\begin{eqnarray} u(x) = \begin{cases} e^{-\frac{1}{1-\|x\|^2}}& \text{ if } \|x\| < 1\\ 0& \text{ if } \|x\|\geq 1 \end{cases} \end{eqnarray} I am trying to show the integral of the mollifier function $u$ equals $1$. Help please. Thank you.
The integral of mollifier function is $1 $
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calculus
real-analysis
pde
regularity-theory-of-pdes
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1If you look up 'mollifier' in wikipedia you'll find this function with a normalizing factor $1/I_n$ as an example. The integral of your function is not $=1$, but you can just divide it by it's own integral to acchieve this. – 2017-02-25
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0but i could not find it 1 by dividing its integral. Can you explain How to solve the integral ? Since they are same it equals 1 ? – 2017-02-25
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1$\int \frac{|f(x)|}{\int |f(y)|dy }dx= \frac{\int |f(x)|dx}{\int |f(y)|dy}=1$ for every integrable function (the integral of which is nonzero). You don't need to show the integral of your function is $=1$ (and I doubt it is), you just need to normalize it. – 2017-02-25