If $fh\in \mathcal{L}^1(\mu )$ for every $h\in \mathcal{L}^{p'}(\mu )$ then $f\in \mathcal{L}^p(\mu)$

Question

I have an exercise that I dont know how to handle, it says

Let $(X,\mathcal{S},\mu )$ a $\sigma $-finite measure space and $f:X\to \mathbb{F}$ an $\mathcal{S}$-measurable function. Suppose that $p\in[1,\infty ]$. Show that if $fh\in \mathcal{L}^1(\mu )$ for every $h\in \mathcal{L}^{p'}(\mu )$ then $f\in \mathcal{L}^p(\mu)$.

Here $\mathbb{F}$ is $\Bbb R $ or $\Bbb C $. My first idea was trying to use the following theorem

Theorem: if $f\in \mathcal{L}^p(\mu)$ in a $\sigma $-finite space for $p\in[1,\infty ]$ then $$\|f\|_p=\sup\left\{\left|\int fh \,\mathrm d \mu \right|:\|h\|_{p'}\leqslant 1\right\}$$

But the identity above is just stated when we already knows that $f\in \mathcal{L}^p(\mu )$, so it is not clear if I can use it in this context (it seems that I cant do it).

Maybe I must use the PUB (aka Principle of Uniform Boundedness). However I didn't find a way to use the PUB because it needs a set of bounded linear maps who sets of norms of the image is bounded.

Another idea: setting $\varphi _f(h):=\int fh\,\mathrm d \mu $ if I show that it is a continuous linear functional in $\mathcal{L}^{p'}(\mu )$ then we are done (what is almost equivalent to the application of the theorem above stated). However I dont see a way to show this.

Setting $h:=\frac{\bar f}{|f|}\sum_{k\geqslant 1}c_k\chi_{A_k}$ for appropiate values of $c_k$ and where $(A_k)_k$ is a cover of $X$ where each set have finite measure, I shown that $h\in \mathcal{L}^1\cap \mathcal{L}^{p'}(\mu )$ and therefore $\int_{A_k}|f|\,\mathrm d \mu <\infty $ for each $A_k$. Im not sure if this result could help, I just had shown that $f$ is integrable "locally".

I dont know what to do, some help will be appreciated, thanks.

Answer 1

Let $f_{n}=f\chi_{X_{n}\cap|f|\leq n}$, where $X=\displaystyle\bigcup_{n}X_{n}$, $(X_{n})$ increasing, $X_{n}$ is of finite measure. Then $|f_{n}g|\leq|fg|$, where $g\in L^{q}$, $|fg|\in L^{1}$, and hence $\lim_{n}\displaystyle\int|f_{n}g|=\int|fg|$ by Lebesgue Dominated Convergence Theorem, then so is $u_{n}(g)=\displaystyle\int f_{n}g$, $u(g)=\displaystyle\int fg$, that $u_{n}(g)\rightarrow u(g)$. By Uniform Boundedness, $\|u_{n}\|\leq M$.

On the other hand, $\|f\|_{L^{p}}=\lim_{n\rightarrow\infty}\|f_{n}\|_{L^{p}}$ by Monotone, and we have $\|f_{n}\|_{L^{p}}=|u_{n}(g)|$ for some $\|g\|_{L^{q}}\leq 1$ by Riesz Representation Theorem and the fact that $L^{p}$ is reflexive for $p>1$, but then $|u_{n}(g)|\leq M$ and hence $\|f\|_{L^{p}}\leq M<\infty$.

Edit:

$\|f\|_{L^{p}}=\lim_{n\rightarrow\infty}\|f_{n}\|_{L^{p}}$ by Monotone and we have \begin{align*} \|f_{n}\|_{L^{p}}=\sup\left\{\int |f_{n}||g|: \|g\|_{L^{q}}\leq 1\right\} \end{align*} since $f_{n}\in L^{p}$.

But then $\displaystyle\int|f_{n}||g|\rightarrow\int|f||g|$ and \begin{align*} \sup\left\{\int |f||g|: \|g\|_{L^{q}}\leq 1\right\}=\sup\left\{\left|\int fg\right|: \|g\|_{L^{q}}\leq 1\right\}\leq M, \end{align*} so we deduce that $\|f_{n}\|_{L^{p}}\leq M$.