1 Introduction

Our motivation comes from a very basic question about the Hilbert transform \({\mathcal {H}}^{\mathbb {T}}\) on the unit circle \({\mathbb {T}}\simeq (-\pi ,\pi ]\) equipped with a normalized uniform measure \(m\). Recall that this operator is given by the singular integral

$$\begin{aligned} {\mathcal {H}}^{\mathbb {T}}f(x)=\text{ p.v. }\int _{-\pi }^\pi f(t)\cot \frac{x-t}{2}m(\text{ d }t),\qquad x\in {\mathbb {T}}, \end{aligned}$$

when \(f\in L^1({\mathbb {T}})\). A classical result of Riesz [13] states that for any \(1<p<\infty \) there is a finite universal constant \(C_p\) such that

$$\begin{aligned} \left| \left| {\mathcal {H}}^{\mathbb {T}}f\right| \right| _{L^p({\mathbb {T}})} \le C_p||f||_{L^p({\mathbb {T}})},\qquad f\in L^p({\mathbb {T}}). \end{aligned}$$
(1.1)

For \(p=1\) the above estimate does not hold with any \(C_1<\infty \), but, as Kolmogorov showed in [11], there is an absolute \(c_1<\infty \) such that

$$\begin{aligned} \left| \left| {\mathcal {H}}^{\mathbb {T}}f\right| \right| _{L^{1,\infty }({\mathbb {T}})}{:=}\sup _{\lambda >0}\bigg [\lambda \, m\left( \left\{ x\in {\mathbb {T}}:|{\mathcal {H}}^{\mathbb {T}}f(x)|\ge \lambda \right\} \right) \bigg ] \le c_1||f||_{L^1({\mathbb {T}})}, \end{aligned}$$
(1.2)

whenever \(f\in L^1({\mathbb {T}})\). The optimal values of the constants \(C_p\) and \(c_1\) were determined in 1970s: Pichorides [12] and Cole (unpublished: see Gamelin [9]) proved that the best constant in (1.1) equals \(\cot \frac{\pi }{2p^*}\), where \(p^*=\max \{p,p/(p-1)\}\), and Davis [6] showed that the optimal choice for the constant \(c_1\) in (1.2) is

$$\begin{aligned} \left( \frac{1}{\pi }\int _{\mathbb {R}}\frac{\left| \frac{2}{\pi }\log |t|\right| }{t^2+1}\text{ d }\!t\right) ^{-1}=\frac{1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\cdots }{1-\frac{1}{3^2}+\frac{1}{5^2}-\frac{1}{7^2}+\cdots }=1.347\ldots . \end{aligned}$$

The above results are of fundamental importance to harmonic analysis. Furthermore, the methods developed by Riesz [13] have had a profound influence on the shape of the contemporary mathematics. For numerous extensions and applications of the above statements, consult e.g. the works of Burkholder [3], Calderón and Zygmund [5], Essén [8], Gohberg and Krupnik [10], Stein [14] and Zygmund [15], and many more.

We will continue the research in this direction. We will be interested in a “dual” version of Kolmogorov’s result, i.e., in a weak-\(L^\infty \) estimate for \({\mathcal {H}}^{\mathbb {T}}\). To explain what the weak-\(L^\infty \) space is, we need more notation. For a given measurable function \(f:{\mathbb {T}}\rightarrow {\mathbb {R}}\), we define \(f^*\), the decreasing rearrangement of \(f\), by

$$\begin{aligned} f^*(t)=\inf \left\{ \lambda \ge 0:m\left( \left\{ x\in {\mathbb {T}}:|f(x)|>\lambda \right\} \right) \le t\right\} . \end{aligned}$$

Then \(f^{**}:(0,1]\rightarrow [0,\infty )\), the maximal function of \(f^*\), is given by the formula

$$\begin{aligned} f^{**}(t)=\frac{1}{t}\int _0^t f^*(s)\text{ d }s,\qquad t\in (0,1]. \end{aligned}$$

One easily verifies that \(f^{**}\) can alternatively be defined by

$$\begin{aligned} f^{**}(t)=\frac{1}{t}\sup \left\{ \int _E |f| \text{ d }m\,:\,E\subseteq {\mathbb {T}},\,m(E)=t\right\} . \end{aligned}$$

