Characterization of extendible distributions I asked this question on Mathematics Stackexchange, but got no answer.
I found the following question which characterize the extension of a distribution in $\mathbb{R}$:

Let $f \in L_{\text{loc}}^{1}(\mathbb{R}_{>0})$ such that $f \geq 0$ a.e. Show that $f$ extends to a distribution $F$ in $\mathbb{R}$ if and only if there exists $k \geq 0$ $$\int_{\varepsilon}^{1}f(x)dx=O(\varepsilon^{-k}),$$
for $\varepsilon \rightarrow 0^{+}$.

I would like to know if someone could give me some hint to solve this problem. The only thing I could think was evaluate the distribution $T_f$ generated by $f$ in functions like $\varphi_\varepsilon(x)$ satisfying $\operatorname{supp}(\varphi_\varepsilon) \subset (\varepsilon,1)$, but without success.
Another question I have is:

Is there any generalization of this result to higher dimensions?

 A: $\newcommand{\R}{\mathbb R}$Without loss of generality, $k$ is a positive integer.
First, the "if" part. For $t\in(0,1]$, let
\begin{equation*}
    (Jf)(t):=\int_t^1 dx\,f(x). 
\end{equation*}
Here it is assumed that
\begin{equation*}
    (Jf)(t)=O(t^{-k}) 
\end{equation*}
for $t\in(0,1]$.
As suggested in the comment by paul garrett, this implies
\begin{equation*}
h(t):=(J^kf)(t)=O(\ln\tfrac1t)
\end{equation*}
for $t\in(0,1]$. So, $h$ extends to a distribution $H$ (by the formula $H\phi:=\int_{(0,1]} h\phi$ for smooth $\phi$ with compact support $S_\phi$) and hence $f$ extends to the distribution $(-1)^k H^{(k)}$, because $f=(-1)^k h^{(k)}$ on $(0,1]$.
For the "only if" part, we can use the characterization of distributions as generalized multiple derivatives of continuous functions.
Suppose that $f$ extends to a distribution $F$. Then, according to Theorem 6.26, there exist a continuous function $g$ and a nonnegative integer $k$ such that for all smooth functions $\phi$ with support $S_\phi\subseteq(0,2]$
\begin{equation*}
    F\phi=(-1)^k\int_{\R} g\phi^{(k)} 
\end{equation*}
and hence
\begin{equation*}
    \int_{(0,\infty)}f\phi=(-1)^k\int_{\R} g\phi^{(k)}. \tag{1}\label{1}
\end{equation*}
Let $\Phi$ be any nonnegative smooth function such that
\begin{equation*}
    \text{$\Phi=1$ on $[1,\infty)$ and $\Phi=0$ on $(-\infty,1/2]$} \tag{2}\label{2}
\end{equation*}
and let $\psi$ be any nonnegative smooth function with $S_\psi\subset[-1,2]$ such that
\begin{equation*}
    \text{$\psi=1$ on $[0,1]$.}
\end{equation*}
For $t\in(0,1]$ and real $x$, let
\begin{equation*}
    \phi_t(x):=\Phi(x/t)\psi(x),
\end{equation*}
so that $\phi_t$ is a nonnegative smooth function with $S_{\phi_t}\subseteq[t/2,2]\subset(0,2]$ such that $\phi_t=1$ on $[t,1]$.
Then for $t\in(0,1]$ we have $\phi_t^{(k)}=O(t^{-k})$ on $\R$ (see Detail at the end of this answer) and hence,
by the condition $f\ge0$ a.e., \eqref{1}, and the continuity of $g$
\begin{equation*}
    0\le\int_t^1 f\le\int_{(0,\infty)}f\phi_t=(-1)^k\int_{\R} g\phi_t^{(k)}
    =(-1)^k\int_0^2 g\phi_t^{(k)}=O(t^{-k}),
\end{equation*}
as desired. $\quad\Box$
Detail: Because $\Phi$ is smooth and in view of \eqref{2}, we see that $\Phi$ and all its derivatives, of any order, are bounded. Because $\psi$ is smooth and has a compact support,  we see that $\psi$ and all its derivatives, of any order, are bounded. So, by the Leibniz and chain rules of differentiation,
\begin{equation*}
    \phi_t^{(k)}(x)=\sum_{j=0}^k\binom kj \Phi^{(j)}(x/t)t^{-j}\psi^{(k-j)}(x) =O(t^{-k})  
\end{equation*}
for $t\in(0,1]$.
