Proving Density of $S$ in $L^{p'}(E)$ for $g \in L^p(E)$

[joypav]

Problem:
$E$ is a measurable set and $1 \leq p < \infty$. Let $p′$ be the conjugate of $p$, and $S$ is a dense subset of $L^{p′}(E)$. Show that if $g \in L^p(E)$ and $\int_{E}fg = 0$ for all $f \in S$, then $g= 0$.

Definition of Density:
$S$ is dense in $L^{p'}(E)$ if $\forall h \in L^{p'}(E), \forall \epsilon > 0, \exists f \in S$ s.t. $\left| \left| f-h \right| \right|_{p'} < \epsilon$
or equivalently
$\exists (f_n)$ in $S$ s.t. $\lim_{{n}\to{\infty}}f_n=h$ a.e. on $E$.

Idea?:
As p and p' are conjugates, I was thinking to use Holder's Inequality.
$\int_{E}\left| fg \right| \leq \left| \left| f \right| \right|_{p'} \left| \left| g \right| \right|_p$

 

[Euge]

Hi joypav,

By density of $S$ in $L^{p'}(E)$, the integral $\int_E fg = 0$ for all $f \in L^{p'}(E)$. Construct $f\in L^{p'}(E)$ such that $fg = \lvert g\rvert^p$ and $\int_E \lvert f\rvert^{p'} \le \int_E \lvert g\rvert^p$. Deduce that $\int_E \lvert g\rvert^p = 0$, i.e., $\|g\|_{L^p(E)} = 0$. Then $g = 0$ in $L^p(E)$.