[proofplan]
We prove both parts of the exponential convergence theorem for Fourier approximation. For part 1, we establish the exponential decay of Fourier coefficients by shifting the contour of integration into the complex strip: the integral defining $\widehat{f}_n$ is rewritten along a horizontal line at height $\pm a'$ using Cauchy's theorem and 2-periodicity, then bounded using $|f(z)| \leq M$ on the strip. Letting $a' \to a$ gives $|\widehat{f}_n| \leq Mc^{|n|}$. For part 2, we bound the tail of the Fourier series $f(x) - \phi_N(x) = \sum_{|n| > N/2} \widehat{f}_n e^{i\pi nx}$ term by term using part 1, and sum the resulting geometric series.
[/proofplan]
[step:Shift the contour of the Fourier coefficient integral into the complex strip]
Fix $n \in \mathbb{Z}$ and $0 < a' < a$. The Fourier coefficient is defined by
\begin{align*}
\widehat{f}_n = \frac{1}{2}\int_{-1}^{1} f(t) \, e^{-i\pi n t} \, dt.
\end{align*}
Define the auxiliary function
\begin{align*}
F: \{z \in \mathbb{C} : -a < \operatorname{Im} z < a\} &\to \mathbb{C}, \quad z \mapsto f(z) \, e^{-i\pi n z}.
\end{align*}
Since $f$ is analytic on this strip and the exponential is entire, $F$ is analytic on the strip. Consider the positively oriented rectangle $\mathcal{R}$ with vertices $-1, 1, 1 + i\sigma a', -1 + i\sigma a'$, where $\sigma = \operatorname{sgn}(n)$ for $n \neq 0$ (and $\sigma = 1$ for $n = 0$, though the bound is immediate in that case). By Cauchy's theorem applied to $F$ on $\mathcal{R}$:
\begin{align*}
\oint_{\mathcal{R}} F(z) \, dz = 0.
\end{align*}
The two vertical sides of the rectangle contribute integrals at $z = 1 + iy$ and $z = -1 + iy$ for $y \in [0, \sigma a']$. Since $f$ is 2-periodic (so $f(1 + iy) = f(-1 + iy)$) and $e^{-i\pi n(1+iy)} = e^{-i\pi n} \cdot e^{-i\pi n(-1+iy)} \cdot e^{-i\pi n \cdot 2} = e^{-i\pi n(1+iy)}$, while $e^{-i\pi n(-1+iy)} = e^{i\pi n} e^{-i\pi n iy}$ — in fact, $e^{-i\pi n \cdot 1} = (-1)^n$ and $e^{-i\pi n \cdot (-1)} = (-1)^n$, so $F(1 + iy) = f(1+iy) e^{-i\pi n} e^{\pi ny}$ and $F(-1+iy) = f(-1+iy) e^{i\pi n} e^{\pi ny}$. By 2-periodicity of $f$, the integrands on the two vertical sides cancel (they are traversed in opposite directions with equal integrands). Therefore
\begin{align*}
\int_{-1}^{1} f(t) e^{-i\pi nt} \, dt = \int_{-1}^{1} f(t + i\sigma a') \, e^{-i\pi n(t + i\sigma a')} \, dt.
\end{align*}
[guided]
The strategy is a standard contour shift: we move the horizontal contour from the real axis to height $\sigma a'$ in the direction that makes the exponential $e^{-i\pi n z}$ decay. At $z = t + i\sigma a'$, we have $|e^{-i\pi n(t + i\sigma a')}| = e^{\pi n \sigma a'}$. By choosing $\sigma = \operatorname{sgn}(n)$, we get $n\sigma = |n|$, so the modulus is $e^{-\pi |n| a'}$ — this is where the exponential decay comes from.
The vertical sides cancel because of the 2-periodicity of $f$. Explicitly, $F(z)$ at $z = 1 + iy$ and $z = -1 + iy$ differ by the factor $e^{-i\pi n \cdot 2} = 1$ (since $n$ is an integer), so $F(1+iy) = f(1+iy)e^{-i\pi n(1+iy)}$ and $F(-1+iy) = f(-1+iy)e^{-i\pi n(-1+iy)} = f(1+iy)e^{i\pi n}e^{-i\pi n iy}$. The vertical integrals are traversed in opposite directions and have the same integrand (after accounting for the $e^{-i\pi n \cdot 2}$ factor), so they cancel. Thus
\begin{align*}
\int_{-1}^{1} f(t) e^{-i\pi nt} \, dt = \int_{-1}^{1} f(t + i\sigma a') \, e^{-i\pi n(t + i\sigma a')} \, dt.
