Let \(f\) be a rational function with no poles in the closed upper half-plane. Prove that \[ \sup\{|f(z)|:\operatorname{Im}z\ge0\} = \sup\{|f(z)|:\operatorname{Im}z=0\}. \]
Solution
Set \[ M=\sup\{\abs{f(x)}:x\in\RR\}. \]
1. If \(f\) has a pole at infinity, then both suprema in the statement are\(+\infty\).
Write \(f=p/q\) with relatively prime polynomials. A pole at infinity means \(\deg p>\deg q\), so \(\abs{f(x)}\to\infty\) as \(\abs{x}\to\infty\) along the real axis. Hence \(M=+\infty\). Since the real axis is contained in the closed upper half-plane, the supremum there is also \(+\infty\).
2. Suppose \(f\) has no pole at infinity. Then the finite limit\[ L=\lim_{z\to\infty}f(z) \]exists and satisfies \(\abs{L}\leq M\).
For a rational function \(p/q\) with \(\deg p\leq\deg q\), the limit at infinity exists and is finite. Taking the limit along the real axis gives \[ \abs{L} = \lim_{\abs{x}\to\infty}\abs{f(x)} \leq M. \]
3. Under the hypothesis of step 2, for every \(z\) with\(\operatorname{Im}z\geq0\),\[ \abs{f(z)}\leq M. \]
Fix such a \(z\) and let \(\varepsilon>0\). By step 2 and the definition of the limit at infinity, there is \(R_0\) such that \[ \abs{f(w)-L}<\varepsilon \] whenever \(\abs{w}\geq R_0\). Choose \(R>\max\{R_0,\abs{z}\}\). On the semicircular part of the boundary of the upper half-disk \[ D_R=\{w:\abs{w}<R,\ \operatorname{Im}w>0\}, \] the triangle inequality and step 2 give \[ \abs{f(w)} \leq \abs{L}+\varepsilon \leq M+\varepsilon. \] On the diameter \([-R,R]\), the definition of \(M\) gives \(\abs{f(w)}\leq M\). Since \(f\) has no poles in the closed upper half-plane, it is holomorphic on a neighborhood of the closed upper half-disk. The maximum modulus principle therefore yields \[ \abs{f(z)}\leq M+\varepsilon. \] As \(\varepsilon>0\) is arbitrary, \(\abs{f(z)}\leq M\).
4.\[ \sup\{\abs{f(z)}:\operatorname{Im}z\geq0\} = \sup\{\abs{f(z)}:\operatorname{Im}z=0\}. \]
If \(f\) has a pole at infinity, step 1 proves the equality. Otherwise, step 3 shows that the left-hand supremum is at most \(M\), while inclusion of the real axis in the closed upper half-plane shows that it is at least \(M\). The right-hand supremum is \(M\) by definition.
5. Q.E.D.
Step 4 is the required equality.