Am reading the proof from here reproduced below:
Let $X$ be any real or complex Banach space of functions or equivalence classes of functions such that with $f\in X$ also $|f|\in X$ and $\||f|\| = \|f\|$ and let $\phi$ be a positive linear functional on $X$. It is easily seen that $|\phi(f)|\le 2\phi(|f|)$ for all $f\in X$.
Suppose that $\phi$ is unbounded. Then there exists a sequence $(f_n)\subset X$ with $\|f_n\|=1$ and $|\phi(f_n)|\ge 2\cdot 4^n$. Put $g_n := |f_n|\ge 0$. Then $\|g_n\|=1$ and $\phi(g_n)\ge 4^n$ for $n\in\mathbb N$. Define the function $h := \sum_{n=1}^\infty2^{-n}g_n\in X$. Then $h\ge 2^{-n}g_n$ for each $n\in\mathbb N$ and so $\phi(h)\ge 2^{-n}\phi(g_n)\ge 2^n$ for all $n\in\mathbb N$, which is impossible.
Question: Why do we have $h\ge 2^{-n}g_n$ for each $n\in \mathbb N$? This doesn't seem to follow from the assumptions made. Do we need $X$ to be an $L^p$ space with $1\leq p< \infty$?
It looks like we are using the property that if $f_n\geq 0$ for each $n$ and $f_n\to f$ with respect to $\|\cdot \|$ then $f\geq 0$. But why is this true?
Am I missing something? Thanks a lot for your help.
EDIT:
For the real case it seems that the natural requirement is to let $X$ be a Banach lattice . This will ensure that $|x|:=\sup\{x,-x\}\in X$ whenever $x\in X$, that $\||x|\| = \|x\|$, and that the positive cone $P := \{ x \in X : x\geq 0\}$ is closed (see here).
In the case where $X=L^p$ for some $1\leq p< \infty$ I believe we can argue as follows:
Choose a representative for $h$ and $\sum_{i=1}^n2^{-i}g_i$ for each $n$. Then $\sum_{i=1}^n2^{-i}g_i \to h$ in $L^p$ as $n\to \infty$. We can extract a subsequence $(n_k)$ with $\sum_{i=1}^{n_k}2^{-i}g_i \to h$ a.s. as $k\to \infty$. This implies that $h\ge 2^{-n}g_n$ a.s. for each $n\in \mathbb N$.