Doubt in Rudin's Proof:

Question

Once I go through the proof of the below theorem, I could encounter that he used dominated convergence theorem to prove $(f)$, in that how they claim that $$\frac{e^{-ix(s-t)}-1}{s-t}\leq |x|$$ Kindly explain.

Answer 1

$$ \frac{e^{ixu}-1}{u}=\left.\frac{1}{u}e^{ixv}\right|_{v=0}^{u}=\frac{1}{u}\int_{0}^{u}ixe^{ixv}dv. \\ \left|\frac{e^{ixu}-1}{u}\right| \le \frac{1}{|u|}\int_{0}^{|u|}|x|du=|x|. $$

Answer 2

$\phi(x,u)=(e^{-ixu}-1)/u$ so that $$ |\phi(x,u)|=2\,\frac{|\sin(ux/2)|}{|u|}\le2\,\frac{\min(1,|\tfrac12xu|)}{|u|}=\min\left(\frac{2}{|u|},|x|\right) $$ which implies the inequality.

---

The critical step is $$e^{iy}-1=e^{iy/2}(e^{iy/2}-e^{-iy/2})=2ie^{iy/2}\sin(y/2)$$ and $|ie^{iy/2}|=1$