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I read this question, and I would like to ask the opposite: Assume that I have a sequence of smoothing operators $(P_n)$ with (hence smooth) kernels $(p_n)$ converging strongly to some smoothing operator $P$ with kernel $p$. Does this imply convergence of $(p_n)$ to $p$, and if yes, in what sense?

Assume that I have $P_n \longrightarrow P$ strongly in $H^s$ for all $s \in \mathbb{R}$. Or conversely, which kind of convergence do I need for the operators to conclude that $p_n \longrightarrow p$ in $C^\infty$?

/Edit: As Liviu asked me to, I will be a bit more specific. It is ok for me to start with a compact manifold (everything is Riemannian), but I would like to have the results as well on a complete manifolds, with bounded curvature (or even with bounded geometry, if necessary, i.e. that all derivatives of the curvature tensor are uniformly bounded). It is enough to have the results on the kernel to be local though.

Regarding the convergence of my operator family, let us say that $\|P_n\|, \|P\| \leq C$ in $H^s_0(M)$ and that $P_nu \longrightarrow Pu$ for each $u \in C^\infty_c(M)$ (then we have the same statement for all $u \in H^s_0(M)$ be the Banach-Steinhaus Theorem).

Now the question is: If each $P_n$ is smoothing, as well as $P$, and the above holds for each $s \in \mathbb{R}$, can we conclude that the kernels converge $p_n$ converge to $p$ in $C^\infty(M \times M)$, at least locally?

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  • $\begingroup$ You need to be a bit more precise. Could you rewrite the question so we know whether the kernels are defined on a compact manifold or a noncompact manifold? (In the compact case the various locally convex topologies are better behaved in the compact case.) Additionally, can you define precisely the meaning of ``strong'' convergence? In any case, I believe that Treves' book on distributions has the answer to your question. Another good source is vol. 4 of Gelfand & co opus on Generalized Functions. $\endgroup$ Apr 25 '14 at 10:47
  • $\begingroup$ I deleted my answer because too many misunderstandings went into it. $\endgroup$ Apr 26 '14 at 14:36
  • $\begingroup$ @Kofi, you'll need some kind of condition to prevent e.g. convergence of convolution kernels to a delta function: if $p_n(x) = n p(nx)$, with $p\in C_c^\infty(\mathbb R)$, then convolution with $p_n$ satisfies all your assumptions, with the exception (I assume) of the limit operator being 'smoothing'. That suggests that you need to quantify what 'smoothing' means, and how the 'smoothing' nature of $p_n$ is conserved in the limit. Can you say something about that? $\endgroup$ Apr 26 '14 at 14:40
  • $\begingroup$ Smoothing means for me that the operator maps distributions to smooth functions. If we have a non-compact manifold, it is not really clear which distribution spaces are suitable for this, but on a compact manifolds there is basically only one such space. In particular, a smoothing operator is one with a smooth Schwartz kernel. $\endgroup$ Apr 26 '14 at 21:40
  • $\begingroup$ Also, I actually do think that your answer was helpful and regret that you deleted it! $\endgroup$ Apr 26 '14 at 21:41

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