Finding an $L^2$-$L^2$ bound on the Bogovski Operator While looking at the Stokes operator in an online course I'm taking, we came across what is known as the Bogovski operator. It is used to show that, for $\Omega\subset\mathbb{R}^n$ open, bounded, connected and with Lipschitz boundary, for every $f\in L^2(\Omega)$ with $\int f=0$ there exists $u\in H_0^1(\Omega;\mathbb{K}^n)=H_0^1(\Omega)^n$ such that $\text{div }u=f$.
The operator is defined likewise: let $\rho\in C^\infty_c(\mathbb{R}^n)_+$ be a fixed non-negative test function such that $\int\rho=1$. Then for $f\in C^\infty_c(\mathbb{R}^n)$ we define
$$Bf(x):=\int_{\mathbb{R}^n}\int_1^\infty f(y)\rho(y+r(x-y))r^{n-1}(x-y)\ \mathrm{d}r\ \mathrm{d}y.$$
Using a change of variables one can show
$$Bf(x)=\int_{\mathbf{R}^n}\int_0^\infty f(x-z)\rho\left(x+s\frac{z}{|z|}\right)(s+|z|)^{n-1}\frac{z}{|z|^n}\ \mathrm{d}s\ \mathrm{d}z$$
from which we can deduce $Bf\in C^\infty(\mathbb{R}^n)$ (we can use $g(s,z)=M\mathbf{1}_{B(0,2R)}(z)\mathbf{1}_{(0,2R)}(s)|z|^{1-n}$ as dominating function, where $M>0$ is a suitably large constant and $R>0$ is large enough that $f$ and $\rho$ both have support in $B(0,R)$). It is also possible to show that $Bf$ has compact support and that if $\int f=0$ then $\text{div }Bf=f$.
The exercise I am trying to attempt is to prove that there exists $c>0$ such that, if $f\in C^\infty_c(\Omega)$ satisfies $\int f=0$, then $\|Bf\|_2\le c\|f\|_2$. I am given the hint to use the dominating function $g$ mentioned above and either prove $L^1$-$L^1$ and $L^\infty$-$L^\infty$ bounds and interpolate using the Riesz-Thorin theorem, or use the fact that $\|\varphi_k* h\|_p\le\|h\|_p$ and $\varphi_k*h\to h$ in $L^p$, where $(\varphi_k)$ is an approximate identity of convolution, $1\le p<\infty$ and $h\in L^p(\mathbb{R}^n)$.
I have tried both methods but without much luck. A hurdle I've been unable to overcome is that the dominating function $g$ necessarily depends on $f$ - I have been unable to find a suitable bound for the integrand of the second form of $Bf$ which does not depend on the support of $f$. I also am unsure where to use the condition that $\int f=0$. Does anyone have any ideas?
EDIT: Crucially, I had misread the question as $f\in C^\infty_c(\mathbb{R}^n)$, not $f\in C_c^\infty(\Omega)$. I have come up with a solution, but if anyone would like to have a go before I share it that is welcome.
 A: In case anyone is interested in this problem, I have written a solution. It ended up not being all that difficult once I realised the question specified $f\in C^\infty_c(\Omega)$ (where $\Omega$ is open and bounded) rather than $f\in C^\infty_c(\mathbb R^n)$.
Let $R>0$ be large enough that both $\Omega$ and the support of $\rho$ are contained in $B(0,R)$. For any $x\in\Omega$, if $s\ge2R$ then $\rho(x+s\frac{z}{|z|})=0$ and if $|z|\ge2R$ then $f(x-z)=0$. For $s\in(0,2R)$ and $z\in B(0,2R)$ we have $(s+|z|)^{n-1}\le2^{n-2}(s^{n-1}+|z|^{n-1})\le(2R)^{n-1}$, so we find
\begin{align*}
|Bf(x)|&\le\int\int_0^\infty|f(x-z)|\rho\left(x+s\frac{z}{|z|}\right)\mathbf1_{(0,2R)}(s)\mathbf1_{B(0,2R)}(z)(s+|z|)^{1-n}\ \mathrm d s\ \mathrm d z\\
&\le(2R)^n\|\rho\|_\infty\int_{B(0,2R)}|f(x-z)||z|^{1-n}\ \mathrm d z.
\end{align*}
Setting $M:=(2R)^n\|\rho\|_\infty$, this gives us an $L^\infty$-$L^\infty$ bound:
$$\|Bf\|_\infty\le\left(M\int_{B(0,2R)}|z|^{1-n}\ \mathrm d z\right)\|f\|_\infty$$
and integrating over $x\in\Omega$ gives us an $L^1$-$L^1$ bound:
\begin{align*}
\|Bf\|_1&\le M\int_\Omega\int_{B(0,2R)}|f(x-z)||z|^{1-n}\ \mathrm d z\ \mathrm d x\\
&=M\int_{B(0,2R)}\left(\int_\Omega|f(x-z)|\ \mathrm d x\right)|z|^{1-n}\ \mathrm d z\\
&\le\left(M\int_{B(0,2R)}|z|^{1-n}\ \mathrm d z\right)\|f\|_1.
\end{align*}
To verify that $c:=M\int_{B(0,2R)}|z|^{1-n}\ \mathrm d z$ is finite, convert to polar coordinates:
\begin{align*}
z_1&=r\cos\phi_1,\\
z_2&=r\sin\phi_1\cos\phi_2,\\
\vdots\\
z_{n-1}&=r\sin\phi_1\ldots\sin\phi_{n-2}\cos\phi_{n-1},\\
z_n&=r\sin\phi_1\ldots\sin\phi_{n-2}\sin\phi_{n-1}
\end{align*}
and note that the Jacobian of the transformation is $r^{n-1}F(\phi_1,\ldots,\phi_{n-2})$ for some continuous (and bounded) function $F$. It hence follows by the Riesz-Thorin Theorem that $\|Bf\|_2\le c\|f\|_2$ for all $f\in C_c^\infty(\Omega)$.
