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Prove that the following: $$\int_{\mathbb{R}^{n}}\eta_{\epsilon}(x)dx=1~\forall\epsilon>0$$ Where $\eta_{\epsilon}(x)$ is the Friedrichs' mollifier: $$\eta_{\epsilon}(x)= \frac{1}{\epsilon^n}\eta(\frac{x}{\epsilon}) =\left\{\begin{matrix} \frac{1}{\epsilon^n}e^{\frac{1}{|\frac{x}{\epsilon}|^2-1}} & x<1\\ \\ 0 & |x|\geq 1\\ \end{matrix}\right. $$ and $\eta(x)$ is: $$\eta(x) =\left\{\begin{matrix} e^{\frac{1}{|x|^2-1}} & x<1\\ \\ 0 & |x|\geq 1\\ \end{matrix}\right. $$ Im kind of lost as I do not know where to begin, the only thing that I did prove is that $\eta_{\epsilon}(x)$ has a compact support on $\overline{B_{\epsilon}(0)}$ and that it is $C^{\infty}$. Other than that I am kind of lost.

AdrinMI49
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    Do you know why it suffices to check that $\eta(x)$ has integral equal to $1$? – Elchanan Solomon Nov 07 '21 at 01:33
  • I would suppose that it is sufficient as $\eta_{\epsilon}(x)$ is an escalation (I don't know if that is correctly said, though I don't know how to translate the word I have in my mother language) of the function $\eta(x)$. In other words, it would be practically the same function but with a change of variables and a scalar multiplying it – AdrinMI49 Nov 07 '21 at 01:38
  • Where do you see this definition? Do you miss a normalization constant in $\eta$? See: https://en.wikipedia.org/wiki/Mollifier#Concrete_example –  Nov 07 '21 at 01:50
  • I saw the definition of $\eta_{\epsilon}(x)$ on my course of Spectral Theory, And I'm not missing a normalization constant in $\eta$. I have been thinking and reading my notes, and the thing is that $\eta$ is just a function in $C_{0}^\infty(\mathbb{R}^n)$ with a compact support, but it is not a mollifier yet. When I compose it to the function $\eta_{\epsilon}(x)$ then it becomes a mollifier and the normalization constant is $\frac{1}{\epsilon^n}$. So I'm still stuck, as I don't know where to begin to prove the proposition, as the function does not have an obvious antiderivative. – AdrinMI49 Nov 07 '21 at 02:19
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    The important part is that $\eta$ is smooth, positive, has the stated compact support, and has finite integral. Also your $\eta_\epsilon$ is incorrect … its support is $(-\epsilon, \epsilon)$. – A rural reader Nov 07 '21 at 03:02
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    @adrinmajestuosso: $\eta$ is a special case of $\eta_\epsilon$ where $\epsilon=1$. It must be that $\int \eta=1$ if you have $\int \eta_\epsilon=1$ for all $\epsilon$. Moreover, $\frac{1}{\epsilon^n}$ is not the normalization constant, but due to a change of variables. I would not see your reply in time unless you ping me using @Louis Pan in the comment box. –  Nov 07 '21 at 14:37
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    I dint think this is true without a normalizing constant. Note that $0\leq \eta \leq 1$ and that $\eta$ is supported in the n-dimensional unit ball. The measure of this ball goes to zero with $n\to\infty$. – PhoemueX Nov 07 '21 at 21:22
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    The mollifier needs to be defined with a constant in front obviously, which is actually some sort of hypergeometric function, see here. Is the calculation of this integral of interest to you? – DinosaurEgg Nov 08 '21 at 15:31
  • @DinosaurEgg Thank you very much. I do have an interest in the calculation of this integral, though I think I can calculate it with a change to spherical coordinates. I haven't tried it, but today I was going to try and solve it. – AdrinMI49 Nov 08 '21 at 20:58
  • I think that I have the answer and will update the post at a later date, as I am busy right now. – AdrinMI49 Nov 30 '21 at 01:01

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