- #1
bruno67
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A function f is both integrable and infinitely differentiable, i.e. f\in L_1(\mathbb{R}) \cap C^{\infty}(\mathbb{R}). Is it correct to say that this implies that the derivatives of f are also in L_1(\mathbb{R})? My reasoning: we have I<\infty, where
I=\int_{-\infty}^{\infty} f(x) dx = [x f(x)]_{-\infty}^{\infty} - \int_{-\infty}^{\infty} xf'(x) dx = - \int_{-\infty}^{\infty} xf'(x) dx
where the boundary term disappears because, since f is integrable, we must have f(x) =O(x^{-1-\alpha}) for |x|\to \infty, for some \alpha>0. Hence f'(x)=O(x^{-2-\alpha}), and in general f^{(n)}(x)=O(x^{-n-1-\alpha}), for |x|\to \infty.
I=\int_{-\infty}^{\infty} f(x) dx = [x f(x)]_{-\infty}^{\infty} - \int_{-\infty}^{\infty} xf'(x) dx = - \int_{-\infty}^{\infty} xf'(x) dx
where the boundary term disappears because, since f is integrable, we must have f(x) =O(x^{-1-\alpha}) for |x|\to \infty, for some \alpha>0. Hence f'(x)=O(x^{-2-\alpha}), and in general f^{(n)}(x)=O(x^{-n-1-\alpha}), for |x|\to \infty.
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