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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\} $.

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

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

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