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Suppose $f,g$ are $a$-Hoelder continuous real-valued functions on some domain $\Omega \subset \mathbb{R}^n$ satisfying $$ \|f\|_{C^{0,a}(\bar\Omega)},\|g\|_{C^{0,a}(\bar\Omega)}<\infty. $$ Then the following inequality holds

$$ \|(fg)*\varphi_{\ell} - (f*\varphi_{\ell})(g*\varphi_{\ell})\|_{C^r(\bar\Omega)}\leq C \ell^{2 a-r}\|f\|_{C^{0,a}(\bar\Omega)}\|g\|_{C^{0,a}(\bar\Omega)} $$

where $\varphi \in C^{\infty}_c(\mathbb{R}^n)$ is standard mollifier and

$$ \varphi_{\ell}(x)=\frac{1}{\ell^n}\varphi(x/\ell). $$ The same estimate holds true on a compact smooth manifold. Apparently the proof should be straightforward using a partition of unity argument, but the details are not clear to me. Is this explained somewhere in detail?

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Could you say where you learned that the same estimate is true on a compact manifold and the proof is straightforward? Maybe you're misinterpreting something you read? It's not even clear how to define "convolution" on a compact manifold. – Deane Yang Jul 2 '12 at 10:51
The statement can be found in on page 17. – Luca Scolari Jul 2 '12 at 12:11
Thanks for providing the reference. The convolution is being done on a fixed co-ordinate chart with respect to the co-ordinates. The functions being convolved are compactly supported on the chart. So you are taking two global functions, writing them as a sum of functions that are compactly supported on co-ordinate charts. Each is smoothed using convolution of each term with a fixed mollifier on each co-ordinate chart. I encourage you to work everything out carefully yourself. – Deane Yang Jul 2 '12 at 12:44

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