Is $\sup_{f\in \mathcal{F}}\left|\int _Xfg \, d\mu\right|<\infty$ true for all $g\in L^\infty _\mathbb{C}(\mu )$?

Question

Suppose that $(X,\mathcal{A},\mu )$ is a finite measure space. Let $\mathcal{F}\subseteq L^1_\mathbb{C}(\mu )$. If $\sup_{f\in \mathcal{F}}\left|\int _Xf\varphi \, d\mu\right|<\infty$ for all simple function $\varphi :X\to \mathbb{C}$, can we conclude that $\sup_{f\in \mathcal{F}} \left|\int _Xfg \, d\mu\right|<\infty$ for all $g\in L^\infty _\mathbb{C}(\mu )$.

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In my opinion the answer should be yes since we're dealing with a finite measure, the simple functions are dense in $L^\infty_\mathbb{C}(\mu)$, and $L^\infty_\mathbb{C}(\mu)$ is the space of essentially bounded functions. However I don't know how to prove it.

If that inequality is indeed true, please tell me how to prove it!

Answer 1

$\newcommand\vpi\varphi\newcommand\F{\mathcal F}$Suppose that $|g|\le C$ $\mu$-almost everywhere for some real $C>0$. Let $f_1:=\Re f\,1(\Re f>0)$, $f_2:=\Re f\,1(-\Re f>0)$, $f_3:=\Im f\,1(\Im f>0)$, $f_4:=\Im f\,1(-\Im f>0)$, so that the functions $f_1,\dots,f_4$ are $\ge0$ and $f=f_1-f_2+i(f_3-f_4)$.

It suffices to show that $\sup_{f\in\F}\int f_j |g|<\infty$ for each $j=1,2,3,4$. But $\vpi:=1(\Re f>0)$ is a simple function. So, $$\sup_{f\in\F}\int f_1 |g|\le C\sup_{f\in\F}\int f_1 =C\sup_{f\in\F}\int \Re f \vpi \\ =C\sup_{f\in\F}\Re\int f \vpi \le C\sup_{f\in\F}\Big|\int f \vpi\Big| <\infty, $$ so that $\sup_{f\in\F}\int f_1 |g|<\infty$. Similarly, $\sup_{f\in\F}\int f_j |g|<\infty$ for each $j=2,3,4$. $\quad\Box$