Why do we use the convolution and not the cross-correlation to mollify functions?

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Given a function $f$, we want to mollify it. In other words, we want to construct $\rho_n$ with

$\rho_n\in C_c^{\infty}(\mathbb{R^d}),\operatorname{supp}(\rho_n)\subset \overline{B(0,1/n)},\int\rho_n=1,\rho_n\ge 0$ on $\mathbb{R}^d$.

with the properties that $f\ast\rho_n$ is smooth and $f\ast\rho_n\to f$ in $L^p$ when $n\to\infty$ (where $\ast$ is some operator).

For example $\rho(x)=\begin{cases}e^{1/(|x|^2-1)}\quad&\text{ if }|x|<1\\0\quad&\text{ if }|x|>1\end{cases}$

And let $\rho_n(x)=Cn^d\rho(nx)$ with $C=1/\int\rho$. For the operator $\ast$ we may choose I think (and it should work)

(1)$f\ast\rho_n(x)=\int f(x-t)\rho_n(t)\,dt$ (convolution)

(2)$f\ast\rho_n(x)=\int f(t+x)\rho_n(t)\,dt$ (Cross-correlation)

(3)$f\ast\rho_n(x)=\int f(t-x)\rho_n(t)\,dt$

It seems that the convolution is the visually less intuitive procedure among the three and I was wondering why this is in most case the one used to mollify functions.