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## Reverse Riesz Type Problem

Suppose $p \in (1,\infty)$ and consider $q$ such that $1/p+1/q=1$. Take $(X,\mathcal{A},\mu)$ a $\sigma$-finite measure space. Suppose $f:X \to \Bbb{C}$ is a measurable function such that $fg \in L^1(\mu)$ for all $g \in L^q(\mu)$.

Prove that $f \in L^p(\mu)$.

Solution: Define the following functional $\displaystyle \phi : L^q \to \mathbb{R},\ \phi(g)= \int_Xf g d \mu$. This is well defined and therefore, by the theorem of representation of linear functionals on $L^q$ we get that there exists a function $h \in L^p$ such that $\displaystyle\phi(g)=\int_X hg d \mu, \ \forall g \in L^q$. This means that we have $latex \displaystyle \int_X(f-h)gd\mu=0,\ \forall g \in L^q$. Now, take $X=\bigcup X_n$ to be a decomposition of $X$ into countably many measurable spaces with finite measure. For $g= g\chi_{X_k}$ we have that $\displaystyle\int_{X_k}(f-h)gd\mu=0,\ \forall g\in L^q(X_k,\mu)$. Since the space $X_k$ has finite measure, all step functions are contained in $L^q(X_k,\mu)$, and this means that there exists a sequence of step functions $(g_n)\subset L^q(X_k,\mu)$ such that $g_n \to f-h$, proving that $\displaystyle\int_{X_k}(f-h)^2=0$ and therefore $f=h$ almost everywhere on $X_k$. Doing this for all $k$ we see that $f=h$ a.e. on $X$ and therefore $f \in L^p(\mu)$.