$f=g+ih$ is $L^p$ if and only if $g$ and $h$ are $L^p$?

Question

Let $X$ be a measure space, $f:X\to\mathbb{C}$ a measurable function, and $1<p<\infty$. $f$ is said to be $L^p$ if $|f|^p$ is integrable. Now, let $f=g+ih$, where $g$ and $h$ are real-valued. Is $f$ $L^p$ if and only if $g$ and $h$ are $L^p$ ? If not, is there any relationship between these two conditions?

Answer 1

Suppose that $\frac1p+\frac1q=1$. By Holder's inequality https://en.wikipedia.org/wiki/H%C3%B6lder%27s_inequality#Notable_special_cases, for all $x \in X$, we have $$ \max\{|g(x)|^p, |h(x)|^p \} \le |f(x)|^p \le (|g(x)|+|h(x)|)^p \le 2^{p/q}\cdot (|g(x)|^p+|h(x)|^p) \,.$$ Integrating over $x \in X$ proves the desired equivalence.