U − 4|cb − ad|^{2}(1 − π/ω)^{2}).

In particular, if U > 0 but U − 4|cb − ad|^{2}(1 − π/ω)^{2} < 0, then for every ρ ∈ (−1/2, 1/2) which satisfies the conditions of Theorem 22, the index of κ(A_{1}^{ρ,ω}) is not equal to zero. Conversely, if U > 0 and U − 4|cb − ad|^{2}(1 − π/ω)^{2} > 0, then there always exists a ρ such that all conditions of Theorem 22 are satisfied and that κ(A_{1}^{ρ,ω}) = 0.

Proof of Corollary 23. To prove the claim requires describing the range of the function

ρ ↦ (Re Φ(1/2)(ρ))^{2} = (\frac{sin((ω − π)(1/2 − ρ))}{cos(πρ)})^{2}

or equivalently, after substituting y := π(1/2 − ρ) ∈ (0, π) and δ := ω/2π, that of the function

ψ_{δ}^{2}(y) := (\frac{sin((2δ − 1)y)^{2})}{sin y})^{2} .

The one-sided limits of ψ_{δ} at 0 and π are

lim_{y→+0} ψ_{δ}(y) = 2δ − 1 and lim_{y→π−0} ψ_{δ}(y) = \{ \begin{matrix}−∞ if 2δ − 1 < 0 \\ +∞ if 2δ − 1 > 0, \end{matrix} , (42)

respectively, the derivative of ψ_{δ}(y) is

ψ′_{δ}(y) = \frac{(δ − 1) sin(2δy) + 2δ(sin(1 − δ)y)}{sin^{2} y},

and the derivative of the nominator η_{δ}(y) of that fraction is given by

η′_{δ}(y) = 4δ(1 − δ) sin y sin((2δ − 1)y).

The following considerations depend on the location of δ in (0, 1).

Let δ ∈ (0, 1/4). Then η′_{δ}(y) < 0 for all y ∈ (0, π), hence, η_{δ} is monotonically decreasing on [0, π]. Since η_{δ}(0) = 0, the function η_{δ} is negative on (0, π), which implies that ψ′_{δ}(y) < 0 for all y ∈ (0, π). Thus, ψ_{δ} is monotonically decreasing on this interval, which together with (42) shows that I_{ψ_{δ}}(0, π) = (−∞, 2δ − 1) and

I_{ψ_{δ}}(0, π) = ((2δ − 1)^{2}, ∞) if δ ∈ (0, 1/4). (43)

If δ = 1/4, then ψ_{1/4}(y) = −1/(2 cos(y/2)), which immediately gives

I_{ψ_{1/4}^{2}}(0, π) = (1/4, ∞). (44)

In case δ ∈ (1/4, 1/2) we have η′_{δ}(y) < 0 for all y ∈ (0, π) again. However, now we have to restrict ourselves to those y for which the curve Γ_{0} has no more than two points in common with the real line (since otherwise, we have no information about the location of the zeros of the symbol). Thus, according to Theorem 22, we get y ∈ (0, π/(4δ)), and therefore I_{ψ_{δ}}(0, π/(4δ)) is the interval (cot \frac{π}{4δ}, 2δ − 1), whence follows

I_{ψ_{δ}^{2}}(0, π/(4δ)) = ((2δ − 1)^{2}, cot^{2} \frac{π}{4δ}) if δ ∈ (1/4, 1/2). (45)

Analogously we obtain

I_{ψ_{δ}^{2}}(0, π/(4(1 − δ))) = ((2δ − 1)^{2}, cot^{2} \frac{π}{4(1 − δ)}) if δ ∈ (1/2, 3/4), (46)

I_{ψ_{3/4}^{2}}(0, π) = (1/4, ∞), (47)

I_{ψ_{δ}^{2}}(0, π) = ((2δ − 1)^{2}, ∞) if δ ∈ (3/4, 1). (48)

A comparison of (43) - (48) with (41) indicates that the conditions of Corollary 23 indeed allow to choose the weight parameter ρ in such a way that the product t_{1} · t_{2}(ρ) becomes positive and, hence the index of A_{1}^{ρ,ω} becomes zero..

5 Modification of the quadrature method by cutting off corner singularities

In case the operator A is invertible, and the local operators A_{1,τ} are Fredholm operators with index 0 but not invertible, then the quadrature rule (7) can be modified in such a way that the necessary and sufficient stability conditions become essentially weaker than for the original system. The modification under consideration was proposed by Chandler and Graham [12] and consists in cutting off the quadrature rule in the neighbourhood of the corner points.

