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Exercise from Stein with partial differential operator

Let $S$ be a linear partial differential operator in $\mathbb{R}^{n}$, $n \geq 2$ and consider the vector space of $f \in C^{\infty}(\mathbb{R}^{n})$ such that $Sf = 0$. Why is this vector space not finite dimensional?

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marked as duplicate by Davide Giraudo, Leonid Kovalev, Asaf Karagila, t.b., Chris Eagle Aug 17 '12 at 10:45

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A function can be constant in one direction and non-constant in another.

For instance, any smooth function of $y$ like $e^y$ or $\sin(y)$ can be thought of as a function on $\mathbb{R}^2$ which is smooth and annihilated by $\frac{\partial}{\partial x}$, and that's an infinite dimensional vector space of two-variable smooth functions.

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How does this help answer the question for a general linear partial differential operator? –  Nate Eldredge Apr 28 '12 at 21:56

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