Constructing an $L^2$ space on the unit ring $\mathcal{S^1}$ Revised Question:
Starting with $L^2[0,2\pi]$, does the canonical map $$[0,2\pi)\ni\theta\mapsto e^{i\theta}\in\mathcal{S^1}$$(with functions going across in the obvious way) turn $L^2[\mathcal{S^1}]$ into a bona fide Hilbert space?
In particular, does the difference in topology between $[0,2\pi]$ and ${S^1}$ have any nasty implications?
Original Question: 
Is the Hilbert Space of $L^2$ functions on $[0,2\pi]$ with $f(0)=f(2\pi)$ equivalent to a Hilbert Space of $L^2$ functions defined on the unit ring $\mathcal{S^1}$? Can I even construct the latter?
The reason I ask is that I'm uncertain whether the difference in the measure and topologies between $[0,2\pi]$ and $\mathcal{S^1}$ 'bubbles' its way up into functional analytic results.
Note: The boundary condition given is shorthand for: "The set of equivalence classes of Lebesgue square measurable functions (modulo sets of measure zero) containing a continuous member satisfying the given boundary condition".

Of course as Hilbert spaces $L^2(S^1)$ and $L^2\bigl([0,1]\bigr)$ are isomorphic, and you could also say that $L^2\bigl([0,1]\bigr)$ is the prime example of a Hilbert space arising from Lebesgue theory.
But note that $L^2(S^1)$ is one of the most important Hilbert spaces in the world, and there definitively is an essential difference between $L^2\bigl([0,1]\bigr)$ and $L^2(S^1)$ which has not been exposed so far: While $[0,1]$ is just a measure space the ground set $S^1$ carries additional structure, namely a transitive group of translations. It is not necessary to go into details here. We all have heard of Fourier theory. E.g., there is a famous theorem saying that any function $f\in L^2(S^1)$ is represented (meaning "equal to") almost everywhere by its Fourier series.