We are ready to introduce the weak-\(L^\infty \) space. Following Bennett, DeVore and Sharpley [1], we let

$$\begin{aligned} ||f||_{W({\mathbb {T}})}=\sup _{t\ge 0}(f^{**}(t)-f^*(t)) \end{aligned}$$

and define \(W({\mathbb {T}})=\{f:{\mathbb {T}}\rightarrow {\mathbb {R}}:||f||_{W({\mathbb {T}})}<\infty \}\). Some words explaining the meaning of this space are in order. For each \(1\le p<\infty \), the usual weak space \(L^{p,\infty }\) properly contains \(L^p\), but for \(p=\infty \), the two spaces coincide. Thus, there is no Marcinkiewicz interpolation theorem between \(L^1\) and \(L^\infty \) for operators which are unbounded on \(L^\infty \). The reason for introducing the space \(W\) was to fill this gap. It can be verified that this space contains \(L^\infty \), can be understood as an appropriate limit of \(L^{p,\infty }\) as \(p\rightarrow \infty \), and enjoys the required interpolation property: if \(T\) is bounded as an operator from \(L^1\) to \(L^{1,\infty }\) and from \(L^\infty \) to \(W\), then it has an extension which is bounded on \(L^p\) spaces, \(1<p<\infty \). See [1] for details. There is a further evidence, again rooted in the interpolation theory, that the space \(W\) can serve as a substitute for weak-\(L^\infty \). Namely, the Peetre \(K\)-functional for the pair \((L^1,L^\infty )\) (cf. [4, p. 184]) is explicitly given by

$$\begin{aligned} K\left( f,t;L^1,L^\infty \right) =\int _0^t f^*(s)\text{ d }s=tf^{**}(t),\qquad t\in (0,1]. \end{aligned}$$

Thus, the weak-\(L^1\) norm can be expressed in terms of the \(K\)-functional by

$$\begin{aligned} ||f||_{L^{1,\infty }({\mathbb {T}})}=\sup _{t\in (0,1]}tf^*(t)=\sup _{t\in (0,1]}t\frac{d}{dt}K\left( f,t;L^1,L^\infty \right) . \end{aligned}$$
(1.3)

Now if we reverse the roles of \(L^1\) and \(L^\infty \), and make use of the identity \(K(f,t;L^\infty ,L^1)=tK(f,t^{-1};L^1,L^\infty )\), we see that the expression on the right of (1.3) is precisely \(\sup _{t\in (0,1]}[f^{**}(t)-f^*(t)]\). Hence this number can be understood as a substitute for the norm in the weak-\(L^\infty \). For more on this interplay, the connections between \(W\) and \(BMO\), as well as other interesting properties of \(W\), we refer the reader to [1] and the monograph [2] by Bennett and Sharpley.

One of our main results is the identification of the norm of \({\mathcal {H}}^{\mathbb {T}}\) as an operator acting from \(L^\infty ({\mathbb {T}})\) to \(W({\mathbb {T}})\). Here is the precise statement.

Theorem 1.1

For any \(f\in L^\infty ({\mathbb {T}})\) we have

$$\begin{aligned} \left| \left| {\mathcal {H}}^{\mathbb {T}}f\right| \right| _{W({\mathbb {T}})}\le ||f||_{L^\infty ({\mathbb {T}})}. \end{aligned}$$
(1.4)

The inequality is sharp: for any \(c<1\) there is a function \(f\in L^\infty ({\mathbb {T}})\) such that \(||{\mathcal {H}}^{\mathbb {T}}f||_{W({\mathbb {T}})}> c||f||_{L^\infty ({\mathbb {T}})}\).

We will also study an analogue of the above result in the nonperiodic case. Recall that the Hilbert transform \({\mathcal {H}}^{\mathbb {R}}\) on the real line is defined by the principal value integral

$$\begin{aligned} {\mathcal {H}}^{\mathbb {R}}f(x)=\frac{1}{\pi }\text{ p.v. }\int _{\mathbb {R}}\frac{f(t)}{x-t}\text{ d }t, \qquad x\in {\mathbb {R}}, \end{aligned}$$

when \(f\in L^1({\mathbb {R}})\). The above strong and weak-type inequalities (1.1), (1.2) can be extended to analogous statements for \({\mathcal {H}}^{\mathbb {R}}\) and the optimal constants remain unchanged (see e.g. [13, 15]). It is natural to ask about a sharp weak-type \((\infty ,\infty )\) inequality in this setting. To study this problem, define the weak space \(W({\mathbb {R}})\) in the same manner as above:

$$\begin{aligned} W({\mathbb {R}})=\bigg \{f:{\mathbb {R}}\rightarrow {\mathbb {R}}\,:\,||f||_{W({\mathbb {R}})}{:=}\sup _{t>0}\big [f^{**}(t)-f^*(t)\big ]<\infty \bigg \}, \end{aligned}$$

where, as previously, \(f^*\) denotes the decreasing rearrangement of \(f\) and \(f^{**}\) stands for the maximal function of \(f^*\). Here is the nonperiodic version of Theorem 1.1. It is well known that some technical problems arise when one defines the action of the Hilbert transform on \(L^\infty ({\mathbb {R}})\); to avoid these, we impose a slightly stronger integrability on functions.

