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.


1 Answer 1


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 $tx\equiv 0 \bmod 2\pi$.

  • $\begingroup$ I'm losing you somewhere between the o's and the integral cut off using h (not sure what that even means). Could you maybe construct a more explicit counterexample using the idea you are describing? $\endgroup$ Jul 3, 2018 at 15:01
  • $\begingroup$ @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. $\endgroup$
    – user138530
    Jul 3, 2018 at 16:27
  • $\begingroup$ What is an example set E that satisfies both of your requirements though? $\endgroup$ Jul 4, 2018 at 19:11

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