This is part of a derivation in a text that I am struggling to follow. It says that if we write $e_k(t) = e^{2 \pi i k t}$ then

$$\langle \sum_{n=- \infty}^{\infty} \langle f, e_n \rangle e_n,\sum_{m=- \infty}^{\infty} \langle f, e_m \rangle e_m \rangle = \sum_{n,m} \langle f, e_n\rangle \overline{\langle f, e_m\rangle} \langle e_n, e_m\rangle.$$

I don't see how this follows. Here $\langle., .\rangle$ is the L2 inner product, so $\langle f, g\rangle = \int_0^1 f(t) \overline{g(t)} dt$.

  • $\begingroup$ You mean $e_n(t) = e^{2\pi i n t}$, right? $\endgroup$ – Paul K Jan 8 '16 at 17:06
  • $\begingroup$ Yes, I fixed it. $\endgroup$ – Fequish Jan 8 '16 at 17:10

This actually holds for any inner product, simply by linearity.

Consider the following simpler case:

$$\langle \langle a,b\rangle c,\langle d,e\rangle f\rangle = \langle a,b\rangle\overline{\langle d,e\rangle} \langle c,f\rangle$$

  • $\begingroup$ Could you show why that holds? $\endgroup$ – Fequish Jan 8 '16 at 17:13
  • $\begingroup$ It's true by definition for any inner product. Is the question to show that the particular integral you've defined is a proper inner product? $\endgroup$ – co9olguy Jan 8 '16 at 17:14
  • $\begingroup$ On also has to argue that you can take the limit out of the inner product. This is allowed due to the continuity of the inner product and Parseval's identity. $\endgroup$ – Peter Jan 8 '16 at 17:19
  • $\begingroup$ I see it now. I think you also have to use conjugate symmetry. $\endgroup$ – Fequish Jan 8 '16 at 17:23

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