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How to show that $$\psi =(x^2\phi)’$$where $\phi$ is a test function, is a test function if and only if $\int_{-\infty}^{\infty} \psi\, dx=\int_{0}^{\infty} \psi\, dx=\psi(0)=0$

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The question is not worded correctly. If $\phi$ is a test function, then so is $\psi$, no additional conditions are needed. The actual question is: given that $\psi$ is a test function, show that the following are equivalent:

- $\psi$ is of the form $\psi=(x^2\phi)’$ (with $\phi$ a test function).
- $\int_{-\infty}^{\infty} \psi dx=\int_{0}^{\infty} \psi dx=\psi(0)=0$

That 1 implies 2 is explained in Daniel Fischer’s comment. In the converse direction, let $\Psi$ be an antiderivative of $\psi$ such that $\Psi(0)=0$. The first two conditions in 2 imply that $\Psi$ vanishes at $\pm \infty$, hence is a test function. The last condition says that $\Phi'(0)=0$. From L’Hôpital’s rule it follows that the function $\phi(x)=\Psi(x)/x^2$ has a finite limit at $0$, namely $\Psi”(0)/2$. Extending it by $\phi(0)=\Psi”(0)/2$, we get a continuous function on $\mathbb R$. It remains to show that $\phi$ is $C^\infty$ smooth in a neighborhood of $0$, for which I refer to Quotient of two smooth functions is smooth.

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