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Let $M$ be a smooth compact Riemannian manifold of dimension $n$, and let $H^s_p(M)$ for $s\in \mathbb{R}$ be the fractional Sobolev space of order $s$ on the manifold (defined for instance through the Laplace-Beltrami operator). Let $k:\mathbb{R}\to \mathbb{R}$ be a radial kernel and let $k_h=h^{-n}k(\cdot/h)$ for $h>0$. For $f:M\to \mathbb{R}$, one define $k_h*f$ by $$ k_h*f(x) = \int_{T_x M} k_h(|v|)f(\exp_x(v)) d v.$$ Consider the following assertion: For $s<r\in \mathbb{R}$, one can choose $k$ (with sufficiently zero moments) such that there exists a constant $C$ depending on the manifold, $r$ and $s$ with, $\forall f\in H_p^r(M)$, $$\|f-k_h*f\|_{H_p^s(M)} \leq Ch^{r-s} \|f\|_{H_p^r(M)}.$$

If one replace $M$ by $\mathbb{R}^n$, a simple proof of this assertion is possible by using the Fourier characterization of the Sobolev spaces. Such a construction does not generalize well to manifolds. I suspect this result to be standard. Any reference would be helpful.

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  • $\begingroup$ It seems to me that if you restrict to functions compactly supported on a sufficiently small neighborhood of a point and use exponential coordinates, you might be able to adapt the $\mathbb{R}^n$ proof. There will be error terms due to the metric not being flat, but you should be able to make them negligible by making the neighborhood small enough. I'm unfortunately not familiar enough with $L^p$ estimates for singular integrals or pseudodifferential operators to say anything more. However, the $L^2$ case should be pretty straightforward. $\endgroup$
    – Deane Yang
    Nov 19 '19 at 23:22
  • $\begingroup$ I modified my answer. Does it answer your question? $\endgroup$ Nov 21 '19 at 21:16
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For $s>0$ the easiest way to obtain convolution estimates on manifolds is described below.

First of all, by the Whitney embedding theorem (or by Nash theorem if you want to preserve the Riemannian metric) we may assume that $M$ is $n$-dimensional submimanifold of $\mathbb{R}^k$ for some $k>n$.

The trace operator is bounded as $T:H^{s+1/p}_p(\mathbb{R}^k)\to H^{s}_p(\mathbb{R}^{k-1})$ and there is an extension operator $E:H^{s}_p(\mathbb{R}^{k-1})\to H^{s+1/p}_p(\mathbb{R}^k)$. The same applies to compact submanifolds of codimension $1$. By induction, if $M\subset \mathbb{R}^k$ s a minifold of dimension $n$, there is a trace operator $T:H^{s+(k-n)/p}_p(\mathbb{R}^k)\to H^s_p(M)$ and the extension operator $E:H^s_p(M)\to H^{s+(k-n)/p}_p(\mathbb{R}^k)$. Now for $f\in H^s_p(M)$ we define convolution on $M$ as follows $$ K_h f=T(k_h*(Ef)). $$ That is, we extend $f$ to $\mathbb{R}^k$, we apply this desired convolution on $\mathbb{R}^k$, and then we restrict the resulting function back to $M$. That usually gives all estimates you want.

Assuming that the convolution in $\mathbb{R}^k$ satisfies your estimate, the same estimate will hold on $M$. Indeed, \begin{equation*} \begin{split} &\Vert f-K_h f\Vert_{H^s_p(M)}= \Vert T(Ef-K_h*(Ef))\Vert_{H^s_p(M)} \leq C_1\Vert Ef-K_h*(Ef)\Vert_{H^{s+(k-n)/p}(\mathbb{R}^k)}\\ &\leq C_2 h^{r-s}\Vert Ef\Vert_{H^{r+(k-n)/p}_p(\mathbb{R}^k)}\leq C_3 h^{r-s}\Vert f\Vert_{H^r_p(M)}. \end{split} \end{equation*}

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  • $\begingroup$ Thank you for your very clear answer. I however think that this does not really answer my question, as I think that your definition of convolution does not match with mine (or maybe I am missing something). The definition I gave is really the one I need in my work. $\endgroup$
    – Vincent
    Nov 25 '19 at 16:02
  • $\begingroup$ @Vincent You are right, my convolution is different, but it was not clear to me if you need to work with your particular convolution or if you just wanted to have a convolution with the given decay estimate. $\endgroup$ Nov 25 '19 at 16:13
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Since your manifold is compact, $H^s_{loc}$ regularity will be equivalent to $H^s$ regularity. To check the local regularity, you can use cutoff functions and the coordinate charts.

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  • $\begingroup$ I agree that it is a local question as the manifold is compact. However, I do not know how to bound the distance in a chart. If for instance one wants to use the Fourier approach, then it is false that the Fourier transform of a convolution as defined here is equal to the product of the Fourier transforms. To conclude, one would probably need to measure how far this equality is from being true, but this is non trivial for me. Could you be more explicit? $\endgroup$
    – Vincent
    Nov 17 '19 at 17:34
  • $\begingroup$ I guess that your estimate holds true for $h$ small enough (smallness depending on the geometry) and with a constant $C$ depending also on the geometry: the scalar curvature and the injectivity radius must play a rôle, but if you do not care about the precise dependence of the constants, you can use the charts. $\endgroup$
    – Bazin
    Nov 17 '19 at 20:12

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