Let \(f\) be holomorphic on the open unit disk \[ \mathbb D=\{z:|z|<1\}. \] Prove that there is a sequence \((z_n)\) in \(\mathbb D\) such that \[ |z_n|\to1 \] and \((f(z_n))\) is bounded.
Solution
1. If \(f\equiv0\), the conclusion holds.
Take \[ z_n=1-\frac1n \] for \(n\geq2\). Then \(z_n\in\DD\), \(\abs{z_n}\to1\), and \(f(z_n)=0\) for every \(n\).
2. Assume henceforth that \(f\) is not identically zero. If the desiredsequence does not exist, then for every \(M>0\) there is \(r_M<1\) suchthat\[ \abs{z}>r_M \quad\Longrightarrow\quad \abs{f(z)}>M. \]
Suppose the displayed assertion failed for some \(M>0\). Then for every integer \(n\geq2\) there would be a point \(z_n\in\DD\) with \[ \abs{z_n}>1-\frac1n \] and \[ \abs{f(z_n)}\leq M. \] This would give \(\abs{z_n}\to1\) while \((f(z_n))\) remained bounded, contrary to the assumption that the desired sequence does not exist.
3. Under the assumption of step 2, the function \(f\) has onlyfinitely many zeros in \(\DD\).
Apply step 2 with \(M=1\). There is \(r_1<1\) such that \[ \abs{z}>r_1 \quad\Longrightarrow\quad \abs{f(z)}>1. \] Thus every zero of \(f\) lies in the compact disk \[ \{z:\abs{z}\leq r_1\}. \] Since \(f\) is holomorphic and not identically zero, its zeros are isolated. An infinite set of zeros in this compact disk would have an accumulation point there, contradicting the identity theorem. Hence there are only finitely many zeros.
4. Let \(a_1,\ldots,a_k\) be the zeros of \(f\) in \(\DD\), withmultiplicities \(m_1,\ldots,m_k\), and define\[ P(z)=\prod_{j=1}^k(z-a_j)^{m_j}, \]with \(P=1\) if \(f\) has no zeros. Then\[ h(z)=\frac{P(z)}{f(z)} \]extends to a holomorphic function on \(\DD\) that is not identicallyzero.
At each zero \(a_j\), the functions \(P\) and \(f\) vanish to the same order \(m_j\). Their quotient therefore has a removable singularity at \(a_j\), so \(h\) extends holomorphically across all the zeros. At every point where \(f\) is nonzero, both \(P\) and \(f\) are nonzero, so \(h\) is nonzero there. Hence \(h\) is not identically zero.
5. For every \(\varepsilon>0\), there is \(r_\varepsilon<1\) such that\[ r_\varepsilon<\abs{z}<1 \quad\Longrightarrow\quad \abs{h(z)}<\varepsilon. \]
The polynomial \(P\) is bounded on the closed unit disk. Put \[ C=\max_{\abs{z}\leq1}\abs{P(z)}. \] Since \(P\) is not the zero polynomial, \(C>0\). Apply step 2 with \[ M=\frac{C}{\varepsilon}. \] For \(\abs{z}>r_M\), \[ \abs{h(z)} = \frac{\abs{P(z)}}{\abs{f(z)}} < \frac{C}{C/\varepsilon} = \varepsilon. \] Take \(r_\varepsilon=r_M\).
6. The function \(h\) is identically zero on \(\DD\).
Fix \(w\in\DD\) and \(\varepsilon>0\). By step 5, choose \(r_\varepsilon<1\) as there. Select \[ \rho \quad\text{with}\quad \max\{\abs{w},r_\varepsilon\}<\rho<1. \] The maximum-modulus principle on the disk \(\abs{z}\leq\rho\) gives \[ \abs{h(w)} \leq \max_{\abs{z}=\rho}\abs{h(z)} < \varepsilon. \] Since \(\varepsilon>0\) is arbitrary, \(h(w)=0\). As \(w\) was arbitrary, \(h\equiv0\) on \(\DD\).
7. The assumption that no desired sequence exists is impossible.
Step 4 says that \(h\) is not identically zero, whereas step 6 says that \(h\) is identically zero. This contradiction arose from the assumption in step 2 that the desired sequence does not exist.
8. Therefore there is a sequence \((z_n)\) in \(\DD\) such that\[ \boxed{\abs{z_n}\to1} \]and \((f(z_n))\) is bounded.
1. If \(f\) is not identically zero,step 7 proves the conclusion by contradiction.
9. Q.E.D.
Step 8 is the required conclusion.