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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 $F=f \ast 1_{x>0}\frac{ x^m}{m!}$ is continuous and $$f = F^{(m+1)}$$

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 $$f = (F')^{(m)}$$

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 $$f = (F'')^{(m)}$$

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 $F=f \ast 1_{x>0}\frac{ x^m}{m!}$ is continuous and $$f = F^{(m+1)}$$

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 $[a,b]$, 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 $$f = (F')^{(m)}$$

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 $F=f \ast 1_{x>0}\frac{ x^m}{m!}$ is continuous and $$f = F^{(m+1)}$$

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 $$f = (F')^{(m)}$$

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 $F=f \ast 1_{x>0}\frac{ x^m}{m!}$ is continuous and $$f = F^{(m+1)}$$

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 $[a,b]$, 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 $0$.

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 $F=f \ast 1_{x>0}\frac{ x^m}{m!}$ is continuous and $$f = F^{(m+1)}$$

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 $[a,b]$, 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 $$f = (F')^{(m)}$$