Abstract

Let \(D\) be a connected bounded domain in \({\mathbb R}^2\), \(S\) be its boundary which is closed, connected and smooth, or \(S=(-\infty, \infty)\). Let \(f=B\phi=\frac 1 {\pi i}\int_S\frac{\phi(s)ds}{s-t}ds\). Here \(\phi, f\in X\), where \(X\) is a set of distributions, that is, \(\phi, f\) are bounded linear functionals on the Banach space \(H^\mu\) of Hölder-continuous functions on \(S\). A definition of the operator \(B\) on \(X\) is given. The equation \(B\phi=f\) is studied. It is proved that this equation has a solution in \(X\), and this solution is unique in \(X\).

Introduction

Let \(D\) be a connected bounded domain on \({\mathbb R}^2\), \(S\) be its boundary, which is closed and \(C^{1,a}\)-smooth, \(0<a\le 1\) or \(S=(-\infty,\infty)\). Consider the singular integral equation (see [1]): \[\label{e1} f=B\phi=\frac 1 {i\pi}\int_S\frac{\phi(s)}{s-t}ds.------(1.1)\] We assume that \(f\in X\). This is the basic new assumption. Our goal is to give a new definition of the operator \(B\). This definition makes the proof of the existence of the unique solution to the equation \(B\phi=f\) in \(X\) very simple.

Let \(H^\mu:=H^\mu(S)\) be the space of Hölder-continuous functions, \(0<\mu<1\), with the usual norm, and \(X\) be the Banach space of the bounded linear functionals \(h\) on \(H^\mu\) with the standard norm: \[\label{e2} \|h\|_X=\sup_{\|\psi\|_{H^\mu}=1} |h(\psi)|, \quad \psi\in H^\mu.------(1.2)\] One has \[\label{e3} |h(\psi)|\le \|h\|_X \|\psi\|_{H^\mu}.------(1.3)\] It is known, see [1], that on \(H^\mu\) the operator \(B\) is bounded, has a bounded inverse, the range \(R(B)\) of \(B\) is equal to \(H^\mu\), \(R(B)=H^\mu\), and \(B^2=I\), where \(I\) is the identity operator. Therefore, the equation \[\label{e4} B\phi=f------(1.4)\] is solvable in \(H^\mu\) for any \(f\in H^\mu\) and its solution \(\phi=Bf\) is unique. One has \(\|B\psi\|_{H^\mu}\le c_\mu \|\psi\|_{H^\mu}\), where \(c_\mu\) is a constant.

The aim of this paper is to generalize this result to the space \(X\) which contains distributions. By \(D(B)\) and \(R(B)\) the domain of definition of \(B\) and, respectively, the range of \(B\) are denoted. Our result is the following Theorem.

Theorem 1. The \(D(B)=X\), \(R(B)=X\), the operators \(B\) and \(B^{-1}\) are bounded.

Theorem 1 is proved in Section 2.

Proofs

Let us define \(B\) on \(X\):

Definition 1. \[\label{e5} (B\phi, \psi)=-(\phi, B\psi) \quad \forall \psi\in H^\mu(S), \,\, 0<\mu<1.------(2.1)\] Here \((h,\psi)\) is the value of the functional \(h\in X\) on the element \(\psi\in H^\mu\). This definition is similar to the one used in . We prove that \(B\) is defined on all of \(X\) and is bounded on \(X\).

Lemma 1. The operator \(B:X\to X\) is bounded, \(D(B)=X\).

Proof. Let \(\phi\in X\). Then, using Definition 1, one gets: \[\label{e6} |(B\phi, \psi)|=|(\phi, B\psi)|\le \|\phi\|_X \|B\psi\|\le c_{\mu}\|\phi\|_X \|\psi\|_{H^\mu}, \quad \forall \psi\in H^\mu(S), \,\, 0<\mu<1.------(2.2)\] Thus, \(D(B)=X\) and \(||B||_X\le c_{\mu}\).

Lemma 1 is proved. \(\Box\)

Lemma 2. Relation \(B^2=I\) holds in \(X\).

Proof. By definition (2.1) one gets: \[\label{e7} (B^2\phi, \psi)=-(B\phi, B\psi)=(\phi, B^2\psi)=(\phi, \psi) \quad \forall \psi\in H^\mu,------(2.3)\] where the last equation holds because \(B^2=I\) on \(H^\mu\).

Therefore, Lemma 2 is proved. \(\Box\)

From equation (1.4) and from Lemma 2 one gets: \[\label{e4a} \phi=Bf.------(2.4)\] Lemma 3. Equation (1.4) has a solution in \(X\) and this solution is unique.

Proof. First, let us prove the uniqueness. If \(B\phi_1=f=B\phi_2\), then \(B\phi=0\), where \(\phi:=\phi_1-\phi_2\). By (2.1), one has \((\phi, B\psi)=0, \forall \psi\in H^\mu.\) Since the set \(B\psi=H^\mu\) when \(\psi\) runs through all of \(H^\mu\), it follows that \(\phi=0\). The uniqueness of the solution in \(X\) is proved.

Let us prove the existence of the solution. By Lemma 2 and equation (2.4), one obtains: \[\label{e8} B^2\phi=\phi=Bf.------(2.5)\] Therefore, Lemma 3 is proved. \(\Box\)

The conclusions of Theorem 1 follows from Lemmas 1–3. \(\Box\)

Remark 1. It does not follow that \(B\) maps \(L^1(S)\) into itself.

Example 1. Let us show that there is an \(f\in L^1(S)\) such that \(Bf\notin L^1(S)\). Let \(S=(-\infty, \infty)\), \(\mathcal{F}(f):=\tilde{f}\), \(\tilde{f}:=\int_Se^{i\xi s}f(s)ds\), \(\mathcal{F}(Bf)=\mathcal{F}(f) \mathcal{F}(s^{-1})=\tilde{f} i\pi sgn (\xi)\). We have used the known formula, see : \(\mathcal{F}(s^{-1})= i\pi sgn (\xi)\), where \(sgn (\xi)=1\) if \(\xi>0\), \(sgn (\xi)=-1\) if \(\xi<0\). The Fourier transform of \(f\in L^1(S)\) is a continuous uniformly bounded function. Therefore, \(\tilde{f} sgn (\xi)\) is not, in general, a continuous function at \(\xi=0\). Thus, if \(\tilde{f}|_{\xi=0}\neq 0\), then the function \(Af\notin L^1(S)\).

Conclusion

A new definition of the singular integral operator in a special space \(X\) of distributions is given.

Conflict of interest

There is no conflict of interest.

References

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2. Gel’fand, I. and Shilov, G. (1959). Generalized functions, Vol. 1, Gos. Izdat. Fiz.-Math. Lit., Moscow, (In Russian). View

3. Ramm, A. G. (2023). Boundary values of analytic functions, Far East Journal of Appl. Math., 116, N3, 215-227. View