Is $F(x)=\int_{\mathbb{R}}f(t)\frac{\sin(xt)}{t}dt$ differentiable for $f\in L^1$
The problem asks to show that $F$ is differentiable a.e, but I do not see why it would be a.e
By MVT and DCT (Leibniz rule) we get that $F'(x)=\int_{\mathbb{R}}f(t)cos(xt)dt$ everywhere. Is it not true? Also this would imply that $F$ is actually lipshitz as $F'(x)$ is bounded.
Look at the difference quotient.
\begin{align*} \left| \frac{F(x+h)-F(x)}{h} \right| &\leq \int_{\mathbb{R}} \left| \frac{f(t)}{ht} \right| \left| \sin((x+h)t) - \sin(xt) \right| \ dt \\ & \leq \int_{\mathbb{R}} \left| \frac{f(t)}{ht} \right| \left| xt + ht - xt \right| \ dt \\ &\leq \int_{\mathbb{R}} | f(t) | \ dt \\ &< \infty \end{align*}
where we used that $|\sin x - \sin y| \leq |x - y|$ for each $x, y$ (the sine is Lipschitz). The difference quotient is bounded. Now let $h = \frac{1}{n},$ and define $F_n(x) = |n (F(x + \frac{1}{n}) - F(x))|$. Then the sequence is bounded, and by the Dominated Convergence Theorem, the limit exists. Thus $F$ is a.e. differentiable.