Peetre's Inequality - not strict?












1












$begingroup$



(Peetre's inequality)
Let $x,y in Bbb R^n$ and $s in Bbb R$. Then
$$ frac{(1+|x|^2)^s}{(1+|y|^2)^s} le 2^{|s|} (1+|x-y|^2)^{|s|}.$$




Proof: By switching roles of $x,y$ we may suppose $s ge 0$, and taking $s$th root may assume $s=1$. Then the argument i found online:



begin{align*}
(1+|x|^2) &= 1 + |x-y|^2 + |y|^2 +2(x-y)y \
& le 1 + |x-y|^2 + |y|^2 + (|x-y|^2+|y|^2) \
& le 2(1+|y|^2 + |x-y|^2 +|y|^2|x-y|^2 ) \
& = 2(1+|y|^2)(1+|x-y|^2).
end{align*}




What I don't understand is that on the third inequality, isn't this clearly a strict inequality when $snot= 0$? (at least by 1)






real-analysis functional-analysis ordinary-differential-equations inequality







share|cite|improve this question











$endgroup$

















    1












    $begingroup$



    (Peetre's inequality)
    Let $x,y in Bbb R^n$ and $s in Bbb R$. Then
    $$ frac{(1+|x|^2)^s}{(1+|y|^2)^s} le 2^{|s|} (1+|x-y|^2)^{|s|}.$$




    Proof: By switching roles of $x,y$ we may suppose $s ge 0$, and taking $s$th root may assume $s=1$. Then the argument i found online:



    begin{align*}
    (1+|x|^2) &= 1 + |x-y|^2 + |y|^2 +2(x-y)y \
    & le 1 + |x-y|^2 + |y|^2 + (|x-y|^2+|y|^2) \
    & le 2(1+|y|^2 + |x-y|^2 +|y|^2|x-y|^2 ) \
    & = 2(1+|y|^2)(1+|x-y|^2).
    end{align*}




    What I don't understand is that on the third inequality, isn't this clearly a strict inequality when $snot= 0$? (at least by 1)






    real-analysis functional-analysis ordinary-differential-equations inequality







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$



      (Peetre's inequality)
      Let $x,y in Bbb R^n$ and $s in Bbb R$. Then
      $$ frac{(1+|x|^2)^s}{(1+|y|^2)^s} le 2^{|s|} (1+|x-y|^2)^{|s|}.$$




      Proof: By switching roles of $x,y$ we may suppose $s ge 0$, and taking $s$th root may assume $s=1$. Then the argument i found online:



      begin{align*}
      (1+|x|^2) &= 1 + |x-y|^2 + |y|^2 +2(x-y)y \
      & le 1 + |x-y|^2 + |y|^2 + (|x-y|^2+|y|^2) \
      & le 2(1+|y|^2 + |x-y|^2 +|y|^2|x-y|^2 ) \
      & = 2(1+|y|^2)(1+|x-y|^2).
      end{align*}




      What I don't understand is that on the third inequality, isn't this clearly a strict inequality when $snot= 0$? (at least by 1)






      real-analysis functional-analysis ordinary-differential-equations inequality







      share|cite|improve this question











      $endgroup$





      (Peetre's inequality)
      Let $x,y in Bbb R^n$ and $s in Bbb R$. Then
      $$ frac{(1+|x|^2)^s}{(1+|y|^2)^s} le 2^{|s|} (1+|x-y|^2)^{|s|}.$$




      Proof: By switching roles of $x,y$ we may suppose $s ge 0$, and taking $s$th root may assume $s=1$. Then the argument i found online:



      begin{align*}
      (1+|x|^2) &= 1 + |x-y|^2 + |y|^2 +2(x-y)y \
      & le 1 + |x-y|^2 + |y|^2 + (|x-y|^2+|y|^2) \
      & le 2(1+|y|^2 + |x-y|^2 +|y|^2|x-y|^2 ) \
      & = 2(1+|y|^2)(1+|x-y|^2).
      end{align*}




      What I don't understand is that on the third inequality, isn't this clearly a strict inequality when $snot= 0$? (at least by 1)






      real-analysis functional-analysis ordinary-differential-equations inequality




      real-analysis functional-analysis ordinary-differential-equations inequality






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 23 '18 at 6:55







      CL.

















      asked Dec 23 '18 at 6:50









      CL.CL.

