$L^2$ Cauchy sequence of convolution functions For $f\in L^2(\mathbb{R}^2)$ lets define $g_n(x):=(\phi*f)(x)$ where $\phi_n(x)=n^4x_1x_2e^{-n|x|}$ with $x=(x_1, x_2)\in \mathbb{R}^2$ and $|x|=\sqrt{x_1^2+x_2^2}$. I have to prove that $(g_n)_n$ is a Cauchy sequence in $L^2(\mathbb{R}^2)$ and eventually also compute the limit.
I'm stuck trying to prove that it is a Cauchy sequence. By using Minkowski's integral inequality I end up with
$$\int |g_n-g_m|^2dx\leq ||f||_2^2\left(\int_{\mathbb{R}^2}\int_{\mathbb{R}^2} |(x_1-y_1)(x_2-y_2)|^2\cdot |n^4e^{-n|x-y|}-m^4e^{-m|x-y|}|^2dydx\right)$$
and my first idea was spliting up the last integral in two regions: $|x-y|<1$ and $|x-y|\geq 1$. At a first glance, I don't see any problem working with the region where $|x-y|<1$, but for the other one I have no idea how to treat it.
I would really appreciate any suggestions or hints!
 A: Consider the constant $c=\int_{\mathbb{R}^2} x_1x_2 e^{-|x|}\, dx$. We claim that $g_n\to cf$ as $n\to \infty$. The basic idea is the following:
Step 1: Case of $f$ continuous and compactly supported. Here we have that $g_n \to cf$ pointwise as follows
$$
|g_n(x)-cf(x)|= \left| \int_{\mathbb{R}^2} (f(x-y)- f(x))\phi_n(y)\, dy\right|  \leq \left| \int_{|y|<\delta} \cdots \right| + \left| \int_{|y|\geq \delta} \cdots \right|=: I+II.
$$
The first terms can be made small due to the continuity of $f$ (by choosing $\delta$ small enough). The second one we use the boundedness of $f$ and the fact that $\int_{|y|\geq \delta} \phi_n\, dy\to 0$ for any $\delta >0$.
To get the convergence in $L^2$ we need to control the tail of the convolution (notice that $g_n$ is pointwise uniformly bounded in $n$ by H"older's inequality). Suppose that the support of $f$ is contained in $B_R(0)$ for some $R>1$. Notice that $|\phi_1(x)|\leq e^{-|x|/2}$ for any $x$ and $\phi_n(x)=n^2\phi_1(nx)$ and so for $|x|>2R$
$$
|g_n(x)|\leq \| f\|_\infty \int_{B_R(0)} n^2e^{-n|x-y|/2}\, dy\leq \| f\|_\infty \int_{B_R(0)} n^2e^{-n|x|/4}\, dy = \| f\|_\infty \pi R^2 n^2e^{-n|x|/4} \leq \| f\|_\infty \dfrac{16R^2}{|x|^2},
$$
where we used that $|x|\leq 2|x-y|$ if $|x|>2R$ and $|y|<R$; as well as $e^{-z}\leq z^{-2}$ for any $z>0$. Combining all these we get
$$
|g_n(x)|\leq \| f\|_\infty ( \chi_{B_{2R}(0)}(x) + \frac{16\pi R^2}{|x|^2}\chi_{\mathbb{R}^2\setminus B_{2R}(0)}(x)),
$$
and this last is an $L^2$ (in fact any $L^p$, with $p\geq 1$) function. We conclude by the dominated convergence theorem. (the $L^2$ convergence argument here can be simplified, I just took the first thing that worked)
Step 2: For general $f$ we approximate it by continuous, compactly supported functions $f_k$ and call $h_{k,n}=f_k*\phi_n$.
$$
g_n-g_m= (g_n-h_{k,n})+ (h_{k,n}-h_{k,m})+ (h_{k,m}-g_m)= (f-f_k)*\phi_n + f_k*(\phi_m-\phi_n)+ (f-f_k)*\phi_m.
$$
The $L^2$ norms of the first and third term can be made arbitrarily small, uniformly in $n,m$, by choosing $k$ large enough. Having fixed such $k$ pick $n,m$ large so that $h_{k,n}$ and $h_{k,m}$ are both very close (in $L^2$) to $cf_k$, controlling the second term.
