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I want to use a theorem which states that if $\{\phi_n\}_1^{\infty}$ is an orthonormal basis of $L^2(a,b)$ then for all $f\in L^2(a,b)$ we have $\parallel f \parallel^2 = \sum_1^\infty|\langle f,\phi_n\rangle|^2$. The problem is that my ON basis is $\{\phi_n\}_{-\infty}^{\infty}$.

However, I am thinking that if I define a new set $\{\Phi_n\}_1^{\infty}$, where $\Phi_1 = \phi_0$ and $\Phi_n = \phi_{(-1)^n\lfloor\frac{n}{2}\rfloor}$ for $n\geq 2$, then the two sets contain the same elements, so $\{\Phi_n\}_1^{\infty}$ is also an ON basis of $L^2(a,b)$. In this way I could use the theorem anyway.

In general, is there something that could possibly go wrong with such a "re-indexing"?

jock.marshall
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  • may I know which theorem it is exactly ? is it Parseval's identity ? – the_firehawk Jan 29 '18 at 13:28
  • Your method counts all of the frequencies I think. The indexes are important integers. – user76568 Jan 29 '18 at 14:42
  • The theorem does not come with a name: it is theorem 3.4 in Fourier Analysis and its Applications by Folland. But the equation $\parallel f \parallel^2 = \sum_1^{\infty}|\langle f, \phi_n \rangle|^2$ is called Parseval's equation. – jock.marshall Jan 29 '18 at 14:44
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    It could have gone wrong due to Riemann if it was conditionally convergent. But it is absolutely convergent obviously and doesn't touch the value. – user76568 Jan 29 '18 at 15:02

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