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Let $B_1$ be the open ball with radius $1$ around $0\in\mathbb{R}^n$ and $\phi:\mathbb{R}^n\to [0,\infty)$ with

  1. $\phi\in C_0^\infty(B_1)$, i.e. $\phi$ is infinitely many differentiable in $B_1$ and vanishes at infinity
  2. $\phi\equiv 0$ in $\mathbb{R}^n\setminus\overline{B}_1$
  3. $\int\phi d\lambda^n=1$, where $\lambda^n$ is the Lebesgue measure on the Borel $\sigma$-algebra in $\mathbb{R}^n$

Now, let $$\phi_\varepsilon(x):=\frac 1{\varepsilon^n}\phi\left(\frac x\varepsilon\right)\;\;\;\text{for }x\in\mathbb{R}^n$$ for $\varepsilon>0$. I want to show, that $$\phi_\varepsilon\to 0\;\text{in }L_\text{loc}^\infty\left(B_1\setminus\left\{0\right\}\right)\text{ as }\varepsilon\to 0\tag{1},$$ where $L^p_\text{loc}(\Omega)$ is the space of all $p$-times locally integrable functions.


I've got problems to understand what exactly is meant by $(1)$. From the definition of local integrability and $L^\infty$, we should have $f_n\stackrel{L_\text{loc}^\infty(\Omega)}{\to}0$ iff $$\forall K\subseteq\Omega\text{ compact }:\operatorname{ess sup}\limits_{K}|f_n|\stackrel{n\to\infty}{\to}0$$ for any $(f_n:\Omega\to\mathbb{R})_{n\in\mathbb{N}}$, where $\operatorname{ess sup}$ is the essential supremum.

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Let $K\subset\mathbb{R}^n\setminus\{0\}$ be compact. Let $$ d=\min_{x\in K}|x| $$ be the distance from the origin to $K$; then $d>0$. From the properties of $\phi$, $\phi_\epsilon(x)=0$ if $|x|>\epsilon$. If $0<\epsilon<d$, then $\phi_\epsilon(x)=0$ for all $x\in K$. Thus $$ \lim_{\epsilon\to0}\operatorname{ess sup}\limits_{K}|\phi_\epsilon|=0. $$

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  • $\begingroup$ $\phi(x)=0$, if $|x|>1$. So, shouldn't we have $\phi_\varepsilon(x)=0$, if $|x|>\varepsilon$? $\endgroup$ – 0xbadf00d Jun 4 '15 at 10:02
  • $\begingroup$ Yes. Typo corected. $\endgroup$ – Julián Aguirre Jun 4 '15 at 10:17

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