Theorem 1.2

If \(f\) belongs to \(L^\infty ({\mathbb {R}})\cap L^p({\mathbb {R}})\) for some \(1\le p<\infty \), then we have the sharp bound

$$\begin{aligned} \left| \left| {\mathcal {H}}^{\mathbb {R}}f\right| \right| _{W(\mathbb {{\mathbb {R}}})}\le ||f||_{L^\infty ({\mathbb {R}})}. \end{aligned}$$
(1.5)

The paper is organized as follows. In the next section we establish Theorem 1.1. In the proof of (1.4) we make use of Bellman function method: the estimate is deduced from the existence of a certain special superharmonic function. In the final part of the paper we present the proof of Theorem 1.2, which follows from Theorem 1.1 by certain transference-type arguments.

2 Periodic Case

For any \(c\ge 0\), define the function \(V^{(c)}:[-1,1]\times [0,\infty )\rightarrow {\mathbb {R}}\) by \(V^{(c)}(x,y)=(y-c)\chi _{\{y> 0\}}\) (here and below, \(\chi _A\) denotes the indicator function of a set \(A\)). Furthermore, let \(U^{(c)}:(-1,1)\times (0,\infty ) \rightarrow {\mathbb {R}}\) be given by the formula

$$\begin{aligned} U^{(c)}(x,y)=&\, y-c+\frac{2c}{\pi }\arctan \left( \frac{e^{-\pi y/2}}{\cos (\pi x/2)}-\tan \frac{\pi x}{2}\right) \\&+ \frac{2c}{\pi }\arctan \left( \frac{e^{-\pi y/2}}{\cos (\pi x/2)}+\tan \frac{\pi x}{2}\right) . \end{aligned}$$

It is easy to check that \(U^{(c)}\) is a harmonic function. Actually, it can be regarded as a harmonic lift of \(V^{(c)}\), in the sense explained in the first part of the lemma below.

Lemma 2.1

The function \(U^{(c)}\) has the following properties.

  1. 1.

    If \(Y>0\), then \(\lim _{(x,y)\rightarrow (\pm 1,Y)} U^{(c)}(x,y)=V^{(c)}(\pm 1,Y)\); if \(X\in (-1,1)\), then \(\lim _{(x,y)\rightarrow (X,0)} U^{(c)}(x,y)=V^{(c)}(X,0)\).

  2. 2.

    For any \(x\in (-1,1)\), we have

    $$\begin{aligned} \lim _{y\downarrow 0} U^{(c)}(x,y)/y=1-c\left( \cos \frac{\pi x}{2}\right) ^{-1}. \end{aligned}$$
  3. 3.

    For any \((x,y)\in (-1,1)\times (0,\infty )\), we have \(U^{(c)}(x,y)\ge V^{(c)}(x,y)\).

Proof

The properties (1) and (2) are straightforward and left to the reader. The majorization (3) is also easy: we must show that

$$\begin{aligned} \arctan \left( \frac{e^{-\pi y/2}}{\cos (\pi x/2)}-\tan \frac{\pi x}{2}\right) + \arctan \left( \frac{e^{-\pi y/2}}{\cos (\pi x/2)}+\tan \frac{\pi x}{2}\right) \ge 0. \end{aligned}$$

This follows from the estimate

$$\begin{aligned} \left( \frac{e^{-\pi y/2}}{\cos (\pi x/2)}-\tan \frac{\pi x}{2}\right) +\left( \frac{e^{-\pi y/2}}{\cos (\pi x/2)}+\tan \frac{\pi x}{2}\right) \ge 0 \end{aligned}$$

and the fact that the arctangent function is odd and increasing on the real line. \(\square \)

It will be convenient for us to extend \(U^{(c)}\) to the halfstrip \([-1,1]\times [0,\infty )\) by the requirement that \(U^{(c)}\) and \(V^{(c)}\) match at the boundary of this set. Then \(U^{(c)}\) becomes a harmonic majorant of \(V^{(c)}\) on the whole \([-1,1]\times [0,\infty )\), and it is continuous except for the points \((\pm 1,0)\). In addition, part (2) of the above lemma implies that for \(c\ge 1\), the one-sided partial derivative \(U^{(c)}_{y+}\) satisfies \(U^{(c)}_{y+}(x,0)\le 0\) for all \(x\in (-1,1)\).

