Brezis’s book gives an example of a sequence of mollifiers. Considering the following smooth function:
$$ p(x)=\left\{\begin{array}{rc} e^{\frac{1}{\vert x\vert^2-1}},&\mbox{if}\quad \vert x\vert< 1,\\ 0, &\mbox{if}\quad \vert x\vert>1. \end{array}\right. $$
Then he says: "we obtain a sequence of mollifiers by letting $p_n(x)=Cn^N p(nx)$ with $C=\frac{1}{\int p}$.
-> My question is as follows:
How to prove that $\int p_n=1$? Being more punctual: I am having difficulties in changing variables. I thank you in advance for your help.
Of course you would agree with me that $$\lambda\int_a^b f(\lambda x)dx = \int_a^b f(\lambda x)d(\lambda x)= \int_{\lambda a}^{\lambda b}f(y)dy \tag{CoV}\label{CoV}$$ a To calculate $\int p_n$ we can note that $p_n$ is zero if $|x|>1/n$ i.e. $x$ is outside the ball $B_{1/n}(0)$ and therefore zero if $\max(|x_1|,|x_2|,\dots,|x_n|) > 1/n$ i.e. outside the $N$-dimensional 'cube around $0$' $X$ of sidelength $2/n$. So
$$ \int_{\mathbb R^N} p_n dx= \int_{\mathbb B_{1/n}(0)} p_n dx= \int_X p_n dx = \int_{-1/n}^{1/n}\dots\int_{-1/n}^{1/n} p_n(x_1,\dots,x_N) dx_1\dots dx_N$$ Apply definition of $p_n$ $$ \int_{\mathbb R^N} p_n dx = Cn^N\int_{-1/n}^{1/n}\dots\int_{-1/n}^{1/n} p(nx_1,\dots,nx_N) dx_1\dots dx_N$$ Now apply \ref{CoV} $N$ times. Each time uses up one copy of $n$ on the outside. So the $n^N$ disappears exactly, leaving you with $$ \int_{\mathbb R^N} p_n dx = C\int_{-1}^1 \dots \int_{-1}^1 p(y_1,\dots,y_N)dy_1\dots dy_N= C\int_{B_1(0)}p(y)dy = C\int_{\mathbb R^N} p(y) dy$$ now we see $C$ was chosen specifically for this to be equal to 1.
For functions that are not compactly supported you can prove a more general multivariate change of variables theorem that avoids talking about boxes.