Sub-Gaussian decay of convolution of $L^1$ function with Gaussian kernel

Question

I think it might be helpful to put the new statement at the beginning and put the original post at the end. This new statement is more mathematically elegant.

Let $f\geq0$ be in $L^1(\mathbb{R}^d)$ and $g(x)=\exp(-\|x\|^2)$. Let $1_{B_1}$ be the indicator function of the unit ball centered at origin. Let $*$ be the convolution operation. Does the condition $$(f*g)(x)\leq C_1\exp(-C_2\|x\|^2)$$ for some $C_1,C_2>0$ imply $$\lim_{n\to+\infty}\frac{(f*1_{B_1})(\mu_n)}{(f*g)(\mu_n)}=0$$ for some sequence $\mu_n\in\mathbb{R}^d$? If this is not true, what additional regularity conditions on $f$ do we need? Any idea or possibly useful reference would be appreciated! The result can be verified easily when $f$ is another Gaussian function as well as some linear combination of Gaussian functions.

----------------Original post---------------------

Let $X$ be a random vector in $\mathbb{R}^d$ satisfying the following property: there exists $C_1,C_2>0$ such that $$\int_0^{+\infty}\mathbb{P}(\|X-\mu_0\|\leq\sqrt{t})\exp(-t)dt\leq C_1\exp(-C_2\|\mu_0\|^2)$$ for any $\mu_0\in\mathbb{R}^d$. Here $\|\|$ is the Euclidean norm in $\mathbb{R}^d$. If the above property holds, is the following statement true: there exists a sequence of vectors $\mu_n$ in $\mathbb{R}^d$ and a sequence of real numbers $t_n\to+\infty$ ($t_n$ may depend on $\mu_n$ for example $t_n=\|\mu_n\|^2/4$) such that: $$\lim_{n\to+\infty}\frac{\mathbb{P}(\|X-\mu_n\|\leq1)}{\mathbb{P}(\|X-\mu_n\|\leq \sqrt{t_n})\exp(-t_n)}=0$$

If this is not true, is there a counter example? Or is the the following result true? $$\lim_{n\to+\infty}\frac{\mathbb{P}(\|X-\mu_n\|\leq1)}{\int_0^{+\infty}\mathbb{P}(\|X-\mu_n\|\leq\sqrt{t})\exp(-t)dt}=0$$

Answer 1

We assume that $f$ is not identically zero, whence $f*g(x)>0$ for all $x \in \mathbb{R}^d$. The answer is positive, and this is also true when the density f is replaced by an arbitrary positive finite measure (or equivalently, a probability measure, as in the original formulation). Moreover, for any fixed nonzero vector $v \in \mathbb{R}^d$, the sequence $\{\mu_n\}$ can be taken as a subsequence of the positive integer multiples $\{k v\}_{k \ge 0}$ of $v$. Indeed, if this fails, then there exists $\alpha>0$ and an integer $k_0>0$ such that $$\frac{(f*1_{B_1})(kv)}{(f*g)(kv)} \ge \alpha $$ for all $k \ge k_0$. Let $$\beta=\beta(v):=\inf \{g(x) \, : \, x \in v+B_1\}>0. $$ Then $$(f*g)((k+1)v) \ge \beta (f*1_{B_1})(kv),$$ so the first display gives $$(f*g)((k+1)v) \ge \alpha \beta (f*g)(kv) \,. $$ We infer inductively that $$(f*g)(kv) \ge (\alpha\beta)^{k-k_0} (f*g)(k_0 v) \,. $$ This exponentially decaying lower bound contradicts the hypothesis $$(f*g)(x)\leq C_1\exp(-C_2\|x\|^2).$$