Simple proposition about sequences with certain properties












0












$begingroup$


Edit: In the paper it says $c_1:= c^k/(k-1)^2$, but that is propably a typo, since it works out with $c_1:= c^{k/(k-1)^2}$. From there on, it is not to tough.



I found a remark in a paper where I couldn't follow the proof, despite the fact, that it looks fairly simple.



Claim: Take a sequence $(Phi_m)$ with $Phi_0 >0$ and
$0< Phi_m leq c^n Phi^k_{n-1}$ for $m in mathbb{N}$; $k>1$ and for a constant $c>0$. Then the assertion



$limsup_{m to infty} Phi_m^{k^{-m}} leq c_1 Phi_0$



holds where $c_1:= c^{k/(k-1)^2}$.



As the proof it is given:



Define the sequence $ Psi_m:=c_1^{m+1-k^{-1}m} Phi_m.$
Then using the assumption, one obtains



$0 < Psi_m leq Psi_{m-1}^{k}$.



By iteration we then get



$Psi_m leq Psi_{0}^{k^m}.$



This already shows the claim.



So the last step is clear to me. In opposite, I really have no clue how to derive the first inequality from the assumption. I further don't know where that factor $(k-1)^{-2}$ could be coming from.



Thank all of you in Advance.



The problem is from Jürgen Moser's "A New Proof of de Giorgi's Theorem Concerning the Regularity Problem for Elliptic Differential Equations" (which can be found here: https://onlinelibrary.wiley.com/doi/abs/10.1002/cpa.3160130308). It starts at equation (17).






real-analysis ordinary-differential-equations







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Edit: In the paper it says $c_1:= c^k/(k-1)^2$, but that is propably a typo, since it works out with $c_1:= c^{k/(k-1)^2}$. From there on, it is not to tough.



    I found a remark in a paper where I couldn't follow the proof, despite the fact, that it looks fairly simple.



    Claim: Take a sequence $(Phi_m)$ with $Phi_0 >0$ and
    $0< Phi_m leq c^n Phi^k_{n-1}$ for $m in mathbb{N}$; $k>1$ and for a constant $c>0$. Then the assertion



    $limsup_{m to infty} Phi_m^{k^{-m}} leq c_1 Phi_0$



    holds where $c_1:= c^{k/(k-1)^2}$.



    As the proof it is given:



    Define the sequence $ Psi_m:=c_1^{m+1-k^{-1}m} Phi_m.$
    Then using the assumption, one obtains



    $0 < Psi_m leq Psi_{m-1}^{k}$.



    By iteration we then get



    $Psi_m leq Psi_{0}^{k^m}.$



    This already shows the claim.



    So the last step is clear to me. In opposite, I really have no clue how to derive the first inequality from the assumption. I further don't know where that factor $(k-1)^{-2}$ could be coming from.



    Thank all of you in Advance.



    The problem is from Jürgen Moser's "A New Proof of de Giorgi's Theorem Concerning the Regularity Problem for Elliptic Differential Equations" (which can be found here: https://onlinelibrary.wiley.com/doi/abs/10.1002/cpa.3160130308). It starts at equation (17).






    real-analysis ordinary-differential-equations







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Edit: In the paper it says $c_1:= c^k/(k-1)^2$, but that is propably a typo, since it works out with $c_1:= c^{k/(k-1)^2}$. From there on, it is not to tough.



      I found a remark in a paper where I couldn't follow the proof, despite the fact, that it looks fairly simple.



      Claim: Take a sequence $(Phi_m)$ with $Phi_0 >0$ and
      $0< Phi_m leq c^n Phi^k_{n-1}$ for $m in mathbb{N}$; $k>1$ and for a constant $c>0$. Then the assertion



      $limsup_{m to infty} Phi_m^{k^{-m}} leq c_1 Phi_0$



      holds where $c_1:= c^{k/(k-1)^2}$.



      As the proof it is given:



      Define the sequence $ Psi_m:=c_1^{m+1-k^{-1}m} Phi_m.$
      Then using the assumption, one obtains



      $0 < Psi_m leq Psi_{m-1}^{k}$.



      By iteration we then get



      $Psi_m leq Psi_{0}^{k^m}.$



      This already shows the claim.



      So the last step is clear to me. In opposite, I really have no clue how to derive the first inequality from the assumption. I further don't know where that factor $(k-1)^{-2}$ could be coming from.



      Thank all of you in Advance.



