Why is this function continuous in the strip and analytic in it's interior? (Rubel's proof of Holder's Inequality) Rubel's proof of Holder's Inequality is done in the following way: 
Given the measure space $(\Omega, \mathcal{F}, \mu)$, functions $f \in L^p$, $g \in L^q$ and conjugate exponents $p,q$, define
$\phi(z)= \frac{ \int_{\Omega} |f|^{p (1-z)} |g|^{qz} d\mu}{(\int_{\Omega} |f|^p d\mu)^{1-z} (\int_{\Omega} |g|^q d\mu)^{z}}$, where $z$ is a complex variable taking values in the strip $\{ z \in \mathbb{C} : 0 \leq \mathcal{R}(z) \leq 1\}$. 
He claims this map $\phi$ is defined and continuous in the whole strip and analytic in it's interior. Why is that so?
The rest of the proof follows easily, since it's clear that $|\phi(\sigma+it)| \leq \phi(\sigma)$ for $0 \leq \sigma \leq 1$ and that $|\phi(it)|, |\phi(1+it)|$ are bounded by $1$ (so by the maximum modulus principle, $\phi(z) \leq 1$ in the whole strip and thus Holder's inequality holds (pun unintended) by taking $z=\frac{1}{q}$.

Continuity follows from a dominated  convergence argument. Indeed, we take $(z_n)_n$ contained in the strip such that $z_n\to z$. We define 
$$h_n(x):=|f(x)|^{pz_n}|g(x)|^{q(1-z_n)}.$$
Then $h_n(x)\to h(x):=|f(x)|^{pz}|g(x)|^{q(1-z)}$. Let $a_n$ be the real part of $z_n$. We have 
$$|h_n(x)|=|f(x)|^{pa_n}|g(x)|^{q(1-a_n)},$$
and we can bound it by an integrable function using Young's inequality. 
For analyticity, we can use Morera's theorem. It's indeed, sufficient to prove that the numerator is analytic. To this aim, we integrate since over a triangle contained in the strip $\{0\lt \Re z\lt 1\}$ and we use Fubini's theorem. (Morera's theorem requires the triangle is contained in the considered open set)
