What sequences in $Bbb Z[frac16]$ converge to the fixed point $0$ in $lvertcdotrvert_2$ while being fixed in $lvert cdotrvert_3$?

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Suppose some uniformly continuous (under the 2-adic topology) function $f:Bbb Z[frac16]toBbb Z[frac16]$ converges under composition with itself to $0$ (in the 2-adic topology) for all inputs, and also satisfies $f(0)=0$.



Also let $f$ preserve the $3$-adic valuation, i.e. $lvert f(x)rvert_3=lvert xrvert_3$




Is there some reason why such a function can only approach $0$ through some sequence terminating in $2^m3^n:mtoinfty$?




I can't find some counterexample, but then my skills are sadly lacking.




convergence p-adic-number-theory fixed-point-theorems




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    Suppose some uniformly continuous (under the 2-adic topology) function $f:Bbb Z[frac16]toBbb Z[frac16]$ converges under composition with itself to $0$ (in the 2-adic topology) for all inputs, and also satisfies $f(0)=0$.



    Also let $f$ preserve the $3$-adic valuation, i.e. $lvert f(x)rvert_3=lvert xrvert_3$




    Is there some reason why such a function can only approach $0$ through some sequence terminating in $2^m3^n:mtoinfty$?




    I can't find some counterexample, but then my skills are sadly lacking.




    convergence p-adic-number-theory fixed-point-theorems




    share|cite|improve this question





















      up vote
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      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Suppose some uniformly continuous (under the 2-adic topology) function $f:Bbb Z[frac16]toBbb Z[frac16]$ converges under composition with itself to $0$ (in the 2-adic topology) for all inputs, and also satisfies $f(0)=0$.



      Also let $f$ preserve the $3$-adic valuation, i.e. $lvert f(x)rvert_3=lvert xrvert_3$




      Is there some reason why such a function can only approach $0$ through some sequence terminating in $2^m3^n:mtoinfty$?




      I can't find some counterexample, but then my skills are sadly lacking.




      convergence p-adic-number-theory fixed-point-theorems




      share|cite|improve this question











      Suppose some uniformly continuous (under the 2-adic topology) function $f:Bbb Z[frac16]toBbb Z[frac16]$ converges under composition with itself to $0$ (in the 2-adic topology) for all inputs, and also satisfies $f(0)=0$.



      Also let $f$ preserve the $3$-adic valuation, i.e. $lvert f(x)rvert_3=lvert xrvert_3$




      Is there some reason why such a function can only approach $0$ through some sequence terminating in $2^m3^n:mtoinfty$?




      I can't find some counterexample, but then my skills are sadly lacking.





      convergence p-adic-number-theory fixed-point-theorems





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      share|cite|improve this question




      share|cite|improve this question









      asked 5 hours ago









      Robert Frost

      3,877836




      3,877836




















          1 Answer
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          You have $|2^m3^n|_2to 0$ as $mto+infty$, and since $f$ is continuous at $0$ respect to $2$-adic topology, then
          beginalign
          0
          &=f(0)\
          &=fleft(lim_mtoinfty2^m3^nright)\
          &=lim_mto+inftyf(2^m3^n)
          endalign
          for all $ninBbb Z$.






          share|cite|improve this answer





















          • Sorry, are you saying that's the only such sequence, i.e. that $2^ncdot15$ can't approach zero? Which is what I ask.
            – Robert Frost
            3 hours ago











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          1 Answer
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          up vote
          0
          down vote













          You have $|2^m3^n|_2to 0$ as $mto+infty$, and since $f$ is continuous at $0$ respect to $2$-adic topology, then
          beginalign
          0
          &=f(0)\
          &=fleft(lim_mtoinfty2^m3^nright)\
          &=lim_mto+inftyf(2^m3^n)
          endalign
          for all $ninBbb Z$.






          share|cite|improve this answer





















          • Sorry, are you saying that's the only such sequence, i.e. that $2^ncdot15$ can't approach zero? Which is what I ask.
            – Robert Frost
            3 hours ago















          up vote
          0
          down vote













          You have $|2^m3^n|_2to 0$ as $mto+infty$, and since $f$ is continuous at $0$ respect to $2$-adic topology, then
          beginalign
          0
          &=f(0)\
          &=fleft(lim_mtoinfty2^m3^nright)\
          &=lim_mto+inftyf(2^m3^n)
          endalign
          for all $ninBbb Z$.






          share|cite|improve this answer





















          • Sorry, are you saying that's the only such sequence, i.e. that $2^ncdot15$ can't approach zero? Which is what I ask.
            – Robert Frost
            3 hours ago













          up vote
          0
          down vote










          up vote
          0
          down vote









          You have $|2^m3^n|_2to 0$ as $mto+infty$, and since $f$ is continuous at $0$ respect to $2$-adic topology, then
          beginalign
          0
          &=f(0)\
          &=fleft(lim_mtoinfty2^m3^nright)\
          &=lim_mto+inftyf(2^m3^n)
          endalign
          for all $ninBbb Z$.






          share|cite|improve this answer













          You have $|2^m3^n|_2to 0$ as $mto+infty$, and since $f$ is continuous at $0$ respect to $2$-adic topology, then
          beginalign
          0
          &=f(0)\
          &=fleft(lim_mtoinfty2^m3^nright)\
          &=lim_mto+inftyf(2^m3^n)
          endalign
          for all $ninBbb Z$.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered 4 hours ago









          Fabio Lucchini

          5,55411025




          5,55411025











          • Sorry, are you saying that's the only such sequence, i.e. that $2^ncdot15$ can't approach zero? Which is what I ask.
            – Robert Frost
            3 hours ago

















          • Sorry, are you saying that's the only such sequence, i.e. that $2^ncdot15$ can't approach zero? Which is what I ask.
            – Robert Frost
            3 hours ago
















          Sorry, are you saying that's the only such sequence, i.e. that $2^ncdot15$ can't approach zero? Which is what I ask.
          – Robert Frost
          3 hours ago





          Sorry, are you saying that's the only such sequence, i.e. that $2^ncdot15$ can't approach zero? Which is what I ask.
          – Robert Frost
          3 hours ago













           

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