I asked this question on [Mathematics Stackexchange](https://math.stackexchange.com/questions/4505957/characterization-of-extendible-distributions), but got no answer.


I found the following [question](http://www.cmls.polytechnique.fr/perso/golse/MAT431-10/pc10.pdf) 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?
> 
> 
> 



Characterization of extendible distributions
$\newcommand{\R}{\mathbb R}$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](https://rads.stackoverflow.com/amzn/click/com/0070542368), 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 $t\in(0,1]$.