      2,2172825




      2,2172825






















          1 Answer
          1






          active

          oldest

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          2












          $begingroup$

          Yes, you are correct. We have that for $x,yin mathbb{R}^n$,
          begin{align*}
          (1+|x|^2) &=1+|(x-y)+y|^2\
          &= 1 + |x-y|^2 + |y|^2 +2langle (x-y), y rangle \
          &leq 1 + |x-y|^2 + |y|^2 +2|x-y| |y| \
          & le 1 + |x-y|^2 + |y|^2 + (|x-y|^2+|y|^2) \
          & = 1+ 2|y|^2 + 2|x-y|^2 \
          & < 2+ 2|y|^2 +2|x-y|^2 +2|y|^2|x-y|^2 \
          & = 2(1+|y|^2)(1+|x-y|^2).
          end{align*}






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            if you have time I hope you don't mind giving me some comments on my recent question on elliptic operator.
            $endgroup$
            – CL.
            Dec 23 '18 at 7:21










          • $begingroup$
            I read the other question but I don't know much about elliptic operators. I did not know even Peetre's Inequality, but as far as I understand, in its applications in PDE, a "strict" inequality is not so important. Am I wrong?
            $endgroup$
            – Robert Z
            Dec 23 '18 at 7:29












          • $begingroup$
            Unfortunately I am new to this, so I do not know ...
            $endgroup$
            – CL.
            Dec 23 '18 at 8:20










          • $begingroup$
            I have simplified the problem now - does this seem easier?
            $endgroup$
            – CL.
            Dec 23 '18 at 23:51











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          1 Answer
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          1 Answer
          1






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          active

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          2












          $begingroup$

          Yes, you are correct. We have that for $x,yin mathbb{R}^n$,
          begin{align*}
          (1+|x|^2) &=1+|(x-y)+y|^2\
          &= 1 + |x-y|^2 + |y|^2 +2langle (x-y), y rangle \
          &leq 1 + |x-y|^2 + |y|^2 +2|x-y| |y| \
          & le 1 + |x-y|^2 + |y|^2 + (|x-y|^2+|y|^2) \
          & = 1+ 2|y|^2 + 2|x-y|^2 \
          & < 2+ 2|y|^2 +2|x-y|^2 +2|y|^2|x-y|^2 \
          & = 2(1+|y|^2)(1+|x-y|^2).
          end{align*}






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            if you have time I hope you don't mind giving me some comments on my recent question on elliptic operator.
            $endgroup$
            – CL.
            Dec 23 '18 at 7:21










          • $begingroup$
            I read the other question but I don't know much about elliptic operators. I did not know even Peetre's Inequality, but as far as I understand, in its applications in PDE, a "strict" inequality is not so important. Am I wrong?
            $endgroup$
            – Robert Z
            Dec 23 '18 at 7:29












          • $begingroup$
            Unfortunately I am new to this, so I do not know ...
            $endgroup$
            – CL.
            Dec 23 '18 at 8:20










          • $begingroup$
            I have simplified the problem now - does this seem easier?
            $endgroup$
            – CL.
            Dec 23 '18 at 23:51
















          2












          $begingroup$

          Yes, you are correct. We have that for $x,yin mathbb{R}^n$,
          begin{align*}
          (1+|x|^2) &=1+|(x-y)+y|^2\
          &= 1 + |x-y|^2 + |y|^2 +2langle (x-y), y rangle \
          &leq 1 + |x-y|^2 + |y|^2 +2|x-y| |y| \
          & le 1 + |x-y|^2 + |y|^2 + (|x-y|^2+|y|^2) \
          & = 1+ 2|y|^2 + 2|x-y|^2 \
          & < 2+ 2|y|^2 +2|x-y|^2 +2|y|^2|x-y|^2 \
          & = 2(1+|y|^2)(1+|x-y|^2).
          end{align*}






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            if you have time I hope you don't mind giving me some comments on my recent question on elliptic operator.
            $endgroup$
            – CL.
            Dec 23 '18 at 7:21










          • $begingroup$
            I read the other question but I don't know much about elliptic operators. I did not know even Peetre's Inequality, but as far as I understand, in its applications in PDE, a "strict" inequality is not so important. Am I wrong?
            $endgroup$
            – Robert Z
            Dec 23 '18 at 7:29












          • $begingroup$
            Unfortunately I am new to this, so I do not know ...
            $endgroup$
            – CL.
            Dec 23 '18 at 8:20










          • $begingroup$
            I have simplified the problem now - does this seem easier?
            $endgroup$
            – CL.
            Dec 23 '18 at 23:51














