Bounding a function with second moments 
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.
 A: 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 $\sigma=2/n$. 
