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

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

• Am I right that the first display in your question is just an obscure way to say that the convolution of the distribution of $X$ with a fixed centred Gaussian is dominated by a multiple of another centred Gaussian? May 30, 2019 at 23:11
• Sorry I cannot see where the convolution comes from. Could you please elaborate a little bit about this? @MateuszKwaśnicki May 30, 2019 at 23:51
• \begin{aligned}\int_0^\infty P(|X-\mu|>\sqrt{t})e^{-t}dt&=\int_0^\infty E[\mathbb{1}_{(|X-\mu|^2,\infty)}(t)e^{-t}]dt\\&=E\biggl[\int_{|X-\mu|^2}^\infty e^{-t}dt\biggr]=E[\exp(-|X-\mu|^2)]\end{aligned} May 31, 2019 at 7:35
• Yes it is. Does this convolution have any implication on the distribution of $X$? May 31, 2019 at 14:25

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 $$(f*g)(x)\leq C_1\exp(-C_2\|x\|^2).$$
• This is just the power of proof by contradiction- If your conclusion did not hold, then the convolutions $f*1_{B_1}$ and $f*g$ would be comparable, but the first is completely local while the second takes into account neighboring values, so for them to have bounded ratio implies that the decay can be no faster than exponential. Jun 3, 2019 at 7:43