[proofplan]
We prove pointwise convergence of the Fourier partial sums $\phi_N(x) \to f(x)$ for a 2-periodic Lipschitz continuous function by the classical Dirichlet kernel argument. We express $\phi_N(x) - f(x)$ as an integral against the Dirichlet kernel, use the 2-periodicity and the reproducing property of the kernel to rewrite the integrand in terms of the difference $f(x - t) - f(x)$, and then show this integral vanishes as $N \to \infty$ by combining the Lipschitz condition (which controls the integrand near $t = 0$) with the Riemann-Lebesgue lemma (which kills the oscillatory contributions away from $t = 0$).
[/proofplan]
[step:Express $\phi_N(x) - f(x)$ as a convolution with the Dirichlet kernel]
The Fourier partial sum is $\phi_N(x) = \sum_{n=-N/2}^{N/2} \widehat{f}_n e^{i\pi n x}$. Substituting the definition $\widehat{f}_n = \frac{1}{2}\int_{-1}^{1} f(t) e^{-i\pi n t} \, dt$ and exchanging the finite sum with the integral:
\begin{align*}
\phi_N(x) = \frac{1}{2}\int_{-1}^{1} f(t) \sum_{n=-N/2}^{N/2} e^{i\pi n(x-t)} \, dt = \frac{1}{2}\int_{-1}^{1} f(t) \, D_{N/2}(x - t) \, dt,
\end{align*}
where $D_M(s) := \sum_{n=-M}^{M} e^{i\pi n s}$ is the Dirichlet kernel of order $M = N/2$. Since $D_M$ integrates to $\frac{1}{2}\int_{-1}^{1} D_M(s) \, ds = 1$ (all terms except $n = 0$ integrate to zero by orthogonality), we write
\begin{align*}
\phi_N(x) - f(x) = \frac{1}{2}\int_{-1}^{1} \bigl[f(t) - f(x)\bigr] \, D_{N/2}(x - t) \, dt.
\end{align*}
Substituting $t = x - s$ and using 2-periodicity to keep the domain as $[-1, 1]$:
\begin{align*}
\phi_N(x) - f(x) = \frac{1}{2}\int_{-1}^{1} \bigl[f(x - s) - f(x)\bigr] \, D_{N/2}(s) \, ds.
\end{align*}
[guided]
The goal is to express the error $\phi_N(x) - f(x)$ in a form amenable to asymptotic analysis. We start from the definition of $\phi_N$ and insert the integral formula for each Fourier coefficient $\widehat{f}_n$. Since the sum over $n$ is finite (from $-N/2$ to $N/2$), exchanging the sum and integral is justified without any convergence hypotheses — this is simply linearity of a finite sum.
The resulting kernel $D_M(s) = \sum_{n=-M}^{M} e^{i\pi n s}$ is the Dirichlet kernel. Its key property for us is the reproducing identity: $\frac{1}{2}\int_{-1}^{1} D_M(s) \, ds = 1$, which follows because $\frac{1}{2}\int_{-1}^{1} e^{i\pi n s} \, ds = 0$ for $n \neq 0$ and equals $1$ for $n = 0$. Using this reproducing property, we write $f(x) = f(x) \cdot 1 = \frac{1}{2}\int_{-1}^{1} f(x) \, D_{N/2}(x-t) \, dt$ and subtract:
\begin{align*}
\phi_N(x) - f(x) = \frac{1}{2}\int_{-1}^{1} \bigl[f(t) - f(x)\bigr] \, D_{N/2}(x - t) \, dt.
\end{align*}
The substitution $s = x - t$ (so $t = x - s$, $dt = -ds$) reverses the limits, and 2-periodicity of both $f$ and $D_M$ allows us to re-centre the integration domain back to $[-1, 1]$:
\begin{align*}
\phi_N(x) - f(x) = \frac{1}{2}\int_{-1}^{1} \bigl[f(x - s) - f(x)\bigr] \, D_{N/2}(s) \, ds.
\end{align*}
This form is what we need: the integrand is the product of a difference quotient (controlled by Lipschitz continuity near $s = 0$) and a highly oscillatory kernel (controlled by the Riemann-Lebesgue lemma as $N \to \infty$).
[/guided]
[/step]
[step:Rewrite the Dirichlet kernel in closed form using the geometric series]
The Dirichlet kernel $D_M(s) = \sum_{n=-M}^{M} e^{i\pi n s}$ is a finite geometric series with ratio $e^{i\pi s}$. Summing:
\begin{align*}
D_M(s) = \frac{\sin\bigl((M + \tfrac{1}{2})\pi s\bigr)}{\sin(\tfrac{\pi s}{2})}
\end{align*}
for $s \neq 0$ (with $D_M(0) = 2M + 1$ by continuity). This closed form reveals that $D_M(s)$ oscillates at frequency proportional to $M$ with an envelope decaying like $1/\sin(\pi s/2)$.
