Given a Lipschitz function $f(x)$ with $f(0)=0$, its derivative exists and is in $L^{\infty}$ almost everywhere under Lebesgue measure by Rademacher theorem. However, it is not continuous. Applying Fourier transform to the derivative, I wish to obtain somehow a $C^1$-version of $f$. Then I will do some calculation with the transformed version and then transform back.
I will work with the integral $$f(t)=\int_0^tf'(x)\,dx$$ and its transformed version is thus $$F(t)=\int_0^t\int_0^tf'(\xi)e^{-i\xi x} \,d\xi \,dx,$$ where I perform the fourier transform on the bounded domain $[0,t]$. However, I wish express the transformed version $F(t)$ as some operation of $f(t)$, i.e. rewritting into $F(t)=O(f(t))$, where $O$ is some mapping from space of $f(t)$ into space where $F(t)$ lives. Is it possible to find an expression for such mapping?
My Goal: Once such mapping is found, I will investigate the continuity of the mapping. Then I will be able to approximate $F(t)$ by choosing an approximate sequence of $f(t)$ and an appropriate topology.
Integration by parts: $$ \begin{split} F(t) &=\int_0^t\int_0^tf'(\xi)e^{-i\xi x} \,d\xi \,dx\\ &=f(t)\int_0^te^{-ixt}dx+\int_0^t\int_0^tixf(\xi)e^{-i\xi x} \,d\xi \,dx\\ &=\frac{1-e^{-it^2}}{it}f(t)+\int_0^t\frac{(i-\xi)e^{-i\xi t}-1}{\xi^2}\cdot f(\xi)\,d\xi . \end{split} $$