Let \(f\) be analytic on a neighborhood of the closed unit disk, suppose \[ f(-\log2)=0, \] and assume \[ |f(z)|\le |e^z| \] for every \(|z|=1\). How large can \[ |f(\log2)| \] be?
Solution
Set \[ r\coloneqq\log2. \] Then \[ 0<r<1. \]
1. The function\[ g(z)\coloneqq e^{-z}f(z) \]is analytic on a neighborhood of the closed unit disk, satisfies\[ g(-r)=0, \]and obeys\[ \abs{g(z)}\leq1 \]for \(\abs z=1\).
The exponential never vanishes, so \(g\) has the same domain of analyticity as \(f\). The prescribed zero gives \[ g(-r) = e^r f(-r) =0. \] On \(\abs z=1\), \[ \abs{g(z)} = e^{-\Re z}\abs{f(z)} \leq e^{-\Re z}\abs{e^z} =1. \]
2. Define\[ B(z)\coloneqq\frac{z+r}{1+rz}. \]Then \(B\) is analytic on a neighborhood of the closed unit disk,\[ B(-r)=0, \]and\[ \abs{B(z)}=1 \]for \(\abs z=1\).
The only pole of \(B\) is at \[ z=-\frac1r, \] whose modulus is greater than \(1\) because \(0<r<1\). Thus \(B\) is analytic near the closed unit disk. Its zero at \(-r\) is immediate.
For \(\abs z=1\), \[ \abs{z+r}^2 = 1+r(z+\overline z)+r^2 \] and \[ \abs{1+rz}^2 = 1+r(z+\overline z)+r^2. \] Hence the two moduli are equal, so \(\abs{B(z)}=1\).
3. The quotient\[ q(z)\coloneqq\frac{g(z)}{B(z)} \]extends holomorphically across \(z=-r\) and satisfies\[ \abs{q(z)}\leq1 \]throughout the closed unit disk.
Both \(g\) and \(B\) vanish at \(-r\), while \(B\) has a simple zero there. Therefore \(g/B\) has at worst a removable singularity at \(-r\), and it extends holomorphically.
On \(\abs z=1\), steps 1 and 2 give \[ \abs{q(z)} = \frac{\abs{g(z)}}{\abs{B(z)}} \leq1. \] The maximum modulus principle then gives the same bound throughout the unit disk.
4. One has\[ \abs{g(r)} \leq \frac{2r}{1+r^2}. \]
Step 3 gives \[ \abs{g(r)} = \abs{B(r)}\abs{q(r)} \leq \abs{B(r)}. \] Since \(r\) is real, \[ B(r) = \frac{2r}{1+r^2}. \]
5. Therefore\[ \abs{f(\log2)} \leq \boxed{ \frac{4\log2}{1+(\log2)^2} }. \]
Since \(f(z)=e^zg(z)\) and \(e^r=2\), \[ \abs{f(r)} = 2\abs{g(r)}. \] Apply step 4 and substitute \(r=\log2\).
6. The bound in step 5 is attained.
For any constant \(C\) with \(\abs C=1\), define \[ f_C(z) \coloneqq C e^z B(z). \] By step 2, this function is analytic on a neighborhood of the closed unit disk and satisfies \[ f_C(-r)=0. \] On \(\abs z=1\), \[ \abs{f_C(z)} = \abs{e^z}\abs{B(z)} = \abs{e^z}. \] At \(z=r\), \[ \abs{f_C(r)} = 2\frac{2r}{1+r^2}, \] which is exactly the bound in step 5.
7. The largest possible value is\[ \boxed{ \frac{4\log2}{1+(\log2)^2} }. \]
Step 5 gives the upper bound and step 6 shows it is sharp.
8. Q.E.D.
Step 7 answers the question.