Relationship between decay of Fourier transform and smoothness in $L^2$ This is Ex $2.3.6$ in Dym and Mckean's Fourier Series and Integrals.
One can define the operator $\hat{f}(\gamma)$ for $f \in L^2(\mathbb{R})$ as $\lim_{b \rightarrow \infty, \, a \rightarrow -\infty} \int_b^a f(x) e^{-2 \pi i x \gamma} dx$ where the limit takes place in $L^2$. 
It turns out this is a length preserving isomorphism in $L^2$, with inverse operator $\check{f}(x) = \lim_{b \rightarrow \infty, \, a \rightarrow -\infty} \int_b^a f(\gamma) e^{2 \pi i x \gamma} d\gamma$. 
I am now asked to show that for $f \in L^2$, if $\gamma^n \hat{f} (\gamma) \in L^2$ then $f$ is $C^\infty$ and all derivatives $D^n f \in L^2$.  
Here is my proof which doesn't work yet:
Let $g = \check{[(2 \pi i \gamma) \hat{f}]}$ and the idea is to show that $g = f'$ via $f(x) = f(0) + \int_0^x g(s) ds$. First you can see that $\int_0^x g(s) dx \lt \infty $ a.e. in $x$ since $g \in L^2$ by assumption and the integral can be expressed as the inner product $(g, \mathbb{1}(0, x))$ . 
Then you can substitute $g$ to get $f(0) + \int_0^x \lim_{b \rightarrow \infty, \, a \rightarrow -\infty} \int_b^a 2 \pi i \gamma \,\hat{f}(\gamma) e^{2 \pi i t \gamma} d\gamma\, dt$. 
Now the bad part is that the inner integral is an $L^2$ limit whereas I want to be able to say $f(0) + \int_0^x \int_{-\infty}^{\infty} 2 \pi i \gamma \,\hat{f}(\gamma) e^{2 \pi i t \gamma} d\gamma\, dt = f(0) + \int_{-\infty}^{\infty} \int_{0}^{x} 2 \pi i \gamma \,\hat{f}(\gamma) e^{2 \pi i t \gamma} dt \,d\gamma = f(0) + \int_{-\infty}^{\infty} \hat{f}(\gamma)(e^{2 \pi i x \gamma} - 1) \,d\gamma = \check{\hat{f}}(x) - \check{\hat{f}}(0) + f(0) = f(x)$ which would complete the proof.
I don't know how to justify these operations though. 
Edit:
Here is something I overlooked. Inner product with $\mathbb{1}(0,x)$ is a continuous operator on $L^2$ and therefore we can pull the inner limits out to get the simpler expression $f(0) + \lim_{b \rightarrow \infty, \, a \rightarrow -\infty} \int_0^x \int_{a}^{b} 2 \pi i \gamma \,\hat{f}(\gamma) e^{2 \pi i t \gamma} d\gamma\, dt $. Interchange of integrals is justified because of Fubini and I think this completes the proof.
 A: Let $y, \xi \in \mathbb{R}$ be fixed. Note
$$
|e^{2\pi i\xi y} \left( \frac{e^{2\pi i \xi h} - 1}{h} \right) \hat{f}(\xi)| \leq |L \xi \hat{f}(\xi)|
$$
where $L$ is a Lipschitz constant for $\exp$. We can prove that the RHS is integrable in $\xi$.
We have
$$
\int | \xi \hat{f}(\xi) | d\xi = \int_{[-1,1]} |\xi \hat{f}(\xi)|d\xi + \int_{[-1,1]^c}|\xi \hat{f}(\xi)|d\xi.
$$
Using Cauchy-Schwarz we have
$$
\int_{[-1,1]} |\xi \hat{f}(\xi)|d\xi \leq ||\xi \hat{f}||_{L^2} || 1 ||_{L^2(-1,1)} = \sqrt{2}||\xi \hat{f}||_{L^2}
$$
which is finite by assumption. We have by Cauchy-Schwarz
$$
\int_{[-1,1]^c} \xi \hat{f}(\xi)d\xi = \int_{[-1,1]^c} \frac{1}{\xi} \cdot \xi^2\hat{f}(\xi)d\xi \leq || \xi^2 \hat{f}||_{L^2} || \xi^{-1} ||_{L^2([-1,1]^c)}.
$$
This quantity is also finite since $\xi^2 \hat{f} \in L^2$ by assumption.
Consider the difference quotient
$$
\frac{f(y+h) - f(y)}{h} = \frac{1}{h} \int e^{2\pi i \xi y} \left( e^{2\pi i \xi h} - 1 \right) \hat{f}(\xi) d\xi.
$$
Now use dominated convergence theorem to pass a limit in $h$ inside the integral. Make sure you explain why Fourier inversion works in this case.
