A function f is both integrable and infinitely differentiable, i.e. [itex]f\in L_1(\mathbb{R}) \cap C^{\infty}(\mathbb{R})[/itex]. Is it correct to say that this implies that the derivatives of f are also in [itex]L_1(\mathbb{R})[/itex]? My reasoning: we have [itex]I<\infty[/itex], where(adsbygoogle = window.adsbygoogle || []).push({});

[tex]I=\int_{-\infty}^{\infty} f(x) dx = [x f(x)]_{-\infty}^{\infty} - \int_{-\infty}^{\infty} xf'(x) dx = - \int_{-\infty}^{\infty} xf'(x) dx[/tex]

where the boundary term disappears because, since f is integrable, we must have [itex]f(x) =O(x^{-1-\alpha})[/itex] for [itex]|x|\to \infty[/itex], for some [itex]\alpha>0[/itex]. Hence [itex]f'(x)=O(x^{-2-\alpha})[/itex], and in general [itex]f^{(n)}(x)=O(x^{-n-1-\alpha})[/itex], for [itex]|x|\to \infty[/itex].

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# Regularity condition

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