A sequence of random variables $X\_n$ converges in distribution to $X$, if there is pointwise convergence of its characteristic functions, i.e. $\lim\_{n\rightarrow\infty}\phi\_{X\_n}(\lambda) = \phi\_X(\lambda)$ for all $\lambda \in \mathbb R$.


Imagine we do not only have pointwise, but uniform convergence of the characteristic functions ($\lim\_{n\rightarrow\infty}\|\phi\_{X\_n}-\phi\_X\|\_\infty = 0$). Can we estimate the error in the distributions by knowing $\|\phi\_{X\_n}-\phi\_X\|\_\infty$, i.e. do we have $$ |P(X\_n \le z ) - P(X < z)| \le \alpha\left(\|\phi\_{X\_n}-\phi\_X\|\_\infty\right)$$ for a function $\alpha(\cdot)$ and any $z \in \mathbb R$? What is the form of $\alpha(\cdot)$?



Error for the convergence by distribution
Assuming that you want $\alpha(0+)=0$, there is no such inequality. In fact, by Proposition 11.7.6 in Dudley's Real Analysis and Probability, the optimal $\alpha$ in your question, if taken literally, is identical to $1$ for strictly positive arguments. 


As Dudley states in the end-of-chapter notes, the result of his Proposition 11.7.6 is due to Dyson (1953). If I recall it correctly, Dyson's original example is more complicated than Dudley's.


