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Exercise 5: If $f\in L^1$ and $\int |t\hat{f}(t)|<\infty$, prove that $f$ coincide a.e. with a differentiate function whose derivative is $i\int_{-\infty}^{\infty}t\hat{f}(t)e^{ixt}dt$

I know a theorem which claims If $f\in L^1$,$\exists g\in L^1$ such that $\hat{g}(t)=t\hat{f}(t)$ then $f(x)=\int_{-\infty}^{x}g(t)dt$ a.e. I think it may have some relation between them, who can give me some suggestion?

Thank you very much!

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If $t\hat{f} \in L^{1}$, define $$ g(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}s\hat{f}(s)ie^{isx}\,ds, \;\;\; x \in\mathbb{R}. $$ The function $g$ is continuous by the Lebesgue dominated convergence theorem. Using Fubini's theorem to switch order of integration gives $$ \int_{0}^{t}g(x)\,dx = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\hat{f}(s)(e^{ist}-1)\,ds = f(t)-C,\;\;\; a.e. t\in\mathbb{R}. $$ In other words, $f(t) = C+\int_{0}^{t}g(x)\,dx$ a.e.. The function $C+\int_{0}^{t}g(x)\,dx$ is continuously differentiable on $\mathbb{R}$ with derivative $g$, which has the stated form (up to a constant multiple of $1/\sqrt{2\pi}$ that one of us has wrong.)

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  • $\begingroup$ Excellent proof! Thank you very much. $\endgroup$ Nov 7, 2014 at 0:43
  • $\begingroup$ @Ylath : Thank you, and you're welcome. $\endgroup$ Nov 7, 2014 at 1:01

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