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Inspired by one of the answers to this post, I constructed my own proof. In addition, I constructed a basis of the solution space. I'd like to know if the following is correct.

Let $(a_0,\ldots,a_n)\in\mathbf{C}^{n+1}$ ($a_n\neq 0$). If $I\subset\mathbf{R}$ is an open interval and $f\colon I\to\mathbf{C}$ has the property $a_0f^{(0)}+...+a_nf^{(n)}=0$, then $f\in C^{\infty}$$.^1$ This means that "solving" the differential equation is equivalent to determining the set \begin{equation} X=\left\{f\in C^{\infty}:a_0f^{(0)}+...+a_nD^nf^{(n)}=0\right\}. \end{equation} If we introduce \begin{align} D^k\colon C^{\infty}&\to C^{\infty}\\ f&\mapsto f^{(k)} \end{align} for all $k\in\mathbf{N}_0$ and \begin{equation} O:=a_0D^0+...+a_nD^n, \end{equation} then \begin{equation} X=\ker O. \end{equation} Like for all polynomials, there is a $(c_1,\ldots,c_n)\in\mathbf{C}^n$ with \begin{equation} O=a_n\prod_{k=1}^nD^1-c_kD^0=a_n\prod_{k=1}^nD-c_kI \end{equation} Thus, if we define $D-a:=D-aI$ ($a\in\mathbf{C}$), \begin{equation} X=\ker\prod_{k=1}^nD-a_k. \end{equation} $D-a$ and $D-b$ commute for all $a,b\in\mathbf{C}$, thus we can reorder the product however we want. Therefore we can concentrate on finding the kernel of an operator that has the form \begin{equation}\tag{1} (D-a_1)^{k_1}\cdots(D-a_m)^{k_m}. \end{equation} (This will allow us to construct a basis later on.)

$D-a$ has a $1$-dimensional kernel and is surjective (here is the proof), meaning that $\dim\ker O=n$ according to this post.

We can even construct a basis: \begin{align} e_n(a)\colon I&\to\mathbf{C}\\ x&\mapsto\frac{x^n}{n!}e^{ax} \end{align} Thanks to the factorial, we have $De_n(a)=e_{n-1}(a)+ae_n(a)$ for $n\in\mathbf{N}$ and $De_0(a)=ae_0(a)$. This means $(D-c)e_n(a)=e_{n-1}(a)+(a-c)e_n(a)$ for $n\in\mathbf{N}$ and $(D-c)e_0(a)=(a-c)e_0(a)$.

Thus, \begin{equation} \{e_0(a_1),\ldots,e_{k_1-1}(a_1),\ldots,e_0(a_m),\ldots,e_{k_m-1}(a_m)\} \end{equation} is a basis of the operator in equation $(1)$.

$^1$ $a_0f^{(0)}+...+a_nf^{(n)}=0$ is equivalent to \begin{equation} f^{(n)}=\frac{-a_0f^{(0)}\ldots-a_{n-1}f^{(n-1)}}{a_n} \end{equation} and that is equivalent to \begin{equation} f^{(m)}=\frac{-a_0f^{(m-n)}\ldots-a_{n-1}f^{(m-1)}}{a_n} \end{equation} for all $n\leq m\in\mathbf{N}$ (you can prove this using induction).

Filippo
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  • The factorization of the polynomial differential operator holds formally, and it factors as if it was an ordinary polynomial. –  Nov 10 '20 at 08:31
  • @YvesDaoust Thank you very much for the confirmation :) – Filippo Nov 10 '20 at 08:34
  • Please stop editing this question. Making repeated edits confuses the intent of your question, has the chance of invalidating existing answers, and bumps the question to the top of the front page over and over again. Please decide what it is that you want to ask, ask that question, and then leave it alone as much as possible. – Xander Henderson Dec 20 '20 at 13:00
  • @XanderHenderson The questions I originally had weren't answered (I think I asked too many things at once, since I basically wanted a completion of the proof sketched under the first link), so I asked more specific questions to fill the gaps - which was succesfull. Then I thought that it could be useful for others if I added that information. But I guess that it would have been much better to add an answer, instead of editing the question (I have to admit that it isn't really a question anymore). – Filippo Dec 20 '20 at 13:59

1 Answers1

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In order to get (2), you need your caracteristic polynom : $\chi(X)=\sum_{i=0}^na_iX^i $ to have simple roots

If you work in $\mathbb{C}$, you have the following result :

$$\ker\left({O=a_n\prod_{i=1}^{k}(D-c_i1)^{m_i}} \right)=\bigoplus_{i=1}^k\ker\left((D-c_i1)^{m_i}\right)$$

NHL
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  • Is $\chi$ meant to be the map $\chi\colon\mathbb C\to\mathbb C,\ X\mapsto\sum_{i=0}^na_iX^i$? – Filippo Nov 09 '20 at 14:41
  • You can see that your differential equation is a composition of $D$ and the polynom $\chi$. For a polyno, having simple roots mean that it can be decomposed into a product of polynoms of degree 1. For example the polynom $(X^2+1)^2 $ is of degree 4, but it has only 2 roots of degree 2 – NHL Nov 09 '20 at 14:41
  • I indeed made a mistake, the result for polynoms of endomorphisms is that if P and Q are co-prime, then $\ker((P\circ Q)(f))=\ker(P(f))\oplus\ker(Q(f)) $ – NHL Nov 09 '20 at 14:45
  • What do you mean with the composition of $D$ and $\chi$? How is $\chi$ defined? – Filippo Nov 09 '20 at 14:57
  • $\chi$ is defined as th polynom such that $\chi\circ D=0$, where you replace X by D and the power become composition. You can check the formalism on wikipedia on "polynoms of endomorphisms" – NHL Nov 09 '20 at 14:58