Friday, 7 September 2012

rt.representation theory - closed irreducible subspaces in L²(R)

Taking Fourier transforms, you get the action by multiplication. (Really, x acts by multiplying with eix.) The closed invariant subspaces will thus be of the form finL2(mathbbR)colonhatf|E=0 for measurable sets E.



But you were asking for closed irreducible subspaces: There aren't any, because the Lebesgue measure has no atoms.

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