I am reading Rudin's "Principles of Mathematical Analysis" and I am stuck at the following theorem:
Theorem 8.14:
If, for some $x$, there are constants $\delta > 0$ and $M < \infty$ such that: $\left| f(x + t) - f(x) \right| \leq M|t|$ for all $t \in (-\delta, \delta)$, then $\lim_{N\rightarrow\infty} s_N(f; x) = f(x)$.
A few Definitions that needed:
Dirichlet kernel: $$D_N(x) = \sum_{n=-N}^Ne^{inx} = \frac{\sin\left( (N+\frac12)x \right)}{\sin(x/2)} \qquad (77)$$ and $$s_N(f; x) = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x - t) D_N(t) dt\qquad (78)$$
The proof goes as follows: Define $g(t) = \frac{f(x-t) - f(x)}{\sin(t/2)}$ for $0 < |t| \leq \pi$, and put $g(0) = 0)$. By the definition (77), $\frac{1}{2\pi} \int_{-\pi}^{\pi}D_N(x) dx = 1$. Hence (78) shows that: $$ \begin{aligned} s_N(f; x) - f(x) &= \frac{1}{2\pi} \int_{-\pi}^{\pi} g(t) \sin((N +\frac12)t)\\ & = \frac{1}{2\pi} \int_{-\pi}^{\pi} \left[ g(t)\cos\frac{t}2 \right]\sin(Nt) \space dt + \frac{1}{2\pi} \int_{-\pi}^{\pi} \left[ g(t)\sin\frac{t}2 \right]\cos(Nt) \space dt. \end{aligned} $$
$g(t)\cos(t/2)$ and $g(t)\sin(t/2)$ are bounded. The last two integrals thus tend to $0$ as $N \rightarrow \infty$.
Can anyone explain why they tend to $0$?
Since $f$ is bounded, for all $t\in (-\pi,\pi)$, there exists $M_1<\infty$ such that $|f(x-t)-f(x)|<M_1|t|<M_1\pi$, that is, $|g(t)\;\sin(t/2)|<M_1\pi$.
That is $g(t)\;\sin(t/2)$ is bounded.
Analogously there is $M_2<\infty$ such that $|g(t)\;\cos(t/2)|<M_2\pi$.
That is, $g(t)\;\cos(t/2)$ is bounded.
By Equation (74),$S_N(f;x)$ converges as $N\to\infty$.
Thus, the integrals on the right exist as $N\to\infty$.
Then, choosing $M=max(M_1,M_2)$, we get $\int_{-\pi}^\pi{[g(t) \;\cos(t/2)]\;\sin (Nt) dt}$
$\le M\delta\int_{-\pi}^\pi{\sin (Nt) dt}=0$ as $N\to\infty$.
Analogously we can show that $\int_{-\pi}^\pi{[g(t)\;\sin(t/2)] \cos (Nt) dt}=0$ as $N\to\infty$.