Inequality regarding difference of characteristic functions We want to show if $X, Y$ are random variables defined on a common probability space, with characteristic functions $f, g$ respectively, then the following inequality is valid:
$$\sup |f(x)-g(x)| \le 2P(X\neq Y).$$
This is from an old qualifying exam and I cannot solve it.  I tried to analyze the difference in probability measures, but to no avail. Any help would be greatly appreciated.
 A: We have for two real numbers $a$ and $b$ that 
$$|e^{ia}-e^{ib}|=|e^{ia/2}-e^{i(b-a/2)}|=|e^{\frac{a-b}2i}-e^{\frac{b-a}2i}|=2\left|\sin\left(\frac{a-b}2\right)\right|.$$
This gives 
\begin{align}
|f(x)-g(x)|&\leq \int_{\Omega}|e^{iX(\omega)x}-e^{iY(\omega)x}|d\mathbb P\\
&\leq 2\int_{\Omega}\left|\sin\left(\frac{X(\omega)-Y(\omega)}2x\right)\right|d\Bbb P(\omega)\\
&=2\int_{\{X\neq Y\}}\left|\sin\left(\frac{X(\omega)-Y(\omega)}2x\right)\right|d\Bbb P(\omega)\\
&\leq 2\mathbb P(X\neq Y).
\end{align}
The constant $2$ can't be improved. This can be seen taking Dirac measures at distinct points. 
Note that the bound can be taken quite straightforwardly considering the first line, integrating over $\{X\neq Y\}$ and bounding the integrand by $2$. But the equality in the first displayed equation helps us to see why $2$ is optimal.
A: Almost content-free, since once the inequality in 1. is stated the rest is routine:


*

*For every real numbers $a$ and $b$, $\color{red}{|\mathrm e^{\mathrm ia}-\mathrm e^{\mathrm ib}|\leqslant2\cdot[a\ne b]}$.

*Apply 1. to $a=xX$ and $b=xY$, for some real $x$. This yields $|\mathrm e^{\mathrm ixX}-\mathrm e^{\mathrm ixY}|\leqslant2\cdot[X\ne Y]$ almost surely.

*Integrate 2. This yields $|\mathrm E(\mathrm e^{\mathrm ixX})-\mathrm E(\mathrm e^{\mathrm ixY})|\leqslant2\cdot\mathrm P(X\ne Y)$.

*The inequality in 3. holds for every $x$ uniformly hence 
$$
\sup\limits_x|\mathrm E(\mathrm e^{\mathrm ixX})-\mathrm E(\mathrm e^{\mathrm ixY})|\leqslant2\cdot\mathrm P(X\ne Y).$$