For, choose a non-negative integer i_{0}, define a set of subscripts I_{n,i_{0}} as the set of all k in {0, 1, …, n − 1} such that

|k/n − j/n_{0}| ≥ i_{0}/n for every j = 1, 2, …, n_{0} − 1,

and replace (7) by the system

[a(τ_{k}^{(n)}) − b(τ_{k}^{(n)})i cot(π(ε − δ))]ξ_{k}^{(n)} + \frac{b(τ_{k}^{(n)})}{πi} ∑_{j∈I_{n,i_{0}}} \frac{Δt_{j}^{(n)}}{t_{j}^{(n)} − τ_{k}^{(n)}} ξ_{j}^{(n)}

+ [c(τ_{k}^{(n)}) − d(τ_{k}^{(n)})i cot(π(ε − δ))]\overline{ξ_{k}^{(n)}} + \frac{d(τ_{k}^{(n)})}{πi} ∑_{j∈I_{n,i_{0}}} \frac{Δt_{j}^{(n)}}{t_{j}^{(n)} − τ_{k}^{(n)}} \overline{ξ_{j}^{(n)}}

+ ∑_{j∈I_{n,i_{0}}} k_{1}(t_{k}^{(n)}, t_{j}^{(n)})Δt_{j}^{(n)}ξ_{j}^{(n)} + ∑_{j∈I_{n,i_{0}}} k_{2}(t_{k}^{(n)}, t_{j}^{(n)})Δt_{j}^{(n)}\overline{ξ_{j}^{(n)}}

= f(τ_{k}^{(n)}),  k ∈ I_{n,i_{0}}.  (49)

Observe that the choice i_{0} = 0 yields the original system (7).

The modified method can be analysed in a similar way as the original one (compare Sections 4.2.4 – 4.2.6 in [14] for details), and what results is the following theorem where r_{τ} = (α − 1)/2 − ρ_{τ}α again, where P_{i} stands for the projection operator

P_{i} : l_{r_{τ}}^{2} → l_{r_{τ}}^{2} , (x_{0}, x_{1}, …) ↦ (x_{0}, …, x_{i−1}, 0, 0, …)

in case i ≥ 1, and where Q_{i} := I − P_{i}. The same notations P_{i} and Q_{i} will be used for the diagonal operators diag (P_{i}, P_{i}, P_{i}, P_{i}) and diag (Q_{i}, Q_{i}, Q_{i}, Q_{i}), respectively.

Theorem 24 The modified quadrature method (49) is stable if and only if (a) the operator A is invertible,

(b) the local operators A_{1,τ} are invertible for all τ ∈ Γ \ {γ(u_{1}), …, γ(u_{n_{0}})}, and

(c) the sequences (P_{i}+Q_{i}A_{1,τ}Q_{i})_{i≥1} are stable for all τ ∈ {γ(u_{1}), …, γ(u_{n_{0}})}.

Observe that, if the sequence (P_{i} + Q_{i}A_{1,τ}Q_{i})_{i≥1} is stable, then the operators P_{i} + Q_{i}A_{1,τ}Q_{i} are invertible for sufficiently large i, hence, A_{1,τ} is necessarily a Fredholm operator with index 0 in that case (due to the compactness of the P_{i}). On the other hand, the invertibility of A_{1,τ} is not necessary for the stability of that sequence since the strong limit of the operators P_{i} + Q_{i}A_{1,τ}Q_{i} as i → ∞ is the identity operator (which evidently is invertible). In that sense, (P_{i} + Q_{i}A_{1,τ}Q_{i})_{i≥1} is an approximation method for the identity operator.

The operator A_{1,τ} is of the form Toeplitz operator + discretized Mellin convolution; thus Theorem 4.4 in [14] applies to give the following stability result for the sequence (P_{i} +Q_{i}A_{1,τ}Q_{i})_{i≥1} (compare also Example 4.7 in [14]). In this theorem, we again make use of the abbreviations a := a(τ), b := b(τ), c := c(τ), d := d(τ) introduced in Section 3.

Theorem 25 The sequence (P_{i} + Q_{i}A_{1,τ}Q_{i})_{i≥1} is stable if and only if the following conditions are satisfied:

(a) the operator A_{1,τ} is Fredholm on l_{r_{τ}}^{2} ⊕ … ⊕ l_{r_{τ}}^{2} ,

(b) the 4 × 4 block Toeplitz operator T with generating function

( \begin{matrix}a + bf^{(ε−δ)} & 0 & c + df^{(ε−δ)} & 0 \\ 0 & a − bf^{(δ−ε)} & 0 & c − df^{(δ−ε)} \\ c̄ + d̄(f^{(ε−δ)})^{∗} & 0 & ā + b̄(f^{(ε−δ)})^{∗} & 0 \\ 0 & c̄ − d̄(f^{(δ−ε)})^{∗} & 0 & ā − b̄(f^{(δ−ε)})^{∗} \end{matrix} )

is invertible on l_{0}^{2} ⊕ … ⊕ l_{0}^{2}, and

(c) the singular integral operator

χ_{Γ_{τ} \Γ_{τ,1}}(aI + bS + (cI + dS)M)χ_{Γ_{τ} \Γ_{τ,1}}I

with Γ_{τ} := e^{iω_{τ}} [0, ∞) ∪ [0, ∞) and Γ_{τ,1} := e^{iω_{τ}} [0, 1) ∪ [0, 1) is invertible on L^{2}(Γ_{τ} \ Γ_{τ,1}, r_{τ}) ⊆ L^{2}(Γ_{τ}, r_{τ}).

