1 Background and statement of results

In this paper, we consider the behavior of Lipschitz functions which are analytic on a bounded open subset of the complex plane and how much analyticity extends to the boundary of the domain. Let U be an open subset in the complex plane and let \(0< \alpha < 1\). A function \(f : U \rightarrow \mathbb {C}\) satisfies a Lipschitz condition with exponent \(\alpha \) on U if there exists \(k>0\) such that for all \(z,w \in U\)

$$\begin{aligned} |f(z) - f(w)| \le k |z-w|^{\alpha } \end{aligned}$$
(1)

Let Lip\(\alpha (U)\) denote the space of functions that satisfy a Lipschitz condition with exponent \(\alpha \) on U. Lip\(\alpha (U)\) is a Banach space with norm given by \(||f||_{Lip\alpha (U)} = \sup _{U} |f| + k(f)\), where k(f) is the smallest constant that satisfies (1). If we let \(||f||_{Lip\alpha (U)}' = k(f)\) then \(||f||_{Lip\alpha (U)}'\) is a seminorm on Lip\(\alpha (U)\).

An important subspace of Lip\(\alpha (U)\) is the little Lipschitz class, lip\(\alpha (U)\), which consists of those functions in Lip\(\alpha (U)\) that also satisfy the additional property that for each \(\epsilon >0\), there exists \(\delta >0\) such that for all z, w in U, \(|f(z)-f(w)| \le \epsilon |z-w|^{\alpha }\) whenever \(|z-w| < \delta \).

The importance of lip\(\alpha (U)\) is illustrated by the following result of De Leeuw [1]. Let \(\varDelta \) be a closed disk. Then the restriction spaces Lip\(\alpha (\varDelta ) = \{f|\varDelta : f \in \text{ Lip } \alpha (\mathbb {C})\}\) and lip\(\alpha (\varDelta ) = \{f|\varDelta : f \in \text{ lip } \alpha (\mathbb {C})\}\) are Banach spaces and lip\(\alpha ^{**}(\varDelta )\) is isometrically isomorphic to Lip\(\alpha (\varDelta )\). Thus the weak-star topology can be applied to Lip\(\alpha (\varDelta )\) as the dual of lip\(\alpha ^{*}(\varDelta )\).

Let U be a bounded open subset of the complex plane. We will restrict our study to those functions in lip\(\alpha (\mathbb {C})\) which are analytic on U. Let \(A_{\alpha }(U) = \{f \in \text{ lip }\alpha : \overline{\partial } f = 0 \text{ on } U\}\), where \(\overline{\partial }f = \dfrac{1}{2} \left( \dfrac{\partial f}{\partial x} + i \dfrac{\partial f}{\partial y} \right) \). For an arbitrary set \(E \subset \mathbb {C}\), let \(A_{\alpha }(E) = \bigcup \{A_{\alpha }(U): U \text{ open } , E \subset U\}\).

While the functions in \(A_{\alpha }(U)\) are differentiable on the interior of U, they need not be differentiable on the boundary of U. In this paper, we consider the question of how close the functions in \(A_{\alpha }(U)\) come to being differentiable at boundary points of U. To answer this question we will make use of the concept of a bounded point derivation. For \(x_0 \in \mathbb {C}\), it is known that \(A_{\alpha }(U \cup \{x_0\})\) is dense in \(A_{\alpha }(U)\). [4, Lemma 1.1] Thus we say that \(A_{\alpha }(U)\) admits a bounded point derivation at \(x_0\) if the map \(f\rightarrow f'(x_0)\) extends from \(A_{\alpha }(U \cup \{x_0\})\) to a bounded linear functional on \(A_{\alpha }(U)\). Equivalently, \(A_{\alpha }(U)\) admits a bounded point derivation at \(x_0\) if and only if there exists a constant \(C >0\) such that

$$\begin{aligned} |f'(x_0)| \le C ||f||_{Lip\alpha (\mathbb {C})}, \end{aligned}$$
(2)

for all f in \(A_{\alpha }(U \cup \{x_0\})\).