          2












          2








          2





          $begingroup$

          Yes, you are correct. We have that for $x,yin mathbb{R}^n$,
          begin{align*}
          (1+|x|^2) &=1+|(x-y)+y|^2\
          &= 1 + |x-y|^2 + |y|^2 +2langle (x-y), y rangle \
          &leq 1 + |x-y|^2 + |y|^2 +2|x-y| |y| \
          & le 1 + |x-y|^2 + |y|^2 + (|x-y|^2+|y|^2) \
          & = 1+ 2|y|^2 + 2|x-y|^2 \
          & < 2+ 2|y|^2 +2|x-y|^2 +2|y|^2|x-y|^2 \
          & = 2(1+|y|^2)(1+|x-y|^2).
          end{align*}






          share|cite|improve this answer











          $endgroup$



          Yes, you are correct. We have that for $x,yin mathbb{R}^n$,
          begin{align*}
          (1+|x|^2) &=1+|(x-y)+y|^2\
          &= 1 + |x-y|^2 + |y|^2 +2langle (x-y), y rangle \
          &leq 1 + |x-y|^2 + |y|^2 +2|x-y| |y| \
          & le 1 + |x-y|^2 + |y|^2 + (|x-y|^2+|y|^2) \
          & = 1+ 2|y|^2 + 2|x-y|^2 \
          & < 2+ 2|y|^2 +2|x-y|^2 +2|y|^2|x-y|^2 \
          & = 2(1+|y|^2)(1+|x-y|^2).
          end{align*}







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 23 '18 at 7:12

























          answered Dec 23 '18 at 7:05









          Robert ZRobert Z

          98.5k1068139




          98.5k1068139












          • $begingroup$
            if you have time I hope you don't mind giving me some comments on my recent question on elliptic operator.
            $endgroup$
            – CL.
            Dec 23 '18 at 7:21










          • $begingroup$
            I read the other question but I don't know much about elliptic operators. I did not know even Peetre's Inequality, but as far as I understand, in its applications in PDE, a "strict" inequality is not so important. Am I wrong?
            $endgroup$
            – Robert Z
            Dec 23 '18 at 7:29












          • $begingroup$
            Unfortunately I am new to this, so I do not know ...
            $endgroup$
            – CL.
            Dec 23 '18 at 8:20










          • $begingroup$
            I have simplified the problem now - does this seem easier?
            $endgroup$
            – CL.
            Dec 23 '18 at 23:51


















          • $begingroup$
            if you have time I hope you don't mind giving me some comments on my recent question on elliptic operator.
            $endgroup$
            – CL.
            Dec 23 '18 at 7:21










          • $begingroup$
            I read the other question but I don't know much about elliptic operators. I did not know even Peetre's Inequality, but as far as I understand, in its applications in PDE, a "strict" inequality is not so important. Am I wrong?
            $endgroup$
            – Robert Z
            Dec 23 '18 at 7:29












          • $begingroup$
            Unfortunately I am new to this, so I do not know ...
            $endgroup$
            – CL.
            Dec 23 '18 at 8:20










          • $begingroup$
            I have simplified the problem now - does this seem easier?
            $endgroup$
            – CL.
            Dec 23 '18 at 23:51
















          $begingroup$
          if you have time I hope you don't mind giving me some comments on my recent question on elliptic operator.
          $endgroup$
          – CL.
          Dec 23 '18 at 7:21




          $begingroup$
          if you have time I hope you don't mind giving me some comments on my recent question on elliptic operator.
          $endgroup$
          – CL.
          Dec 23 '18 at 7:21












          $begingroup$
          I read the other question but I don't know much about elliptic operators. I did not know even Peetre's Inequality, but as far as I understand, in its applications in PDE, a "strict" inequality is not so important. Am I wrong?
          $endgroup$
          – Robert Z
          Dec 23 '18 at 7:29






          $begingroup$
          I read the other question but I don't know much about elliptic operators. I did not know even Peetre's Inequality, but as far as I understand, in its applications in PDE, a "strict" inequality is not so important. Am I wrong?
          $endgroup$
          – Robert Z
          Dec 23 '18 at 7:29














          $begingroup$
          Unfortunately I am new to this, so I do not know ...
          $endgroup$
          – CL.
          Dec 23 '18 at 8:20




          $begingroup$
          Unfortunately I am new to this, so I do not know ...
          $endgroup$
          – CL.
          Dec 23 '18 at 8:20












          $begingroup$
          I have simplified the problem now - does this seem easier?
          $endgroup$
          – CL.
          Dec 23 '18 at 23:51




          $begingroup$
          I have simplified the problem now - does this seem easier?
          $endgroup$
          – CL.
          Dec 23 '18 at 23:51


















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