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I am wondering the link as the title implies. The Spring 87 problem in Berkeley Problems in Mathematics is as follows:

Let $V$ be a finite dimensional linear subspace of $C^{\infty}(\mathbb{R})$. Assume that $V$ is closed under differentiation. Prove that there is a constant coefficient operator $L=\sum^{n}_{k=0}a_{k}D^{k}$ such that $V=\{f:Lf=0\}$.

I am wondering why the finite dimensional condition given would imply such a strong result. Because if we assume the required relationship the reverse is not necessarily true(the whole space is obviously closed under any differential operator), I feel something deeper may be buried in this problem I do not know. A hint or an illuminating example would be mostly welcome. I just do not know how to attack this problem.

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Let $\{f_1,\dots,f_N\}$ a basis of $V$. Since $f'_k\in V$ for all $k$, we can write $f'_k=\sum_{l=1}^Na_{k,l}f_l,$ and using matrices $\pmatrix{f'_1\\\vdots\\ f'_N}=A\pmatrix{f_1\\\vdots\\ f_N}.$ Let $P$ the minimal polynomial of $A$ over $\Bbb R$. We can check that the differential operator $L$ associated to it is such that $V\subset \{f\mid Lf=0\}$. The general theory of differential equations ensures the converse: if $Lf=0$, then $f$ is in the vector space generated by $x^pe^{\lambda x}$, where $\lambda$ are eigenvalues of $A$, $m$ is the multiplicity of the root $\lambda$ in the minimal polynomial and $0\leq p\leq m$. Each of these maps are in $V$.

Note that if $V=\ker L$ for $L$ of the form $\sum_{k=0}^n a_kD^k$, where $a_k$ are constant, then $V$ is necessarily finite dimensional.

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    Davide, if so then we would have $0,1...m$ together $m+1$ basis vectors.2012-07-22