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Given $f$ be a $2\pi$-periodic complex-valued function which is integrable on $[−\pi, \pi]$. Write $$f(x) \sim \sum_{n=-\infty}^{\infty}c_ne^{inx}$$ and $$\overline {f(x)} \sim \sum_{n=-\infty}^{\infty}d_ne^{inx}$$

But even if I can prove $f$ and $\bar f$ have the same fourier series, what I can conclude is that they are equal almost everywhere except on a null set. $f$ may still be complex-valued on a null set.

How can we actually prove this?

mathshungry
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1 Answers1

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Suppose $c_{-n} = \overline{c_n}$. Then \begin{align*} \sum_{n = -\infty}^\infty c_n \mathrm{e}^{\mathrm{i} n x} &= c_0 \mathrm{e}^{\mathrm{i} 0 x} + \sum_{n = 1}^\infty c_n \mathrm{e}^{\mathrm{i} n x} + c_{-n} \mathrm{e}^{-\mathrm{i} n x} \\ &= c_0 + \sum_{n = 1}^\infty c_n \mathrm{e}^{\mathrm{i} n x} + \overline{c_{n}} \mathrm{e}^{-\mathrm{i} n x} \\ &= c_0 + \sum_{n = 1}^\infty c_n \mathrm{e}^{\mathrm{i} n x} + \overline{c_{n} \mathrm{e}^{\mathrm{i} n x}} \text{.} \end{align*}

First, $c_0 = \overline{c_0}$, so $c_0 \in \mathbb{R}$. Then, the summand is $z_n + \overline{z_n} \in \mathbb{R}$ for each $n$.

Converse: read bottom to top.

Technical detail: To justify the reorganization of the sum in the first line, sneak up on the summations through the sequence of trigonometric polynomials $\left(\sum_{n=-N}^{N} c_n \mathrm{e}^{\mathrm{i} nx}\right)_{N=0}^\infty$. The trigonometric polynomials are dense in the continuous functions on the circle (under the uniform norm), which are dense in the integrable functions on the circle. Moving diagonally through the resulting sequence of sequences, we get a sequence of trigonometric polynomials approaching $f(x)$.

roi_saumon
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Eric Towers
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  • I understand we can get a sequence of trigonometric polynomials approaching $f$ but I don't understand how you use this to reorganize the terms of the sum. – roi_saumon Aug 12 '20 at 13:17
  • @roi_saumon : Since the sequence of partial sums on the left of the first equality is unlikely to match the sequence of partial sums on the right, we need something to show that the limits of the sequences of partial sums agree. Contrast with $\int_{-1}^1 1/x ,\mathrm{d}x$ diverges, but $\int_0^1 (1/x +1/(-x)) ,\mathrm{d}x = 0$. We are always permitted to reorganize finitely many terms or we find some other criterion that allows arbitrary reorganization (such as absolute convergence). Any polynomial has finitely many terms, so we can reorder to put the complex conjugates adjacent. – Eric Towers Aug 12 '20 at 17:52