Let $b \in (0,\infty)$ and, for $n \in \mathbb{Z}$ $e_n:\mathbb{R} \to \mathbb{C}$ by $e_n(t) = e^{int}$. In $L^2([0,b])$ let $V = $ span$\{e_n : n \in \mathbb{Z}\}$. My question is, for which $b$ is $V^{\perp} = 0$?
I suspect that $V^{\perp} = 0$ if and only if $b \leq 2\pi$. If $b \leq 2\pi$ then $L^2([0,b])$ embeds into $L^2([0,2\pi])$ by extending a function $f:[0,b] \to \mathbb{C}$ to a function $\tilde{f}:[0,2\pi] \to \mathbb{C}$ defined to be zero on $[b,2\pi]$. We can observe that
$$ \langle f , e_n \rangle_{L^2([0,b])} = \langle \tilde{f}, e_n \rangle_{L^2([0,1])}. $$
Thus if $f \in V^{\perp}$ then $\tilde{f}$ is orthogonal to every $e_n$ and since $V$ is dense in $L^2([0,2\pi])$ $\tilde{f} = 0$ almost everywhere and thus so does $f$.
If $b \geq 2$ we can define a function $v(t) = \chi_{[0,2\pi]} - \chi_{[2\pi,4\pi]}$. Then such a function will be orthogonal to every $e_n$.
It remains to find a non-zero element in $V^{\perp}$ for $2\pi < b < 4\pi$. In essence this amounts to finding two functions $f$ and $g$ so that, for every $n \in \mathbb{Z}$,
$$ \int_{0}^{2\pi}fe^{inx}dx = \int_{2\pi}^{b}ge^{inx}dx.$$
You can do some change of variables to get that this is equivalent to finding $f$ and $g$ such that
$$\int_0^{2\pi} fe^{inx} - ge^{in \frac{x}{b}}dx = 0$$
for each $n \in \mathbb{Z}$. Both of these formulations are intractable to me so far.
If $0\le b<2\pi$ then the span of the $e^{nit}$ are dense in the $L^\infty$-norm on $C[0,b]$ by the Stone-Weierstrass theorem, and so dense in the $L^2$ norm too, and so dense too in $L^2([0,b])$. A bit more effort, and one gets that for $b=2\pi$ too. But for $b>2\pi$ periodicity scuppers that.