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Let $\theta$ be a given fixed test function ($\mathscr{D}(\mathbb{R})=C_c^\infty(\mathbb{R}))$ such that $\theta(a)=1$ for $a \in \mathbb{R}$. Then why is every test function $\chi \in \mathscr{D}(\mathbb{R})$ of the form: $\chi = \lambda \theta + \psi$ with $\psi = (x-a)\phi \in \mathscr{D}(\mathbb{R})$, $\phi \in \mathscr{D}(\mathbb{R})$ such that $\lambda = \chi(a), \psi(a)=0$?

This is equivalent to saying that all test functions that vanish at $x=a$ is given by $(x-a)\phi$ for some $\phi \in \mathscr{D}(\mathbb{R})$, but I cannot see why this is so.

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This is really just Taylor's theorem. Consider $$\frac{\chi(x)-\theta(x)\chi(a)}{x-a}.$$ This is readily seen to be a test function (or extend to one), and it vanishes at $x=a$ by applying Taylor's theorem. Hence, $$\frac{\chi(x)-\theta(x)\chi(a)}{x-a}=R_a(x)$$ for some test function $R_a$, or $$\chi=\lambda\theta+\psi,$$ where $\psi(x)=(x-a)R_a(x).$ It is helpful to view Taylor's theorem as a statement on division, and you'll see this approach in most distribution theory textbooks.

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  • $\begingroup$ I thought about Taylor's formula but I cannot see how it applies here directly. Could you elaborate this please? So in this case, we have $f \in C_c^\infty(\mathbb{R})$ with $f(a)=0$ then $f(x)=f'(a)(x-a) + R_a(x)(x-a)$ where $R_a(x) \to 0$ as $x \to a$. So we would like to set $\psi(x) = (x-a)(f'(a)+R_a(x))$, but how do we know that $R_a$ is also a test function? I cannot see any conditions that give this from Taylor's theorem. $\endgroup$ Jan 15 at 10:56
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    $\begingroup$ Think of it more in this way: if you Taylor expand, then the $x-a$ terms cancel, so you know that your function on the right extends smoothly ($x=a$ is not a problem). To see that it's compactly supported, well, the numerator is compactly supported ($x=a$ has nothing to do with that, especially is we know that it's not a problematic point). That cancellation wouldn't affect the support. $\endgroup$
    – cmk
    Jan 15 at 14:16

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