This post about characteristic functions asks about absolutely integrable $\phi(t)$. This answer mentions "we can find $R$ such that $\int_{\mathbb R\setminus[-R,R]}\lvert\phi(t)\rvert dt<2\pi\varepsilon$".
The argument I don't get is how $\frac 1{2\pi}\left\lvert\int_{-R}^R\phi(t)\left(e^{-itx}-e^{-ity} \right)dt\right\rvert \leqslant \frac{R^2}{\pi}\lvert x-y\rvert$; in particular, the $|x - y|$ bit. It seems to me that $|e ^ {-itx} - e ^ {-ity}|$ gets upper-bounded by $|-tx - (-t)y| = |t||x - y|$. Why is that?
Or, $|e ^ {-i\theta}| = 1$, so if you only deal with $\int_{-R}^{R} |\phi(t)||t| \mathrm{d}t$ (in this comment), where does the $|x - y|$ come from?
$|e^{-itx}-e^{-ity}|=|-it \int_x^{y} e^{-its}ds|\le |t||\int_x^{y} 1ds|=|t| |x-y|$.