# Boundedness of functions in complex interpolation method

In the method of complex interpolation one evaluates traces of suitable holomorphic functions on the strip. I have looked in the book of Lunardi and the one of Bergh/Löfström and in both this "suitable" meant for a function $f: S\to X+Y$ that $f$ is continuous, bounded, holomorphic on the interior of the strip and $f(it)\in \mathrm{C}_b(\mathbb{R},X)$, $f(1+it)\in \mathrm{C}_b(\mathbb{R},Y)$.

My question is, whether it is redundant to require that $f$ is bounded. Using the maximum principle for holomorphic functions, the continuous embeddings $X,Y \hookrightarrow X+Y$ and the boundedness of the restriction of $f$ to the lines bordering the strip this already is clear?

## 1 Answer

No, the assumption of boundedness is not redundant. On an unbounded domain, the maximum principle fails.

A typical counterexample: let $f(z) = \exp(\exp(\pi i (z-1/2)))$. When $\operatorname{Re} z \in \{0,1\}$, the imaginary part of $w = \pi i (z-1/2)$ is $\pm \pi /2$, hence $\exp(w)$ is purely imaginary, hence $\exp(\exp(w))$ has modulus $1$. Yet, at $z=1/2 - in$ we get $f(z) = \exp(\exp(\pi n))$ which grows to infinity rather rapidly.

The maximum principle can be repaired by adding an additional assumption about the asymptotic behavior at infinity, at which point it becomes known as Phragmén–Lindelöf principle.

• I would apply the maximum principle to intervals in the strip and not the whole strip – Sebastian Bechtel Jul 10 '17 at 19:44
• Try it. I gave an explicit counterexample that shows it will not work. If you cut a rectangle out of the strip, what to do about its top and bottom sides? – user357151 Jul 10 '17 at 19:58
• Found my mistake... – Sebastian Bechtel Jul 10 '17 at 20:03