      The problem is from Jürgen Moser's "A New Proof of de Giorgi's Theorem Concerning the Regularity Problem for Elliptic Differential Equations" (which can be found here: https://onlinelibrary.wiley.com/doi/abs/10.1002/cpa.3160130308). It starts at equation (17).






      real-analysis ordinary-differential-equations







      share|cite|improve this question











      $endgroup$




      Edit: In the paper it says $c_1:= c^k/(k-1)^2$, but that is propably a typo, since it works out with $c_1:= c^{k/(k-1)^2}$. From there on, it is not to tough.



      I found a remark in a paper where I couldn't follow the proof, despite the fact, that it looks fairly simple.



      Claim: Take a sequence $(Phi_m)$ with $Phi_0 >0$ and
      $0< Phi_m leq c^n Phi^k_{n-1}$ for $m in mathbb{N}$; $k>1$ and for a constant $c>0$. Then the assertion



      $limsup_{m to infty} Phi_m^{k^{-m}} leq c_1 Phi_0$



      holds where $c_1:= c^{k/(k-1)^2}$.



      As the proof it is given:



      Define the sequence $ Psi_m:=c_1^{m+1-k^{-1}m} Phi_m.$
      Then using the assumption, one obtains



      $0 < Psi_m leq Psi_{m-1}^{k}$.



      By iteration we then get



      $Psi_m leq Psi_{0}^{k^m}.$



      This already shows the claim.



      So the last step is clear to me. In opposite, I really have no clue how to derive the first inequality from the assumption. I further don't know where that factor $(k-1)^{-2}$ could be coming from.



      Thank all of you in Advance.



      The problem is from Jürgen Moser's "A New Proof of de Giorgi's Theorem Concerning the Regularity Problem for Elliptic Differential Equations" (which can be found here: https://onlinelibrary.wiley.com/doi/abs/10.1002/cpa.3160130308). It starts at equation (17).






      real-analysis ordinary-differential-equations




      real-analysis ordinary-differential-equations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 12 at 12:48







      Max

















      asked Jan 11 at 18:13









      MaxMax

      586




      586






















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          $begingroup$

          Regarding the Edit, it has now worked out for the edited constant, assuming it was just a typo.



          Proof:



          Define the sequence



          $Psi_m:=c_1^{m+1-k^{-1}m} Phi_m$,



          with $c_1:= c^{k/(k-1)^2}$.
          Using the assumption, we get to



          $ 0 < Psi_m
          =c_1^{m+1-k^{-1}m} Phi_m
          leq c_1^{m+1-k^{-1}m} c^m Phi_{m-1}^k
          = c^{frac{km+k-m}{(k-1)^2}+m} Phi_{m-1}^k
          =(c_1^{m-k^{-1}(m-1)}Phi_{m-1})^k
          = Psi_{m-1}^{k}.$



          Via Induction we obtain



          $Psi_m leq Psi_{0}^{k^m}=(c_1 Phi_0)^{k^m}.$



          That implies



          $Phi_m^{k^{-m}}
          leq Psi_{0} c_1^{-((m+1)k^{-m}+-k^{-m-1}m)}
          =c_1 Phi_0 c_1^{-((m+1)k^{-m}+-k^{-m-1}m)} .$



          Since $k>1$, the following holds



          $ lim_{mto infty} c_1^{-frac{(m+1)}{k^{m}}+frac{m}{k^{m+1}}}=1$



          This finally yields



          $limsup_{m to infty} Phi_{m}^{1/{kappa^{m}}} leq c_1 Phi_0 .$






          share|cite|improve this answer









          $endgroup$













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            0












            $begingroup$

            Regarding the Edit, it has now worked out for the edited constant, assuming it was just a typo.



            Proof:



            Define the sequence



            $Psi_m:=c_1^{m+1-k^{-1}m} Phi_m$,



            with $c_1:= c^{k/(k-1)^2}$.
            Using the assumption, we get to



            $ 0 < Psi_m
            =c_1^{m+1-k^{-1}m} Phi_m
            leq c_1^{m+1-k^{-1}m} c^m Phi_{m-1}^k
            = c^{frac{km+k-m}{(k-1)^2}+m} Phi_{m-1}^k
            =(c_1^{m-k^{-1}(m-1)}Phi_{m-1})^k
            = Psi_{m-1}^{k}.$



            Via Induction we obtain



            $Psi_m leq Psi_{0}^{k^m}=(c_1 Phi_0)^{k^m}.$



            That implies



            $Phi_m^{k^{-m}}
            leq Psi_{0} c_1^{-((m+1)k^{-m}+-k^{-m-1}m)}
            =c_1 Phi_0 c_1^{-((m+1)k^{-m}+-k^{-m-1}m)} .$



            Since $k>1$, the following holds



            $ lim_{mto infty} c_1^{-frac{(m+1)}{k^{m}}+frac{m}{k^{m+1}}}=1$



            This finally yields



            $limsup_{m to infty} Phi_{m}^{1/{kappa^{m}}} leq c_1 Phi_0 .$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Regarding the Edit, it has now worked out for the edited constant, assuming it was just a typo.



