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Given a shift invariant probability measure $\mu$ on a mixing subshift of finite type.

What are the Lipschitz functions with zero integral with respect to the measure $\mu?$

Clearly any $\phi\in\{-u+u\circ \sigma: u \mbox{ Lipschitz continuous}\}$ is solution. More general, any $\phi\in\{\sum _{i=1}^m (-1)^iu\circ \sigma^{a_i}: u \mbox{ Lipschitz continuous}, m \mbox{ even},(a_i)_{i=1}^m\in \mathbb{Z}^m\}$ is solution.

Are there other solutions? How to solve this, is there a general technique?

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  • $\begingroup$ I guess that on some spaces of smooth functions it is a trivial problem in harmonic analysis. $\endgroup$
    – user39115
    Commented Apr 30, 2014 at 14:35
  • $\begingroup$ What's the equation you are talking about? Are you just looking for functions with zero integral with respect to the measure $\mu$? $\endgroup$
    – R W
    Commented Apr 30, 2014 at 14:48
  • $\begingroup$ Yes. I am just looking for functions with zero integral with respect to the measure $\mu.$ I will take your question in consideration and I will rewrite my question as you suggest. $\endgroup$
    – user39115
    Commented Apr 30, 2014 at 15:27
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    $\begingroup$ To my mind this is not very different to the question "Given a probability measure $\mu$ on a metric space, what are the Lipschitz functions with zero integral with respect to the measure $\mu$?" They are simply the functions of the form $f - \int f\,d\mu$ and that is about all you can say. It is straightforward to see that there are solutions other than those you mention: take $f$ with the property $\int f\,d\mu=0$, but such that $\int f\,d\nu \neq 0$ for a different shift-invariant measure $\nu$ (e.g. a Dirac measure on a fixed point). Clearly $f$ is not one of these examples. $\endgroup$
    – Ian Morris
    Commented May 6, 2014 at 15:25
  • $\begingroup$ The "more general" class of solutions defined in the questions is not genuinely more general. Clearly both classes of solutions are linear spaces. Since we may write $u - u \circ \sigma^m = v - v \circ \sigma$ where $v:=\sum_{i=0}^{m-1}u \circ \sigma^i$ it is not difficult to see that a solution in the "more general" class actually belongs to the first class of solutions. $\endgroup$
    – Ian Morris
    Commented May 6, 2014 at 18:24

2 Answers 2

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Even for integer-valued continuous $f: X \to $ Z, there are generically lots of solutions which are not coboundaries (of either integer or real-valued ct functions). For example, suppose that $\mu$ is ergodic, and is supported on a uniquely ergodic minimal subset of the mSFT. Then the GPS (Giordano, Putnam, Skau) results apply, and the dimension group invariant gives plenty of examples with nontrivial infinitesimals. Every nontrivial infinitesimal gives rise to an integer-valued continuous function $f$ which vanishes under integration at $\mu$. On the other hand, if the minimal subset has more than one ergodic measure, then there must be non-coboundaries that integrate to zero (although there need not be any integer-valued examples of these).

I suppose there are some that must be Lipschitz continuous ...

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  • $\begingroup$ Many Thanks. I will see the GPS results. $\endgroup$
    – user39115
    Commented Apr 30, 2014 at 15:45
  • $\begingroup$ I have not done anything yet, I will, but my immediate instinct says me that those functions are not going to be very smooth. $\endgroup$
    – user39115
    Commented Apr 30, 2014 at 18:29
  • $\begingroup$ I modified the answer, as there was an inaccuracy. There is another result that might be of interest: if $f:X \to $ R is continuous and $\int f \, d\mu = 0$ for all ergodic measures, and $X$ is an mSFT, then $f$ is coboundary. $\endgroup$
    – David Handelman
    Commented May 1, 2014 at 0:10
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I am not quite sure what kind of answer you are looking for, but perhaps this helps ...

Of course every function of the form $g-g\circ\sigma$ (a coboundary) has integral $0$ with respect to every shift-invariant measure. So does every function in the closed linear span of coboundaries. Let us call this close linear span $B$. Are there functions outside $B$ with integral $0$ with respect to $\mu$?

As David mentions above, it can be shown that the property of having $0$ integral with respect to all shift-invariant measures characterizes $B$:

  • a continuous function $f$ is in $B$ if and only if $\pi(f)=0$ for all shift-invariant measures $\pi$.

On the other hand, if you have a continuous function $f$, then $f'(\cdot)\triangleq f(\cdot)-\mu(f)$ is a function with integral $0$ with respect to $\mu$.

So, if you are interested in continuous (of Lipschitz) functions with $0$ integral with respect to $\mu$ that are not in $B$, you can look for any continuous (or Lipschitz) function $f$ such that $\pi_1(f)\neq\pi_2(f)$ for two (arbitrary) shift-invariant measures $\pi_1$ and $\pi_2$, and form $f'(\cdot)\triangleq f(\cdot)-\mu(f)$. This would give you all the possible solutions.

In fact, since you are working with a mixing shift of finite type, you can use the following simpler characterization of Lipschitz functions in $B$:

  • $f$ is in $B$ if and only if $\overline{f}(x)=0$ for every periodic point $x$,

where $\overline{f}(x)$ denotes the average of $f$ over one period of $x$. So, you can look for any function $f$ that has different averages over two periodic points.

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  • $\begingroup$ Thanks, but I m fixing the measure $\mu$ At the beginning. $\endgroup$
    – user39115
    Commented May 2, 2014 at 14:42
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    $\begingroup$ Sure! That's what I am suggesting how you could try to find: non-trivial solutions of $\mu(f)=0$ (non-trivial in the sense that $\pi(f)\neq 0$ for at least one other shift-invariant measure $\pi$). $\endgroup$
    – Algernon
    Commented May 2, 2014 at 15:37

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