Maybe I'm missing something, but I'm not sure why it would be any more complicated than
$$
\mathcal{P}(\vec{x};\vec{x_0},r) = \frac{1}{V_n} \Theta(r - ||\vec{x} - \vec{x_0}||)
$$
for the interior of a hyperball, and
$$
\mathcal{P}(\vec{x};\vec{x_0},r) = \frac{1}{S_n} \delta(||\vec{x} - \vec{x_0}|| - r)
$$
for the surface of a hypersphere.
The notation here is that
EDIT: to show that these have the correct normalization, we use polar coordinates centered at $\vec{x}_0$; in other words, we define $\rho = \| \vec{x} - \vec{x}_0 \|$ and let the other $n-1$ coordinates be angular coordinates $\theta_1, \theta_2, \dots$ on the $n$-sphere. For the surface of a hypersphere, we then have
$$
\int_{\mathbb{R}^n} \mathcal{P}(\vec{x};\vec{x}_0,r) d^n x = \frac{1}{S_n} \int_0^\infty \delta(\rho - r) (\rho^{n-1} \, d\rho \, d\Omega)
$$
where $d^{n-1} \Omega$ stands for the solid angle element in $\mathbb{R}^n$. This is then equal to
$$
\frac{1}{S_n} \left[ \int_0^\infty \delta(\rho - r) \rho^{n-1} d\rho \right] \left[ \int d \Omega \right]
$$
The integral of the solid angle in $n$ dimensions is a [well-known result], while the "picking property" of the delta function means that the radial integral is equal to $r^{n-1}$; so the integral becomes
$$
\frac{1}{S_n} \left[ r^{n-1} \right] \left[ \frac{2 \pi^{n/2}}{\Gamma(n/2)} \right] = 1.
$$
The proof for the interior of the hyperball proceeds similarly, except this time the integral over $\rho$ is
$$
\int_0^{\infty} \Theta(r - \rho) \rho^{n-1} \, d \rho = \int_0^r \rho^{n-1} \, d\rho = \frac{r^n}{n}.
$$
This factor and the angular factor then cancel out with the factor of $1/V_n$. (In the first equality above, we use the fact that $\Theta(r - \rho) = 0$ for $r > \rho$ and 1 for $r < \rho$.)
$\|\mathbf{J}\|_{\infty}$ can be considered as a function of $\{\mathbf{J}_{i,j}\}_{i,j\in [n]^2}$, denoted it by $f(\mathbf{J}_{1,1},\dots,\mathbf{J}_{n,n})$. Let $\mathbf{J}_{vec}=(\mathbf{J}_{1,1},\dots,\mathbf{J}_{n,n})\in \mathbb{R}^{n^2}$ represents the unrolled version of $\mathbf{J}$.
Based on Theorem 5.8 in lecture note [1], if we manage to show that $\|\mathbf{J}\|_{\infty}$ is $1$-Lipschitz in the sense that
$$
\|\mathbf{J}\|_{\infty} - \|\mathbf{J'}\|_{\infty} \leq \| \mathbf{J}_{vec}- \mathbf{J'}_{vec} \|_2
$$
Then, we are done. Consider
$$
\sup_{||u||=1}<u, \mathbf{J}u>-\sup_{||v||=1}<v, \mathbf{J'}v> \\
\leq \sup_{||u||=1}<u, \mathbf{J}u>-<u, \mathbf{J'}u>
$$
Then, $\sup_{||u||=1}<u, \mathbf{J}u>-<u, \mathbf{J'}u>=\sup_{||u||=1}<u, (\mathbf{J}-\mathbf{J'})u> = \|\mathbf{J}-\mathbf{J'}\|_2$
where $\|\mathbf{J}-\mathbf{J'}\|_2$ is the matrix $\ell_2$ norm. Then, by a standard result (for a proof see [2]) we have $\|\mathbf{J}-\mathbf{J'}\|_2 \leq \|\mathbf{J}-\mathbf{J'}\|_F$ where $\|\mathbf{J}-\mathbf{J'}\|_F$ is the Frobenius norm. Finally note that $\|\mathbf{J}-\mathbf{J'}\|_F = \| \mathbf{J}_{vec}- \mathbf{J'}_{vec} \|_2$
[1]https://www.stat.cmu.edu/~arinaldo/Teaching/36755/F16/Scribed_Lectures/LEC0914.pdf
[2] Why is the Frobenius norm of a matrix greater than or equal to the spectral norm?
