Let \(\ts{f_n}_{n=1}^\infty\) is a sequence of holomorphic functions on \(\DD\) and \(f\) is also holomorphic on \(\DD\). Show that the following are equivalent:
- \(f_n\to f\) uniformly on compact subsets of \(\DD\).
- For \(0 < r < 1\), \[ \int_{\abs z = r} \abs{f_n(z) - f(z)} \abs{dz} \convergesto{n\to\infty}0 .\]
Note: \(\abs{\dz} = \abs{\gamma'(t)}\dt\) for \(\gamma\) a parameterization of the curve.
Solution
\(\implies\):
- Fix \(r \in (0, 1)\) and let \(\gamma = \ts{\abs{z} = r}\). This is compact, so \(f_n\to f\) uniformly on \(\gamma\): \[ \int_\gamma \abs{f_n(z) - f(z) } \dz &\leq\int_\gamma \sup_{w\in \gamma } \abs{f_n(w) - f(w) } \dz \\ &\leq\int_\gamma \norm{f_n(w) - f(w) }_{\infty} \dz \\ &= \norm{f_n(w) - f(w) }_{\infty} \int_\gamma \dz \\ &= \norm{f_n(w) - f(w) }_{\infty} \length(\gamma) \\ &\convergesto{n\to\infty} 0 .\]
\(\impliedby\):
Let \(K\) be compact, then choose \(\gamma\) enclosing but not intersecting \(K\).
Since \(\gamma, K\) are disjoint compact sets, define \(M \da \inf \ts{\abs{z-\xi} \st z\in K, \xi\in \gamma}\), the \(0<M<\infty\).
Apply Cauchy’s formula to the function \(F_n(z) \da f_n(z) - f(z)\), where we want to show \(\abs{F_n(z)} < \eps\): \[ F_n(z) &= {1\over 2\pi i} \int_\gamma { F_n(\xi) \over z-\xi} \dxi \\ \implies \abs{f_n(z) - f(z) } &\leq {1\over 2\pi }\int_\gamma \abs{f_n(\xi) - f(\xi) \over z-\xi} \dxi \\ &\leq {1\over 2\pi} \int_\gamma {\abs{ f_n(\xi) - f(\xi) } \over M} \dxi \\ &\leq {1\over 2\pi M} \int_\gamma {\abs{ f_n(\xi) - f(\xi) } } \abs{\dxi} \\ ,\] where by hypothesis we can bound this integral by an \(\eps\). So given \(\eps\), choose \(n\) large enough to bound the integral as above by some \(\eps\) depending only on \(n\) and not on \(z\). Taking \(\sup\) of both sides yields \(\norm{f_n - f}_{\infty, K} \leq {\eps\over 2\pi M}\), so \(f_n\to f\) uniformly on \(K\).