Suppose $f \in L^2(\mathbb{R})$ and assume $\|\gamma \, \hat{f} \|_2 < \infty$.
Pick $~f_n \in C^{\infty}_{\downarrow}(\mathbb{R})$ such that $$\|f_n-f \|_2 + \|f_{n}^{'}-f'\|_2 = \int(1+4\pi^2\gamma^2)\, | \, \hat f_n-\hat f|^2 \, d\gamma \rightarrow 0 ~~ \text{as}~n\rightarrow\infty,$$ where $f' := [2\pi i \gamma \hat f]^{\check{}}.$
(I've checked that we can always pick such $f_n$ under the assumption $ \| \gamma \, \hat{f} \|_2 < \infty$.)
I also know that $$|f_n(x) - f(x) | \leq \|\hat{f}_n - \hat{f} \|_1 \leq \left[\int(1+4\pi^2 \gamma^2 )^{-1} \, d\gamma \right]^{1/2}\left[\int(1+4\pi^2 \gamma^2) \, |\hat{f}_n-\hat{f}|^2 \, d\gamma \right]^{1/2}. $$ So, $f_n \rightarrow f$ uniformly.
What I want to prove is that $[2\pi i \gamma \, \hat{f}_n ]^{\check {}}(x) \rightarrow [2\pi i \gamma \, \hat{f} ]^{\check {}}(x)$ as $n\rightarrow \infty$ for each $x$.
I tried to prove by considering the following:
$$|[2\pi i \gamma \, \hat{f} ]^{\check {}}(x)- [2\pi i \gamma \, \hat{f} ]^{\check {}}(x)| \leq 2\pi\int |\gamma \, \hat{f}_n - \gamma \, \hat{f}| \, d\gamma. $$
Any help will be appreciatd!
I proved that it suffices to show that $\gamma \, \hat{f}(\gamma)$ is integrable. But I failed to prove this.