# How show that $\lim_{\varepsilon \rightarrow 0}\int_A h_\varepsilon(x)dx =0$, whenever $\bigg|\int_I h\bigg|\leq |I|^{1/2}$?

Let $h$ be a bounded, measurable function, such that, for any interval $I$ $$\bigg|\int_I h\bigg|\leq |I|^{1/2}.$$

I want show that, for any $A$, with $|A|<\infty$, $$\int_A h_\varepsilon(x)dx\rightarrow 0,\ \ \mathrm{as}\ \varepsilon \rightarrow 0,$$

where $h_\varepsilon(x) = h(\frac x\varepsilon)$.

My idea was to take $\eta>0$, and a open set $O$ such that $|O-A|<\eta$. How $O$ is open in $\mathbb{R}$, there exist disjoint intervals, $\{I_k\}$, such that $O = \cup I_k$. Then

$$\bigg| \int_{O-A} h_\varepsilon(x)dx \bigg|\leq \|h\|_{\infty}\eta$$ and $$\bigg| \int_{O} h_\varepsilon(x)dx \bigg| \leq \sum_k\bigg| \int_{I_k} h_\varepsilon(x)dx \bigg| \leq \varepsilon^{1/2} \sum_k|I_k|^{1/2}.$$

But, for instance, if $|I_k|=1/n^2$, then $|O| = \sum_k|I_k|=\pi/6$, however $\sum_k|I_k|^{1/2}=\infty$.

How can I solve this problem?

Thank you!

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Let $h=\chi_{[0,1]}$, then $h$ satisfies your condition. But $\int_{[0,1]} h_{\epsilon}(x)=1$ for any $\epsilon\leq 1$. –  JSchlather Jan 29 '13 at 5:35
Sorry @Jacob, but $\int_0^1 h_\varepsilon(x)dx = \varepsilon \int_0^{1/\varepsilon} h(x)dx = \varepsilon$, for $\varepsilon\ll 1$.Thank you anyway. –  Kelson Vieira Jan 29 '13 at 5:59
Hm, well it seems like you could try something like bounding the tail of your series by $\eta$ and then splitting the sum there. But it's not clear to me if that gives you enough control to get it to go to zero, you do ensure a finite bound at least. –  JSchlather Jan 29 '13 at 6:35
@JacobSchlather, it gives you enough control indeed. First pick $\eta$, this will fix the number of intervals $I_k$ too, then choose a small enough $\epsilon$. –  user27126 Jan 29 '13 at 7:46
Do you know about the Hardy-Littlewood maximal function? –  AD. Jan 29 '13 at 8:13

1. As $h$ is bounded and measurable, we can find simple functions $\{h_n\}$ such that $\sup_{x\in\Bbb R}|h(x)-h_n(x)|\leqslant 1/n$.
2. Write $h_n(x):=\sum_{j=1}^{N_n}a_{n,j}\chi_{B_{n,j}}$. We have $$\left|\int_Ah_\varepsilon(x)dx\right|\leqslant \frac 1n+\sum_{j=1}^{N_n}|a_{n,j}|\int_{\Bbb R}\chi_{B_{n,k}}(x\varepsilon^{-1})\chi_A(x)dx.$$ Use the substitution $t=x\varepsilon^{—1}$, take the $\limsup_{\varepsilon \to 0}$ to get what we want.