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Projection to Invariant Functions:

  1. Jan 30, 2014 #1
    Context:
    [itex]T : X \rightarrow X[/itex] is a measure preserving ergodic transformation of a probability measure space [itex]X[/itex]. Let [itex]V_n = \{ g | g \circ T^n = g \} [/itex] and [itex]E = span [ \{g | g \circ T = \lambda g, [/itex] for some [itex]\lambda \} ][/itex] be the span of the eigenfunctions of the induced operator [itex]T : L^2 \rightarrow L^2[/itex], [itex]Tf = f \circ T[/itex].

    Problem:
    I was reading this paper by Fursternberg and Weiss where they implicitly claim if [itex]f \perp E[/itex] then [itex] f \perp V_n[/itex]. However, I don't see how this isso.

    Some help would be greatly appreciated. : )
     
  2. jcsd
  3. Jan 30, 2014 #2

    Office_Shredder

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    I think you can do an inductive argument. For example f us perpendicular to V1 obviously. If g is in V2, then g+gT is in E, as is g-gT. So both of these are perpendicular to f, and therefore their sum is as well. You can probably keep working your way up.
     
  4. Feb 3, 2014 #3
    Ooh! I like it! Awesome, thanks!

    One more question:

    They also makes a claim as follows. Let [itex]P : L^2 \rightarrow V_{n}[/itex] be the projection operator. Then, it can be represented as an integral operator with kernel [itex]K(x,y) = l\sum_{i=1}^{l} 1_{A_i} (x) 1_{A_i} (y) [/itex] where [itex]\cup A_i = X[/itex] are [itex]T^n[/itex]-invariant sets. I don't see how this is even possible, nor where the [itex]l[/itex] comes into play, or how they can have only finitely many. Any ideas with this one?
     
  5. Feb 4, 2014 #4
    Nevermind, got it! Thanks anyways!
     
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