With the question in the title I mean this: Pick an $f \in L^2 (\mathbb{R}^3)$. Can one then find positive constants $a$, $\varepsilon$, and $\gamma$ such that $$ |f(x)| \leq a |x|^{-3/2 + \gamma} $$ for almost all $x \in B_\varepsilon (0)$?
I am trying to turn the folklore statement "$f \in L^2 (\mathbb{R}^3)$ behaves in the worst case like $a |x|^{-3/2 + \gamma}$ around $0$" into a rigorous one.
In order to answer my question positively, I aimed at a contradiction but didn't succeed. However, I also couldn't prove that $$ \frac{e^{-x^2}}{|\sin\frac{1}{|x|}|^{-3/2 + \delta}}, $$ what seems to be a counterexample if $\delta > 0$ is sufficiently small, lies in $L^2 (\mathbb{R}^3)$ at all.
Any help concerning my specific question but also references or general remarks are highly appreciated!
As a direct counterexample you can take $$f(x) = |x|^{-3/2} \, (\log|x|)^\alpha$$ for small $x$ and a suitable $\alpha$.
Before you have further guesses, one can find a function in $L^2(\mathbb R^3)$ with a singularity in any open set. Indeed, fix a nonnegative function $f$ in $L^2(\mathbb R^3)$ with a singularity at $0$. For an enumeration $\{q_i\}_{i \in \mathbb N}$ of $\mathbb Q^3$, consider $$ g(x) = \sum_{i=1}^\infty \frac1{2^i} f(x - q_i).$$ This function belongs to $L^2(\mathbb R^3)$ and has a singularity at each rational number.