# Prove that if $\int_E f \cdot g = 0$ then $f=0$

Let $E$ be a measurable set, $1\le p < \infty$, $q$ the conjugate of $p$, and $\mathcal S$ a dense subset of $L^q(E)$. show that if $f \in L^p(E)$ and $\int_E f \cdot g = 0 \; \forall \; g \in \mathcal S$, then $f=0$

I previously did a similar problem where I showed that if this integral was $0$ for all $g \in L^q(E)$ then $f=0$ and I used the fact that $\|f\|_p = Max_{g \in L^q(E), \|g\|_q \le 1} \int_E f \cdot g$

Heres my attempt at this problem but I know there is a mistake:

Suppose that $\int_E f \cdot g = 0 \; \forall \; g \in \mathcal S$ . Let $h \in L^q(E)$ be such that $\|f\|_p = Max_{g \in L^q(E), \|g\|_q \le 1} \int_E f \cdot g = \int_E f \cdot h \;$. Then $|h| \in L^q(E)$

Then there exists nonnegative ${g_n} \in \mathcal S$ such that $\|g_n - |h|\|_q \rightarrow 0$. So we get that $\|f\|_p = Max_{g \in L^q(E), \|g\|_q \le 1} \int_E f \cdot g = \int_E f \cdot h \le \int_E |f| \cdot |h| \le \lim_{n \rightarrow \infty}\int_E |f| \cdot g_n =\lim_{n \rightarrow \infty}0 = 0$. The last inequality comes from Fatou's and that each $g_n \in \mathcal S$

I pretty sure this is incorrect because in order to use Fatou's lemma I am assuming that $g_n \rightarrow |h|$ but we only know that $\|g_n - |h|\|_q \rightarrow 0$

• How do you know that the $g_n$ are non-negative...? – saz Feb 27 '16 at 19:13
• I don't know if your proof is correct, but in any case, if $g_n \to |h|$ in $L^q$ then you can find a subsequence $g_{n_k}$ which converges to $|h|$ almost everywhere, so this in itself is not a problem. – Mark Feb 27 '16 at 19:15

I don't think that your argumentation works. Not because of the missing pointwise convergence (that's something we could fix), but because the functions $g_n$ do, in general, not need to be non-negative (unless $\mathcal{S}$ has the additional property that $u \in \mathcal{S}$ implies $|u| \in \mathcal{S}$) ...
$$\left| \int f \cdot g \right| = \left| \int f \cdot (g-g_n) \right| \leq \|f\|_p \|g-g_n\|_q$$
for any $g \in L^q(\mu)$ and $(g_n)_{n \in \mathbb{N}} \subseteq \mathcal{S}$.