              Proof:



              Define the sequence



              $Psi_m:=c_1^{m+1-k^{-1}m} Phi_m$,



              with $c_1:= c^{k/(k-1)^2}$.
              Using the assumption, we get to



              $ 0 < Psi_m
              =c_1^{m+1-k^{-1}m} Phi_m
              leq c_1^{m+1-k^{-1}m} c^m Phi_{m-1}^k
              = c^{frac{km+k-m}{(k-1)^2}+m} Phi_{m-1}^k
              =(c_1^{m-k^{-1}(m-1)}Phi_{m-1})^k
              = Psi_{m-1}^{k}.$



              Via Induction we obtain



              $Psi_m leq Psi_{0}^{k^m}=(c_1 Phi_0)^{k^m}.$



              That implies



              $Phi_m^{k^{-m}}
              leq Psi_{0} c_1^{-((m+1)k^{-m}+-k^{-m-1}m)}
              =c_1 Phi_0 c_1^{-((m+1)k^{-m}+-k^{-m-1}m)} .$



              Since $k>1$, the following holds



              $ lim_{mto infty} c_1^{-frac{(m+1)}{k^{m}}+frac{m}{k^{m+1}}}=1$



              This finally yields



              $limsup_{m to infty} Phi_{m}^{1/{kappa^{m}}} leq c_1 Phi_0 .$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Regarding the Edit, it has now worked out for the edited constant, assuming it was just a typo.



                Proof:



                Define the sequence



                $Psi_m:=c_1^{m+1-k^{-1}m} Phi_m$,



                with $c_1:= c^{k/(k-1)^2}$.
                Using the assumption, we get to



                $ 0 < Psi_m
                =c_1^{m+1-k^{-1}m} Phi_m
                leq c_1^{m+1-k^{-1}m} c^m Phi_{m-1}^k
                = c^{frac{km+k-m}{(k-1)^2}+m} Phi_{m-1}^k
                =(c_1^{m-k^{-1}(m-1)}Phi_{m-1})^k
                = Psi_{m-1}^{k}.$



                Via Induction we obtain



                $Psi_m leq Psi_{0}^{k^m}=(c_1 Phi_0)^{k^m}.$



                That implies



                $Phi_m^{k^{-m}}
                leq Psi_{0} c_1^{-((m+1)k^{-m}+-k^{-m-1}m)}
                =c_1 Phi_0 c_1^{-((m+1)k^{-m}+-k^{-m-1}m)} .$



                Since $k>1$, the following holds



                $ lim_{mto infty} c_1^{-frac{(m+1)}{k^{m}}+frac{m}{k^{m+1}}}=1$



                This finally yields



                $limsup_{m to infty} Phi_{m}^{1/{kappa^{m}}} leq c_1 Phi_0 .$






                share|cite|improve this answer









                $endgroup$



                Regarding the Edit, it has now worked out for the edited constant, assuming it was just a typo.



                Proof:



                Define the sequence



                $Psi_m:=c_1^{m+1-k^{-1}m} Phi_m$,



                with $c_1:= c^{k/(k-1)^2}$.
                Using the assumption, we get to



                $ 0 < Psi_m
                =c_1^{m+1-k^{-1}m} Phi_m
                leq c_1^{m+1-k^{-1}m} c^m Phi_{m-1}^k
                = c^{frac{km+k-m}{(k-1)^2}+m} Phi_{m-1}^k
                =(c_1^{m-k^{-1}(m-1)}Phi_{m-1})^k
                = Psi_{m-1}^{k}.$



                Via Induction we obtain



                $Psi_m leq Psi_{0}^{k^m}=(c_1 Phi_0)^{k^m}.$



                That implies



                $Phi_m^{k^{-m}}
                leq Psi_{0} c_1^{-((m+1)k^{-m}+-k^{-m-1}m)}
                =c_1 Phi_0 c_1^{-((m+1)k^{-m}+-k^{-m-1}m)} .$



                Since $k>1$, the following holds



                $ lim_{mto infty} c_1^{-frac{(m+1)}{k^{m}}+frac{m}{k^{m+1}}}=1$



                This finally yields



                $limsup_{m to infty} Phi_{m}^{1/{kappa^{m}}} leq c_1 Phi_0 .$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 12 at 12:53









                MaxMax

                586




                586






























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