Suppose \(f\) is holomorphic on \(\DD\smz\) and there exist \(M, k\) such that \[ \abs{f^{(k)}(z)} \leq {M\over \abs{z}^k} && \forall 0 < \abs z < 1 .\]
Show that if \(f\) has a singularity at \(z=0\), then it must be removable.
- \(\dd{}{z}\) is a left-shift on power series, \(z^m\) is a right-shift.
- \(f'\) has the same poles as
\(f\), possibly with worse order due to
the left-shift.
- In general, if \(z_0\) is an order \(\ell\) pole of \(f\), then it is at least an order \(\ell + m\) pole of \(f^{(m)}\).
Solution
Define \(F(z) \da z^k f^{(k)}(z)\) and note that \(\abs{F(z)} \leq M\) on \(\DD\smz\).
If \(f\) has an essential singularity at \(z=0\), then so does \(F\) by considering power series expansions: \[ f(z) = \sum_{k\in \ZZ} c_k z^k \implies z^m f^{(m)}(z) = \sum_{k\leq 1} \tilde c_k z^{-k} + \sum_{k\geq m}\tilde c_{k}z^{k} ,\] which will still have infinitely many terms in its principal part at \(0\). However, if \(F\) had an essential singularity, the image of \(F\) in a neighborhood of \(0\) would be dense in \(\CC\) by Casorati-Weierstrass, contradicting that its image is bounded (by \(M\)).
Suppose instead \(z=0\) is a pole of order \(\ell\) of \(f\), so \(\abs{f(z)}\to \infty\) as \(z\to 0\). Then again by considering power series expansions, \(z=0\) remains a pole of \(F\), now of order at worst \(\ell\): \[ f(z) = \bigo(z^{\ell}) \implies z^m f^{(m)}(z) \approx z^m \cdot \bigo(z^{\ell - m}) = \bigo(z^\ell) .\] But if this is an order \(\ell\) pole of \(F\), then \(\lim_{z\to 0} \abs{F(z)} = \infty\) and \(\lim_{z\to 0} z^\ell F(z))\) is finite and nonzero. Apply the assumed bound yields the last contradiction: \[ z^{\ell}F(z) = z^{\ell + m}f(z) \leq z^{\ell + m} \cdot Mz^{-m} = z^{\ell} \convergesto{z\to 0} 0 .\] \(\contradiction\)