Suppose that $f$ is a $2\pi$-periodic function that satisfies the estimate $$|f(x)-f(y)|\leq M|x-y|^\alpha$$ for an $0<\alpha<1,$ and let $$\hat{f}(n)=\frac{1}{2\pi}\int_0^{2\pi}f(x)e^{-inx}dx.$$ Show that $$S_N(x)=\sum_{n=-N}^N\hat{f}(n)e^{inx}$$ converges uniformly to $f(x)$ for all real $x.$
Hint: $$S_N(x)=\int_0^{2\pi}f(y)\frac{\sin\left((N+\frac{1}{2})(x-y)\right)}{2\pi\sin\left(\frac{1}{2}(x-y)\right)}dy$$
My attempt:
After substituting $\hat{f}(n)$ into $S_N(x)$ and exchanging the integral and summation we get an expression similar to that in the hint and $$\frac{\sin\left((N+\frac{1}{2})(x-y)\right)}{\sin\left(\frac{1}{2}(x-y)\right)}=1+\cos t+\cos 2t+\ldots+\cos Nt$$ but what next?
The standard trick is to notice that $$ \int_{0}^{2\pi}\frac{\sin\left((n+\frac{1}{2})(x-y)\right)}{2\pi\sin\left(\frac{1}{2}(x-y)\right)}\,dy = 1. $$ That allows you to write $$ S_{N}(x)-f(x) = \int_{0}^{2\pi}(f(y)-f(x))\frac{\sin\left((n+\frac{1}{2})(x-y)\right)}{2\pi\sin\left(\frac{1}{2}(x-y)\right)}\,dy $$ Given $\epsilon > 0$, you can find $\delta > 0$ such that $$ \int_{-\delta}^{\delta}\frac{|t|^{\alpha}}{2\pi\left|\sin\left(\frac{1}{2}t\right)\right|}\,dt < \frac{\epsilon}{2}. $$ That should get your started. The remaining part has to do with the uniform convergence of $$ \lim_{n\rightarrow\infty} \int_{0}^{2\pi}h(t)\sin\left(\left(n+\frac{1}{2}\right)t\right)\,dt=0. $$