How can i contruct a smooth bump function $F$ on $\Bbb{R}^n$ such that $F(0)=1$ and with integral $1$?
I have tried to manipulate the function $f(x)=e^{-\frac{1}{x^2}}$ if $x>0$ and $f(x)=0$ if $x \leq 0$, ta find a such function $g$, but i only get one of the 2 conditions.
Then, of course,we can define $F(x)=g(x_1)* g(x_2)***g(x_n)$ on $\Bbb{R}^n$
Thank you in advance.
Consider $$g_0(x)=f\left(\frac12 + x\right)f\left(\frac12-x\right)$$ This is a smooth bump function centred at the origin, but $g_0(0)\neq1$. We can fix that with $g_1(x)=\frac{g_0(x)}{g_0(0)}$.
Now for the multidimensional aspect. That's easy. You have one solution, with $$g_2(x_1,x_2,\ldots,x_n)=g_1(x_1)g_1(x_2)\cdots g_1(x_n)$$Another option is $$g_2(x_1,x_2,\ldots,x_n)=g_1(x_1^2+x_2^2+\cdots+x_n^2)$$.
Finally, the normalisation. Note that $$ \int_{\Bbb R^n}g_2(c\mathbf x)dV=\frac 1{c^n}\int_{\Bbb R^n}g_2(\mathbf x)dV $$ Setting $c=\sqrt[n]{\int_{\Bbb R}g_2(\mathbf x)dV}$, then $g_3(\mathbf x)=g_2(c\mathbf x)$ finally gives you what you want.