Whereas it seems to be quite hard to analyse condition (c) further (it is actually equivalent to the invertibility of the Wiener-Hopf operator acting on the space L^{2}([0, ∞), r_{τ}) ⊕ … ⊕ L^{2}([0, ∞), r_{τ}) with generating function

( \begin{matrix}a + bs & −bn_{2π−β} & c + ds & −dn_{2π−β} \\ bn_{β} & a − bs & dn_{β} & c − ds \\ c̄ − d̄s & d̄n_{β} & ā − b̄s & b̄n_{β} \\ −d̄n_{2π−β} & c̄ + d̄s & −b̄n_{2π−β} & ā + b̄s \end{matrix} )

where β := ω_{τ} and where s(z) := coth π(z + i(1/2 + r_{τ})) and n_{β}(z) := e^{(z+i(1/2+r_{τ}))(π−β)}/ sinh π(z + i(1/2 + r_{τ}))), we have effective criteria to check the invertibility of the Toeplitz operator T in condition (b).

Indeed, as in the proof of Lemma 4 one easily gets that the operator T is invertible if and only if the Toeplitz operator with generating function

( \begin{matrix}a + bf^{(ε−δ)} & c + df^{(ε−δ)} & 0 & 0 \\ c̄ + d̄(f^{(ε−δ)})^{∗} & ā + b̄(f^{(ε−δ)})^{∗} & 0 & 0 \\ 0 & 0 & a − bf^{(δ−ε)} & c − df^{(δ−ε)} \\ 0 & 0 & c̄ − d̄(f^{(δ−ε)})^{∗} & ā − b̄(f^{(δ−ε)})^{∗} \end{matrix} )

is invertible and, hence, if and only if the 2 × 2 block Toeplitz operators T_{1} and T_{2} with generating functions

( \begin{matrix}a + bf^{(ε−δ)} & c + df^{(ε−δ)} \\ c̄ + d̄(f^{(ε−δ)})^{∗} & ā + b̄(f^{(ε−δ)})^{∗} \end{matrix} ) , ( \begin{matrix}a − bf^{(δ−ε)} & c − df^{(δ−ε)} \\ c̄ − d̄(f^{(δ−ε)})^{∗} & ā − b̄(f^{(δ−ε)})^{∗} \end{matrix} ) ,(50)

respectively, are invertible.

Making use of the identity (f^{(ν)})^{∗} = −f^{(ν)} (again from the proof of Lemma 4) we can rewrite the operator T_{1} as

( \begin{matrix}a + bT (f^{(ε−δ)}) & c + dT (f^{(ε−δ)}) \\ c̄ − d̄T (f^{(ε−δ)}) & ā − b̄T (f^{(ε−δ)}) \end{matrix} )

and since all entries in this operator matrix commute with each other, we conclude that the operator T_{1} is invertible if and only if the operator

(a + bT (f^{(ε−δ)}))(ā − b̄T (f^{(ε−δ)})) − (c + dT (f^{(ε−δ)}))(c̄ − d̄T (f^{(ε−δ)}))
= (aā − cc̄)I + (āb − ab̄ − c̄d + cd̄)T (f^{(ε−δ)}) − (bb̄ − dd̄)(T (f^{(ε−δ)}))^{2}

is invertible. This clearly happens if and only if

(aā − cc̄)I + (āb − ab̄ − c̄d + cd̄)λ − (bb̄ − dd̄)λ^{2} ≠ 0 (51)

for all λ belonging to the spectrum of the Toeplitz operator T (f^{(ε−δ)}).

So what we still need is the spectrum of that Toeplitz operator considered as acting on the Hilbert space l^{2}. For, abbreviate ε − δ by ν and observe that, for ν ≠ 0, the piecewise continuous function f^{(ν)} maps the unit circle into a circular arc C_{ν} in the complex plane which joins -1 to 1 and runs through the point i tan(πν/2) (the center of that circular arc is −i cot πν, and its radius is equal to 1/| sin πν|). The well known criterion for the invertibility of Toeplitz operators with piecewise continuous generating function (see, e.g., Corollary 2.40 and Theorem 2.74 in [1]) states that the spectrum of T (f^{(ν)}) just coincides with that compact region in the complex plane which is bounded by the circular arc C_{ν} and the interval [−1, 1]. Thus, denoting this region by D_{ν}, we get

Lemma 26 The Toeplitz operator T_{1} is invertible if and only if (51) is satisfied for all λ ∈ D_{ε−δ}.

An analogous result holds for the invertibility of T_{2}.

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Authors’ addresses:

V. S. Didenko University of Brunei Darussalam Department of Mathematics Bandar Seri Begawan 2028 Brunei

B. Silbermann Fakultät für Mathematik Technische Universität Chemnitz D 09107 Chemnitz Germany

S. Roch Fachbereich Mathematik Technische Universität Darmstadt Schlossgartenstrasse 7 D 64289 Darmstadt Germany