Understanding Test functions

Question

Now I'm studying distributions and I have a quesition. One of conditions of function $\varphi$ to be a test function is that it has to be in $C^{\infty}(\mathbb{R})$. So, what are the methods to prove it? For example, $$ \varphi = \begin{cases} e^{\frac{1}{x^2-1}}, \text{if} \, |x|< 1, \\ 0, \text{if} \, |x| \geq 1. \end{cases} $$

Answer 1

You need to prove that $\varphi$ and all its derivatives are continuous.

So first check that $\lim_{x \nearrow 1} \varphi(x) = \varphi(1)=0$ and $\lim_{x \searrow -1} \varphi(x) = \varphi(-1)=0$.

Afterwards prove that the derivatives are continuous.
For $|x|>1$, we have $\varphi'(x)=0$ and for $|x|<1$, we have $\varphi'(x)=\frac d {dx} e^{\frac 1 {x^2-1}}$. Thus we need to check $$\ \lim_{x \nearrow 1} \frac d {dx} e^{\frac 1 {x^2-1}}= \lim_{x \nearrow 1} \varphi'(x) = \lim_{x \searrow 1} \varphi'(x)=0,$$ and the corresponding statement for $x\to -1$.

Then continue for the derivatives of higher (arbitrary) order. Here you need to use the fact that the exponential function grows faster than any polynomial.

Answer 2

There are two parts of the proof. Second part is to prove that it's infinitely differentiable at $\pm 1$ and on the first is to prove it to be everywhere else.

The first part we see obviously it's true for $|x|>1$. For $|x|<1$ we use properties of rational functions under derivation. We have that if we have a class of functions on the form $r(x)e^{q(x)}$ where $r$ and $q$ are rational and continuous in $(-1,1)$ we have it's derivative $(r'(x)+r(x)q'(x))e^{r(x)}$ is also in that class and it will be indefinitely differentiable. Note that the derivative $r'(x)$ of a rational does not add any poles/discontinuitis as $D_x f(x)/g(x) = (f'(x)g(x)-g'(x)f(x))/(g(x))^2$ so it's only discontinuous when $g(x)=0$ which was the poles of the original rational function.

Now for the point $\pm 1$ we have to consider limits of $r(x)e^{-1/({x^2-1})}/(x-1)$ as $x\to 1$ which are $0$. So this will lead it to be indefinitely differentiable at $\pm 1$ as well. The limit is the "in"-side limit and the "out"-side limit is obviously $0$ as well. The important point here is that $e^{-1/(x^2-1)}$ approaches $0$ so fast it will dominate any pole of any rational function (ie that $e^{-1/(x^2-1)}/(1-x)^n \to 0$ for any $n$).