Ratio of Gaussian measure over Euclidean balls

Question

Let $\nu\in\mathcal P(\mathbf R^d)$ be the standard Gaussian distribution $\mathcal N(0,I_d)$. Denote by $\mathscr B$ the class of Euclidean balls $B_r(x)$ (centered in $x\in\mathbf R^d$ with radius $r>0$) of $\mathbf R^d$.

Fix $x_0\in \mathbf R^d$. I am interested by the quantity $$F(x_0):=\sup_{B\in\mathscr B}\frac{\nu(B)}{\nu(B+x_0)}.$$ More precisely, I would like to have a simpler expression of $F$, ideally depending only on $x_0$.

I have two questions:

1° I have the intuition that the supremum is atteined when considering balls centered in $0\in\mathbf R^d$, i.e.,
$$ F(x_0 ) = \sup_{r>0}\frac{B_r(0)}{B_r(x_0)}.$$ Is it true and how can we prove that statement ?

2° Can we use some monotonicity argument to prove that $$\sup_{r>0}\frac{B_r(0)}{B_r(x_0)} = \lim_{r\to 0}\frac{B_r(0)}{B_r(x_0)} \quad ? $$ In that case we would obtain by Lebesgue's Differentiation Theorem, that $$\frac{B_r(0)}{B_r(x_0)} = \frac{B_r(0)/|B_r|}{B_r(x_0)/|B_r|}\to_{r\to 0}\frac{f(0)}{f(x_0)} = 1/f(x_0), $$ where $f$ denotes the density of $\mathcal N(0,I_d)$.

If points 1° and 2° are correct, we would have proved that $$F(x_0) = 1/f(x_0).$$

Thank you for your help.

Answer 1

We have \begin{equation*} F(x_0)=\infty \tag{1}\label{1} \end{equation*} for any nonzero $x_0$.

Indeed, by spherical symmetry, without loss of generality \begin{equation*} x_0=(2a,0,\dots,0) \end{equation*} for some real $a>0$.

Let $b_t:=B_{2a}((t,0,\dots,0))$ for real $t>0$. Let \begin{equation*} C_{d,a}:=\nu_{d-1}\big(B^{d-1}_a\big), \end{equation*} where $\nu_k$ is the standard Gaussian measure over $\Bbb R^k$ and $B^k_r$ is the ball in $\Bbb R^k$ of radius $r$ centered at the origin. Note that $b_t\supset[t-a,t-a/2]\times B^{d-1}_a$, if $\Bbb R^d$ is identified with $\Bbb R\times\Bbb R^{d-1}$. So, if $t>a$, then \begin{equation*} \nu(b_t)\ge\frac1{\sqrt{2\pi}}\int_{t-a}^{t-a/2} dx_1\,e^{-x_1^2/2} \;C_{d,a} \ge c_d\, e^{-(t-a/2)^2/2}, \end{equation*} where $c_d:=\frac{C_{d,a}}{\sqrt{2\pi}}\frac a2>0$.

On the other hand, \begin{equation*} \nu(b_t+x_0)\le\nu([t+2a-2a,\infty)\times\Bbb R^{d-1})= \frac1{\sqrt{2\pi}}\int_t^\infty dx_1\,e^{-x_1^2/2}\le e^{-t^2/2}. \end{equation*}

So, if $t>a$, then \begin{equation*} \frac{\nu(b_t)}{\nu(b_t+x_0)}\ge c_d\,\frac{e^{-(t-a/2)^2/2}}{e^{-t^2/2}} \to\infty \end{equation*} as $t\to\infty$. So, \eqref{1} is proved. $\quad\Box$

Answer 2

Here's another proof that $F(x_0)=\infty$. I work in dimension 1 for simplicity.
Fix $x_0\in\mathbf R$ and introduce $A(x,r) = \nu(B_r(x))/\nu(B_r(x+x_0))$. By Lebesgue differentiation theorem, it holds, for all $x\in\mathbf R$,
$$A(x,r) = \frac{\nu(B_r(x))/|B_r(x)|}{\nu(B_r(x+x_0))/|B_r(x+x_0)|}\to_{r\to0}\frac{e^{-x^2/2}}{e^{-(x+x_0)^2/2}}$$ Hence, for all $x\in\mathbf R$, $$\sup_{r>0}A(x,r)\ge\frac{e^{-x^2/2}}{e^{-(x+x_0)^2/2}}$$
The right-hand side tends to $+\infty$ when $x$ has the same sign as $x_0$ and $|x|\to\infty$. Hence $F(x_0) = +\infty$.