The above function \(U^{(c)}\) is a “building block” for a larger class of superharmonic functions. For a fixed parameter \(\lambda \ge 0\), introduce the functions \(\mathcal {U}^{(c)}_\lambda \), \(\mathcal {V}^{(c)}_\lambda \) on the strip \([-1,1]\times {\mathbb {R}}\) by the formulas

$$\begin{aligned} \mathcal {U}^{(c)}_\lambda (x,y)=U^{(c)}\left( x,(|y|-\lambda )_+\right) ={\left\{ \begin{array}{ll} U^{(c)}(x,y-\lambda ) &{}\text{ if } y\ge \lambda ,\\ 0 &{}\text{ if } |y|<\lambda ,\\ U^{(c)}(x,-\lambda -y) &{}\text{ if } y<-\lambda \end{array}\right. } \end{aligned}$$

and \(\mathcal {V}^{(c)}_\lambda (x,y)=V^{(c)}(x,(|y|-\lambda )_+)=(|y|-\lambda )_+-c\chi _{\{|y|> \lambda \}}\).

Lemma 2.2

For each \(\lambda \ge 0\) and \(c\ge 1\), the function   \(\mathcal {U}_\lambda ^{(c)}\) is a superharmonic majorant of \(\mathcal {V}_\lambda ^{(c)}\).

Proof

Assume first that \(c>1\). The inequality \(\mathcal {U}_\lambda ^{(c)}\ge \mathcal {V}_\lambda ^{(c)}\) follows immediately from the majorization \(U^{(c)}\ge V^{(c)}\) established above; hence all we need is the superharmonicity of \(\mathcal {U}_\lambda ^{(c)}\). Observe that this function is harmonic on each of the domains \((-1,1)\times (-\infty ,-\lambda )\), \((-1,1)\times (-\lambda ,\lambda )\) and \((-1,1)\times (\lambda ,\infty )\). Consequently, it is enough to check that \(\mathcal {U}_\lambda ^{(c)}\) satisfies the mean value property at each point of the form \((x,\pm \lambda )\). But this follows at once from the inequality \(U_{y+}^{(c)}(x,0)<0\) (here the strictness is due to \(c>1\)). To get the claim for \(c=1\), note that \(U^{(1)}\) is a pointwise limit of \(U^{(c)}\) as \(c\downarrow 1\). \(\square \)

In the next lemma we establish an intermediate result which is of its own interest.

Lemma 2.3

For any \(f:{\mathbb {T}}\rightarrow [-1,1]\) and any \(\lambda \ge 0\), we have

$$\begin{aligned} \int _{\mathbb {T}} \left( \left| {\mathcal {H}}^{\mathbb {T}}f\right| -\lambda \right) _+\text{ d }m\le m\left( \left\{ x\in {\mathbb {T}}:\left| {\mathcal {H}}^{\mathbb {T}}f(x)\right| > \lambda \right\} \right) . \end{aligned}$$
(2.1)

Proof

Let \(u\), \(v\) denote the harmonic extensions of \(f\) and \({\mathcal {H}}^{\mathbb {T}}f\) to the unit disc, obtained via the Poisson kernel. Then \(u\), \(v\) satisfy Cauchy-Riemann equations and \(v(0,0)=0\) (cf. Riesz [13]). Consequently, the function \(\mathcal {U}^{(1)}_\lambda (u,v)\) is superharmonic (being the composition of a superharmonic \(\mathcal {U}^{(1)}\) and the analytic \(u+iv\)) and it majorizes \(\mathcal {V}^{(1)}_\lambda (u,v)\). Therefore, by the mean value property,

$$\begin{aligned} \int _{\mathbb {T}} \mathcal {V}^{(1)}_\lambda (u,v)\text{ d }m&\le \int _{\mathbb {T}} \mathcal {U}^{(1)}_\lambda (u,v)\text{ d }m\\&\le \mathcal {U}^{(1)}_\lambda (u(0,0), v(0,0))= \mathcal {U}_\lambda ^{(1)}(u(0,0),0)=0. \end{aligned}$$

This is precisely (2.1). \(\square \)

We turn our attention to Theorem 1.1.

