Let $h:\mathbb{R}\rightarrow\mathbb{R}$ be a measurable function such that

$$\left|\int_I h\right|\leq c \sqrt{|I|}$$

for each interval $I$. Then $h_\epsilon(x)=h(x/\epsilon)$ satisfies

$$\int_Ah_\epsilon(x)dx\rightarrow0 $$ as $\epsilon$ goes to zero, for each Borel set $A$ such that $|A|<\infty$.

  • 3
    $\begingroup$ Do a change of variables. $\endgroup$ – Najib Idrissi Mar 2 '12 at 10:15
  • $\begingroup$ I edited the question based on chessmath's comments on a currently deleted answer. $\endgroup$ – Willie Wong Mar 2 '12 at 11:15
  • $\begingroup$ Could you provide an example of $f$ which satisfy the hypothesis but which is not square integrable? $\endgroup$ – Davide Giraudo Jul 26 '12 at 19:50
  • $\begingroup$ @DavideGiraudo $x^{-1/2}$ $\endgroup$ – Alexander Shamov Oct 3 '12 at 0:55
  • $\begingroup$ What are the precise integrability assumptions on $h$? $L^1$, $L^1_\mathrm{loc}$, or what? $\endgroup$ – Alexander Shamov Oct 3 '12 at 1:04

The answer depends on the integrability assumptions on $h$:

  1. If $h \in L^1$, or indeed $L^q,q<\infty$, this is trivial, since $\Vert h_\epsilon \Vert \to 0$.

  2. If $h \in L^\infty$, then this is true, since this holds for "elementary" sets, that is, fininte unions of intervals, and hence for any other sets by means of $L^1$ approximation of their indicators.

  3. If $h \in L^1_\mathrm{loc}$, this is just not true, and I have a counterexample.

So here it is. Let's denote

$s(x) := \begin{cases} 1, & x\in[2k,2k+1)\\ -1, & x\in[2k+1,2k+2) \end{cases}$

$h(x):=\begin{cases} 0, & x\le 1\\ 2^{n/2}s(2^{n}x), & x\in(2^{n},2^{n+1}] \end{cases}$

I will have to show that it satisfies the square-root hypothesis later, but first let's construct a set $A$ and see why it violates the claim.

$A := \bigsqcup_n A_n$, where $A_n := \bigcup_k [2^{-2n^2} \cdot 2k, 2^{-2n^2} \cdot (2k+1)] \cap [1+2^{-n-1}, 1+2^{-n})$.

Then $h(2^{n^2} x) = 2^{n^2/2} s(2^{2n^2} x), x \in [1,2]$, so it will only "resonate" with $A_n$:

$\displaystyle \intop_A h_{2^{-n^2}} dx = \intop_{A_n} h_{2^{-n^2}} dx$.

The latter integral equals $2^{n^2/2-n}$ up to a constant.

Finally, let's estimate $\intop_I h \, dx$. Clearly, we may assume that $I \subset [0,+\infty)$. I claim that if $[2^m,2^n] \subset I \subset [2^{m-1},2^{n+1}]$, then $|\intop_I h \, dx| \le 2^{-m/2}$, up to a constant, since an integral over $[2^m,2^n]$ is zero, and when changing the left or right limit of integration within the abovementioned segments the integral varies within $2^{-m/2} + 2^{-n/2}$. So for sufficiently long segments (that is, longer than $2^{-m}$) everything is fine. For the ones shorter than $2^{-m}$, within $[2^{m-1}, 2^m]$, $|\intop_I h \, dx| \le 2^{m/2} |I| \le |I|^{1/2}$.

  • 1
    $\begingroup$ Nice counter-example! You deserved the bounty. $\endgroup$ – Davide Giraudo Oct 6 '12 at 10:01

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.