1 Introduction

All the functions considered in this letter are real-valued. Recall that the Hilbert transform H of a function f on \({{\mathbb {R}}}\) (in proper function spaces) is defined by

$$\begin{aligned} H(f)(x)=\text {p.v.}\frac{1}{\pi }\int _{{{\mathbb {R}}}}\frac{f(y)}{x-y}dy, \end{aligned}$$

where “p.v." stands for the principal value. This singular integral operator has a local version, say on the interval \(I=(-1,1)\), and is given by

$$\begin{aligned} H_I(f)(x)=\text {p.v.}\frac{1}{\pi }\int _{-1}^1\frac{f(y)}{x-y}dy. \end{aligned}$$

It is called the finite Hilbert transform and arises naturally in applied science. In particular, the resolution of the following airfoil equation from aerodynamics,

$$\begin{aligned} H_I(f)(x)=\phi (x)\quad (-1<x<1), \end{aligned}$$

involves the inversion of \(H_I\) in proper function spaces. The airfoil equation is approached by Tricomi in [7] via establishing some convolution theorems for \(H_I\); these convolution identities are motivated by his earlier study on mixed type equations. A byproduct from [7] is the following arcsine distribution uniqueness.

Theorem 1.1

(Tricomi 1951) Let \(f(x)(1-x^2)^{\frac{1}{4}}\in L^2_I=L^2(-1,1)\). If H(f) vanishes identically on \(I=(-1,1)\), then for some real-valued constant c,

$$\begin{aligned} f(x)=\frac{c}{\sqrt{1-x^2}}\chi _{I}(x). \end{aligned}$$

Here, \(\chi _{I}\) is the indicator function of I.

Remark 1.2

For an application of this to Erdös-Turán inequality, see [1].

Recently, this uniqueness result is revisited by Coifman and Steinerberger in [2]. They further observed the following localized Parseval identity for \(H_I\).

Theorem 1.3

(Coifman and Steinerberger 2019) Let \(f(x)(1-x^2)^{\frac{1}{4}}\in L^2_I\). If the mean value of \(f(x)(1-x^2)^{\frac{1}{2}}\) on I is 0, then

$$\begin{aligned} \int _{-1}^1H_I(f)^2(x)\sqrt{1-x^2}dx=\int _{-1}^1f^2(x)\sqrt{1-x^2}dx. \end{aligned}$$
(1.1)

This complements the global \(L^2\)-isometry for the standard Hilbert transform H:

$$\begin{aligned} \Vert H(f)\Vert _{L^2({{\mathbb {R}}})}=\Vert f\Vert _{L^2({{\mathbb {R}}})}. \end{aligned}$$

Two proofs for the localized Parseval identity (1.1) are offered in [2]: one is by Chebychev orthogonal expansion (see e.g. [6]), another by working on the unit circle and using the formula of conjugate functions as carried out in [3]. In this letter we point out that the identity (1.1) admits the following dual weight version.

Theorem 1.4

Let \(f(x)(1-x^2)^{-\frac{1}{4}}\in L^2_I\). Then

$$\begin{aligned} \int _{-1}^1H_I(f)^2(x)\frac{dx}{\sqrt{1-x^2}}=\int _{-1}^1f^2(x)\frac{dx}{\sqrt{1-x^2}}. \end{aligned}$$
(1.2)

Remark 1.5

Such a dual weight mechanism is indeed quite common in boundary value problems, see for example Rosén [5] on Euclidean upper half-space.

2 Proof of Theorem 1.4

Our proof of (1.2) adapts the Chebychev orthogonal expansion arguments in [2]. Consider f so that \(\frac{f(x)}{\sqrt{1-x^2}}\) is a polynomial. For some \(N\in {{\mathbb {N}}}\) we can write

$$\begin{aligned} \frac{f(x)}{\sqrt{1-x^2}}=\sum _{k=0}^N a_kU_k(x), \end{aligned}$$

where \(\{U_k\}\) denotes the family of Chebychev polynomials of the second kind. Thereby,

$$\begin{aligned} \begin{aligned} \int _{-1}^1f^2(x)\frac{dx}{\sqrt{1-x^2}}&=\int _{-1}^1\left( \frac{f(x)}{\sqrt{1-x^2}}\right) ^2\sqrt{1-x^2}dx\\&=\frac{\pi }{2}\sum _{k=0}^N a_k^2. \end{aligned} \end{aligned}$$

Furthermore, we have

$$\begin{aligned} \begin{aligned} H_I(f)&=\text {p.v.}\frac{1}{\pi }\int _{-1}^1\frac{f(y)/\sqrt{1-y^2}}{x-y}\sqrt{1-y^2}dy\\&=\text {p.v.}\frac{1}{\pi }\int _{-1}^1\sum _{k=0}^N \frac{a_kU_k(y)}{x-y}\sqrt{1-y^2}dy. \end{aligned} \end{aligned}$$

After using the crucial formulae (see for example [4, p. 187])

$$\begin{aligned} \text {p.v.}\frac{1}{\pi }\int _{-1}^1 \frac{U_k(y)}{x-y}\sqrt{1-y^2}dy=-T_{k+1}(x), \end{aligned}$$

where \(\{T_k\}\) denotes the family of Chebychev polynomials of the first kind, we get

$$\begin{aligned} \int _{-1}^1H_I(f)^2(x)\frac{dx}{\sqrt{1-x^2}}=\frac{\pi }{2}\sum _{k=0}^N a_k^2. \end{aligned}$$

This proves the identity (1.2) for f such that \(\frac{f(x)}{\sqrt{1-x^2}}\) is a polynomial. Note that the subspace of such functions is dense in \(L^2\left( I,(1-x^2)^{-\frac{1}{2}}dx\right) \), so \(H_I\) extends to an isometry \(\widetilde{H_I}\) on \(L^2\left( I,(1-x^2)^{-\frac{1}{2}}dx\right) \). Moreover, this extension agrees with \(H_I\) since \(L^2\left( I,(1-x^2)^{-\frac{1}{2}}dx\right) \) embeds into \(L^2_I\). This finishes the proof of Theorem 1.4.