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I am getting confused with a very easy things. Here it comes.

let $f\in L^2([-\pi, \pi])$ and consider its Fourier Series

$$f(x)\sim \frac{a_0}{2}+\sum_{k=1}^\infty (a_k\cos(kx)+b_k\sin(kx)),$$

then we also know that $a_k=(f\vert u_{2k-1})$ and $b_k=(f\vert u_{2k})$, where $(\cdot\vert\cdot)$ stands for $L^2$-inner product, where $u_0=\frac{1}{2\sqrt{\pi}}$, $u_{2k-1}=\frac{cos(kx)}{\sqrt{\pi}}$, and $u_{2k}=\frac{\sin(kx)}{\sqrt{\pi}}$.

Then we can write

$$f(x)\sim (f\vert u_0)u_0 +\sum_{k=1}^\infty ((f\vert u_{2k-1}) u_{2k-1} +(f\vert u_{2k})u_{2k})$$ and so we have

$$f(x)\sim\sum_{k=0}^\infty (f\vert u_{k})u_k. $$

Then, my question is, to pass from the second to the third formula for the Fourier Series of $f$, don't one need to know something about absolute convergence of the odd and even series in $u_k$? To me it looks we are summing two infinite series but for this to be done we need to know about absolute convergence. What am I missing?

Thanks for your help!

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    $\begingroup$ In writing $$\sum_{k=1}^{+\infty}\left(a_{2k-1}+a_{2k}\right) = \sum_{k\geq 1}a_k $$ you are not rearranging anything. $\endgroup$ – Jack D'Aurizio Jun 20 '17 at 15:17
  • $\begingroup$ @JackD'Aurizio The OP might be questioning the "grouping" of terms, not the rearrangement. $\endgroup$ – Mark Viola Jun 20 '17 at 15:25

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