Suppose $g\in L^1[0,1]$ satisfies $\int_{[0,1]} fg~dx=0$ for all $f\in C[0,1]$. (Here, we are considering Lebesgue measure on $[0,1]$. ) Then do we necessarily have $g=0$?
I am trying to show that "not" every bounded linear functional on $L^\infty[0,1]$ is of the form $f\mapsto \int_{[0,1]} fh~dx$ for some $h\in L^1[0,1]$, and if the above question is true, then I am done, but I can't see whethere it is indeed true.
This is not yet an answer but rather a proof idea. To make notation a little simpler we will simply write $L^1$ instead of $L^1[0,1]$. So suppose now, that $g \in L^1$ with $$\int fg dx = 0$$ for all $f \in \mathcal{C}[0,1]$. The key to prove the result is to prove, that by denseness of $\mathcal{C}[0,1]$ in $L^1$ there exists a bounded sequence $\{f_n\} \subset \mathcal{C}[0,1]$, i.e. $\sup \|f_n\|_\infty < \infty$, such that $f_n \xrightarrow{L^1} sgn(g)$. Indeed, if that is the case, then from the proof of the Riesz-Fischer completeness theorem for $L^p$ spaces, we know, that there exists a subsequence $\{f_{n_k}\}$ so that $f_{n_k} \to sgn(g)$ pointwise. But then by construction we have $$\underbrace{|f_{n_k}g-sgn(g)g|}_{\to 0} \leq |g| \sup \|f_n\|_\infty + |g|$$ So by the Dominated convergence Theorem $$\int |g| dx = \int sgn(g)g dx = \lim \int f_{n_k}g dx = 0$$ Thus $g = 0$ a.e.