Proof of (1.4)

By homogeneity, we may and do assume that \(||f||_{L^\infty ({\mathbb {T}})}=1\). By the definition of \(({\mathcal {H}}^{\mathbb {T}}f)^{**}\), we may write

$$\begin{aligned}&\left( {\mathcal {H}}^{\mathbb {T}}f\right) ^{**}(t)-\left( {\mathcal {H}}^{\mathbb {T}}f\right) ^*(t)\\&\quad =\sup \left\{ \frac{1}{m(E)}\int _E \left[ \left| {\mathcal {H}}^{\mathbb {T}}f(x)\right| -\left( {\mathcal {H}}^{\mathbb {T}}f\right) ^*(t)\right] m(\text{ d }x)\,:\,E\subseteq {\mathbb {T}},\,m(E)=t\right\} \!. \end{aligned}$$

It is clear that when computing this supremum, we may restrict ourselves to those \(E\), which satisfy

$$\begin{aligned} \left\{ x\in {\mathbb {T}}:\left| {\mathcal {H}}^{\mathbb {T}}f(x)\right| >\lambda \right\} \subseteq E \subseteq \left\{ x\in {\mathbb {T}}:\left| {\mathcal {H}}^{\mathbb {T}}f(x)\right| \ge \lambda \right\} \end{aligned}$$

for some \(\lambda \ge 0\). Actually, since \(m(E)=t\), this \(\lambda \) must be equal to \(({\mathcal {H}}^{\mathbb {T}}f)^*(t)\). For such \(E\), it is clear that

$$\begin{aligned}&\frac{1}{m(E)}\int _E \left[ \left| {\mathcal {H}}^{\mathbb {T}}f(x)\right| -\left( {\mathcal {H}}^{\mathbb {T}}f\right) ^*(t)\right] m(\text{ d }x)\\&\quad \le \frac{1}{m\left( \left\{ x\in {\mathbb {T}}:\left| {\mathcal {H}}^{\mathbb {T}}f(x)\right| >\lambda \right\} \right) }\int _{\left\{ \left| {\mathcal {H}}^{\mathbb {T}}f\right| >\lambda \right\} } \left[ \left| {\mathcal {H}}^{\mathbb {T}}f\right| -\lambda \right] \text{ d }m\le 1, \end{aligned}$$

where the latter bound is due to (2.1). This establishes (1.4). \(\square \)

Sharpness

Let \(c \in (0, 1)\) and fix an arbitrary \(\lambda \in (0,c/2)\). Consider the region \(\mathcal {C}=[-1,1]\times [-\lambda ,\infty )\) and let \(F\) be the conformal mapping which sends the unit disc \(\mathbb {D}\) onto \(\mathcal {C}\) and \((0,0)\in \mathbb {D}\) to \((0,0)\in \mathcal {C}\). Then \(F\) transports \(m\), the harmonic measure on \({\mathbb {T}}\) with respect to \((0,0)\), to \(\mu \), the harmonic measure on \(\partial \mathcal {C}\) with respect to \((0,0)\). Finally, put \(u={\text {Re}}F\) and \(v={\text {Im}}F\); clearly, the restrictions \(f=u|_{{\mathbb {T}}}\) and \(g=v|_{{\mathbb {T}}}\) satisfy the relation \(g={\mathcal {H}}^{\mathbb {T}}f\). The function \((x,y)\mapsto U^{(c)}(x,y+\lambda )\) is harmonic in the interior of \(\mathcal {C}\), so by the mean-value property,

$$\begin{aligned} U^{(c)}(0,\lambda )&=\int _{\partial \mathcal {C}} U^{(c)}(x,y+\lambda )\text{ d }\mu (x,y)\\&=\int _{{\mathbb {T}}} U^{(c)}(u,v+\lambda )\text{ d }m\\&=\int _{{\mathbb {T}}}(v+\lambda -c)\chi _{\{v+\lambda >0\}}\text{ d }m\\&\le \int _{{\mathbb {T}}}(v+\lambda -c)\chi _{\{v-\lambda >0\}}\text{ d }m\\&= \int _{{\mathbb {T}}}{\mathcal {H}}^{\mathbb {T}}f\chi _{\left\{ {\mathcal {H}}^{\mathbb {T}}f>\lambda \right\} }\text{ d }m-(c-\lambda )m\left( \left\{ x\in {\mathbb {T}}:{\mathcal {H}}^{\mathbb {T}}f(x)>\lambda \right\} \right) . \end{aligned}$$

