Let $f$ be a smooth function defined on $\mathbb{R}^3$ with $f(x) = O(1/|x|^{2+\epsilon})$, and let $g(x)=\frac{1}{|x|}$. I want to show that $f\ast g (x) \to 0$ as $|x| \to \infty$. (Maybe this is false, but I think this should be intuitively true.)
I tried to use the Riemann-Lebesgue Lemma so that the problem reduces to proving $\frac{\hat f}{|x|^2} \in L^1$. However, I cannot prove that $\hat f (x)$ has a sufficiently tame singularity at $x=0$, so that the product indeed lies in $L^1$.
My mathematics skill are quite rusty; any helps are appreciated.
I also couldn't show whether $\frac{\hat{f}}{|x|^2}\in L^1$, so I came up with another proof that's a bit more involved, but only uses the convolution's original expression.
Let's choose some arbitrary $R\geq1$ for now, and split the convolution as such: $$ f*g(x) = \underbrace{\int_{|y|\leq R} f(x-y) g(y)\,dy}_{(i)} + \underbrace{\int_{|y|>R} f(x-y) g(y)\,dy}_{(ii)} $$
For $(i)$, we can take out the supremum of $f$ over the ball $B(x,R)$: $$ |(i)| \leq \|f\|_{L^\infty(B(x,R))} \int_{|y|\leq R} g(y)\,dy, $$ and since $\int_{|y|\leq R} \frac{1}{|y|}\,dy = CR^2$ for some constant $C>0$, we have: $$ |(i)| \leq CR^2\,\|f\|_{L^\infty(B(x,R))}. $$
For $(ii)$, let's write: \begin{align*} |(ii)| &= \left|\int_{|y|>R} f(x-y) \cdot\frac{1}{|y|^{1-\epsilon/2}}\cdot \frac{1}{|y|^{\epsilon/2}}\,dy\right| \\ &\leq \frac{1}{R^{\epsilon/2}} \int_{|y|>R} |f(x-y)| \cdot\frac{1}{|y|^{1-\epsilon/2}}\,dy \\ &\leq \frac{1}{R^{\epsilon/2}}\,|f|*\bar{g}(x), \end{align*} with $\bar{g}(x) := \frac{1}{|x|^{1-\epsilon/2}} \boldsymbol{1}_{|x|>1}$ (we're also assuming $\epsilon<1/2$ without loss of generality).
We have that $|f|*\bar{g}$ is independant of $R$, but we also want to bound it uniformly over $x$. We could do that if there were some $p,p^*\geq1$ such that $\frac{1}{p}+\frac{1}{p^*}=1$ and $f\in L^p$, $\bar{g}\in L^{p^*}$, since we could then use Young's convolution inequality.
We know that $f\in L^p$ when $p(2+\epsilon)>3$, and $\bar{g}\in L^{p^*}$ when $p^*(1-\epsilon/2)>3$, so we just need to verify if those conditions are compatible.
After some algebraic rearranging, we get what we want whenever $p$ verifies $1\leq\frac{3}{2+\epsilon}<p<\frac{3}{2+\epsilon/2}$, which is definitely possible since $0<\epsilon<1/2$. Choosing such a $p$, we then have: $$ |f|*\bar{g}(x) \leq \||f|*\bar{g}\|_{L^\infty} \leq \|f\|_{L^p}\,\|\bar{g}\|_{L^{p^*}} < \infty. $$ Taking all our estimates, we get: $$ |f*g(x)| \leq CR^2\,\|f\|_{L^\infty(B(x,R))} + \frac{1}{R^{\epsilon/2}} \|f\|_{L^p}\,\|\bar{g}\|_{L^{p^*}}. $$ Now we can prove the convergence. Choose any $\delta>0$, and take $R>0$ large enough such that: $$ \frac{1}{R^{\epsilon/2}} \|f\|_{L^p}\,\|\bar{g}\|_{L^{p^*}} \leq \frac{\delta}{2}. $$ Since $f(x)\to0$ as $|x|\to\infty$, we can define $M>0$ (which depends on $R$) large enough so that, whenever $|x|>M-R$, we have: $$ CR^2\,|f(x)|<\frac{\delta}{2}. $$ In particular, this is uniformly true over $B(x,R)$ whenever $|x|>M$.
Thus, for all $\delta>0$, we can define $M>0$ such that, for all $|x|>M$: $$ |f*g(x)|<\delta, $$ hence the convergence to $0$.