Best Answer
Let $F$ be the cumulative distribution function of $\left\|Z\right\|$ then , $$F'(x) = \alpha_n x^{n-1} e^{-\frac12x^2}\mathbf 1_{x > 0}.$$
Let $h : \mathbb R^n \to \mathbb B^n$, $$h(z) = \frac1{\left\|z\right\|}F(\left\|z\right\|) z$$
If $z_1, z_2\in \mathbb R^n$ such $\left\|z_2\right\| \le \left\|z_1\right\|$,
\begin{align} \left\|h\left(z_1\right) - h\left(z_2\right)\right\| &= \left\|\frac1{\left\|z_1\right\|}F\left(z_1\right)z_1 - \frac1{\left\|z_2\right\|}F\left(z_2\right)z_2\right\|\\ &=\left\| \frac{z_1\left\|z_2\right\|F\left(\left\|z_1\right\|\right) - z_2\left\|z_1\right\|F\left(\left\|z_2\right\|\right)}{\left\|z_1\right\|\left\|z_2\right\|} \right\|\\ &=\left\| \frac{z_1\left\|z_2\right\|F\left(\left\|z_1\right\|\right) - z_1\left\|z_2\right\|F\left(\left\|z_2\right\|\right) + z_1\left\|z_2\right\|F\left(\left\|z_2\right\|\right) - z_2\left\|z_1\right\|F\left(\left\|z_2\right\|\right)}{\left\|z_1\right\|\left\|z_2\right\|} \right\|\\ &\le \left|F\left(\left\|z_1\right\|\right) - F\left(\left\|z_2\right\|\right)\right| + \frac{\left|F\left(\left\|z_2\right\|\right)\right|}{\left\|z_1\right\|\left\|z_2\right\|}\left\|z_1 \left\|z_2\right\| - z_2\left\|z_1\right\|\right\|\\ &= \left|F\left(\left\|z_1\right\|\right) - F\left(\left\|z_2\right\|\right)\right| + \frac{\left|F\left(\left\|z_2\right\|\right)\right|}{\left\|z_1\right\|\left\|z_2\right\|}\left\|z_1 \left\|z_2\right\| - z_2 \left\|z_2\right\| + z_2 \left\|z_2\right\| - z_2\left\|z_1\right\|\right\|\\ &\le \left|F\left(\left\|z_1\right\|\right) - F\left(\left\|z_2\right\|\right)\right| + \frac{\left|F\left(\left\|z_2\right\|\right)\right|}{\left\|z_1\right\|\left\|z_2\right\|}\left(\left\|z_2\right\|\left\|z_1 - z_2\right\| + \left\|z_2\right\|\left|\left\|z_2\right\| - \left\|z_1\right\|\right|\right)\\ &= \left|F\left(\left\|z_1\right\|\right) - F\left(\left\|z_2\right\|\right)\right| + \frac{\left|F\left(\left\|z_2\right\|\right)\right|}{\left\|z_1\right\|}\left(\left\|z_1 - z_2\right\| + \left|\left\|z_2\right\| - \left\|z_1\right\|\right|\right)\\ &\le \left|F\left(\left\|z_1\right\|\right) - F\left(\left\|z_2\right\|\right)\right| + \frac{\left|F\left(\left\|z_2\right\|\right)\right|}{\left\|z_2\right\|}\left(\left\|z_1 - z_2\right\| + \left|\left\|z_2\right\| - \left\|z_1\right\|\right|\right)\\ &\le \sup_{x\ge 0}\left|F'(x)\right|\left|\left\|z_2\right\| - \left\|z_1\right\|\right| + \frac{\sup\limits_{x\ge 0}\left|F'(x)\right|\left\|z_2\right\|}{\left\|z_2\right\|}\left(\left\|z_1 - z_2\right\| + \left|\left\|z_2\right\| - \left\|z_1\right\|\right|\right)\\ & \le 3 \sup\limits_{x\ge 0} \left|F'(x)\right| \left\|z_1 - z_2\right\| \end{align}
Proof: Look at this post
Proof: Look at this post.
This is a direct consequence of the claims 2 and 3.