Let $0<α<1$. Show that if x and y are positive real numbers, then $|x^α-y^α|≤|x-y|^α$.











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I'm having a hard time proving this statement. I do know that d(x,y)=$|x-y|^α$ defines a metric on R. Any help would be greatly appreciated.






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    I'm having a hard time proving this statement. I do know that d(x,y)=$|x-y|^α$ defines a metric on R. Any help would be greatly appreciated.






    real-analysis







    share|cite|improve this question


























      up vote
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      down vote

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      up vote
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      favorite











      I'm having a hard time proving this statement. I do know that d(x,y)=$|x-y|^α$ defines a metric on R. Any help would be greatly appreciated.






      real-analysis







      share|cite|improve this question















      I'm having a hard time proving this statement. I do know that d(x,y)=$|x-y|^α$ defines a metric on R. Any help would be greatly appreciated.






      real-analysis




      real-analysis






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      edited Nov 21 at 17:30

























      asked Nov 21 at 17:10









      Amanda Varvak

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          Observe the inequality is symmetric in $x,y$ and satisfied with equality for $x=y$. Thus we may restrict to $0<x< y$. Define
          $$t=x/yin (0,1)$$
          then the inequality is
          $$ 1-t^alpha le (1-t)^alpha iff 1 le (1-t)^alpha + t^alpha$$
          this is true because for $tin (0,1)$ and $alpha in (0,1)$, $t^alpha ge t$ and $(1-t)^alpha ge (1-t)$.



          Alternatively $|d(x,0) - d(y,0)| le d(x,y) $ purely from triangle inequality (follow the proof for the usual euclidean norm).






          share|cite|improve this answer





















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            Observe the inequality is symmetric in $x,y$ and satisfied with equality for $x=y$. Thus we may restrict to $0<x< y$. Define
            $$t=x/yin (0,1)$$
            then the inequality is
            $$ 1-t^alpha le (1-t)^alpha iff 1 le (1-t)^alpha + t^alpha$$
            this is true because for $tin (0,1)$ and $alpha in (0,1)$, $t^alpha ge t$ and $(1-t)^alpha ge (1-t)$.



            Alternatively $|d(x,0) - d(y,0)| le d(x,y) $ purely from triangle inequality (follow the proof for the usual euclidean norm).






            share|cite|improve this answer

























              up vote
              2
              down vote













              Observe the inequality is symmetric in $x,y$ and satisfied with equality for $x=y$. Thus we may restrict to $0<x< y$. Define
              $$t=x/yin (0,1)$$
              then the inequality is
              $$ 1-t^alpha le (1-t)^alpha iff 1 le (1-t)^alpha + t^alpha$$
              this is true because for $tin (0,1)$ and $alpha in (0,1)$, $t^alpha ge t$ and $(1-t)^alpha ge (1-t)$.



              Alternatively $|d(x,0) - d(y,0)| le d(x,y) $ purely from triangle inequality (follow the proof for the usual euclidean norm).






              share|cite|improve this answer























                up vote
                2
                down vote










                up vote
                2
                down vote









                Observe the inequality is symmetric in $x,y$ and satisfied with equality for $x=y$. Thus we may restrict to $0<x< y$. Define
                $$t=x/yin (0,1)$$
                then the inequality is
                $$ 1-t^alpha le (1-t)^alpha iff 1 le (1-t)^alpha + t^alpha$$
                this is true because for $tin (0,1)$ and $alpha in (0,1)$, $t^alpha ge t$ and $(1-t)^alpha ge (1-t)$.



                Alternatively $|d(x,0) - d(y,0)| le d(x,y) $ purely from triangle inequality (follow the proof for the usual euclidean norm).






                share|cite|improve this answer












                Observe the inequality is symmetric in $x,y$ and satisfied with equality for $x=y$. Thus we may restrict to $0<x< y$. Define
                $$t=x/yin (0,1)$$
                then the inequality is
                $$ 1-t^alpha le (1-t)^alpha iff 1 le (1-t)^alpha + t^alpha$$
                this is true because for $tin (0,1)$ and $alpha in (0,1)$, $t^alpha ge t$ and $(1-t)^alpha ge (1-t)$.



                Alternatively $|d(x,0) - d(y,0)| le d(x,y) $ purely from triangle inequality (follow the proof for the usual euclidean norm).







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 21 at 17:40









                Calvin Khor

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                11.1k21437






























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