Find a function holomorphic on the closed unit disk, of absolute value \(1\) on the unit circle, whose only zero in the disk is \(1/2\).
Let \(f\) be holomorphic on the closed disk with \(f(1/2)=0\) and \(|f(z)|\le|e^z|\) for \(|z|=1\). How large can \(|f(0)|\) be?
Solution
1. Define\[ g(z) \coloneqq \frac{z-\frac12}{1-\frac z2}. \]Then \(g\) is holomorphic on a neighborhood of the closed unit disk andits only zero there is\[ z=\frac12. \]
The denominator vanishes only at \(z=2\), which lies outside the closed unit disk. Thus \(g\) is holomorphic on a neighborhood of that disk. Its numerator has the unique zero \(z=1/2\), and the denominator is nonzero there.
2. For every \(|z|=1\),\[ |g(z)|=1. \]
If \(|z|=1\), then \(\overline z=z^{-1}\) and \[ \begin{aligned} \left|z-\frac12\right|^2 &= \left(z-\frac12\right) \left(\overline z-\frac12\right)\\ &= \frac54-\frac12(z+\overline z), \end{aligned} \] while \[ \begin{aligned} \left|1-\frac z2\right|^2 &= \left(1-\frac z2\right) \left(1-\frac{\overline z}{2}\right)\\ &= \frac54-\frac12(z+\overline z). \end{aligned} \] The numerator and denominator of \(|g(z)|^2\) are therefore equal. Together with step 1, this proves part (a).
3. For a function \(f\) as in part (b), the quotient\[ h(z) \coloneqq \frac{f(z)}{g(z)e^z} \]has a removable singularity at \(z=1/2\) and extends holomorphically tothe unit disk.
The exponential has no zeros. The function \(g\) has a simple zero at \(1/2\), while \(f(1/2)=0\). Hence one may write locally \[ f(z) = \left(z-\frac12\right)u(z) \] with \(u\) holomorphic, and \[ g(z) = \left(z-\frac12\right) \frac1{1-z/2}. \] Thus near \(1/2\), \[ h(z) = \frac{u(z)(1-z/2)}{e^z}, \] which is holomorphic. This gives the desired extension.
4. On the unit circle,\[ |h(z)|\le1. \]
For \(|z|=1\), step 2 and the hypothesis on \(f\) give \[ |h(z)| = \frac{|f(z)|}{|g(z)|\,|e^z|} \le \frac{|e^z|}{1\cdot|e^z|} = 1. \]
5. One has\[ |f(0)|\le\frac12. \]
By step 3, \(h\) is holomorphic on the unit disk and continuous on its boundary. Step 4 and the maximum modulus principle imply \[ |h(0)|\le1. \] Since \[ g(0)=-\frac12 \qquad\text{and}\qquad e^0=1, \] we have \[ |h(0)| = \frac{|f(0)|}{|g(0)|} = 2|f(0)|. \] Therefore \[ |f(0)|\le\frac12. \]
6. The bound in step 5 is attained.
Take \[ f(z)=g(z)e^z. \] Then \(f\) is holomorphic on the closed unit disk and \[ f\left(\frac12\right)=0. \] For \(|z|=1\), step 2 gives \[ |f(z)| = |g(z)|\,|e^z| = |e^z|. \] Finally, \[ |f(0)| = |g(0)| = \frac12. \]
7. Hence the largest possible value is\[ \boxed{\frac12}. \]
Step 5 gives the upper bound and step 6 realizes it.
8. Q.E.D.
Steps <1>1–<1>2 prove part (a), and steps <1>3–<1>7 prove part (b).