5
$\begingroup$

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.

$\endgroup$

1 Answer 1

0
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.