Proximity in terms of characteristic functions for $n$-dimensional distributions

Question

Let $X\in \mathbb{R}^n$ and $Y\in \mathbb{R}^n$ be random variables with characteristic functions $\phi_X(t)$ and $\phi_Y(t)$, respectively.

Suppose that \begin{align} \sup_{t \in \mathbb{R}^n} \frac{|\phi_X(t)-\phi_Y(t)|}{\|t\|} \le \epsilon. \quad (*) \end{align}

Question: Can we say something about how close the distributions of $X$ and $Y$ are in some metric over probability spaces? In other words, suppose that $(*)$ small what other distance would be small too.

For example, for $n=1$. There exists the following inequality \begin{align} L^2(P_X,P_Y) \le 2 \sup_{t \ge 0} \frac{|\phi_X(t)-\phi_Y(t)|}{t} \end{align} where $L$ is the L\'evy distance.

However, I was not able to locate similar inequality for $n>1$. One distance that I have taken a look at extensively is the Levy-Prokhorov metric. However, whenever I find anything it always depends on the derivatives of characteristic functions.

Answer 1

There is a good reason why you cannot get anything for the standard Levy-Prokhorov distance in high dimensions. Let's consider the uniform distribution on the sphere of radius $R$ in $\mathbb R^3$ and the mixture (with weight $\frac 12$ for each) of the uniform distributions on the spheres of radii $R-r$ and $R+r$ where $R>2r$. Then the difference of the characteristic functions is $|F(R|t|)-\frac 12[F((R-r)|t|)+F((R+r)|t|)]|\le r^2|t|^2\max_{[(R-r)|t|,(R+r)|t|]}|F''|$ where $F(u)=\frac{\sin u}{u}$. However, $|F''(u)|\le \frac Cu$ for $u>0$, so to have your condition, it suffices to ensure that $$ \frac {Cr^2|t|^2}{R|t|/2}\le\varepsilon |t|, $$ i.e. $2Cr^2/R\le\varepsilon$, which still allows $r$ to grow without bound as $R\to+\infty$ for any fixed $\varepsilon>0$.

So, you'll have to either settle for some cruder distance that gives you less control at infinity, or find a way to get that control by some alternative means from other assumptions you may have in your problem.