I am proving that $$ \lim_{t \to \infty} \left \lvert \mathbb{E}[e^{itX}] \right \rvert = 0$$ in the case in whic $X$ has a density function $f(x)$ (wrt the Lebesgue measure).
It seems to good to be true to proceed as follows...
$$ \lim_{t \to \infty} \int_{-\infty}^{\infty} f(x) e^{itx} \mathrm{d} x = \lim_{t \to \infty} \int_{-\infty}^{\infty} \frac{1}{t} f\left(\frac{y}{t}\right) e^{iy} \mathrm{d} x $$ thanks to the change of variable $y=tx$. $$ \lim_{t \to \infty} \int_{-\infty}^{\infty} \frac{1}{t} f\left(\frac{y}{t}\right) e^{iy} \mathrm{d} x = \int_{-\infty}^{\infty} \lim_{t \to \infty} \frac{1}{t} f\left(\frac{y}{t}\right) e^{iy} \mathrm{d} x $$ By dominated convergence ($f$ on any compact around 0 has a maximum).
And $ \lim_{t \to \infty} \frac{1}{t} f\left(\frac{y}{t}\right)$ goes to 0 as $t \to \infty$.
I guess that is it wrong since I am assuming continuity of $f$, right?