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Question: What can you conclude about a tempered distribution $G\ \in\ S'(R^n)$ that is concentrated in some k-dimensional manifold $M\ \subset\ R^n$ (for k < n)? More specifically, is there a result analogous to the following n=1 result?

$n=1$ result (hope I remember it correctly):

Let $\ S(R)\ $ be the set of Schwartz functions ($C^\infty$ functions $f:\ R\ \to\ C\ $ s.t. $\ f^{(n)}$ goes to 0 at infinity faster than any inverse power of x (for n=0, 1, ...)). Let $\ S'(R)\ $ be the set of tempered distributions. $G\ \in\ S'(R)$ is said to be concentrated in a set $A\ \subset\ R\ $ iff $\forall\ \phi\ \in\ S(R)$ that vanishes on some open set $B\ \supset\ A\ $, $G(\phi)\ =\ 0$.

Suppose $\ G\ $ is concentrated in {$\ x\ $}, for some $x\ \in\ R$. Then $\exists\ c_0,\ ...\ c_L\ \in\ C\ $ s.t. $\ G\ $ = $\sum_{j=0}^L\ c_j\ \delta_x^{(j)}$.

[where $\delta_x^{(j)}\ (\phi)\ \equiv\ (-1)^j\ \phi^{(j)}(x)\ $]

  • 1
    That's a wrong definition of "concentrated". For example, you would like $\delta'$ to be concentrated on $\{0\}$, but there are functions (e.g. $\phi(x) = x \exp(-x^2)$) with $\phi(0) = 0$ but $\delta'(\phi) = \phi'(0) \ne 0$.2011-03-23
  • 0
    Instead, you want to say that $\phi$ is supported in $R \backslash A$, i.e. it is 0 on a neighbourhood of $A$.2011-03-23

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