I seem to have lost the ability to do u-substitutions as the following question will show: I am trying to construct a smooth compactly supported function such that it is equal to one on a given open ball $B(x, r)\subset \mathbb{R}^n$. To this end, I was thinking that $g(y) = \chi_{B(x, 2r)}\ast f(y)$ would suffice with $f(y) = r^n\varphi(\frac{y - x}{r})$ and $\varphi$ is the $L^1$-normalized smooth bump function
$$\varphi(x) = \begin{cases}c\exp\left(\left(|x|^2 - 1\right)^{-1}\right) &: |x| < 1\\\ 0 &: |x|\geq 1\end{cases}$$
with $\int_{\mathbb{R}^n}\varphi(x)dx = 1$.
However, I can't seem to be able to conclude that $g(z) = 1$ for all $z\in B(x, r)$: for $z\in B(x, r)$, I compute that
$$g(z) = \chi_{B(x, 2r)}\ast f(z) = \int_{B(x, 2r)}f(z - y)dy\leftrightarrow$$
$$g(z) = r^n\int_{B(x, 2r)}\varphi\left(\frac{z - y - x}{r}\right)dy$$
By taking $u(y) = z - y - x$, we get $du = (-1)^ndy$ and new area of integration $u\left[B(x, 2r)\right] = B(z, 2r)$. Thus,
$$g(z) = (-r)^n\int_{B(z, 2r)}\varphi\left(\frac{u}{r}\right)du$$
and by taking $v(u) = \frac{u}{r}$, we have $r^ndv = du$ and $v\left[B(z, 2r)\right] = B\left(\frac{z}{r}, 2\right)$
$$g(z) = (-1)^n\int_{B\left(\frac{z}{r}, 2\right)}\varphi(v)dv$$
and I don't see how to conclude that $g(z) = 1$ from the last equality. All help is appreciated!