However, if \(\lambda \) is sufficiently close to \(0\), then \(U^{(c)}(0,\lambda )>0\): this follows from Lemma 2.1 (2). Hence, for such \(\lambda \),

$$\begin{aligned} \frac{1}{m\left( \left\{ x\in {\mathbb {T}}:{\mathcal {H}}^{\mathbb {T}}f(x)>\lambda \right\} \right) }\int _{\left\{ {\mathcal {H}}^{\mathbb {T}}f>\lambda \right\} }\left| {\mathcal {H}}^{\mathbb {T}}f(x)\right| m(\text{ d }x)\ge c-\lambda . \end{aligned}$$

Now take \(t=m(\{x\in {\mathbb {T}}:{\mathcal {H}}^{\mathbb {T}}f(x)>\lambda \})\). The above inequality implies that

$$\begin{aligned} \left( {\mathcal {H}}^{\mathbb {T}}f\right) ^{**}(t)\ge c-\lambda . \end{aligned}$$
(2.2)

In addition, since \({\mathcal {H}}^{\mathbb {T}}f\ge -\lambda \) on \({\mathbb {T}}\), we actually have \(t=m(\{x\in {\mathbb {T}}:|{\mathcal {H}}^{\mathbb {T}}f(x)|>\lambda \})\). Hence, from the very definition of the decreasing rearrangement, we infer that \(({\mathcal {H}}^{\mathbb {T}}f)^*(t)\le \lambda \). Combining this with (2.2), we obtain

$$\begin{aligned} \left| \left| {\mathcal {H}}^{\mathbb {T}}f\right| \right| _{W({\mathbb {T}})}\ge c-2\lambda . \end{aligned}$$

It remains to observe that the right-hand side can be made arbitrarily close to \(1\), by choosing \(c\) appropriately close to \(1\) and then picking \(\lambda \) sufficiently small. This proves that the constant \(1\) cannot be replaced in (1.4) by a smaller number. \(\square \)

3 The Non-Periodic Case

Proof of (1.5)

To deduce the weak-type estimate for the Hilbert transform on the real line, we use a standard argument known as “blowing up the circle”, which is due to Zygmund ([15], Chapter XVI, Theorem 3.8). Let \(f\in L^p({\mathbb {R}})\cap L^\infty ({\mathbb {R}})\) be a fixed function. For a given positive integer \(n\) and \(x\in {\mathbb {R}}\), put

$$\begin{aligned} g_n(x)=\frac{1}{2\pi n}\text{ p.v. }\int _{-\pi n}^{\pi n}f(t)\cot \frac{x-t}{2n}\text{ d }t. \end{aligned}$$

As shown in [15], we have \(g_n\rightarrow {\mathcal {H}}^{\mathbb {R}}f\) almost everywhere as \(n\rightarrow \infty \). On the other hand, the function

$$\begin{aligned} x\mapsto g_n(nx)=\text{ p.v. }\int _{-\pi }^\pi f(nt)\cot \frac{x-t}{2}m(\text{ d }t) \end{aligned}$$

is precisely the periodic Hilbert transform of the function \(f_n:x\mapsto f(nx)\), \(|x|\le \pi \). Consequently, by (2.1), we may write

$$\begin{aligned} \left| \{x\in (-\pi n, \pi n]: |g_n(x)|>\lambda \}\right|= & {} 2\pi n\, m\left( \left\{ x\in {\mathbb {T}}: \left| {\mathcal {H}}^{\mathbb {T}}f_n(x)\right| >\lambda \right\} \right) \\\ge & {} 2\pi n \int _{\mathbb {T}} \left( \left| {\mathcal {H}}^{\mathbb {T}}f_n(x)\right| -\lambda \right) _+ m(\text{ d }x)\\= & {} \int _{-\pi n}^{\pi n} \left( \left| g_n(x)\right| -\lambda \right) _+\text{ d }x. \end{aligned}$$

Now we let \(n\rightarrow \infty \); using some routine limiting arguments, we get

$$\begin{aligned} \left| \left\{ x\in {\mathbb {R}}: {\mathcal {H}}^{\mathbb {R}}f(x)> \lambda \right\} \right| \ge \int _{\mathbb {R}}\left( \Big |{\mathcal {H}}^{\mathbb {R}} f(x)-\lambda \right) _+\text{ d }x. \end{aligned}$$

