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How would I construct Schwartz functions $f_1^\epsilon$, $f_2^\epsilon$ on $\mathbb{R}$ such that

$f_1^\epsilon(x)\leq\mathbb{1}_{[a,b]}(x)\leq f_2^\epsilon(x)$,

and $f_1^\epsilon\rightarrow\mathbb{1}_{[a,b]}$, $f_2^\epsilon\rightarrow\mathbb{1}_{[a,b]}$ as $\epsilon\rightarrow0$,

where $a<b$ are real numbers and $\mathbb{1}_{[a,b]}$ is the indicator function?

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Hint: $e^{-x^2/2}$ Is this homework? – user17794 Nov 12 '12 at 6:02
no, it isn't homework – Kelly Nov 12 '12 at 7:03
You seem to be new to the site - in general, I would recommend showing the progress you've made on the question so that your answerer can get a better sense of what you under stand - helps prevent pedantry and saves time. For this particular question, it's a fairly lengthy (though standard) construction, so I'd prefer to get to the part that's tripping you up. – user17794 Nov 12 '12 at 7:12

The quickest way is to convolve the characteristic function of another interval with a smooth bump function. Let $\Phi(x)=c\,\exp(1/(1-x^2))\,\chi_{(-1,1)}$ where $c>0$ is chosen to make $\int_{-1}^1 \Phi(x)\,dx=1$. For $\epsilon>0$ define $\Psi_\epsilon(x)=\epsilon^{-1}\Psi(x/\epsilon)$. The convolutions $f_1^\epsilon = 1_{[a+\epsilon,b-\epsilon]}*\Psi_\epsilon$ and $f_2^\epsilon = 1_{[a-\epsilon,b+\epsilon]}*\Psi_\epsilon$ have the required properties, which I leave for you to check.

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