Converge of a sequence in $L^p(\mathbb{R}^3)$ Let $f(x)\in L^p(\mathbb{R}^3)$ for every $p\in [1, \infty]$. Let $B(n)\subset \mathbb{R}^3$ be the ball of radius $n$ centered at the origin. I want to show that the sequence
$$I_n(x)=\int_{\mathbb{R}^3}e^{-\xi^2/n}\int_{B(n)}f(y)e^{i(x-y)\xi}dyd\xi$$
converges in $L^p(\mathbb{R}^3)$ for every $p\in [1, \infty)$.
I know that $L^p$ is complete, so I've tried to show that $I_n(x)$ is a Cauchy sequence. I've written 
$$I_n(x)=\int\int \chi_{B(n)}(y)f(y)e^{i(x-y)\xi}e^{-\xi^2/n}dyd\xi$$
and tried to use some inequalities to estimate $I_n-I_m$, but nothing obtained. Any suggestions?
 A: The $\xi$-integral can be computed as
$$
\int e^{i(x-y)\xi}e^{-|\xi|^2/n}\mathrm{d}\xi
=
e^{-n|x-y|^2}\int e^{-(\xi/\sqrt{n}-i\sqrt{n}(x-y))^2}\mathrm{d}\xi 
= (\pi n)^{3/2} e^{-n|x-y|^2} ,
$$
and hence
$$
I_n(x)=\int\chi_{B(n)}(y)f(y)(\pi n)^{3/2} e^{-n|x-y|^2}\mathrm{d}y
= \int f_n(y)g_n(x-y)\mathrm{d}y ,
$$
where 
$$
f_n(y) = \chi_{B(n)}(y)f(y) ,
$$
and
$$
g_n(x) = (\pi n)^{3/2} e^{-n|x|^2}.
$$
It is obvious that the operator $T_n:f\mapsto f_n$ is bounded uniformly in $n$ as an operator $L^p\to L^p$, for each $1\leq p\leq\infty$, and moreover that $f_n\to f$ in $L^p$, for each $1\leq p<\infty$.
On the other hand, we have
$$
\int g_n(x)\,\mathrm{d}x = \pi^3 ,
$$
$g_n>0$, and $g_n\to0$ uniformly in $\{|x|>\delta\}$ for each fixed $\delta>0$.
Thus by the general theory of approximate identity, we infer that 
the operator $G_n:f\mapsto f*g_n$ is bounded uniformly in $n$ as an operator $L^p\to L^p$, for each $1\leq p\leq\infty$, and moreover that $f*g_n\to \pi^3f$ in $L^p$, for each $1\leq p<\infty$.
Finally, by combining the results about $T_n$ and $G_n$, we conclude that $f_n*g_n\to \pi^3f$ in $L^p$, for each $1\leq p<\infty$.
