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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, $$ \varphi = \begin{cases} e^{\frac{1}{x^2-1}}, \text{if} \, |x|< 1, \\ 0, \text{if} \, |x| \geq 1. \end{cases} $$

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  • $\begingroup$ Related. $\endgroup$ – Pedro May 2 '17 at 3:00
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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 $n$).

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  • $\begingroup$ Sounds pretty convincing. Thanks! $\endgroup$ – Kamil Apr 4 '17 at 12:37
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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.

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  • $\begingroup$ Not sure that this is correct proof. I think we need Taylor series, or may be someting more sophisticated. $\endgroup$ – Kamil Apr 4 '17 at 11:57
  • $\begingroup$ I saw that i didnt write down that $\varphi$ is smooth smooth on $\mathbb R \setminus \{-1,1\}$, but this is obviously true. The rest of the prooof works, but misses some details which need to be filled in by the op. It is the same idea as skyking uses. $\endgroup$ – klirk Apr 4 '17 at 13:21

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