The existence of a bounded point derivation at \(x_0\) shows that the functions in \(A_{\alpha }(U)\) possess some semblance of analytic structure at \(x_0\). If, in addition, U has an interior cone at \(x_0\), a more explicit description of this analytic structure can be obtained. We say that U has an interior cone at \(x_0\) if there is a segment J ending at \(x_0\) and a constant \(k > 0\) such that dist\((x, \partial U) \ge k |x-x_0|\) for all x in J. The segment J is called a non-tangential ray to \(x_0\). It is a result of O’Farrell [5] that if U has an interior cone at a boundary point \(x_0\), then a bounded point derivation on \(A_{\alpha }(U)\) at \(x_0\) can be evaluated by taking the limit of the difference quotient over a non-tangential ray to \(x_0\). To be precise, O’Farrell has proven the following theorem.

Theorem 1

Let \(0< \alpha <1\), and let U be an open set with \(x_0\) in \(\partial U\). Suppose that U has an interior cone at \(x_0\) and that J is a non-tangential ray to \(x_0\). If \(A_{\alpha }(U)\) admits a bounded point derivation D at \(x_0\), then for every f in \(A_{\alpha }(U)\),

$$\begin{aligned} Df = \lim _{x \rightarrow x_0, x \in J} \dfrac{f(x)-f(x_0)}{x-x_0}. \end{aligned}$$

Thus the difference quotient for boundary points that admit bounded point derivations for \(A_{\alpha }(U)\) converges when taken over a non-tangential ray to the point. This illustrates the additional analytic structure of functions in \(A_{\alpha }(U)\) at these points.

O’Farrell comments that the methods used in his proof of Theorem 1 are nonconstructive, involving abstract measures and duality arguments from functional analysis as opposed to using the Cauchy integral formula directly, and suggests that it should be possible to give a proof using constructive techniques. In this paper we present a constructive proof of Theorem 1, which confirms O’Farrell’s conjecture. In Sect. 2 we review some key properties of \(A_{\alpha }(U)\) and in Sect. 3 we prove Theorem 1 using constructive techniques.

2 Preliminary results

We begin by reviewing the Hausdorff content of a set, which is defined using measure functions. A measure function is a monotone nondecreasing function \(h : [0, \infty ) \rightarrow [0, \infty )\). For example, \(r^{\beta }\) is a measure function for \(0 \le \beta < \infty \). If h is a measure function then the Hausdorff content \(M_h\) associated to h is defined by

$$\begin{aligned} M_h(E) = \inf \sum h(\text{ diam } B), \end{aligned}$$

where the infimum is taken over all countable coverings of E by balls and the sum is taken over all the balls in the covering. If \(h(r) = r^{\beta }\) then we denote \(M_h\) by \(M^{\beta }\). The lower \(1+ \alpha \) dimensional Hausdorff content \(M^{1+\alpha }_*(E)\) is defined by

$$\begin{aligned} M^{1+\alpha }_*(E) = \sup M_h(E), \end{aligned}$$

where the supremum is taken over all measurable functions h such that \(h(r) \le r^{1+\alpha }\) and \(r^{-1-\alpha }h(r)\) converges to 0 as r tends to 0. The lower \(1+ \alpha \) dimensional Hausdorff content is a monotone set function; i.e. if \(E \subseteq F\) then \(M^{1+\alpha }_*(E) \le M^{1+\alpha }_*(F)\).

In [4], Lord and O’Farrell gave necessary and sufficient conditions for the existence of bounded point derivations on \(A_{\alpha }(U)\) in terms of Hausdorff contents. There are similar conditions for bounded point derivations defined on other function spaces. ([2, 3]).

Theorem 2

Let U be an open subset of the complex plane with \(x_0\) on the boundary of U. Let \(0<\alpha < 1\). Then \(A_{\alpha }(U)\) has a bounded point derivation at \(x_0\) if and only if

$$\begin{aligned} \sum _{n=1}^{\infty } 4^n M_*^{1+\alpha }(A_n(x_0) {\setminus } U) < \infty . \end{aligned}$$

Another key lemma is the following Cauchy theorem for Lipschitz functions which also appears in the paper of Lord and O’Farrell [4, pg.110].

