For each \(k>0\), let \(X_k\) be the set of analytic functions \(f\) on the open unit disk \(\mathbb D\) such that \[ \sup_{z\in\mathbb D}(1-|z|)^k|f(z)|<\infty. \] Prove that \[ f\in X_k\quad\Longleftrightarrow\quad f'\in X_{k+1}. \]
Solution
1. If \(f\in X_k\), then there is a constant \(C>0\) such that\[ \abs{f(w)} \le \frac{C}{(1-\abs{w})^k} \]for every \(w\in\DD\).
This is exactly the defining boundedness condition for \(X_k\).
2. If \(f\in X_k\), then\[ f'\in X_{k+1}. \]
Fix \(z\in\DD\), and set \[ d=1-\abs{z}, \qquad r=\frac d2. \] If \(\abs{w-z}=r\), then \[ 1-\abs{w} \ge 1-\abs{z}-\abs{w-z} = \frac d2. \] By step 1, \[ \abs{f(w)} \le \frac{2^kC}{d^k} \] on that circle. Cauchy’s derivative estimate therefore gives \[ \abs{f'(z)} \le \frac1r \frac{2^kC}{d^k} = \frac{2^{k+1}C}{d^{k+1}}. \] Thus \[ (1-\abs{z})^{k+1}\abs{f'(z)} \le 2^{k+1}C \] for every \(z\in\DD\), which is precisely \(f'\in X_{k+1}\).
3. Conversely, suppose \(f'\in X_{k+1}\). Then there is a constant\(C>0\) such that\[ \abs{f'(w)} \le \frac{C}{(1-\abs{w})^{k+1}} \]for every \(w\in\DD\).
This is the defining boundedness condition for \(X_{k+1}\) applied to \(f'\).
4. For every \(z\in\DD\),\[ \abs{f(z)-f(0)} \le \frac{C}{k} \left( \frac1{(1-\abs{z})^k}-1 \right). \]
Let \(r=\abs{z}\). Integrating along the radial segment from \(0\) to \(z\) and using step 3 gives \[ \begin{aligned} \abs{f(z)-f(0)} &\le \int_0^1 \abs{z}\,\abs{f'(tz)}\,dt\\ &\le Cr \int_0^1 \frac{dt}{(1-tr)^{k+1}}\\ &= C \int_0^r \frac{ds}{(1-s)^{k+1}}\\ &= \frac{C}{k} \left( \frac1{(1-r)^k}-1 \right). \end{aligned} \] This is the claimed inequality.
5. If \(f'\in X_{k+1}\), then\[ f\in X_k. \]
By step 4, \[ \begin{aligned} (1-\abs{z})^k\abs{f(z)} &\le (1-\abs{z})^k\abs{f(0)} +\frac{C}{k} \left(1-(1-\abs{z})^k\right)\\ &\le \abs{f(0)}+\frac{C}{k}. \end{aligned} \] The right-hand side is independent of \(z\), so the defining supremum for \(X_k\) is finite.
6. Consequently,\[ \boxed{ f\in X_k \quad\Longleftrightarrow\quad f'\in X_{k+1} }. \]
Step 2 proves the forward implication, and step 5 proves the reverse implication.
7. Q.E.D.
Step 6 is the required equivalence.