$Antiderivative of a Schwartz-function$

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Let $\mathcal{S}(\mathbb{R})$ be the Schwartz-Space. Prove that if $f \in \mathcal{S}(\mathbb{R})$ and $\int_{-\infty}^\infty f(t)dt=0$, then there is $F \in \mathcal{S}(\mathbb{R})$ such that $f=F'$.

Fourier Analysis

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Daniel90

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Let $F(x) = -\int_x^\infty f(t)\,dt$. Then $F(-\infty)=0$, since $\int_{\mathbb R}f(t)\,dt=0$. The decay at $+\infty$ follows from $$|F(x)| \le \int_x^\infty C(n)\, t^{-n}\,dt = O(x^{1-n})$$ where $n$ can be arbitrarily large. The decay at $-\infty$ follows similarly, by using $F(x) = \int_{-\infty}^x f(t)\,dt$ which holds since the integral of $f$ is $0$: $$|F(x)| \le \int_{-\infty}^x C(n)\, |t|^{-n}\,dt = O(|x|^{1-n})$$

Thus $F$ is a Schwartz function.

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$Antiderivative of a Schwartz-function$

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