Let \(E\subset(-1,1)\) be a compact set of zero linear measure, and let \[ f:\Delta\setminus E\to\mathbb C \] be a bounded analytic function. Show that \(f\) extends to an analytic function on the whole unit disk \(\Delta\).
Solution
Let \(|f|\le M\) on \(\Delta\setminus E\). We prove that every point of \(E\) has a neighborhood on which \(f\) is the restriction of a holomorphic function.
1. Fix \(x_0\in E\). There is a disk\[ D=D(x_0,r),\qquad \overline D\subset\Delta, \]such that \(\partial D\cap E=\varnothing\).
The circle centered at the real point \(x_0\) meets the real axis only at \(x_0-r\) and \(x_0+r\). Since \(E\) has linear measure zero, we may choose \(r>0\) small enough that \(\overline{D(x_0,r)}\subset\Delta\) and neither \(x_0-r\) nor \(x_0+r\) belongs to \(E\). Hence \(\partial D\) is disjoint from \(E\).
2. Put \(K=E\cap\overline D\). For every \(\varepsilon>0\) sufficientlysmall, \(K\) can be covered by finitely many pairwise disjoint closed squares\(Q_1,\ldots,Q_N\subset D\) whose boundaries avoid \(E\) and whose totalperimeter is less than \(4\varepsilon\).
Because \(K\subset\mathbb R\) has linear measure zero, it has a finite cover by open intervals with total length less than \(\varepsilon\). Taking the connected components of their union gives finitely many pairwise disjoint open intervals \(I_j\) with \[ K\subset\bigcup_j I_j, \qquad \sum_j |I_j|<\varepsilon. \] Their endpoints may be chosen outside \(E\), since \(\mathbb R\setminus E\) is dense. As \(K\) is a positive distance from \(\partial D\), after taking \(\varepsilon\) small we may replace each \(I_j\) by the closed square \(Q_j\) centered on the real axis whose horizontal side has length \(|I_j|\). The squares remain inside \(D\), are disjoint because their horizontal projections are disjoint, and their boundaries do not meet \(E\): their horizontal sides lie off the real axis and their vertical sides meet the real axis only at the endpoints of \(I_j\). Finally, \[ \sum_j\operatorname{length}(\partial Q_j) =4\sum_j|I_j|<4\varepsilon. \]
3. For every \(z\in D\setminus E\),\[ f(z)=\frac{1}{2\pi i}\int_{\partial D} \frac{f(\zeta)}{\zeta-z}\,d\zeta. \]
Fix \(z\in D\setminus E\). Since \(K\) is compact and \(z\notin K\), \[ \delta=\operatorname{dist}(z,K)>0. \] Choose the squares in 2 so small that they all lie in the \(\delta/2\)-neighborhood of \(K\) and none contains \(z\). On \[ \Omega=D\setminus\bigcup_{j=1}^N Q_j \] the function \[ \zeta\longmapsto \frac{f(\zeta)}{\zeta-z} \] is holomorphic except for its simple pole at \(z\), whose residue is \(f(z)\). The residue theorem therefore gives \[ 2\pi i f(z) =\int_{\partial D}\frac{f(\zeta)}{\zeta-z}\,d\zeta -\sum_{j=1}^N \int_{\partial Q_j}\frac{f(\zeta)}{\zeta-z}\,d\zeta. \] For \(\zeta\in Q_j\) we have \(|\zeta-z|\ge\delta/2\), so \[ \left| \sum_{j=1}^N \int_{\partial Q_j}\frac{f(\zeta)}{\zeta-z}\,d\zeta \right| \le \frac{2M}{\delta} \sum_{j=1}^N\operatorname{length}(\partial Q_j) <\frac{8M}{\delta}\varepsilon. \] Letting \(\varepsilon\to0\) proves the formula.
4. The function\[ F_D(w)=\frac{1}{2\pi i}\int_{\partial D} \frac{f(\zeta)}{\zeta-w}\,d\zeta \qquad(w\in D) \]is holomorphic on \(D\) and agrees with \(f\) on \(D\setminus E\).
Because \(\partial D\cap E=\varnothing\), the boundary values of \(f\) in the integral are defined and holomorphic in a neighborhood of \(\partial D\). The Cauchy integral therefore defines a holomorphic function of \(w\in D\). By 3, \(F_D(w)=f(w)\) for every \(w\in D\setminus E\).
5. These local extensions glue to a holomorphic extension of \(f\) to allof \(\Delta\).
Perform <1>1–<1>4 at every point of \(E\). If two resulting disks overlap, their holomorphic extensions both agree with \(f\) on the overlap minus \(E\). That set is dense in the overlap because \(E\) has empty interior, so by continuity the two extensions agree everywhere on the overlap. Hence the local extensions glue with the original \(f\) on \(\Delta\setminus E\) to a single holomorphic function on \(\Delta\).