Leading term in asymptotic expansion on continuous function in $L^2$ Let $f$ be a continuous real function in $L^2(\mathbb{R}^3)$, I think it is true that $f$ must be $O(r^{-(3/2+\epsilon)})$ for some $\epsilon >0$. My argument is as follows:
Assuming that $f$ has an the asymptotic expansion in terms of the radial function $r$, of the form $f \sim \sum_{i \in \mathbb{Z}} a_i(\theta,\varphi) r^i $ where $\theta,\varphi$ are angular variables. Assume $k$ is the leading order in the expansion, i.e. $k$ is the largest of the exponent coefficients.
Then, integration on the complement of a ball of radius $R$ gives: 
$$\int_\Omega \int_R^\infty (\sum_{i \in \mathbb{Z}} a_i(\theta,\varphi) r^i)^2 r^2 dr d\Omega \leq c \int_\Omega \int_R^\infty (a_k(\theta,\varphi) r^k)^2 r^2 dr d\Omega \leq C \int_R^\infty r^{2k} r^2 dr d\Omega        \leq \|f\|^2$$ which should be finite by hypothesis. Therefore $k<-3/2$. Is my argument correct?
 A: This is incorrect.  For one thing, as far as I know an asymptotic expansion of the form you have written need not exist for $f$ (though I may be wrong about that).  For another thing, even if such an expansion does exist, you have ignored the coefficients $a_i(\theta,\varphi)$ in estimating $\|f\|^2$.  You cannot ignore these coefficients by just bounding their integrals below by constants, since they may not always have the same sign and so there may be cancellation.
In any case, the statement you are trying to prove is very false, and indeed you cannot say that $f$ is $O(r^i)$ for any $i$ at all.  Just take $f$ to be $0$ on most of $\mathbb{R}^3$,  but have narrow but tall "spikes" going out to $\infty$.  By making the spikes get taller and taller fast enough, you can make $f\not\in O(r^i)$ for any $i$ (say, near $(n,0,0)$ there is a spike of height $n!$, for each $n\in\mathbb{N}$).  By making the spikes narrow enough (so the integral of $f^2$ on them shrinks fast enough), you can still ensure that $f\in L^2$.
A: To answer a question in a comment: Yes, at least in dimension  one assuming $f,f'\in L^2$ implies that $f$ tends to zero at infinity; in higher dimensions,  or to get faster decay, you need more derivatives.
Speaking very sloppily, if $f'\in L^2(\Bbb R)$ and $x<y$ then $$|f(x)-f(t)|=
\left|\int_x^y f'(t)\,dt\right|\le|x-y|^{1/2}||f'||_2.$$ In particular, $f'\in L^2$ implies that $f$ is uniformly continuous, and it's easy to see that if $f\in L^2$ and $f$ is uniformly continuous then $f$ must vanish at infinity.
You want to learn about the Sobolev Embedding Theorem.