[guided]
To derive the closed form, factor out $e^{-i\pi M s}$ from the geometric sum:
\begin{align*}
D_M(s) = e^{-i\pi M s} \sum_{n=0}^{2M} e^{i\pi n s} = e^{-i\pi M s} \cdot \frac{e^{i\pi(2M+1)s} - 1}{e^{i\pi s} - 1} = \frac{e^{i\pi(M+1)s} - e^{-i\pi M s}}{e^{i\pi s} - 1}.
\end{align*}
Multiplying numerator and denominator by $e^{-i\pi s/2}$ and using $e^{i\theta} - e^{-i\theta} = 2i\sin\theta$:
\begin{align*}
D_M(s) = \frac{2i\sin\bigl((M + \tfrac{1}{2})\pi s\bigr)}{2i\sin(\tfrac{\pi s}{2})} = \frac{\sin\bigl((M + \tfrac{1}{2})\pi s\bigr)}{\sin(\tfrac{\pi s}{2})}.
\end{align*}
This closed form is essential because it separates the high-frequency oscillation in the numerator (which drives the Riemann-Lebesgue argument) from the singular envelope $1/\sin(\pi s/2)$ (which must be controlled by the Lipschitz condition near $s = 0$).
[/guided]
[/step]
[step:Apply the Lipschitz condition and Riemann-Lebesgue lemma to show the integral vanishes]
Since $f$ is Lipschitz continuous with some constant $L > 0$, we have $|f(x - s) - f(x)| \leq L|s|$ for all $s$. Define
\begin{align*}
g: [-1, 1] &\to \mathbb{C}, \quad s \mapsto \frac{f(x-s) - f(x)}{\sin(\tfrac{\pi s}{2})}
\end{align*}
for $s \neq 0$, with $g(0)$ defined by the limit. The Lipschitz bound gives $|f(x-s) - f(x)| \leq L|s|$, and $|\sin(\tfrac{\pi s}{2})| \sim \tfrac{\pi}{2}|s|$ as $s \to 0$, so
\begin{align*}
|g(s)| \leq \frac{L|s|}{|\sin(\tfrac{\pi s}{2})|} \to \frac{2L}{\pi} \quad \text{as } s \to 0.
\end{align*}
Therefore $g$ is bounded on $[-1,1]$ and hence integrable. Using the closed-form Dirichlet kernel, the error becomes
\begin{align*}
\phi_N(x) - f(x) = \frac{1}{2}\int_{-1}^{1} g(s) \sin\bigl((M + \tfrac{1}{2})\pi s\bigr) \, ds.
\end{align*}
By the Riemann-Lebesgue lemma, the Fourier coefficients of any integrable function tend to zero. Since $g \in L^1([-1,1])$ and $\sin((M+\frac{1}{2})\pi s)$ oscillates with frequency $M + \frac{1}{2} \to \infty$, we conclude
\begin{align*}
\phi_N(x) - f(x) \to 0 \quad \text{as } N \to \infty.
\end{align*}
Since $x \in \mathbb{R}$ was arbitrary, this establishes pointwise convergence $\phi_N(x) \to f(x)$ for all $x$.
[guided]
The critical question is: why does the integral vanish as $N \to \infty$? The Dirichlet kernel oscillates faster and faster, but it also has a singularity at $s = 0$ — the envelope $1/\sin(\pi s/2)$ blows up. If the numerator $f(x-s) - f(x)$ did not vanish at $s = 0$, the singularity would prevent the Riemann-Lebesgue lemma from applying.
This is precisely where the Lipschitz condition is consumed. The bound $|f(x-s) - f(x)| \leq L|s|$ provides a zero of order 1 at $s = 0$, which cancels the order-1 pole of $1/\sin(\pi s/2)$. The quotient $g(s) = [f(x-s) - f(x)]/\sin(\pi s/2)$ is therefore bounded near $s=0$ (with limit $2L/\pi$) and bounded away from $s=0$ (since $\sin(\pi s/2)$ is bounded below on any set $\{|s| \geq \delta\}$). Hence $g \in L^1([-1,1])$.
Once integrability of $g$ is established, the Riemann-Lebesgue lemma applies directly: for any $g \in L^1([-1,1])$, $\int_{-1}^{1} g(s) e^{i\omega s} \, ds \to 0$ as $|\omega| \to \infty$. Writing $\sin((M+\frac{1}{2})\pi s) = \operatorname{Im}(e^{i(M+1/2)\pi s})$, the integral is the imaginary part of a Fourier coefficient of $g$ at frequency $(M+\frac{1}{2})\pi \to \infty$, and hence vanishes.
Note that mere continuity of $f$ would not suffice: a continuous function satisfies $f(x-s) - f(x) \to 0$ as $s \to 0$, but the rate might be slower than $|s|$, leaving $g$ unbounded and potentially non-integrable. The Lipschitz condition (or more generally, a Dini condition) is the precise regularity needed.
[/guided]
[/step]
