Setup
Consider the Banach space $X=\mathcal C^1([0,1],\mathbb R^n)$ with norm $$ \|f\|=\sup |f|+\sup |\dot f|. $$ This banach space has a symmetric bilinear positive definite form $$ b(f,g)=\int_0^1\langle f,g\rangle. $$ Here $\langle \bullet,\,\bullet\rangle$ is the standard scalar product $\mathbb R^n \times \mathbb R^n \to \mathbb R$. By the Cauchy-Schwartz inequality, $b$ is also continuous. We set $$ A^\perp=\{ f\in X| b(f,a)=0\,\forall a\in A \} $$ for every subset $A\subset X$.
Question Let $A\subset X$ be a linear subspace. Under what conditions on $A$ do we have the following implication $$ A^\perp = 0 \implies A=X. $$
Ansatz Say that $A$ has property 1 if $A$ is closed and for every linear bounded functional $\lambda:X\to \mathbb R$ that does not extend to a bounded functional on the completion $\bar X =L^2([0,1],\mathbb R^n)$ there exists some $f\in A$ such that $\lambda(f) \neq 0$.
Conjecture: Let $A\subset X$ be a linear subspace that has property 1. Then $A^\perp =0\implies A=X$.
Examples Consider the evaluation maps $ev_t:X\to \mathbb R^n$ for $t\in[0,1]$. They are all bounded so that $X_t:=\ker ev_t$ is closed. These are proper subspaces with $X_t^\perp = 0$. And they do not have property 1.
But... why??: I want to know under which conditions we have splittings $$ X=B\oplus B^\perp $$ for closed supspaces $B\subset X$ for which $B^\perp$ is closed too. Turns out that closedness of both of these sets is not enough (the $X_t$ are counter examples). $B\cap B^\perp=0$ is clear but $B+ B^\perp = X$ is wrong, we only have that $(B+ B^\perp)^\perp=0$. Hence my question: When does $A^\perp = 0$ imply $A=X$ (Especially for $A=B+ B^\perp$)