-1
$\begingroup$

If $f$ a distribution with compact support then they exist $m$ and measures $f_\beta$,$|\beta|\leq m$ such that $$f=\sum_{|\beta|\leq m}\frac{\partial^\beta f_\beta}{\partial x^\beta}$$

how to demonstrate this result ?

Improve this question
$\endgroup$
1
  • 4
    $\begingroup$ This is a standard result, and should probably be on MathStackExchange... $\endgroup$
    – paul garrett
    Commented Dec 15, 2019 at 22:33

3 Answers 3

Reset to default
4
$\begingroup$

That's an easy consequence of the Hahn-Banach theorem and the Riesz theorem characterising the dual of $C^0(\Omega)$.

"of order m" means that the distribution is linear w.r.t. to the $C^m$-norm. It is defined on a dense subspace of $C^m$ and thus extends uniquely to an element of the dual space of $C^m$. Now $C^m(\Omega) \hookrightarrow \prod_{|\beta|\leq m} C^0(\Omega), \phi\mapsto (D^\beta \phi)_\beta$ is an isometric embedding so that we can extend to distribution with Hahn-Banach further to an element of the dual space of this bigger space which is $\prod_{|\beta|\leq m} (C^0(\Omega))'$. With Riesz identify $(C^0(\Omega))'$ with a space of measures on $\Omega$. This gives you the measures $f_\beta$ you're looking for.

Improve this answer
$\endgroup$
1
  • $\begingroup$ @johanes hahn thank you so much $\endgroup$
    – deval sidi
    Commented Dec 16, 2019 at 15:17
2
$\begingroup$

  • If $f\in D'(\Bbb{R})$ is a distribution with compact support $\subset (a,b)$ then it has finite order $m$. Proof : assume it doesn't, for each $k$ take $\phi_k\in C^\infty_c[a,b]$ with $\sum_{l\le k} \|\phi_k^{(l)}\|_\infty\le 2^{-k}$ and $\langle f,\phi_k\rangle=1$ then $\lim_{K\to\infty}\sum_{k\le K} \phi_k$ converges in $C^\infty_c[a,b]$ whereas $\lim_{K\to\infty}\langle f,\sum_{k\le K} \phi_k\rangle=\infty$ contradicting that $f$ is a distribution.

  • $f$ is compactly supported and has order $m$ means that $F=f \ast 1_{x>0}\frac{ x^{m+1}}{(m+1)!}$ is continuous and $$f = F^{(m+2)}$$

  • On $(-\infty,a)$, $F''=0$, on $(b,\infty)$, $F''$ is a polynomial, thus to show that $F''$ is a measure it suffices to show it has order $0$ on $(a-1,b+1)$, assume it is not the case, then $F''\chi$ has order $1$ for some $\chi \in C^\infty_c$, but $(F'' \chi)^{(m)}=f\chi +...$ has order $\le m$, thus $F''\chi$ has order $0$, $F''$ is a measure and $$f = (F'')^{(m)}$$

Improve this answer
$\endgroup$
2
  • $\begingroup$ Hello @rens we pose $\phi(x)=\chi_{x>0}\frac{x^{m+1}}{(m+1)!}$ $$F^{m+2}=(f\ast \phi)^{m+2}=f\ast \phi^{m+2}=0$$ and $f\neq 0$ $\endgroup$
    – deval sidi
    Commented Dec 16, 2019 at 16:16
  • $\begingroup$ No $\phi^{(m+1)} =1_{x > 0}, \phi^{(m+2)}= (1_{x> 0})' = \delta$ and $(f\ast \phi)^{(m+2)} = f\ast \delta=f$ $\endgroup$
    – reuns
    Commented Dec 16, 2019 at 20:47
1
$\begingroup$

In addition to the other two good answers, one can make a somewhat stronger assertion: let $u$ be a distribution on $\mathbb R^n$, in the Sobolev space $H^{-\infty}$ (which contains all compactly-supported distributions, since $H^{+\infty}\subset C^\infty$ by Sobolev imbedding, and compactly-supported distributions are the dual of $C^\infty$). Then for index $s$ such that $u\in H^s$, for sufficiently large positive $k$, $f=(|x|^2+1)^{-k} \widehat{u}$ gives $\widehat f$ inside $H^{{n\over 2}+\epsilon}\subset C^o$, again by Sobolev imbedding. Then (up to normalization) $(\Delta -1)^k\widehat{f}=u$.

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged .