[proofplan]
We verify the four closure properties of the class $\mathcal{A}$ of 2-periodic analytic functions. Linearity and scalar multiplication are immediate from the definition of Fourier coefficients. The product formula follows from the Cauchy product of absolutely convergent series together with the fact that the product of two analytic functions is analytic on the intersection of their strips. The differentiation formula is the most substantive: we justify termwise differentiation of the Fourier series by showing that the differentiated series converges uniformly on $\mathbb{R}$, which holds because the exponential decay $|\widehat{f}_n| \leq M c^{|n|}$ absorbs the linear factor $|n|$ from differentiation.
[/proofplan]
[step:Verify linearity: sums and scalar multiples preserve $\mathcal{A}$]
Let $f, g \in \mathcal{A}$ with respective strip widths $a_f, a_g > 0$, and let $\alpha \in \mathbb{C}$. Both $f + g$ and $\alpha f$ are 2-periodic (since sums and scalar multiples of 2-periodic functions are 2-periodic) and analytic on the strip $\{z : |\operatorname{Im} z| < \min(a_f, a_g)\}$ (respectively $\{z : |\operatorname{Im} z| < a_f\}$), since sums and scalar multiples of analytic functions are analytic. Hence $f + g, \alpha f \in \mathcal{A}$.
For the Fourier coefficients, linearity of the integral gives
\begin{align*}
\widehat{(f+g)}_n = \frac{1}{2}\int_{-1}^{1} (f(t) + g(t)) e^{-i\pi nt} \, dt = \widehat{f}_n + \widehat{g}_n,
\end{align*}
and $\widehat{(\alpha f)}_n = \alpha \widehat{f}_n$. The stated series representations follow by substituting these coefficients into the Fourier expansion.
[/step]
[step:Derive the product formula via the Cauchy product of absolutely convergent series]
Let $f, g \in \mathcal{A}$. The product $fg$ is 2-periodic and analytic on the strip $\{z : |\operatorname{Im} z| < \min(a_f, a_g)\}$, so $fg \in \mathcal{A}$. We compute the Fourier coefficients of $fg$ by multiplying the two absolutely convergent Fourier series:
\begin{align*}
f(x) g(x) = \left(\sum_{m=-\infty}^{\infty} \widehat{f}_m \, e^{i\pi mx}\right)\left(\sum_{k=-\infty}^{\infty} \widehat{g}_k \, e^{i\pi kx}\right).
\end{align*}
Since $f, g \in \mathcal{A}$, their Fourier coefficients decay exponentially: $|\widehat{f}_m| \leq M_f c_f^{|m|}$ and $|\widehat{g}_k| \leq M_g c_g^{|k|}$ for constants $M_f, M_g > 0$ and $c_f, c_g \in (0,1)$. In particular, $\sum_m |\widehat{f}_m| < \infty$ and $\sum_k |\widehat{g}_k| < \infty$, so both series converge absolutely. By the Cauchy product formula for absolutely convergent series, we may multiply term by term and rearrange:
\begin{align*}
f(x)g(x) = \sum_{n=-\infty}^{\infty} \left(\sum_{m=-\infty}^{\infty} \widehat{f}_{n-m} \, \widehat{g}_m\right) e^{i\pi nx},
\end{align*}
where we collected all pairs $(m, k)$ with $m + k = n$ by setting $k = n - m$. The inner sum is the discrete convolution $(\widehat{f} * \widehat{g})_n$, so $\widehat{(fg)}_n = (\widehat{f} * \widehat{g})_n$.
[guided]
Why does the rearrangement of the double series work? We need to reorder $\sum_{m,k} \widehat{f}_m \widehat{g}_k e^{i\pi(m+k)x}$ by grouping all pairs with $m + k = n$. This rearrangement is justified because the double series converges absolutely:
\begin{align*}
\sum_{m,k} |\widehat{f}_m| \, |\widehat{g}_k| = \left(\sum_m |\widehat{f}_m|\right)\left(\sum_k |\widehat{g}_k|\right) < \infty,
\end{align*}
since both factor sums are finite (exponential decay implies summability). By Fubini's theorem for absolutely convergent series (or equivalently, by the Cauchy product theorem), the rearrangement is valid and yields the convolution formula.
