Test functions as a taylor series with integral remainder

**measureman** · May 11th 2011, 10:52 AM

Hi All,

Given a test function $\phi$ (C^infinity function with compact support), I have shown that we can write $\phi = \sum_{j = 0}^{m - 1} \frac{x^j}{j!}\phi^{(j)}(0) + \frac{1}{(m - 1)!}\int_0^x (x - t)^{m - 1}\phi^{(m)}(t) \, \mathrm{d}t.$

Choose a test function $\phi_0$ on $\R$ such that $\phi_0(x) = 1$ on a neighbourhood of $x = 0$ . I have to show the following:
On this neighbourhood, any test function $\phi$ on $\R$ can be written as $\phi = \phi_0(x)\sum_{j = 0}^{m - 1} \frac{x^j}{j!}\phi^{(j)}(0) + \chi(x)$

where $\chi(x)$ is a test function on $\R$ such that $\chi^{(j)}(0) = 0$ for $j = 1, \ldots m - 1$ .

I cannot see where this is coming from.

I have considered the function defined by $\chi(x) = \phi(x) - \phi_0(x)\sum_{j = 0}^{m - 1} \frac{x^j}{j!}\phi^{(j)}(0)$ , but I do not think this has the desired derivative properties.

Can anyone help?

**girdav** · May 11th 2011, 11:35 AM

You can show the derivative properties by the Leibniz's rule (notice that $\phi_0^{(k)}(0)=0$ if $k\geq 1$ .

**measureman** · May 11th 2011, 11:43 AM

Originally Posted by girdav

You can show the derivative properties by the Leibniz's rule (notice that $\phi_0^{(k)}(0)=0$ if $k\geq 1$ .

Ah cheers, I tried this and thought I ended up with \chi'(x) = \phi'(x) - now I notice the term that cancels.

**measureman** · May 13th 2011, 08:53 AM

I now have another problem.

I have shown that if \phi is a test function on R, with the first m - 1 derivatives at the origin being zero, then \phi = x^m\psi for some test function \psi. (*)

So it follows from my first post that $\phi(x) = \phi_0(x)\sum_{j = 0}^{m - 1} \frac{x^j}{j!}\phi^{(j)}(0) + x^m\mu(\phi(x))$ (**)
for some mapping \mu from test functions to test functions.

If v is a distribution, then apparently I can define $u(\phi) = v(\mu(\phi))$ , and u is a distribution. Does anyone know how to prove this?

The hint in the book is to use (*) and (**), but I just can't see it.

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