It's quite trivial to prove that $$ \int_a^b f(t)g(t)dt=0 , \forall g:\mathbb R\to [a,b] \space \space (1) $$ means that $f=0$. What i'm interested in is the following problem: suppose that $(1)$ doesn't hold for all $g$, but for all $g:\mathbb R\to [a,b]$ that satisfy a certain condition. Specifically we have the following-it doesn't come from nowhere, it's physics related (kinda):
Let $$\vec N:\mathbb R \to \mathbb R^n , \space \vec h: \mathbb R \to \mathbb R^n $$ We define the inner product $\bullet$ in $\mathbb R^n$ in the obvious way. Now suppose all the functions $\vec x:\mathbb R \to \mathbb R^n$ such that $\vec h(t) \bullet \vec x(t) =0$. $$ X= \{\vec x:\mathbb R \to \mathbb R^n |\vec h(t) \bullet \vec x(t) =0, \forall t\in [a,b] \} $$
For some $a,b \in R$. Suppose now that $$ \int_a^b \vec x(t) \bullet \vec N(t) dt =0, \space \forall x \in X $$ What can we then say about $\vec N $? What I'd like to show is that $\vec N$ is parallel to $\vec h$ for all $t \in [a,b]$ (I don't know if it's true though). In the physics-case of the problem $\vec x$ are actually virtual displacements $δ \vec x$, but I'm not sure it matters. Hope all makes sense and that no information about the physics problem is needed. Anyhow:
You can read the post if you'd like, but I dont think it's necessary https://physics.stackexchange.com/questions/334029/deriving-the-lagrangian-in-constraint-motion-where-do-multipliers-come-from
Assume that all functions involved are in $L^2 = L^2((a,b), \mathbb{R}^n)$, and that $h\neq 0$. Then, using the Hilbert structure of $L^2$, you have that $X = h^\perp$ (the subspace of functions orthogonals to $h$).
Then your condition says that $(N,x) = 0$ for every $x\in h^\perp$, i.e. $N \in (h^\perp)^\perp = \text{span}(h)$. This means that there exists $\lambda\in\mathbb{R}$ such that $N = \lambda h$.