Fourier transform property(uniformly converges) proof Suppose that f is a 2π-periodic function that satisfies the estimate
\begin{equation}
|f(x)-f(y)|\leqslant M|x-y|^\alpha
\end{equation}
for an 0< $\alpha$ <1
Show that $S_n(x)$ converges uniformly to f (x) for all real x.
\begin{equation}
S_n(x)=\int_0^{2\pi} f(y)\frac{sin((N+\frac{1}{2})(x-y))}{2\pi ~sin(\frac{1}{2}(x-y))} dy.
\end{equation}
My confuse is, $S_n(x)$ should $\longrightarrow$ 0 when n $\longrightarrow$ $\infty$ by using Riemann–Lebesgue lemma. So what mistake that I made?

$$
    f(x)-S_n(x) = \frac{1}{2\pi}\int_{x-\pi}^{x+\pi}\left[\frac{f(x)-f(y)}{\sin{\frac{1}{2}(x-y)}}\right]\sin((n+1/2)(x-y))dy
$$
Because $|f(x)-f(y)| \le M|x-y|^{\alpha}$, the expression enclosed by square brackets is absolutely integrable. So $S_n(x)\rightarrow f(x)$ for all $x$ by the Riemann-Lebesgue Lemma. The uniformity of the Lipshitz condition leads to uniform convergence.
To show how the above gives uniform convergence of $S_n$ to $f$, let $\epsilon > 0$ be given. Then, for small enough $\delta$, the following estimate holds independent of $x$:
$$
    \left|\frac{1}{2\pi}\int_{x-\delta}^{x+\delta}\frac{f(x)-f(y)}{\sin(\frac{1}{2}(x-y))}\sin((n+1/2)(x-y))dy\right| \\
   \le \frac{1}{2\pi}\int_{x-\delta}^{x+\delta}M\frac{|x-y|^{\alpha}}{|\sin\frac{1}{2}(x-y)|}dy \\
   \le \frac{2M}{2\pi}\int_{x-\delta}^{x+\delta}|x-y|^{\alpha-1}dy \\
    = \frac{4M}{2\pi}\int_{0}^{\delta}t^{\alpha-1}dt = \frac{4M}{2\pi\alpha}\delta
$$
So, by choosing $\delta$ small enough, the above is bounded by $\frac{\epsilon}{2}$ for all $x$. To bound the remaining piece by $\epsilon/2$ for large $n$, choose a function $g$ that is $2\pi$ periodic, continuously differentiable and satisfies
$$
       \frac{1}{2\pi\sin(\delta/2)}\int_{\delta \le |y-x| \le \pi}|f(y)-g(y)|dy < \frac{\epsilon}{6} \tag{$\dagger$}
$$
Then
\begin{align}
&\left|\frac{1}{2\pi}\int_{\delta \le |x-y| \le \pi}\left[\frac{f(x)-f(y)}{\sin\frac{1}{2}(x-y)}\right]\sin((n+1/2)(x-y))dy\right| \\
  & \le \left|f(x)\frac{1}{2\pi}\int_{\delta\le|u| \le \pi}\frac{\sin((n+1/2)u}{\sin\frac{1}{2}u}du\right| \\
  & + \left|\frac{1}{2\pi}\int_{\delta \le |x-y|\le \pi}(g(y)-f(y))\frac{\sin((n+1/2)(x-y))}{\sin\frac{1}{2}(x-y)}dy\right| \\
  & + \left|\frac{1}{2\pi}\int_{\delta \le |x-y| \le \pi}\frac{g(y)}{\sin\frac{1}{2}(x-y)}\sin((n+1/2)(x-y)dy\right|
\end{align}
The function $f$ is uniformly bounded because of the Lipschitz condition. Therefore, by the Riemann-Lebesgue lemma, there exists $N_1$ large enough that the first term on the right is bounded uniformly in $x$ by $\frac{\epsilon}{6}$ for $n \ge N_1$. The second term is automatically bounded by $\epsilon/6$ because of $(\dagger)$. The third term can be uniformly bounded for large $N$ by $\epsilon/6$ after a single integration by parts, which is permitted because $g$ is a smooth approximating function:
$$
   \frac{1}{2\pi}\int_{\delta \le |x-y| \le \pi}\frac{g(y)}{\sin\frac{1}{2}(x-y)}\sin((n+1/2)(x-y)dy. \tag{$\dagger\dagger$}
$$
(Integrating the $\sin$ term gives $1/(n+1/2)$, leading to a uniform bound on the above for large enough $n$.) For example, the integral over $\delta < |x-y| \le \pi$ is over $y \in [x+\delta,x+\pi]$ and $y\in [x-\pi,x-\delta]$. The part of the integral over $[x+\delta,x+\pi]$ becomes
\begin{align}
   &\int_{x+\delta}^{x+\pi}\frac{g(y)}{\sin\frac{1}{2}(x-y)}\sin((n+1/2)(x-y))dy\\
  = &-\left.\frac{g(y)}{\sin\frac{1}{2}(x-y)}\frac{\cos((n+1/2)(x-y))}{(n+1/2))}\right|_{y=x+\delta}^{x+\pi} \\
  +&\frac{1}{n+1/2}\int_{x+\delta}^{x+\pi}\frac{\partial}{\partial y}\left(\frac{g(y)}{\sin\frac{1}{2}(x-y)}\right)\cos((n+1/2)(x-y))dy
\end{align}
Now you can see that $(\dagger\dagger)$ tends to $0$ as $n\rightarrow\infty$. The evaluation terms tend to $0$ because of the presence of $1/(n+1/2)$; the integral term on the right tends to $0$ even without the $1/(n+1/2)$. And the convergence is uniform in $x$. Using a smooth $g$ to approximate $f$ in the integral sense simplifies the argument.
