Proof that these are Fourier coefficients

I proved that for $f \in \ell^1 (\mathbb Z)$ its Gelfand transform $\widehat{f}$ is a map $\widehat{f}: S^1 \to S^1$ defined by $$\widehat{f}(z) = \sum_{n \in \mathbb Z}f(n) z^n$$

In Murphy's book it is stated that it is readily verified that the $f(n)$ are the Fourier coefficients of $\widehat{f}$. But what is there to show? Isn't it the definition of a fourier series of any function $g:S^1 \to S^1$ that $$g(x) = \sum_{n \in \mathbb N} c_n e^{inx}$$

for some coefficients $c_n$? What would I have to show if I want to verify that $f(n)$ are the Fourier coefficients?

• The (complex) Fourier series of $h$ is by definition the (complex) trigonometric series whose coefficients are $$c_n = \frac{1}{2\pi} \int_{-\pi}^\pi h(e^{it})\cdot e^{-int}\,dt.$$ Now the orthogonality relations and the uniform convergence of $\sum f(n)z^n$ readily show that indeed $c_n = f(n)$ for all $n$. – Daniel Fischer Apr 14 '14 at 18:28
• @DanielFischer But doesn't the Fourier series usually start at $0$ and sum over $\mathbb N$? But the sum here is over $\mathbb Z$, how can it still be a Fourier series? – Student Apr 15 '14 at 10:43
• That would be the real Fourier series, with sines and cosines, $$f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n\cos (nx) + b_n\sin(nx),$$ using the orthogonal system $\{ \cos (nx) : n \in \mathbb{N}\} \cup \{ \sin(nx) : n \in \mathbb{N}\setminus\{0\}\}$. The complex Fourier series uses the orthogonal system $\{ e^{inx} : n \in \mathbb{Z}\}$ to represent the functions, and thus the series extends over $\mathbb{Z}$. Remember $$\cos (nx) = \frac{1}{2}(e^{inx} + e^{-inx}),\quad \sin (nx) = \frac{1}{2i}(e^{inx} - e^{-inx}),$$ so we need both, exponentials with positive and negative $n$ to get $\cos$ – Daniel Fischer Apr 15 '14 at 10:51
• or $\sin$, generally most functions have nonzero coefficients for both, positive and negative indices. – Daniel Fischer Apr 15 '14 at 10:53
• Exactly. Of course you need to justify the change of order of summation and integration, but $\sum \lvert f(k)\rvert < \infty$ is more than enough for that. – Daniel Fischer Apr 15 '14 at 11:49