Lemma 1

Let \(\varGamma \) be a piecewise analytic curve bounding a region \(\varOmega \in \mathbb {C}\), and suppose that \(\varGamma \) is free of outward pointing cusps. Let \(0< \alpha < 1\) and suppose that \(f \in \) lip\(\alpha (\mathbb {C})\). Then there exists a constant \(\kappa >0\) such that

$$\begin{aligned} \left| \int f(z)dz\right| \le \kappa \cdot M_{*}^{1+\alpha }(\varOmega \cap S) \cdot ||f||'_{Lip\alpha (\varOmega )}. \end{aligned}$$

The constant \(\kappa \) only depends on \(\alpha \) and the equivalence class of \(\varGamma \) under the action of the conformal group of \(\mathbb {C}\). In particular this means that \(\kappa \) is the same for any curve obtained from \(\varGamma \) by rotation or scaling.

3 The proof of the main theorem

To prove Theorem 1, we first note that by translation invariance we may suppose that \(x_0 = 0\). Moreover by replacing f by \(f-f(0)\) if needed, we may suppose that \(f(0) = 0\). In addition, we may suppose that U is contained in the unit disk. Let J be a non-tangential ray to \(x_0\) and for each x in J, define a linear functional \(L_x\) by \(L_x(f) = \dfrac{f(x)}{x} - Df\). Then to prove Theorem 1 it suffices to show that \(L_x\) tends to the 0 functional as \(x \rightarrow 0\) through J. We make the following claim.

Lemma 2

The collection \(\{L_x : x\in J\}\) is a family of bounded linear functionals on \(A_{\alpha }(U)\); that is there exists a constant \(C >0\) that does not depend on x or f such that \(|L_x(f)| \le C ||f||_{Lip \alpha (\mathbb {C})}\) for all f in \(A_{\alpha }(U)\) and all \(x \in J\).

Proof

We will first prove Lemma 2 for the case when f belongs to \(A_{\alpha }(U \cup \{0\})\) and then extend to the general case. It follows from (2) that it is enough to show that \(\left| \dfrac{f(x)}{x}\right| \le C ||f||_{Lip\alpha (\mathbb {C})}\) where the constant C does not depend on f or x. If f belongs to \(A(U \cup \{0\})\), then there is a neighborhood \(\varOmega \) of 0 such that f is analytic on \(\varOmega \). We can further suppose that \(U \subseteq \varOmega \). Let \(B_n\) denote the ball centered at 0 with radius \(2^{-n}\). Then there exists an integer \(N>0\) such that \(\varOmega \) contains \(B_N\) and hence f is analytic inside the ball \(B_N\). In addition, there exists an integer M such that \(\varOmega \subseteq B_M\). Since J is a non-tangential ray to \(x_0\), it follows that there is a sector in \(\mathring{U}\) with vertex at \(x_0\) that contains J. Let C denote this sector. It follows from the Cauchy integral formula that

$$\begin{aligned} \dfrac{f(x)}{x} = \frac{1}{2 \pi i} \int _{\partial (C \bigcup B_N)} \dfrac{f(z)}{z(z-x)}dz \end{aligned}$$
Fig. 1
figure 1

The contour of integration

Full size image

where the boundary is oriented so that the interior of \(C \bigcup B_N\) lies always to the left of the path of integration. (See Fig. 1.) Let \(D_n = A_n {\setminus } C\). Then

$$\begin{aligned} \dfrac{f(x)}{x} = \frac{1}{2 \pi i} \sum _{n=M}^N \int _{\partial D_n} \dfrac{f(z)}{z(z-x)}dz + \frac{1}{2 \pi i} \int _{|z|=2^{-M}} \dfrac{f(z)}{z(z-x)}dz. \end{aligned}$$

Since x lies on J, which is a non-tangential ray to \(x_0\), there exists a constant \(k>0\) such that for \(z \notin U\), \(\dfrac{|x|}{|z-x|} \le k^{-1}\). Thus for \(z \notin U\), \(\dfrac{|z|}{|z-x|} \le 1 + \dfrac{|x|}{|z-x|} \le 1 + k^{-1}\). Hence \(\dfrac{1}{|z|\cdot |z-x|} \le \dfrac{1+ k^{-1}}{|z|^2}\) and therefore

$$\begin{aligned} \dfrac{|f(x)|}{|x|} \le \frac{1}{2 \pi } \sum _{n=M}^N \left| \int _{\partial D_n} \dfrac{f(z)}{z(z-x)}dz\right| + \dfrac{4^M(1+k^{-1})}{2 \pi } ||f||_{\infty }. \end{aligned}$$
(3)

Since \(\dfrac{f(z)}{z(z-x)}\) is analytic on \(D_n {\setminus } U\) for \(M \le n \le N\), an application of Lemma 1 shows that

$$\begin{aligned} \left| \int _{\partial D_n} \dfrac{f(z)}{z(z-x)}dz \right| \le \kappa M_*^{1+\alpha }(D_n {\setminus } U) \cdot \left\| \dfrac{f(z)}{z(z-x)} \right\| _{Lip\alpha (D_n)}'. \end{aligned}$$
(4)

Recall that the constant \(\kappa \) is the same for curves in the same equivalence class. Since the regions \(D_n\) differ from each other by a scaling it follows that \(\kappa \) doesn’t depend on n in (4).

