Suppose \(f\) is holomorphic on the open upper half-plane, satisfies \[ |f(z)|\le1 \] there, and \(f(i)=0\). How large can \(|f(2i)|\) be?
Solution
Define \[ \phi(z)=\frac{z-i}{z+i}. \]
1. The map \(\phi\) is a biholomorphism from the open upperhalf-plane onto the unit disk, with\[ \phi(i)=0. \]
For \(z=x+iy\) with \(y>0\), \[ \abs{z-i}^2=x^2+(y-1)^2 < x^2+(y+1)^2 = \abs{z+i}^2, \] so \(\abs{\phi(z)}<1\). Solving \[ w=\frac{z-i}{z+i} \] for \(z\) gives \[ z=i\frac{1+w}{1-w}, \] and, for \(\abs w<1\), \[ \im\left(i\frac{1+w}{1-w}\right) = \frac{1-\abs w^2}{\abs{1-w}^2} > 0. \] Thus this formula maps the unit disk into the upper half-plane and is the inverse of \(\phi\). Finally, \(\phi(i)=0\).
2. The function\[ g(w)=f\left(i\frac{1+w}{1-w}\right) \]is holomorphic on the unit disk, satisfies \(\abs{g(w)}\leq1\), and\(g(0)=0\).
By step 1, the argument of \(f\) lies in the upper half-plane for \(\abs w<1\), so \(g\) is holomorphic there and inherits the bound \(\abs g\leq1\). Also \[ g(0)=f(i)=0. \]
3. For every \(w\) in the unit disk,\[ \abs{g(w)}\leq\abs w. \]
This is Schwarz’s lemma applied to the function in step 2.
4. Every function satisfying the hypotheses obeys\[ \abs{f(2i)}\leq\frac13. \]
We have \[ \phi(2i) = \frac{2i-i}{2i+i} = \frac13. \] Since \(g(\phi(z))=f(z)\), step 3 gives \[ \abs{f(2i)} = \abs{g(1/3)} \leq \frac13. \]
5. The bound in step 4 is attained.
Take \[ f(z)=\phi(z)=\frac{z-i}{z+i}. \] By step 1, this function is holomorphic on the upper half-plane, has modulus strictly less than \(1\) there, and satisfies \(f(i)=0\). Moreover, \[ \abs{f(2i)}=\frac13. \]
6. The largest possible value is\[ \boxed{\frac13}. \]
Step 4 gives the upper bound, and step 5 shows that equality is possible.
7. Q.E.D.
Step 6 is the requested sharp value.