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? $$\lim_{n\to+\infty}\frac{\mathbb{P}(\|X-\mu_n\|\leq1)}{\int_0^{+\infty}\mathbb{P}(\|X-\mu_n\|\leq\sqrt{t})\exp(-t)dt}=0$$