# Exercise from Stein with partial differential operator

I have again something from Stein-Shakarchi I would really appreciate some help with. Any references are also welcome!

Suppose $L$ is a linear partial differential operator with constant coefficients. Show that when $d \geq 2$, the linear space of solutions $u$ of $L(u)=0$ with $u \in C^{\infty}(\mathbb{R}^{d})$ is not finitely dimensional.

EDIT: $L$ takes the form $$L= \sum_{|\alpha| \leq n}{a_{\alpha}\left(\frac{\partial{}}{\partial{x}}\right)^{\alpha}}$$ with $a_{\alpha} \in \mathbb{C}$ constants.

-
How is a partial differential operator defined? – Davide Giraudo May 2 '12 at 8:02
Edited. Sorry for not being thorough. – Anna May 2 '12 at 23:58

For $p=(p_1,p_2)\in\Bbb C^2$, define $f_p(x)=\exp\left(p_1x_1+p_2x_2\right)$. If $(p_1^{(k)},p_2^{(k)})\in\Bbb C^2$ are distinct (i.e. $(p_1^{(k)},p_2^{(k)})\neq (p_1^{(j)},p_2^{(j)})$ if $j\neq k$, then the set of maps $\{f_{p^{(j)}},1\leq j\leq N\}$ is linearly independent. Indeed, if $\sum_{j=1}^Nc_jf_{p^{(j)}}$, let $J_1,\ldots,J_l$ a partition of $\{1,\ldots,N\}$ such that for all $j_1,j_2\in J_k,p_1^{(j_1)}=p_1^{(j_2)}$. Then we have $$\forall x_2\in\Bbb R,\forall 1\leq k\leq l,\quad \sum_{j\in J_k}c_je^{p_2^{(j)}x_2}=0.$$ Since $p_2^{(j)}$ are distinct, we get that $c_j=0$ for all $j\in J_k$ hence $c_j=0$ for all $j$.
It's a huge contrast with $d=1$ (in this case the space is necessary finite dimensional).