Which collections $\{g_n\}$ of continuous functions satisfy $\int_{0}^{1} f \cdot g_n = 0$ implies $f \equiv 0$ for any continuous $f$?

Question

What is a characterization of (linearly independent) collections $\{g_n\}_{n \geq 1} \subset C[0,1]$ with the following property: $$\left(f \in C[0,1] \ \text{and} \ \int_{0}^{1} f(x)g_n(x) \ dx = 0 \ \text{for each} \ n=0,1,2 \cdots\right) \ \text{implies} \ f \equiv 0$$

A sufficient condition is that $\{g_n\}_{n \geq 1}$ spans a dense subset of $C[0,1]$, but is this also necessary?

This question is motivated by this classic problem.

Answer 1

For a set $A\subset C([0, 1])$, write $$ A^\bot:=\left\{ f\in C([0, 1])\ : \int_0^1 fg\, dx=0, \ \forall g\in A\right\}.$$ Now, let $G=\{g_n\ :\ n\in\mathbb N\}$; your question is rewritten as follows.

What conditions guarantee that $G^\bot=\{0\}$?

This notation strongly hints at Hilbert space theory. Indeed, since $$A^{\bot \bot}= \overline{\operatorname{span } A},$$ where $\overline{\operatorname{span}A}$ denotes the closure in $L^2(0, 1)$ of the linear span of $A$, we have that $$ G^\bot=\{0\} \quad \iff \quad \overline{\operatorname{span } G}= \{0\}^\bot=L^2(0, 1).$$

The sought condition is, thus, that $\operatorname{span}G$ is dense in $L^2(0, 1)$.

Since $C([0, 1])$ is itself dense in $L^2(0, 1)$, it suffices to check that $\operatorname{span}G$ is dense in $C([0, 1])$, but with respect to the $L^2$ norm.

Final remark. Usually, it is more difficult to prove that $\operatorname{span}G$ is dense than to prove that $G^\bot=\{0\}$.