Exercices : Let $\varphi\in\mathbb{S}(\mathbb{R}) $ such that : $\displaystyle\int\limits_\mathbb{R}\varphi (x)dx=0 $
Let defined : $$\psi(t)=\displaystyle\int\limits_{-\infty}^{t}\varphi(x)dx$$ Then prove that : $\psi\in\mathbb{S}(\mathbb{R}) $
My attempts:
To prove $\psi\in\mathbb{S}(\mathbb{R}) $ We need to prove : $$q_{m,n}(\psi)=\sup_{t\in\mathbb{R}}|t^ {n}\psi^{(m)}(t)|<+\infty$$ So : $\psi^{(m)}(t)=\varphi^{(m-1)}(t)$
by " liebniz rule " So :
$$q_{m,n}(\psi)=\sup_{t\in\mathbb{R}}|t^{n}\varphi^{(m-1)}(t)|=q_{m-1,n<p}(\varphi)<+\infty\Rightarrow\psi\in\mathbb{S}(\mathbb{R})$$ Is my solution correct? But why give me
$\displaystyle\int\limits_\mathbb{R}\varphi (x)dx=0 ~~???$
Thanks!