We denote $X\_{T}$ the vector space of all $T$-periodic function with zero mean in $L^2$ ( we know that $X\_{T}$ is spawn by $(e^{2i\pi nt/T})$). Let be $$X=X\_{2\pi}+X\_{3\pi}.$$
I think that $X\_{2\pi}+X\_{3\pi}$ is closed in $L^2(0,4\pi)$ but i can't prove it.



Almost periodic function and closed spaces
I think that it is just a hyperplane $H$ defined by the condition $\int\_0^{\pi}+2\int\_{\pi}^{3\pi}+\int\_{3\pi}^{4\pi}=0$. Obviously all functions in $X\_{2\pi}$ and $X\_{3\pi}$ lie in this hyperplane $H$. Choose any function $f\in H$. Choose functions $a(t),b(t)$ on $[0,\pi)$ so that $f(t+3\pi)-b(t)=f(t)-a(t)$ on $[0,\pi)$ and $\int\_0^\pi (a+b)=0$. Let 
$$
u(t)=\begin{cases}a(t),t\in [0,\pi)\\b(t-\pi),t\in [\pi,2\pi)\end{cases}
$$ 
define a function from $X\_{2\pi}$, then $v(t)=f(t)-u(t)$ is $3\pi$-periodic and belongs to $X\_{3\pi}$ due to $f\in H$.


