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Due to a theorem of Malliavin (an improvement of an erlier theorem of Dixmier and Malliavin, see here) we know that every compactly supported smooth function $f$ on $\mathbb R^n$ can be written as a sum of two convolutions $$f=u_1\circ v_1+u_2\circ v_2$$ where the $u_i,v_i$ are smooth.

Question: Is it known whether we can choose the $u_i,v_i$ to vary continuously with $f$?

The original theorem By Dixmier and Malliavin proves that $f$ is the sum of finitely many convolutions (I believe they prove that $2^n$ is enough). In this case we can choose the functions to depend continuously on $f$, as is proved in this undergraduate thesis.

Malliavin's improvement of the theorem appears in "$C^{\infty}$ parametrix on Lie groups and two steps factorization on convolution algebras", Harmonic analysis (Proc. Conf., Univ. Crete, Iraklion, 1978), pp. 142–156, Lecture Notes in Math., 781, Springer, Berlin, 1980.

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  • $\begingroup$ I changed the two-letter prefixed tag you created, presumably in error, to the IMO intended one. $\endgroup$
    – user9072
    Commented Mar 8, 2013 at 18:41
  • $\begingroup$ Thanks quid! The tag suggestion doesn't work for me, for some reason, and I tagged it by mistake. $\endgroup$
    – Gjergji Zaimi
    Commented Mar 8, 2013 at 18:45
  • $\begingroup$ What can be done for a sum of two convolution products, but isn't possible with a finite sum? $\endgroup$
    – Marc Palm
    Commented Mar 12, 2013 at 10:52
  • $\begingroup$ Marc, are you asking for motivation for my question or Malliavin's theorem? $\endgroup$
    – Gjergji Zaimi
    Commented Mar 12, 2013 at 16:55
  • $\begingroup$ At first sight it seems to me that in Sotirov's thesis only sequential continuity is considered. Since $\mathscr D(\mathbb R^n)$ is very far from being metrizable this can be much weaker than contiunity. $\endgroup$
    – Jochen Wengenroth
    Commented Mar 13, 2013 at 12:27

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