Radial Kernel with Bounded Support and Norm of Gradient Bounded by a Dimension-free Constant I was wondering if it is possible to construct a compactly supported radial kernel function in $\mathbb{R}^d$ such that the norm of the gradient is bounded by some dimension-free constant. That is, the goal is to construct a kernel function  $K\colon \mathbb{R}^d \rightarrow \mathbb{R}$ of the form $K(\mathbf{x}) = \phi( \| \mathbf{x} \|_2)$ such that 
\begin{align}
\int _{\mathbb{R}^d} K(\mathbf{x}) \mathrm{d}\mathbf{x} = 1, ~~\int_{\mathbb{R}^d} \| \nabla K(\mathbf{x}) \|_2 \mathrm{d}\mathbf{x} \leq C,
\end{align}
where $C$ is an absolute constant that does not involve the dimensionality $d$. Moreover, we assume that $K$ is normalized such that $\int_{\mathbb{R}^d} K(\mathbf{x}) \mathrm{d} \mathbf{x} = 1$.
It would be better if $K$ is three-times differentiable and we can control the norms of the higher-order derivatives of $K$. However, for now, having a dimension-free bound on the gradient is already good.
 A: note: This is not a complete answer, but I think it helps clarify the problem.
Since $K$ is radial, the problem can be formulated in terms of the profile function $\phi$.  If we let $r=\|x\|_2$, then the $i$th component of the gradient is 
\begin{equation*}
\frac{\partial K}{\partial x_i}
=
\frac{d\phi}{dr}\frac{\partial r}{\partial x_i}
=
\frac{d\phi}{dr}\frac{x_i}{r}.
\end{equation*}
Hence 
\begin{equation*}
\|\nabla K(x)  \|_2
=
\left| \frac{d\phi}{dr} (r) \right|
\end{equation*}
Using this, the problem can be restated as follows. We want a 1-d function $\phi$ such that 
\begin{equation*}
C_d \int_{0}^{\infty} |\phi^{\prime}(r)| r^{d-1} dr 
\leq
C
\end{equation*}
for a fixed constant $C$ when
\begin{equation*}
C_d \int_{0}^{\infty} \phi(r) r^{d-1} dr = 1.
\end{equation*} 
We could also eliminate the constant $C_d$ (coming from the change to spherical coordinates), and look for a function $\phi$ satisfying
\begin{equation*}
\frac{\int_{0}^{\infty} |\phi^{\prime}(r)| r^{d-1} dr}
{\int_{0}^{\infty} \phi(r) r^{d-1} dr} 
\leq 
C.
\end{equation*} 
As a first attempt at a solution, we could look for a piecewise polynomial
$\phi$.  One possible example would be the Wendland function $\phi(r)=(1-r)_+^4(4r+1)$ with $|\phi^{\prime}(r)|=20r(1-r)_+^3$. If we compute the integrals above, we get
\begin{equation*}
\frac{\int_{0}^{1} |\phi^{\prime}(r)| r^{d-1} dr}
{\int_{0}^{1} \phi(r) r^{d-1} dr} 
=
\frac{5d(d+5)}{2d+5}.
\end{equation*}
As $d\rightarrow \infty$, so does the fraction. Hence this $\phi$ does not work, and if we look at the integrals, we can see why. 
As $d\rightarrow \infty$, the weight function $r^{d-1}$  is weighting the boundary of the support of $\phi$ more heavily. So what we are looking for is a function where $|\phi^{\prime}|$ goes to $0$ at $r=1$ (wlog) as fast as $\phi$. This cannot happen for a polynomial.
Perhaps a Schwartz class $\phi$ is necessary, or maybe a function decaying to zero exponentially fast at $r=1$.
