Difference of Characteristic Functions is Little $o$ Let $\mu, \lambda$ be two zero-mean probability measures with respective characteristic functions $\hat{\mu}, \hat{\lambda}$. How is $\vert \hat\mu(y)-\hat\lambda(y)\vert=o(\vert y \vert)$ as $y\to 0$?
Attempt: As $y\to 0$, $$\Bigg\vert \int e^{iy}d\mu(y)-\int e^{iy}d \lambda(y)\Bigg\vert \to \Bigg\vert \int d\mu (y) - \int d\lambda(y)\Bigg\vert$$
I can't see how to apply FTC here since the measures are defined on events/intervals and I am not sure about the values of $\mu, \lambda$ evaluated at $\lbrace \pm \infty\rbrace$. Further, I don't see how to use the fact that these measures have $0$ mean. How can the little-$o$ be proved?

Hints:


*

*It suffices to show that for any probability measure $\mu$ with mean $0$ it holds that $$|\hat{\mu}(y)-1| = o(|y|) \qquad \text{as} \, \, y \to 0. \tag{1}$$

*Let $\mu$ be a probability measure with mean $0$. Show that $$\hat{\mu}(y)-1 = \int (e^{iy t}-1-iyt) \, \mu(dt). \tag{2} $$

*Show that $$|e^{iyt}-1-iyt| \leq \min \left\{ \frac{1}{2} |y|^2 |t|^2, 2 |y| \, |t| \right\}. \tag{3} $$

*Conclude from $(2)$ and $(3)$ that $$|\hat{\mu}(y)-1| \leq |y| \int |t| \min \left\{ \frac{1}{2} |y| \, |t|, 2 \right\} \, \mu(dt).$$

*Use the dominated convergence theorem to show that $$\lim_{|y| \to 0} \int |t| \min \left\{ \frac{1}{2} |y| \, |t|, 2 \right\}\, \mu(dt)=0.$$

*Conclude that $(1)$ holds.

