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Suppose we have bounded linear maps $F:L^2(A) \to L^2(B)$ and $G:L^2(A) \to L^2(A)$.

Let $f \in L^2(B)$ and $u \in L^2(A)$. In fact suppose $f$ is smooth.

Is $fF(G(u)) = F(G(fu))$?

I want to say yes. But what if eg. $L^2(A) \subset L^2(B)$ or vice versa??

aere
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  • What should $f F(G(u))$ mean? You cannot do pointwise products in $L^2$-space (except trivial cases). Moreover, what should $fu$ mean?? – Giuseppe Negro Jul 11 '13 at 10:16
  • $fu$ is something like multiplying a function by a constant (from point of view of the function). – aere Jul 11 '13 at 10:17
  • @GiuseppeNegro assume $f$ is smooth – aere Jul 11 '13 at 10:19
  • No, I think that you cannot do anything like that. The question is ill-posed, but even if you formulated it correctly it would be false. To wit, observe that in practice, $F$ and $G$ often are integral operators. Then you are basically saying that you can "take a function out of an integral", which as you know is a gross error. – Giuseppe Negro Jul 11 '13 at 10:27

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You are asking whether the operator $F\circ G$ commutes with the multiplication operator $M_f$. This is not true in general. For a simple example, take $L^2$ on a space with two points, where linear operators are $2\times 2$ matrices, and multiplication operators are precisely the diagonal matrices. Clearly, commutativity fails: $$\begin{pmatrix}1 &1 \\1&1\end{pmatrix}\begin{pmatrix}1 &0 \\0&2\end{pmatrix} \ne \begin{pmatrix}1 &0 \\0&2\end{pmatrix} \begin{pmatrix}1 &1 \\1&1\end{pmatrix} $$ In fact, the following is true: if a matrix commutes with all diagonal matrices, it is itself diagonal. A general statement is that multiplication operators form a maximal abelian subalgebra of $B(L^2)$. See Do we have Maximal Abelian Algebras (MAAs)?

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