# Fourier Series and Inner Product

When working with Fourier series, the inner product is defined as $$\int_{-L}^L f(x)g(x)dx$$

I see this definition everywhere and we know that $\rm{sin}\big(\frac{n\pi x}{L}\big)$ and $\rm{cos}\big(\frac{n\pi x}{L}\big)$ will form a orthogonal basis, but not orthonormal.

My question is: why is not more usual to define the inner product as $$\frac{1}{L}\int_{-L}^L f(x)g(x)dx$$ ?

Because with this definition, the previous basis will be orthonormal.

I think that the basis will be precisely $\Big\{ \frac{1}{2},\rm{sin}\big(\frac{\pi x}{L}\big),\rm{cos}\big(\frac{\pi x}{L}\big),\rm{sin}\big(\frac{2\pi x}{L}\big),\rm{cos}\big(\frac{2\pi x}{L}\big),\ldots \Big\}$.

• Actually, people often divide by L: en.wikipedia.org/wiki/Fourier_series – Julien Mar 10 '13 at 16:43
• Well...in that case I think I'll blame my books XD – Integral Mar 10 '13 at 16:45
• @julien If is there some mistake in my english, please feel free to edit. – Integral Mar 10 '13 at 16:46
• I'll let native english speakers do that if needed. – Julien Mar 10 '13 at 16:47

## 1 Answer

An inner product is bilinear form that's symmetric definite positive, so whatever you multiply this inner product by positive real, it's still an inner product.

• Thank you, that's all I want to know. – Integral Mar 10 '13 at 16:47
• You're welcome. – user63181 Mar 10 '13 at 16:49