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We work in $\mathbb R^n$ and denote $r$ the distance to the origin. Define the smooth approximations of the identity $$\phi_\delta = \frac1{\delta^n \int_{\mathbb R^n}\exp\left( \frac{-1}{1-r^2} \right)dx} \cdot \begin{cases} \exp \left( \frac{-1}{1-(r/\delta)^2} \right)dx &: r < \delta \\ 0 &: r \geq \delta \end{cases} $$ so that $\phi_\delta * f \to f$ as $\delta \to 0$, locally uniformly for all smooth $f$. (The factor is just to normalize the $\phi_\delta$.)

Let $D$ be a differential operator on $\mathbb R^n$, say a monomial in the $\partial/\partial x_i$.

What is a useful sufficient condition under which $(D\phi_\delta) * f \to Df$ ? Is there a better choice of the $\phi_\delta$ for which this holds? (It can depend on $f$.)

Fruitless ideas: (1) I'm unable to do something with the usual criterion for converges of derivatives because the $D\phi_\delta$ certainly don't converge uniformly. (2) Power series behave well under differentiation, but there are no power series here.

I'm actually interested in the more general case of convolution on a Riemannian symmetric space $S$, where the approximate identities are functions of the geodesic distance: $$\int_S \phi_\delta(d(y,x))f(x)dx$$ This here is a special case. I hope it will help to do the general case.

Bart Michels
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1 Answers1

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For convolution on $\mathbb R^n$ we have that $D(f*g) = Df * g = f * Dg$ so $$D\phi_\delta * f = \phi_\delta * Df \to Df$$ locally uniformly. I.e. it simply follows from the definition of approximate identity.

The reason why this works on the Riemannian manifold $\mathbb R^n$ is that convolution can be defined by both $$(f*g)(y) = \int f(y-x)g(x)dx\text{ and }\int f(x)g(y-x)dx$$ whereas we no longer have this symmetry for the generalization to Riemannian manifolds, mentioned at the end of the question. It seems I oversimplified the problem by restricting to $\mathbb R^n$.

Bart Michels
  • 26,355