Suppose \(\left( f _ { n } \right)\) is a sequence of functions which are entire (=analytic throughout the complex plane). Suppose \(\left( f _ { n } \right)\) converges pointwise to a function \(g : \mathbb { C } \to \mathbb { C }\) and the convergence is uniform on each line segment in \(\CC\). Show that \(g\) is entire and that \(f _ { n } \to g\) uniformly on each compact subset of \(\mathbb { C }\)
Solution
1. If \(R\) is any closed rectangle in \(\CC\), then\[ f_n\longrightarrow g \]uniformly on \(\partial R\).
The boundary \(\partial R\) is the union of four line segments. By hypothesis, \(f_n\to g\) uniformly on each of those four segments. Given \(\varepsilon>0\), choose an index for each side after which the error is smaller than \(\varepsilon\), and take the maximum of those four indices. That single index works on all of \(\partial R\).
2. If \(z\) lies in the interior of a rectangle \(R\), then\[ g(z) = \frac{1}{2\pi i} \int_{\partial R} \frac{g(\zeta)}{\zeta-z}\,d\zeta. \]
For every \(n\), Cauchy’s integral formula gives \[ f_n(z) = \frac{1}{2\pi i} \int_{\partial R} \frac{f_n(\zeta)}{\zeta-z}\,d\zeta. \] By step 1, \[ \sup_{\zeta\in\partial R} \abs{f_n(\zeta)-g(\zeta)} \longrightarrow0. \] Since \[ \delta_z=\operatorname{dist}(z,\partial R)>0, \] we have \[ \begin{aligned} &\left| \int_{\partial R} \frac{f_n(\zeta)-g(\zeta)}{\zeta-z}\,d\zeta \right|\\ &\qquad\leq \frac{\operatorname{length}(\partial R)}{\delta_z} \sup_{\zeta\in\partial R} \abs{f_n(\zeta)-g(\zeta)} \longrightarrow0. \end{aligned} \] The left side of Cauchy’s formula satisfies \[ f_n(z)\longrightarrow g(z) \] by the pointwise hypothesis. Passing to the limit gives the displayed formula.
3. The function \(g\) is holomorphic in the interior of every rectangle\(R\).
For a fixed rectangle, define \[ G_R(z) = \frac{1}{2\pi i} \int_{\partial R} \frac{g(\zeta)}{\zeta-z}\,d\zeta, \qquad z\in\operatorname{int}R. \] The same difference-quotient argument as for the Cauchy kernel gives \[ G_R'(z) = \frac{1}{2\pi i} \int_{\partial R} \frac{g(\zeta)}{(\zeta-z)^2}\,d\zeta. \] Indeed, for \(h\) small enough the denominators stay uniformly bounded away from zero on the compact boundary \(\partial R\), so the difference-quotient integrands converge uniformly.
Thus \(G_R\) is holomorphic in \(\operatorname{int}R\). Step 2 gives \[ g(z)=G_R(z) \] there, so \(g\) is holomorphic in the interior of \(R\).
4. The function \(g\) is entire.
Fix any \(z_0\in\CC\) and choose a rectangle whose interior contains \(z_0\). Step 3 shows that \(g\) is holomorphic on a neighborhood of \(z_0\). Since \(z_0\) was arbitrary, \(g\) is holomorphic on all of \(\CC\).
5. Let \(K\subseteq\CC\) be compact. There is a closed rectangle \(R\) suchthat\[ K\subseteq\operatorname{int}R \]and\[ \delta=\operatorname{dist}(K,\partial R)>0. \]
Since \(K\) is compact, it is bounded. Choose a rectangle whose sides lie strictly outside a large disk containing \(K\). Then \(K\) lies in its interior. The disjoint compact sets \(K\) and \(\partial R\) have positive distance, which is the stated \(\delta\).
6. For every \(z\in K\),\[ \abs{f_n(z)-g(z)} \leq \frac{\operatorname{length}(\partial R)}{2\pi\delta} \sup_{\zeta\in\partial R} \abs{f_n(\zeta)-g(\zeta)}. \]
By Cauchy’s integral formula for \(f_n\) and step 2 for \(g\), \[ f_n(z)-g(z) = \frac{1}{2\pi i} \int_{\partial R} \frac{f_n(\zeta)-g(\zeta)}{\zeta-z}\,d\zeta. \] For \[ z\in K, \qquad \zeta\in\partial R, \] step 5 gives \[ \abs{\zeta-z}\geq\delta. \] Estimating the contour integral therefore gives the displayed inequality.
7. The convergence\[ f_n\longrightarrow g \]is uniform on every compact subset of \(\CC\).
For the rectangle from step 5, step 1 gives \[ \sup_{\zeta\in\partial R} \abs{f_n(\zeta)-g(\zeta)} \longrightarrow0. \] The constant in step 6 is independent of \(z\in K\). Hence \[ \sup_{z\in K}\abs{f_n(z)-g(z)} \longrightarrow0. \] Since \(K\) was arbitrary, the convergence is uniform on every compact subset of \(\CC\).
8. Q.E.D.
Step 4 proves that \(g\) is entire, and step 7 proves compact-uniform convergence.