We now show that \(\left\| \dfrac{f(z)}{z(z-x)} \right\| _{Lip\alpha (D_n)}'\) can be bounded by a constant independent of f and x. It follows from the definition of the Lipschitz seminorm that

$$\begin{aligned} \begin{aligned} \left\| \dfrac{f(z)}{z(z-x)} \right\| _{Lip\alpha (D_n)}'&= \sup _{z \ne w; z,w \in D_n} \dfrac{\left| \dfrac{f(z)}{z(z-x)} - \dfrac{f(w)}{w(w-x)}\right| }{|z-w|^{\alpha }}\\&= \sup _{z \ne w; z,w \in D_n} \dfrac{|w(w-x)f(z)-z(z-x)f(w)|}{|z|\cdot |z-x|\cdot |w|\cdot |w-x|\cdot |z-w|^{\alpha }}. \end{aligned} \end{aligned}$$

Thus it follows from the triangle inequality that

$$\begin{aligned} \begin{aligned} \left\| \dfrac{f(z)}{z(z-x)} \right\| _{Lip\alpha (D_n)}'&\le \sup _{z \ne w; z,w \in D_n} \dfrac{|w(w-x)f(z)-w(w-x)f(w)|}{|z|\cdot |z-x|\cdot |w|\cdot |w-x|\cdot |z-w|^{\alpha }} \\&\quad + \sup _{z \ne w; z,w \in D_n} \dfrac{|w(w-x)f(w) -z(z-x)f(w)|}{|z|\cdot |z-x|\cdot |w|\cdot |w-x|\cdot |z-w|^{\alpha }}. \end{aligned} \end{aligned}$$
(5)

We first bound the first term on the right of (5)

$$\begin{aligned}&\sup _{z \ne w; z,w \in D_n} \dfrac{|w(w-x)f(z)-w(w-x)f(w)|}{|z|\cdot |z-x|\cdot |w|\cdot |w-x|\cdot |z-w|^{\alpha }} \\&\quad \le \sup _{z \in D_n} \dfrac{1}{|z|\cdot |z-x|} \cdot ||f||'_{Lip\alpha (D_n)}. \end{aligned}$$

Since \(z \notin U\), \(\dfrac{1}{|z|\cdot |z-x|} < \dfrac{1+k^{-1}}{|z|^2}\), and therefore,

$$\begin{aligned} \begin{aligned} \sup _{z \ne w; z,w \in D_n} \dfrac{|w(w-x)f(z)-w(w-x)f(w)|}{|z|\cdot |z-x|\cdot |w|\cdot |w-x|\cdot |z-w|^{\alpha }} \le C 4^n ||f||'_{Lip\alpha (D_n)}. \end{aligned} \end{aligned}$$
(6)

We now bound the second term on the right side of (5). Since \(f(0) =0\) it follows that for \(w \in \mathbb {C}\), \( \dfrac{|f(w)|}{|w|^{\alpha }} \le ||f||'_{Lip\alpha (\mathbb {C})}\). Moreover, a computation shows that \(w(w-x)-z(z-x) = (w-z)(z+w-x)\). Hence

$$\begin{aligned} \begin{aligned}&\sup _{z \ne w; z,w \in D_n} \dfrac{|w(w-x)f(w) -z(z-x)f(w)|}{|z|\cdot |z-x|\cdot |w|\cdot |w-x|\cdot |z-w|^{\alpha }}\\&\quad \le \left( \sup _{z \ne w; z,w \in D_n} \dfrac{|w-z|^{1-\alpha } }{ |z-x|\cdot |w|^{1-\alpha }\cdot |w-x|} + \dfrac{|w-z|^{1-\alpha } }{|z|\cdot |z-x|\cdot |w|^{1-\alpha } }\right) \cdot ||f||'_{Lip\alpha (\mathbb {C})}. \end{aligned} \end{aligned}$$
(7)

