I learned that the functions $e_n = \dfrac{e^{inx}}{\sqrt{2 \pi}}$ are orthonormal basis of $L^2(\mathbb{T})$, the set of square integerable functions on a torus. But then I saw that the same functions are also the orthonormal basis of the space $L^2(0,2\pi)$, the set of square integrable functions that are not necessarily periodic.
Why is that the case? Why is it that we can ignore the condition for periodicity?
This is very important: $L^2(0, 2\pi)$ is not the set of square integrable functions on $(0, 2\pi)$. The reason is that the "inner product" on this set is not an inner product: there are nonzero square-integrable functions $f$ such that
$$\langle f, f \rangle = \int_0^{2\pi} |f(x)|^2 \, dx = 0$$
(for example the indicator function of a point). $L^2(0, 2\pi)$ is the quotient of the vector space of square integrable functions by the subspace of functions with the above property, and as a result its elements are not functions, but equivalence classes of functions. This taking of equivalence classes destroys the information of what value a function takes at the endpoints, and in fact destroys the information of what value a function takes on any particular subset of $(0, 2\pi)$ of measure zero, which is why it doesn't matter whether you impose periodicity or not, or whether you work with $(0, 2\pi)$ or $[0, 2\pi]$.
Similarly, the condition that a bunch of functions (really their equivalence classes) forms an orthonormal basis means that certain sequences converge in the $L^2$ sense, and this is much weaker than, and does not imply, pointwise convergence. So a series of periodic functions can converge in the $L^2$ sense to a function that is not periodic.
