functions acting as linear functionals on their dual space Supposing $f\in L^p$, where p and q are conjugate exponents, what does it mean that "f is completely determined by its action as a linear functional on $L^q$"? (Quoting Folland's Real Analysis here). I know that f becomes a linear functional on the dual space, which happens to be $L^q$, but I am wondering what is meant by completely determined; do we know exactly what f does to its domain? on its domain minus measure zero? How so?
 A: It means that if for $f,g \in L^p$ we have
$$ \int f h = \int gh$$
for all $h \in L^q$, then $f = g$ (in the usual $L^p$ sense).
A: Let $f\in L^p(\Omega)$, $1<p<\infty$.


*

*$f$ is completely determined by the values of $f(x)$ for almost all $x\in\Omega$;

*The values of $f(x)$ for almost all $x\in\Omega$ are completely determined by the values of $\int_{\Omega} f(x)h(x)\;dx$ for all $h\in L^q(\Omega)$;

*The values of $\int_{\Omega} f(x)h(x)\;dx$ for all $h\in L^q(\Omega)$ are completely determined by the action of $f$ as a linear functional on $L^q(\Omega)$.
Conclusion: $f$ is completely determined by its action as a linear functional on $L^q$.
As said in the menag's answer, this means that if $f$ and $g$ are functions which have the same action as linear functionals on $L^q$, then they are also equal as functions in $L^p$. In other words: what the function $f\in L^p(\Omega)$ does to almost all points of its domain $\Omega$ is completely determined by what the functional $f\in(L^p(\Omega))'\cong L^q(\Omega)$ does to all points of its domain $L^p(\Omega)$.
