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Let $U$ be an open bounded set of $\mathbb{R}^n$. Is it possible to approximate $\chi_U$ as almost everywhere limit of increasing sequence of smooth functions?

vonbrand
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chandu1729
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1 Answers1

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Let $$f_n(x)=\frac{n}{\sqrt{\pi}}e^{-(nx)^2}$$ which has integral $1$ and approaches the dirac delta function as $n\to \infty$. Then the convolution $\chi_U*f_n$ is smooth for each $n$ as $f_n$ is smooth and we have $$\frac{d^k}{dx^k}\int_{\mathbb R}\chi_U(t)f_n(x-t)dt=\int_{\mathbb R}\chi_U(t)\frac{d^kf_n}{dx^k}(x-t)dt$$ and converges everywhere to $\chi_U$.

Alex Becker
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  • In the OP there is the requirement that the sequence $f_n$ be increasing (and on $\mathbb{R}^n$, instead of $\mathbb{R}$). This sequence won't be increasing. – Martin Mar 30 '13 at 19:46
  • @Martin Whoops, I missed that. I'll try and think of something, otherwise delete this answer. – Alex Becker Mar 30 '13 at 19:48
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    I'd start with an exhaustion of $U$ by compact sets such that $K_{n} \subseteq \operatorname{int} K_{n+1}$. Take $\varepsilon_n$ so small that the $2\varepsilon_n$-neighborhood of $K_n$ is still contained in $\operatorname{int} K_{n+1}$. Then mollify the characteristic function of $K_{n}$ with a bump function whose support is contained in a ball of radius $\varepsilon_n$ around the origin to get $f_{n}$. Then the sequence $f_{2n}$ should be increasing. I leave the details to you (if you want). – Martin Mar 30 '13 at 19:56
  • http://www.opentradingsystem.com/quantNotes/Convolution_and_smoothing_.html – Bogdan Jan 30 '20 at 18:17