Completeness of $\{ f_n : n \in \mathbb N \} \subset C[0,1]$ in $L^1[0,1]$ Suppose that $\{ f_n : n \in \mathbb N \} \subset C[0,1]$ is a system of continuous functions which is complete in $L^1[0,1]$, i.e. each $g \in L^1[0,1]$ can be approximated arbitrarily well by a finite linear combination of the $f_n$'s. Is it true that if $g \in L^1[0,1]$ is an arbitrary $L^1$-functions such that
$$
\int_0^1g(t)f_n(t) \, dt = 0 \ \ \ \ \forall n \in \mathbb N
$$
then $g=0$ almost everywhere? I know that this is true if $L^1[0,1]$ gets replaced by $L^2[0,1]$ since then we're in a Hilbert space setting.
Thanks for any help!
 A: The answer seems to be no ($g$ need not be zero), if you accept that the $f_n$ are functions in $\mathcal{C}]0,1[ \cap L^1[0,1]$ instead of $\mathcal{C}[0,1]$.
This is based on theorems 2.2 and 2.3 of The Full Müntz theorem in $\mathcal{C}[0,1]$ and $L^1[0,1]$. Let me reproduce these theorems:

Theorem 2.1 (classic Müntz theorem): suppose that $(\lambda_i)_{i \ge 1}$ is a sequence of distinct positive real numbers. Then $\mbox{Span}\big(1, x^{\lambda_1}, x^{\lambda_2}, ...\big)$ is dense in $\mathcal{C}[0,1]$ iff $\sum \limits_{i=1}^{\infty} \frac{\lambda_i}{\lambda_i^2+1} = \infty$.
Theorem 2.2 (in $L^2$): suppose that $(\lambda_i)_{i \ge 1}$ is a sequence of distinct real numbers greater than $-\frac{1}{2}$. Then $\mbox{Span}\big(1, x^{\lambda_1}, x^{\lambda_2}, ...\big)$ is dense in $L^2[0,1]$ iff $\sum \limits_{i=1}^{\infty} \frac{2\lambda_i+1}{(2\lambda_i+1)^2+1} = \infty$.
Theorem 2.1 (in $L^1$): suppose that $(\lambda_i)_{i \ge 1}$ is a sequence of distinct real numbers greater than $-1$. Then $\mbox{Span}\big(1, x^{\lambda_1}, x^{\lambda_2}, ...\big)$ is dense in $L^1[0,1]$ iff $\sum \limits_{i=1}^{\infty} \frac{\lambda_i+1}{(\lambda_i+1)^2+1} = \infty$.

Then if you take $f_n(x) = x^{-\frac{1}{2} + \frac{1}{2n^2}}$ for $n \ge 1$, these $f_n$ are in $L^2$ and the previous criterion guarantee that $(f_n)$ is dense in $L^1$ but not dense in $L^2$.
Now since $L^2$ is a Hilbert space, we know that there exist some non-zero $g \in L^2 \subset L^1$ in the orthogonal complement of $\mbox{Span}\big((f_n)\big)$, that is a non-zero $g \in L^2$ such that for all $n \ge 1$, $\displaystyle{\int_0^1} \frac{g(t)}{t^{\frac{1}{2} - \frac{1}{2n^2}}} dt = 0$.
Since the family $f_n : t \mapsto t^{-\frac{1}{2} + \frac{1}{2n^2}}$ is dense in $L^1$, that's your counterexample.

Remark: I think it should be possible to use simpler tools based on dual spaces rather than the bazooka above, but at least it provides an explicit counterexample
