Factoring out test functions every $\chi \in \mathscr{D}(\mathbb{R})$ is of the form $\chi = \lambda \theta + (x-a)\phi$

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.

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.
• 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. Jan 15 at 10:56
• 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.