Vtyjkyuk

So given a integrable function $f colon mathbbR to mathbbR$, we define the function $$g colon mathbbR to mathbbR, quad x mapsto sup_epsilon>0 dfrac2epsilon$$ and we assume that $g in mathcalL^2(mathbbR)$. I now need to proof that the Fourier transform $mathscrF(f)$ is integrable.

The start of my proof is as follows:

Define $$g_epsilon colon mathbbR to mathbbR, quad x mapsto dfracf(x-epsilon)-f(x+epsilon)2epsilon$$ for every $epsilon > 0$.

We know that $g_epsilon(x) leq g(x) $ for every $x in mathbbR$ and $epsilon > 0$. Then I calculated the Fourier transform of $g_epsilon$
$$ mathscrF(g_epsilon)(t) = dfrac-iepsilon sin(2piepsilon t)mathscrF(f)(t)$$
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 $| mathscrF(g) |_2 = | g |_2$.

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.

Actually, you have already solved the problem, but have not quite noticed it yet :)

Indeed, you have proven that $(xmathcalF(f)(x))^2$ is integrable. But, by a Cauchy-Schwarz inequality,

$$ int_mathbbRbackslash [-1,1] |mathcalF(f)(x)| , dx = int_mathbbRbackslash [-1,1] |xmathcalF(f)(x)| times frac1 , dx $$

$$ le left( int_mathbbRbackslash [-1,1] |xmathcalF(f)(x)|^2 , dx right)^1/2 times left( int_mathbbRbackslash [-1,1] frac1x^2, dx right)^1/2 < +infty.$$

So, it suffices to analyse the interval $[-1,1].$ But, as we assumed that $f in L^1(mathbbR)$, we get that $mathcalF(f)$ 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 $g in L^2(mathbbR)$ is equivalent to $f$ having a (distributional) derivative $f' in L^2(mathbbR).$ Of course, one needs to make sense of $f$ having a distributional derivative, but this happens if, for instance, $f in mathcalS'(mathbbR).$

Proving that $g in L^2(mathbbR)$ implies $f in L^2(mathbbR)$ is standard, whereas the reverse implication follows from the fact that

$$ g(x) le M(f')(x), forall x in mathbbR,$$

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 $mathcalF(g)(x) sim int g$ for $x sim 0,$ then, as $G' = g Rightarrow mathcalF(G)(x) sim fracint gx$ 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 $G in L^1(mathbbR).$