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 $L^p(E)$.

 

[math771]

Yes, using Holder's inequality would be a good idea. Here's how you can use it to prove the statement:

Assume $g \neq 0$. Since $g \in L^p(E)$, there exists a non-zero measurable set $A \subset E$ such that $g|_A \neq 0$. Let $h = \frac{|g|^{p'}}{\left| \left| g \right| \right|_p^{p'-1}}$. Then $h \in L^{p'}(E)$ and $\left| \left| h \right| \right|_{p'} = 1$. Therefore, by the definition of density, there exists a sequence $(f_n)$ in $S$ such that $\lim_{n \to \infty} f_n = h$ a.e. on $E$.

Now, using Holder's inequality, we have:
\begin{align*}
\int_{A} |f_n g| &= \int_{A} |f_n| |g| \\
&\leq \left| \left| f_n \right| \right|_{p'} \left| \left| g \right| \right|_p \\
&= \left| \left| f_n \right| \right|_{p'} \left| \left| g| \right| \right|_p^{p'-1} \left| \left| g \right| \right|_p \\
&= \left| \left| f_n \right| \right|_{p'} \int_{A} h |g| \\
&= \left| \left| f_n \right| \right|_{p'} \int_{A} |g|^{p'} \\
&= \left| \left| f_n \right| \right|_{p'} \left| \left| g \right| \right|_{p'}^{p'} \\
&= \left| \left| f_n \right| \right|_{p'} \\
&\to 1 \text{ as } n \to \infty
\end{align*}

However, since $\lim_{n \to \infty} f_n = h$ a.e. on $E$, we also have $\lim_{n \to \infty} \int