It remains to repeat the reasoning from the periodic case to obtain, for any \(t>0\),

$$\begin{aligned}&\left( {\mathcal {H}}^{\mathbb {R}}f\right) ^{**}(t)-\left( {\mathcal {H}}^{\mathbb {R}}f\right) ^*(t)\\&\quad \le \frac{1}{\left| \left\{ x\in {\mathbb {R}}:\left| {\mathcal {H}}^{\mathbb {R}}f(x)\right| >\lambda \right\} \right| }\int _{\left\{ \left\| {\mathcal {H}}^{\mathbb {R}}f\right| >\lambda \right\} } \left[ \left| {\mathcal {H}}^{\mathbb {R}}f(x)\right| -\lambda \right] \text{ d }x\le 1. \end{aligned}$$

\(\square \)

Sharpness

As we have shown in the previous section, for any \(c\in (0,1)\) and \(\lambda \) sufficiently close to \(0\), there is a function \(\varphi :{\mathbb {T}}\rightarrow [-1,1]\) such that \(\int _{\mathbb {T}}\varphi \text{ d }m=0\) and

$$\begin{aligned} \frac{1}{\left| \left\{ x\in {\mathbb {T}}:\left| {\mathcal {H}}^{\mathbb {T}}\varphi (x)\right| >\lambda \right\} \right| }\int _{\left\{ \left| {\mathcal {H}}^{\mathbb {T}}\varphi \right| >\lambda \right\} }\left| {\mathcal {H}}^{\mathbb {T}}\varphi (x)\right| \text{ d }x\ge c-\lambda . \end{aligned}$$
(3.1)

We will expand this function onto the real line. We will use Davis’ argument from [6]. For the sake of clarity, we have divided the reasoning into three parts.

1. A conformal mapping and its properties Let \(H\) denote the closed upper halfplane of \(\mathbb {C}\) and consider the conformal mapping \(K(z)=-(1 -z)^2/4z\). This function maps the halfdisc \(\mathbb {D}\cap H\) onto \(H\), and the boundary of \(\mathbb {D}\cap H\) onto \({\mathbb {R}}\). Let \(L\) be the inverse of \(K\). Then \(L\) maps \([0, 1]\) onto the halfcircle \(\{e^{i\theta }:0\le \theta \le \pi \}\), and \({\mathbb {R}}{\setminus }[0, 1]\) onto \((-1, 1)\). Specifically, for \(x\in [0,1]\) we have \( L(x)=\exp ({2i\arcsin (\sqrt{x})})\), while for \(x\notin [0,1]\),

$$\begin{aligned} L(x)={\left\{ \begin{array}{ll} 1-2x-2\sqrt{x^2-x} &{} \text{ if } x<0,\\ 1-2x+2\sqrt{x^2-x} &{} \text{ if } x>1. \end{array}\right. } \end{aligned}$$

We will also need the property

$$\begin{aligned} L(z)\rightarrow 0\qquad \text{ as } z\rightarrow \infty . \end{aligned}$$
(3.2)

Next, for a positive integer \(n\), let \(d_n\) be the density of \(L^n([0, 1])\) on \({\mathbb {T}}\) with respect to \(m\), i.e. for any \(-\pi < \alpha <\beta <\pi \),

$$\begin{aligned} \int _\alpha ^\beta d_n\left( e^{i\theta }\right) \,m(\text{ d }\theta )=\left| \left\{ r\in [0,1]:L^n(r)\in \left\{ e^{i\theta }:\alpha <\theta <\beta \right\} \right\} \right| . \end{aligned}$$

Then it is easy to prove that

$$\begin{aligned} d_n\rightarrow 1\,\text{ uniformly } \text{ on }\,{\mathbb {T}}, \end{aligned}$$
(3.3)

see Lemma 3 in [6].

2. Expansion of \(\varphi \) Let \(\Phi \) denote the holomorphic extension of \(\varphi +i{\mathcal {H}}^{\mathbb {T}}\varphi \) to the unit disc. Then \(\Phi \) satisfies \(\Phi (0)=0\): indeed, \({\text {Re}}\Phi (0)=0\) is due to the condition \(\int _{\mathbb {T}}\varphi \text{ d }m=0\), while \({\text {Im}}\Phi (0)=0\) follows from the normalization property of the periodic Hilbert transform. Combining this with (3.2), we see that the analytic function \(F_n=\Phi (L^n(z))\) (\(n=1,\,2,\,\ldots \)), given on the halfplane \(H\), satisfies \(\lim _{z\rightarrow \infty }F_n(z)=0\). Put \(f_n(x)={\text {Re}}F_n(x)\) for any \(x\in {\mathbb {R}}\). This function is bounded in absolute value by \(1\), since so is \(\varphi \). Furthermore, \(f_n\) is integrable when \(n\ge 2\). Indeed, for any \(x\notin [-1,1]\) we have

