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?
$$\lim\_{n\to+\infty}\frac{\mathbb{P}(\|X-\mu\_n\|\leq1)}{\int\_0^{+\infty}\mathbb{P}(\|X-\mu\_n\|\leq\sqrt{t})\exp(-t)dt}=0$$



Sub-Gaussian decay of convolution of $L^1$ function with Gaussian kernel
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
$$(f\*g)(x)\leq C\_1\exp(-C\_2\|x\|^2).$$


