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Andrey Gogolev
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ThePart I: The answer is yes under additional conditions on norm and conorm of $A$ and $\alpha$.

So, if:

  1. Periodic data conditions are satisfied. That is, for any periodic point $p$ $$ \sum_{x\in O(p)}f(x)=A\big(\sum_{x\in O(p)}g(x)\big)-\sum_{x\in O(p)}g(x). $$$$ \sum_{x\in O(p)}f(x)=0. $$
  2. Exponent $\alpha$ is sufficiently close to 1.
  3. Transformation $A$ is dominated by $T$. That is, the map $(x,v)\mapsto(Tx, Av)$ is partially hyperbolic.

Then Walkden's paper "Solutions to the twisted cocycle equation over hyperbolic systems" proves that there existexists an $\alpha$-Holder solution $g$. The result is more general: the target group is any Lie group with a bi-invariant metric and the equation is the cohomological equation for two cocycles rather than just coboundary equation.

Part II: Notice however that if $A\neq Id$ then the periodic conditions may be no longer necessary. Let's restrict to the case when $k=1$ then our equation takes form $$ f=\lambda g\circ T-g, $$ where $\lambda<1$. Direct computation shows that $$ g=-\sum_{i\ge 0} \lambda^if\circ T^i $$ is a solution. It is also clear that $g$ is Holder continuous. Moreover, in this case uniqueness is clear as well since the above formula for $g$ is obtained recurrently from $$ g=-f+\lambda g\circ T. $$ It seems that this generalizes rather straightforwardly to the case when $A$ is hyperbolic. And I think it's worthwhile to see if anything interesting happens in the case then $A$ has some eigenvalues on the unit circle and some off.

The answer is yes under additional conditions on norm and conorm of $A$ and $\alpha$.

So, if

  1. Periodic data conditions are satisfied. That is, for any periodic point $p$ $$ \sum_{x\in O(p)}f(x)=A\big(\sum_{x\in O(p)}g(x)\big)-\sum_{x\in O(p)}g(x). $$
  2. Exponent $\alpha$ is sufficiently close to 1.
  3. Transformation $A$ is dominated by $T$. That is, the map $(x,v)\mapsto(Tx, Av)$ is partially hyperbolic.

Then Walkden's paper "Solutions to the twisted cocycle equation over hyperbolic systems" proves that there exist an $\alpha$-Holder solution $g$. The result is more general.

Part I: The answer is yes under additional conditions:

  1. Periodic data conditions are satisfied. That is, for any periodic point $p$ $$ \sum_{x\in O(p)}f(x)=0. $$
  2. Exponent $\alpha$ is sufficiently close to 1.
  3. Transformation $A$ is dominated by $T$. That is, the map $(x,v)\mapsto(Tx, Av)$ is partially hyperbolic.

Then Walkden's paper "Solutions to the twisted cocycle equation over hyperbolic systems" proves that there exists an $\alpha$-Holder solution $g$. The result is more general: the target group is any Lie group with a bi-invariant metric and the equation is the cohomological equation for two cocycles rather than just coboundary equation.

Part II: Notice however that if $A\neq Id$ then the periodic conditions may be no longer necessary. Let's restrict to the case when $k=1$ then our equation takes form $$ f=\lambda g\circ T-g, $$ where $\lambda<1$. Direct computation shows that $$ g=-\sum_{i\ge 0} \lambda^if\circ T^i $$ is a solution. It is also clear that $g$ is Holder continuous. Moreover, in this case uniqueness is clear as well since the above formula for $g$ is obtained recurrently from $$ g=-f+\lambda g\circ T. $$ It seems that this generalizes rather straightforwardly to the case when $A$ is hyperbolic. And I think it's worthwhile to see if anything interesting happens in the case then $A$ has some eigenvalues on the unit circle and some off.

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Andrey Gogolev
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The answer is yes under additional conditions on norm and conorm of $A$ and $\alpha$. Transformation $A$ should be dominated by $T$. A precise and short way to put it is the following:

The map $(x,v)\mapsto(Tx, Av)$ is partially hyperbolic.So, if

Exponent $\alpha$ must be sufficiently close to 1.

  1. Periodic data conditions are satisfied. That is, for any periodic point $p$ $$ \sum_{x\in O(p)}f(x)=A\big(\sum_{x\in O(p)}g(x)\big)-\sum_{x\in O(p)}g(x). $$
  2. Exponent $\alpha$ is sufficiently close to 1.
  3. Transformation $A$ is dominated by $T$. That is, the map $(x,v)\mapsto(Tx, Av)$ is partially hyperbolic.

Then Walkden's paper "Solutions to the twisted cocycle equation over hyperbolic systems" proves that $g$ is alsothere exist an $\alpha$-Holder solution $g$. HisThe result is more general.

The answer is yes under additional conditions on norm and conorm of $A$ and $\alpha$. Transformation $A$ should be dominated by $T$. A precise and short way to put it is the following:

The map $(x,v)\mapsto(Tx, Av)$ is partially hyperbolic.

Exponent $\alpha$ must be sufficiently close to 1.

Then Walkden's paper "Solutions to the twisted cocycle equation over hyperbolic systems" proves that $g$ is also $\alpha$-Holder. His result is more general.

The answer is yes under additional conditions on norm and conorm of $A$ and $\alpha$.

So, if

  1. Periodic data conditions are satisfied. That is, for any periodic point $p$ $$ \sum_{x\in O(p)}f(x)=A\big(\sum_{x\in O(p)}g(x)\big)-\sum_{x\in O(p)}g(x). $$
  2. Exponent $\alpha$ is sufficiently close to 1.
  3. Transformation $A$ is dominated by $T$. That is, the map $(x,v)\mapsto(Tx, Av)$ is partially hyperbolic.

Then Walkden's paper "Solutions to the twisted cocycle equation over hyperbolic systems" proves that there exist an $\alpha$-Holder solution $g$. The result is more general.

Source Link
Andrey Gogolev
  • 4.2k
  • 1
  • 22
  • 26

The answer is yes under additional conditions on norm and conorm of $A$ and $\alpha$. Transformation $A$ should be dominated by $T$. A precise and short way to put it is the following:

The map $(x,v)\mapsto(Tx, Av)$ is partially hyperbolic.

Exponent $\alpha$ must be sufficiently close to 1.

Then Walkden's paper "Solutions to the twisted cocycle equation over hyperbolic systems" proves that $g$ is also $\alpha$-Holder. His result is more general.