Let f be $L^1(R) \cap C_0(R)$ and satisfies $|\hat{f}(\alpha)|\leq A\frac{1}{|\alpha|}$, for all non zero real $\alpha$, for some positive A.
Then, show that for any $x \in R$,
$f(x) = \lim_{R\rightarrow \infty}\frac{1}{2\pi}\int_{-R}^{R} \hat{f}(\alpha)e^{i\alpha x} d\alpha$
holds.
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Similar to the proof of the inversion formula, I tried to show this by using poisson kernel, $H(x) = \frac{1}{2\pi}\int_{\mathbb R}e^{-|\alpha|}e^{i\alpha x} d\alpha, H_\lambda (x) = \lambda H(\lambda x)$.
(*)$|f(x) - \frac{1}{2\pi}\int_{-R}^{R} \hat{f}(\alpha)e^{i\alpha x} d\alpha| \leq$ $|f(x) - H_\lambda * f(x)| + \frac{1}{2\pi}|\int_{\mathbb R-[-R,R]}\hat{f}(\alpha)e^{i\alpha x}d\alpha| + |\frac{1}{2\pi}\int_{-R}^{R} (e^{-|\frac{\alpha}{\lambda}|}-1)\hat{f}(\alpha)e^{i\alpha x} d\alpha| $.
$|\int_{\mathbb R-[-R,R]}\hat{f}(\alpha)e^{i\alpha x}d\alpha| \leq 2\int_{R}^{\infty}e^{-\frac{\alpha}{\lambda}}\frac{ A}{\alpha}d\alpha \leq \frac{ 2A}{R}\int_{R}^{\infty}e^{-\frac{\alpha}{\lambda}}d\alpha \leq \frac{ 2A}{R}\frac{ 1}{\lambda}e^{-\frac{R}{\lambda}} $
$|\frac{1}{2\pi}\int_{-R}^{R} (e^{-|\frac{\alpha}{\lambda}|}-1)\hat{f}(\alpha)e^{i\alpha x} d\alpha| \leq (e^{-\frac{R}{\lambda}}-1)|\int_{-R}^{R}\hat{f}(\alpha)e^{i\alpha x} d\alpha| \leq (e^{-\frac{R}{\lambda}}-1)\int_{-R}^{R}||f||_1 d\alpha = 2R||f||_1(e^{-\frac{R}{\lambda}}-1) $
However, If i let $\lambda$ -> $\infty$, (*) becomes 0 for every R, if i don't let R to infinity. I think there should be mistake or misunderstanding. How can I show the equality?