
> 
> Let $f(x,y)$ be a non-negative function with $x,y \in \mathbb R^3$ that satisfies
>  $$
> I\_1(f) := \iint\_{\mathbb R^3 \times \mathbb R^3 } f(x,y) \, dx \ dy < \infty
> $$
>  and
>  $$
> I\_2(f) := \iint\_{\mathbb R^3 \times \mathbb R^3 } |y|^2 f(x,y) \, dx \ dy < \infty.
> $$
>  Also, define the sequence by
>  $$
> f\_n(x,y) := \left(\frac n{4\pi} \right)^3 \iint\_{|x'|<n,|y'|<n} e^{-\frac n4 (|x-x'|^2+|y-y'|^2)} f(x',y') \, dx' \, dy'.
> $$
>  Show that
>  $$
> I\_2(f\_n) \le 4I\_2(f) + 16 \Gamma(5/2) \frac 1n I\_1(f),
> $$
>  where $\Gamma$ is the Gamma function.
> 
> 
> 


Honestly, all I can do is rewrite the expression of $I\_2$ as follows:
$$
I\_2(f\_n) = \left(\frac n{4\pi} \right)^3 \iint\_{\mathbb R^3 \times \mathbb R^3} |y|^2 \left(\iint\_{|x'|<n,|y'|<n} e^{-\frac n4 (|x-x'|^2+|y-y'|^2)} f(x',y') \, dx' \, dy'\right) \, dx \, dy.
$$
There's a product-to-sum inequality being used here, I think. 


Also, I'm okay if any answer to this question does not provide the explicit constants $4$ and $16 \Gamma(5/2)$; they are just positive constants.


Remark: This is the third item on the list of assertions of a lemma on page 142 of [this paper](http://www.mathnet.ru/links/2636cfefb008a481d05d4471963fc588/zvmmf6314.pdf).



Bounding a function with second moments
First let me remark that in this case one can make an explicit computation using the explicit values of first and second moments of Gaussian.
 Then note that the $x$ variable does not play any role, as when we integrate by parts, we get the normalised Gaussian which just give $1$. What caught my attention was this $\Gamma(5/2)$ constant. Here is a way how to recover it. Let us make a slight change of notations
 \begin{align\*}
I\_s(f)=\int\_{\mathbb{R}^d}^{}|x|^sf(x)dx
\end{align\*}
 and for all $\sigma>0$ (note that if you are just looking for inequalities, then the cut-off in $x'$ and $y'$ is useless in the estimate \textit{e.g.} it does not change the inequality)
 \begin{align\*}
f\_\sigma(x)=\int\_{\mathbb{R}^d}^{}\frac{e^{-\frac{|y-x|^2}{2\sigma}}}{(2\pi\sigma)^{\frac{d}{2}}}f(y)dy
\end{align\*}
 Then
 \begin{align\*}
I\_s(f\_\sigma)=\int\_{\mathbb{R}^d}|x|^sf\_\sigma(x)dx=\int\_{\mathbb{R}^d}^{}f(y)\left(\int\_{\mathbb{R}^d}^{}|x|^s\frac{e^{\frac{-|x-y|^2}{2\sigma}}}{(2\pi\sigma)^{\frac{d}{2}}}\right)dy.
\end{align\*}
 Let us prove the inequality for $s\geq 1$ (if $s<1$, we just replace $2^{s-1}$ by $1$ in the following inequalities). By Young's inequality and polar coordinates
 \begin{align\*}
\int\_{\mathbb{R}^d}^{}|x|^s\frac{e^{\frac{-|x-y|^2}{2\sigma}}}{(2\pi\sigma)^{\frac{d}{2}}}&=\int\_{\mathbb{R}^d}^{}|y+x|^s\frac{e^{-\frac{|x|^2}{2\sigma}}}{(2\pi\sigma)^{\frac{d}{2}}}dx\leq 2^{s-1}\int\_{\mathbb{R}^d}^{}(|y|^s+|x|^s)\frac{e^{-\frac{|x|^2}{2\sigma}}}{(2\pi\sigma)^{\frac{d}{2}}}dx\\
&\leq 2^{s-1}|y|^s+2^{s-1}\beta(d)\int\_{0}^{\infty}r^{d+s-1}\frac{e^{-\frac{r^2}{2\sigma}}}{(2\pi\sigma)^{\frac{d}{2}}}dr\\
&=2^{s-1}|y|^s+2^{s-1}\beta(d)\frac{(2\sigma)^{\frac{s}{2}}}{2\pi^{\frac{d}{2}}}\Gamma\left(\frac{d+s}{2}\right)\\
&=C(d,s)+2^{s-1}|y|^s
\end{align\*}
 where $\beta(d)$ is the measure of the unit sphere in $\mathbb{R}^d$. So one should get
 \begin{align\*}
C(d,s)=2^{\frac{3}{2}s-1}\sigma^{\frac{s}{2}}\frac{\Gamma\left(\frac{d+s}{2}\right)}{\Gamma\left(\frac{d}{2}\right)}
\end{align\*}
 and we obtain the inequality
 \begin{align\*}
I\_s(f\_\sigma)\leq 2^{s-1}I\_s(f)+C(d,\sigma)I\_0(f).
\end{align\*}
 We note that $C(d,\sigma)=O(\sigma^{\frac{s}{2}})$ so we have the good behaviour when $\sigma\rightarrow 0$ when $d=3$ and $s=2$ and we recover the $\Gamma(5/2)$ constant and a slightly better estimate plugging $\sigma=2/n$. 