Since x lies on J, there exists a constant \(k>0\) such that \(\dfrac{1}{|z-x|} < \dfrac{1+k^{-1}}{|z|}\) and \(\dfrac{1}{|w-x|} < \dfrac{1+k^{-1}}{|w|}\). Hence

$$\begin{aligned} \begin{aligned} \sup _{z \ne w; z,w \in D_n} \dfrac{|w-z|^{1-\alpha } }{ |z-x|\cdot |w|^{1-\alpha }\cdot |w-x|} \le C \dfrac{2^n \cdot (2^n)^{2-\alpha } }{(2^n)^{1-\alpha }} = C 4^n, \end{aligned} \end{aligned}$$
(8)

and

$$\begin{aligned} \begin{aligned} \sup _{z \ne w; z,w \in D_n} \dfrac{|w-z|^{1-\alpha } }{|z|\cdot |z-x|\cdot |w|^{1-\alpha } } \le C \dfrac{4^n \cdot (2^n)^{1-\alpha } }{(2^n)^{1-\alpha }} = C 4^n. \end{aligned} \end{aligned}$$
(9)

Then (7), (8), and (9) yield

$$\begin{aligned} \sup _{z \ne w; z,w \in D_n} \dfrac{|w(w-x)f(w) -z(z-x)f(w)|}{|z|\cdot |z-x|\cdot |w|\cdot |w-x|\cdot |z-w|^{\alpha }} \le C 4^n ||f||'_{Lip\alpha (\mathbb {C})}, \end{aligned}$$
(10)

and it follows from (5), (6), and (10) that

$$\begin{aligned} \left\| \dfrac{f(z)}{z(z-x)} \right\| _{Lip\alpha (D_n)}' \le C 4^n ||f||'_{Lip\alpha (\mathbb {C})}. \end{aligned}$$
(11)

Thus (3), (4), and (11) together yield

$$\begin{aligned} \dfrac{|f(x)|}{|x|} \le C \sum _{n=1}^{\infty } 4^n M_*^{1+\alpha }(D_n {\setminus } U) \cdot ||f||'_{Lip\alpha (\mathbb {C})}. \end{aligned}$$

Since Hausdorff content is monotone, \(M_*^{1+\alpha }(D_n {\setminus } U) \le M_*^{1+\alpha }(A_n {\setminus } U)\) and hence

$$\begin{aligned} \dfrac{|f(x)|}{|x|} \le C \sum _{n=1}^{\infty } 4^n M_*^{1+\alpha }(A_n {\setminus } U) \cdot ||f||_{Lip\alpha (\mathbb {C})}, \end{aligned}$$

and it follows from Theorem 2 that

$$\begin{aligned} \dfrac{|f(x)|}{|x|} \le C ||f||_{Lip\alpha (U)}, \end{aligned}$$

where C does not depend on x or f. Thus \(L_x(f) \le C ||f||_{Lip\alpha (\mathbb {C})}\) for \(f \in A_{\alpha }(U \cup \{0\})\) and since \(A_{\alpha }(U \cup \{0\})\) is dense in \(A_{\alpha }(U)\), it follows that \(L_x\) is a family of uniformly bounded linear functionals on \(A_{\alpha }(U)\). \(\square \)

To complete the proof of Theorem 1, since \(A_{\alpha }(U \cup {0})\) is dense in \(A_{\alpha }(U)\), there exists a sequence \(\{f_j\}\) in \(A_{\alpha }(U \cup {0})\) such that \(f_j \rightarrow f\) in the Lipschitz norm. Since each \(f_j\) is analytic in a neighborhood of 0 and since \(Df_j = f_j'(0)\), it follows that for each j, \(L_x(f_j) \rightarrow 0\) as \(x \rightarrow 0\). It follows from the claim that \(|L_x(f)-L_x(f_j)| \le C ||f-f_j||_{Lip \alpha (U)}\). By first choosing j sufficiently large, the right hand side can be made arbitrarily small. Then by choosing x sufficiently close to 0, \(L_x(f_j)\) can be made arbitrarily close to 0. Thus \(L_x(f) \rightarrow 0\) as \(x\rightarrow 0\) through J, which proves Theorem 1.