I have an issue about a question posted on another forum. The user gatsu on that forum posted (originally in French) that
starting from the following hermitian inner product on periodic functions space:
$$(f|g)=\frac{1}{T}\int_{-T/2}^{T/2}f^{*}(x)\,g(x)\,dx$$
to proove that corresponding Fourier series has a basis, one has to proove this equality :
$$\frac{1}{T}\sum_{n\in N}e^{j2\pi \frac{n}{T}(x-\alpha)} = \delta(x-\alpha)\tag{1}$$
I don't understand very well this reasoning. If we want to prove that Fourier series can be written under: $$x(t)=\sum_{n} c_{n}e^{j2\pi n\,f_{0}t}$$ with $$c_{n}=\frac{1}{T_{0}}\int_{-T_{0}/2}^{T_{0}/2} x(t) e^{-j2\pi nf_{0}t}dt$$ one has to prove that basis vectors $e^{j2\pi n\,f_{0}t}$ are such that inner product: $$(e^{j2\pi k\,f_{0}t}|e^{j2\pi l\,f_{0}t})=\frac{1}{T_{0}}\int_{-T_{0}/2}^{T_{0}/2}e^{j2\pi k\,f_{0}t}\,e^{-j2\pi l\,f_{0}t}\,dt=\delta_{kl}$$
isn't it ?
I can't find the link between the goal of above statement and the demonstration that Fourier series has a basis with Hermitian inner product.
Indeed, in equation $(1)$, there is not an integral but a $\sum$.
If someone could help me to clarify this proof, I do confusions and get all mixed up with this demo.