Proof that a Fourier transform is integrable So given a integrable function $f \colon \mathbb{R} \to \mathbb{R}$, we define the function $$g \colon \mathbb{R} \to \mathbb{R}, \quad x \mapsto \sup_{\epsilon>0} \dfrac{|f(x-\epsilon)-f(x+\epsilon)|}{2\epsilon}$$ and we assume that $g \in \mathcal{L}^2(\mathbb{R})$. I now need to proof that the Fourier transform $\mathscr{F}(f)$ is integrable.
The start of my proof is as follows:
Define $$g_\epsilon \colon \mathbb{R} \to \mathbb{R}, \quad x \mapsto  \dfrac{f(x-\epsilon)-f(x+\epsilon)}{2\epsilon}$$ for every $\epsilon > 0$.
We know that $g_\epsilon(x) \leq g(x) $ for every $x \in \mathbb{R}$ and $\epsilon > 0$. Then I calculated the Fourier transform of $g_\epsilon$ 
$$ \mathscr{F}(g_\epsilon)(t) = \dfrac{-i}{\epsilon} \sin(2\pi\epsilon t)\mathscr{F}(f)(t)$$
But from this point, it get's a little tricky and I don't know if I'm correct, but I think that we can assume that $g$ is an integrable function and then we know that $\| \mathscr{F}(g) \|_2 = \| g \|_2$. 
Now, I want to let $\epsilon \to 0$ and then I think I'm really close to the solution, but I can't figure out how.
 A: Actually, you have already solved the problem, but have not quite noticed it yet :) 
Indeed, you have proven that $(x\mathcal{F}(f)(x))^2$ is integrable. But, by a Cauchy-Schwarz inequality, 
$$ \int_{\mathbb{R}\backslash [-1,1]} |\mathcal{F}(f)(x)| \, dx =  \int_{\mathbb{R}\backslash [-1,1]} |x\mathcal{F}(f)(x)| \times \frac{1}{|x|} \, dx $$
$$ \le \left( \int_{\mathbb{R}\backslash [-1,1]} |x\mathcal{F}(f)(x)|^2 \, dx \right)^{1/2} \times \left(  \int_{\mathbb{R}\backslash [-1,1]} \frac{1}{x^2}\, dx \right)^{1/2} < +\infty.$$
So, it suffices to analyse the interval $[-1,1].$ But, as we assumed that $f \in L^1(\mathbb{R})$, we get that $\mathcal{F}(f)$ is continuous, and therefore bounded, in $[-1,1].$ This implies the required integrability of the Fourier transform. 

On the other hand, we see that the integrability of $f$ is crucial. Indeed, one can show, by using elementary Sobolev space methods, that the condition that $g \in L^2(\mathbb{R})$ is equivalent to $f$ having a (distributional) derivative $f' \in L^2(\mathbb{R}).$ Of course, one needs to make sense of $f$ having a distributional derivative, but this happens if, for instance, $f \in \mathcal{S}'(\mathbb{R}).$ 
Proving that $g \in L^2(\mathbb{R})$ implies $f \in L^2(\mathbb{R})$ is standard, whereas the reverse implication follows from the fact that 
$$ g(x) \le M(f')(x), \forall x \in \mathbb{R},$$
where $M$ stands for the usual Hardy-Littlewood maximal function on the real line. As $M:L^2 \to L^2$ boundedly, we are through. 
To come up with a counterexample is not very difficult then: for any function $g$, continuous, of compact support and such that $\int g \ne 0,$ Let 
$$G(t) = \int_{-\infty}^t g(x) \, dx.$$
This function satisfies the hypotheses of the problem, but clearly, as $\mathcal{F}(g)(x) \sim \int g$ for $x \sim 0,$ then, as $G' = g \Rightarrow \mathcal{F}(G)(x) \sim \frac{\int g}{x}$ for $x \sim 0,$ which does not integrate.
Of course, this all can be made formal, but this is intended as a sketch to just illustrate how important it is that $G \in L^1(\mathbb{R}).$ 
