# Demonstration to check basis for fourier series

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.

## migrated from physics.stackexchange.comAug 8 at 23:13

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• Where is the forum post that you took the quoted text from? There should be some attribution of the source of that text, so if you could edit it in, that would be great. – David Z Aug 8 at 23:16
• -@DavidZ sorry, I forgot to mention the source : it comes from a french forum, I tried to translate it as best as possible : forums.futura-sciences.com/mathematiques-superieur/… . thanks – youpilat13 Aug 8 at 23:29
• OK thanks! I edited in the attribution to show you how it might be done. – David Z Aug 8 at 23:33
• -@DavidZ . did you understand my issue ? – youpilat13 Aug 18 at 16:58