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.


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 $r=1$.


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