\end{align*}
[/guided]
[/step]
[step:Bound the shifted integral to obtain $|\widehat{f}_n| \leq M c^{|n|}$]
Taking absolute values in the identity from the previous step:
\begin{align*}
|\widehat{f}_n| = \frac{1}{2}\left|\int_{-1}^{1} f(t + i\sigma a') \, e^{-i\pi n(t + i\sigma a')} \, dt\right| \leq \frac{1}{2}\int_{-1}^{1} |f(t + i\sigma a')| \cdot |e^{-i\pi n(t + i\sigma a')}| \, dt.
\end{align*}
Since $t + i\sigma a'$ lies in the strip $\{z : |\operatorname{Im} z| < a\}$ (because $|a'| < a$), the hypothesis $|f(z)| \leq M$ applies. For the exponential factor:
\begin{align*}
|e^{-i\pi n(t + i\sigma a')}| = |e^{-i\pi nt}| \cdot |e^{\pi n \sigma a'}| = e^{-\pi |n| a'},
\end{align*}
where we used $n\sigma = |n|$ (by the choice $\sigma = \operatorname{sgn}(n)$). Therefore
\begin{align*}
|\widehat{f}_n| \leq \frac{1}{2} \cdot M \cdot e^{-\pi |n| a'} \cdot 2 = M \, e^{-\pi a' |n|}.
\end{align*}
Since this holds for every $0 < a' < a$, taking $a' \to a$ yields
\begin{align*}
|\widehat{f}_n| \leq M \, e^{-\pi a |n|} = M \, c^{|n|},
\end{align*}
where $c = e^{-a\pi}$. This proves part 1.
[guided]
The bound is obtained in two stages. First, on the shifted contour at height $\sigma a'$, the analyticity bound $|f| \leq M$ controls the function, while the exponential $|e^{-i\pi nz}|$ provides the decay factor $e^{-\pi |n| a'}$. The integral over $t \in [-1, 1]$ contributes a factor of 2 (the length of the interval), which cancels with the $\frac{1}{2}$ in the definition of $\widehat{f}_n$.
Second, we take $a' \to a$ to get the sharpest bound. Strictly speaking, the inequality $|\widehat{f}_n| \leq Me^{-\pi a'|n|}$ holds for every $a' < a$, and since the left side does not depend on $a'$, the bound persists in the limit $a' \to a$. The resulting constant $c = e^{-a\pi} \in (0,1)$ (since $a > 0$) governs the exponential decay rate.
[/guided]
[/step]
[step:Bound the approximation error by summing the tail of the Fourier series]
The approximation error at any $x \in [-1, 1]$ is
\begin{align*}
f(x) - \phi_N(x) = \sum_{\substack{n \in \mathbb{Z} \\ |n| > N/2}} \widehat{f}_n \, e^{i\pi n x}.
\end{align*}
Taking absolute values and applying the triangle inequality, then using $|e^{i\pi nx}| = 1$ and the coefficient bound from part 1:
\begin{align*}
|f(x) - \phi_N(x)| \leq \sum_{|n| > N/2} |\widehat{f}_n| \leq \sum_{|n| > N/2} M \, c^{|n|} = 2M \sum_{k = N/2 + 1}^{\infty} c^k,
\end{align*}
where we grouped positive and negative indices by writing $k = |n|$. This is a geometric series with ratio $c \in (0, 1)$:
\begin{align*}
2M \sum_{k=N/2+1}^{\infty} c^k = 2M \cdot \frac{c^{N/2+1}}{1-c} = \frac{2Mc}{1-c} \, c^{N/2}.
\end{align*}
Since this bound is independent of $x$, we conclude
\begin{align*}
\max_{x \in [-1,1]} |f(x) - \phi_N(x)| \leq \frac{2Mc}{1-c} \, c^{N/2}.
\end{align*}
This proves part 2 and completes the proof.
[guided]
The error $f(x) - \phi_N(x)$ consists of all Fourier modes with $|n| > N/2$ — these are the modes that the partial sum $\phi_N$ truncates. Each such mode has coefficient bounded by $Mc^{|n|}$ from part 1, and contributes $|\widehat{f}_n e^{i\pi nx}| = |\widehat{f}_n|$ since the exponentials are unimodular.
Summing over $n$ with $|n| > N/2$: by symmetry between positive and negative $n$, the sum equals $2M\sum_{k=N/2+1}^{\infty} c^k$. Evaluating the geometric series with first term $c^{N/2+1}$ and ratio $c$:
\begin{align*}
2M \sum_{k=N/2+1}^{\infty} c^k = 2M \cdot \frac{c^{N/2+1}}{1-c} = \frac{2Mc^{N/2+1}}{1-c} = \frac{2Mc}{1-c} \cdot c^{N/2}.
\end{align*}
The bound is uniform in $x$ because only $|\widehat{f}_n|$ appears (not $x$), so the maximum over $[-1,1]$ satisfies the same inequality. The rate $c^{N/2}$ is exponential in $N$, confirming spectral-speed convergence.
[/guided]
[/step]
