# For E of finite Lebesgue measure, is $f(t) = \int_E \cos(tx) dx$

Let $E \subset \mathbb{R}$ be a Lebesgue measurable set with finite measure. Prove or give a counterexample that $$f(t) = \int_E \cos(tx) dx$$ is continuously differentiable.

I proved that f is continuous, and furthermore, f is absolutely continuous on each compact subset and hence is differentiable almost everywhere.

Further, if $E$ is contained in some compact set (up to a null set), then $f(t)$ is analytic and hence infinitely differentiable (as in this question). However, if $E$ is unbounded more generally, I am not sure how to proceed. Any help would be appreciated.

The function need not be differentiable. That is plausible right away because we would expect the derivative to be $-\int_E x \sin tx \, dx$, but this integral need not converge.
For a more precise version, make the additional assumption that $E\subseteq (0,\infty)$ and $$|E\cap (L,\infty)| = o(1/L^4)$$ as $L\to\infty$. I also want to replace the cosine by $e^{itx}$, for convenience. Then $$f(t+h)-f(t) = \int_E e^{itx} (e^{ihx}-1)\, dx .$$ I cut off the integral at $h^{-1/4}$. By our assumption on $E$, the removed part contributes only $o(h)$, so is irrelevant for the computation of the derivative. On $0\le x\le h^{-1/4}$, we have that $e^{ihx}-1=ihx + O(h^{3/2})$. It follows that if $f'(t)$ exists, then $$f'(t) = \lim_{h\to 0} i\int_0^{h^{-1/4}} \chi_E(x)xe^{itx}\, dx .$$
However, it's easy to cook up an $E$ for which this limit fails to exist for a given $t$; just concentrate $E$ near points with $tx\equiv 0 \bmod 2\pi$.
• @user334137: (1) The "integral $\int_0^{\infty} \ldots$ cut off at $a$" refers to $\int_0^a$; (2) the example is fully explicit in its current version. – user138530 Jul 3 '18 at 16:27