$$\begin{aligned} |f_n(x)|=|{\text {Re}}\Phi (L^n(x))|\le \kappa _1|L^n(x)|\le \kappa _2|x|^{-n}, \end{aligned}$$

for some universal constants \(\kappa _1\), \(\kappa _2\). Thus, we may speak of \({\mathcal {H}}^{\mathbb {R}}f_n\). Furthermore, by the aforementioned property \(\lim _{z\rightarrow \infty }F_n(z)=0\), we have \({\mathcal {H}}^{\mathbb {R}}f_n={\text {Im}}F_n|_{\mathbb {R}}\).

3. Computations If \(x\notin [0,1]\), then \(L(x)\in (-1,1)\) and hence \(L^n(x)\rightarrow 0\) as \(n\rightarrow \infty \). Consequently, we have

$$\begin{aligned} \left| \left\{ x\notin [0,1]:\left| {\mathcal {H}}^{\mathbb {R}}f_n(x)\right| >\lambda \right\} \right| \rightarrow 0 \end{aligned}$$

and, by Lebesgue’s dominated convergence theorem,

$$\begin{aligned} \int _{\left\{ x\notin [0,1]: \left| {\mathcal {H}}^{\mathbb {R}}f_n(x)\right| >\lambda \right\} }\left| {\mathcal {H}}^{\mathbb {R}}f_n(x)\right| \text{ d }x\xrightarrow {n\rightarrow \infty } 0. \end{aligned}$$

Next, observe that by (3.3),

and

$$\begin{aligned} \int _{\left\{ x\in [0,1]: \left| {\mathcal {H}}^{\mathbb {R}}f_n(x)\right| >\lambda \right\} }\!\left| {\mathcal {H}}^{\mathbb {R}}f_n(x)\right| \text{ d }x&=\int _{\left\{ x\in [0,1]: \left| {\mathcal {H}}^{\mathbb {T}}\varphi (L^n(x))\right| >\lambda \right\} }\!\left| {\mathcal {H}}^{\mathbb {T}}\varphi (L^n(x))\right| \text{ d }x\\&\quad \,\xrightarrow {n\rightarrow \infty } \int _{\mathbb {T}}\left| {\mathcal {H}}^{\mathbb {T}}\varphi (x)\right| m(\text{ d }x). \end{aligned}$$

Let us put all the above facts together and combine them with (3.1). We get that for an arbitrary \(\eta <1\) we have

provided \(n\) is sufficiently large. Now, set \(t=|\{x\in {\mathbb {R}}:|{\mathcal {H}}^{\mathbb {R}}f(x)|>\lambda \}|\). Arguing as above, we prove that

$$\begin{aligned} \left| \left\{ x\notin [0,1]: \left| {\mathcal {H}}^{\mathbb {R}}f_n(x)\right| >2\lambda \right\} \right| \rightarrow 0 \end{aligned}$$

and

$$\begin{aligned} \left| \left\{ x\in [0,1]: \left| {\mathcal {H}}^{\mathbb {R}}f_n(x)\right| >2\lambda \right\} \right| \rightarrow m\left( \left\{ x\in {\mathbb {T}}:\left| {\mathcal {H}}^{\mathbb {T}}\varphi (x)\right| >2\lambda \right\} \right) <t. \end{aligned}$$

This shows that if \(n\) is sufficiently large, then \(({\mathcal {H}}^{\mathbb {R}}f_n)^*(t)\le 2\lambda \). Hence, for large \(n\),

$$\begin{aligned} \left| \left| {\mathcal {H}}^{\mathbb {R}}f_n\right| \right| _{W({\mathbb {R}})}\ge \left( {\mathcal {H}}^{\mathbb {R}}f_n\right) ^{**}(t)-\left( {\mathcal {H}}^{\mathbb {R}}f_n\right) ^*(t)\ge \eta (c-\lambda )-2\lambda . \end{aligned}$$

The latter constant can be made arbitrarily close to \(1\), by choosing appropriate values for the parameters \(\eta \), \(c\) and \(\lambda \). This proves that the constant \(1\) is indeed the best possible in (1.5). \(\square \)