[/guided]
[/step]
[step:Justify termwise differentiation of the Fourier series to obtain the derivative formula]
Let $f \in \mathcal{A}$ be analytic on the strip $\{z : |\operatorname{Im} z| < a\}$ with $|f(z)| \leq M$ there. Since $f$ is analytic (hence smooth) and 2-periodic, $f'$ is also 2-periodic and analytic on the same strip, so $f' \in \mathcal{A}$. It remains to show that the Fourier series of $f'$ is obtained by termwise differentiation.
We claim that the formally differentiated series
\begin{align*}
\sum_{n=-\infty}^{\infty} (i\pi n) \widehat{f}_n \, e^{i\pi nx}
\end{align*}
converges uniformly on $\mathbb{R}$ to $f'(x)$. By the [Exponential Convergence of Fourier Approximation](/theorems/1379) (part 1), $|\widehat{f}_n| \leq Mc^{|n|}$ where $c = e^{-a\pi} \in (0,1)$. The terms of the differentiated series satisfy
\begin{align*}
|(i\pi n) \widehat{f}_n \, e^{i\pi nx}| = \pi |n| \, |\widehat{f}_n| \leq \pi M |n| \, c^{|n|}.
\end{align*}
Since $c \in (0, 1)$, the series $\sum_{n=-\infty}^{\infty} |n| c^{|n|}$ converges (the exponential decay $c^{|n|}$ dominates the linear growth $|n|$, as verified by the ratio test: $\frac{(|n|+1)c^{|n|+1}}{|n|c^{|n|}} = c \cdot \frac{|n|+1}{|n|} \to c < 1$). Therefore, by the Weierstrass M-test, the differentiated series converges uniformly on $\mathbb{R}$.
Since the original Fourier series $\sum_{n} \widehat{f}_n e^{i\pi nx}$ converges to $f(x)$ (pointwise, and in fact uniformly by the same Weierstrass argument with $Mc^{|n|}$) and the termwise derivative converges uniformly, the classical theorem on termwise differentiation of uniformly convergent series guarantees that $f$ is differentiable and
\begin{align*}
f'(x) = \sum_{n=-\infty}^{\infty} \frac{d}{dx}\left(\widehat{f}_n \, e^{i\pi nx}\right) = i\pi \sum_{n=-\infty}^{\infty} n \, \widehat{f}_n \, e^{i\pi nx}.
\end{align*}
[guided]
The key question is: why can we differentiate the Fourier series term by term? In general, differentiating a convergent series term by term is not valid — the derivative of each term may grow, destroying convergence. What saves us here is the exponential decay of the Fourier coefficients.
Differentiating $\widehat{f}_n e^{i\pi nx}$ introduces a factor of $i\pi n$, so the $n$th term of the differentiated series has size $\pi|n| \cdot |\widehat{f}_n| \leq \pi M |n| c^{|n|}$. The factor $|n|$ is a polynomial in $|n|$, while $c^{|n|}$ is exponential decay. For any $c \in (0,1)$, exponential decay beats polynomial growth: $|n| c^{|n|} \to 0$ as $|n| \to \infty$, and the series $\sum_n |n| c^{|n|}$ converges (by the ratio test: consecutive term ratio tends to $c < 1$).
The Weierstrass M-test then gives uniform convergence of the differentiated series. Combined with uniform convergence of the original series, the standard theorem on termwise differentiation applies: if $\sum g_n$ converges at some point and $\sum g_n'$ converges uniformly, then $(\sum g_n)' = \sum g_n'$.
This is precisely the mechanism the course notes describe: "the coefficients $\widehat{f}_n$ decay exponentially — multiplying by $n$ does not destroy absolute convergence." The result would fail for functions with only algebraic decay $|\widehat{f}_n| = O(|n|^{-k})$, where differentiation loses one power and could destroy convergence.
[/guided]
[/step]
