Proof of the Pseudolocal property of a pseudo differential operator

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Let $ain S^infty, uinmathcalS', Omega=mathbbR^n-singsupp(u)$. Then $phi uin C_0^infty, forall phi in C_0^infty(Omega)$ and for any $psiin C_0^infty(Omega)$ one can find a $phiin C_0^infty(Omega)$ with $phi=1$ near $suppphi$ and write $$phi a(x,D)u = phi a(x,D)(psi u) + phi a(x,D)((1-psi)u)$$



Im struggling to see how the second expression can be written as $b(x,D)u$ where $$b=phi a # (1-psi)$$
Where the # is the compound.



Any help would be great, thanks.




analysis differential localization pseudo-differential-operators microlocal-analysis




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    Let $ain S^infty, uinmathcalS', Omega=mathbbR^n-singsupp(u)$. Then $phi uin C_0^infty, forall phi in C_0^infty(Omega)$ and for any $psiin C_0^infty(Omega)$ one can find a $phiin C_0^infty(Omega)$ with $phi=1$ near $suppphi$ and write $$phi a(x,D)u = phi a(x,D)(psi u) + phi a(x,D)((1-psi)u)$$



    Im struggling to see how the second expression can be written as $b(x,D)u$ where $$b=phi a # (1-psi)$$
    Where the # is the compound.



    Any help would be great, thanks.




    analysis differential localization pseudo-differential-operators microlocal-analysis




    share|cite|improve this question





















      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Let $ain S^infty, uinmathcalS', Omega=mathbbR^n-singsupp(u)$. Then $phi uin C_0^infty, forall phi in C_0^infty(Omega)$ and for any $psiin C_0^infty(Omega)$ one can find a $phiin C_0^infty(Omega)$ with $phi=1$ near $suppphi$ and write $$phi a(x,D)u = phi a(x,D)(psi u) + phi a(x,D)((1-psi)u)$$



      Im struggling to see how the second expression can be written as $b(x,D)u$ where $$b=phi a # (1-psi)$$
      Where the # is the compound.



      Any help would be great, thanks.




      analysis differential localization pseudo-differential-operators microlocal-analysis




      share|cite|improve this question











      Let $ain S^infty, uinmathcalS', Omega=mathbbR^n-singsupp(u)$. Then $phi uin C_0^infty, forall phi in C_0^infty(Omega)$ and for any $psiin C_0^infty(Omega)$ one can find a $phiin C_0^infty(Omega)$ with $phi=1$ near $suppphi$ and write $$phi a(x,D)u = phi a(x,D)(psi u) + phi a(x,D)((1-psi)u)$$



      Im struggling to see how the second expression can be written as $b(x,D)u$ where $$b=phi a # (1-psi)$$
      Where the # is the compound.



      Any help would be great, thanks.





      analysis differential localization pseudo-differential-operators microlocal-analysis





      share|cite|improve this question










      share|cite|improve this question




      share|cite|improve this question









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