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For a continuous function $k:[0,1]^2\to \mathbb{R}$, let $A$ be the generalized Volterra integral operator on $C([0,1],\mathbb{R})$ defined by $$ A(f)(t)\triangleq \int_0^t k(s,t)f(s) ds, \quad t \in [0,1]. $$ Given a finite Borel measure $\mu$ on $C([0,1],\mathbb{R})$ does there always exist a $k$ as above such that $A$ is injective and for every $\epsilon>0$, there exists a Borel subset $B\subseteq C([0,1],\mathbb{R})$ such that

  • for some $n \in \mathbb{N}$ we have $B \cap A^n(B)= \emptyset$,
  • $\mu(C([0,1],\mathbb{R})\setminus B)<\epsilon$?
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  • $\begingroup$ You probably mean "arbitrarily large measure" (as in arbitrarily close to $\mu(C)$), or else $B=\emptyset$ works. $\endgroup$
    – Christian Remling
    Commented Sep 4, 2019 at 19:27
  • $\begingroup$ Though this won't work either if we take $\mu$ as an $A$ invariant measure. $\endgroup$
    – Christian Remling
    Commented Sep 4, 2019 at 19:29
  • $\begingroup$ @ChristianRemling I edited the condition so that it is indeed arbitrarily small. Also, true so suppose we ensure that $\mu$ is not $A$ invariant. If need be, let us assume that $\mu$ is given and that we may choose $k$ (so A). $\endgroup$
    – ABIM
    Commented Sep 4, 2019 at 19:47
  • $\begingroup$ @FabianWirth I added details rigorously described what I'm looking for. So $B$ would be a non-example (unless $\mu=\delta_{f}$ which it need not be in general). $\endgroup$
    – ABIM
    Commented Sep 5, 2019 at 7:21
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    $\begingroup$ I fixe dit for them. $\endgroup$
    – ABIM
    Commented Sep 11, 2019 at 7:11

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