Let $L>0$ and define the space $$ L^2_{per}([0,L]):=\big\{ f: \mathbb{R} \rightarrow \mathbb{R} \; ; \; f \: \text{is periodic with period $L>0$ and} f|_{[0,L]} \in L^2([0,L]) \big\}. $$ The space $L^2_{per}([0,L])=L^2_{per}$ is a Hibert space with the inner product $$ (f,g)_{L^2}=\int_0^L f(x)g(x) \;dx,\; \forall \; f,g \in L^2_{per}. $$
Also define, $$ L^2_{o}:=\{ f \in L^2_{per} \; ; \; f \: \text{is a odd function}\} \quad \text{and} \quad L^2_{e}:=\{ f \in L^2_{per} \; ; \; f \: \text{is a even function}\}. $$
Clearly, $L^2_{o}, L^2_{e} \subset L^2_{per}$ are closed subspaces and hence are Hilbert spaces endowed with the inner product $(\cdot, \cdot)_{L^2}$.
Question. If $f \in L^2_{o}$ and $g \in L^2_{e}$, then $(f,g)_{L^2}=0$ ? If so, can I write the orthogonal decomposition $L^2_{per}=L^2_{o} \oplus L^2_{e}$? Here, the symbol $\oplus$ means 'direct sum'.
Yes. The space of $L^2([0,L])$ functions can be represented as fourier series expansion and hence $$\left\{1,\cos\left(\frac{2\pi nx}{L}\right),\sin\left(\frac{2\pi nx}{L}\right)\; ; \; n \in \mathbb{N}\right\}$$ is an orthonormal basis. Consequentely, the odd functions are spanned by $$\left\{\sin\left(\frac{2\pi nx}{L}\right)\; ; \; n \in \mathbb{N}\right\}$$ and even functions are spanned by $$\left\{1,\cos\left(\frac{2\pi nx}{L}\right)\; ; \; n \in \mathbb{N}\right\}$$ and thus can be written as the direct sum mentioned.
See Theorem $1.3$ in here.