Filling a gap in triangle inequality proof for $d(x,y)= leftlvert frac{|x-y|}{1+|x-y|} rightlvert $.












4












$begingroup$


For the function $d:mathbb{R} times mathbb{R} to mathbb{R}$ defined as $$d(x,y)= leftlvert frac{|x-y|}{1+|x-y|} rightlvert. $$ I want to prove this is a metric but Im having issues proving the triangle inequality. I know this problem has been discused before and the one idea im trying to prove this is using the fact that $f(x)=frac{x}{1+x}$ is increasing and also satisfies the subadditivity property. This is $f(a+b) leq f(a) + f(b)$ for every $a,b geq 0$. Im stuck proving the subadditivity, someone mention this can be done with intermediate value theorem but I dont see how. Thanks for reading and helping.






real-analysis calculus derivatives metric-spaces







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  • $begingroup$
    @Cos Why do you use absolute value signs for non-negative numbers?
    $endgroup$
    – Kavi Rama Murthy
    Jan 9 at 5:50


















4












$begingroup$


For the function $d:mathbb{R} times mathbb{R} to mathbb{R}$ defined as $$d(x,y)= leftlvert frac{|x-y|}{1+|x-y|} rightlvert. $$ I want to prove this is a metric but Im having issues proving the triangle inequality. I know this problem has been discused before and the one idea im trying to prove this is using the fact that $f(x)=frac{x}{1+x}$ is increasing and also satisfies the subadditivity property. This is $f(a+b) leq f(a) + f(b)$ for every $a,b geq 0$. Im stuck proving the subadditivity, someone mention this can be done with intermediate value theorem but I dont see how. Thanks for reading and helping.






real-analysis calculus derivatives metric-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    @Cos Why do you use absolute value signs for non-negative numbers?
    $endgroup$
    – Kavi Rama Murthy
    Jan 9 at 5:50
















4












4








4





$begingroup$


For the function $d:mathbb{R} times mathbb{R} to mathbb{R}$ defined as $$d(x,y)= leftlvert frac{|x-y|}{1+|x-y|} rightlvert. $$ I want to prove this is a metric but Im having issues proving the triangle inequality. I know this problem has been discused before and the one idea im trying to prove this is using the fact that $f(x)=frac{x}{1+x}$ is increasing and also satisfies the subadditivity property. This is $f(a+b) leq f(a) + f(b)$ for every $a,b geq 0$. Im stuck proving the subadditivity, someone mention this can be done with intermediate value theorem but I dont see how. Thanks for reading and helping.






real-analysis calculus derivatives metric-spaces







share|cite|improve this question









$endgroup$




For the function $d:mathbb{R} times mathbb{R} to mathbb{R}$ defined as $$d(x,y)= leftlvert frac{|x-y|}{1+|x-y|} rightlvert. $$ I want to prove this is a metric but Im having issues proving the triangle inequality. I know this problem has been discused before and the one idea im trying to prove this is using the fact that $f(x)=frac{x}{1+x}$ is increasing and also satisfies the subadditivity property. This is $f(a+b) leq f(a) + f(b)$ for every $a,b geq 0$. Im stuck proving the subadditivity, someone mention this can be done with intermediate value theorem but I dont see how. Thanks for reading and helping.






real-analysis calculus derivatives metric-spaces




real-analysis calculus derivatives metric-spaces






share|cite|improve this question













share|cite|improve this question











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asked Jan 9 at 2:10









CosCos

1376




1376












  • $begingroup$
    @Cos Why do you use absolute value signs for non-negative numbers?
    $endgroup$
    – Kavi Rama Murthy
    Jan 9 at 5:50




















  • $begingroup$
    @Cos Why do you use absolute value signs for non-negative numbers?
    $endgroup$
    – Kavi Rama Murthy
    Jan 9 at 5:50


















$begingroup$
@Cos Why do you use absolute value signs for non-negative numbers?
$endgroup$
– Kavi Rama Murthy
Jan 9 at 5:50






$begingroup$
@Cos Why do you use absolute value signs for non-negative numbers?
$endgroup$
– Kavi Rama Murthy
Jan 9 at 5:50












2 Answers
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$begingroup$

Note that $f(x)=frac{x}{1+x}=1-frac{1}{1+x}$. For $a,bgeq 0 $,
$$f(a+b)=1-frac{1}{1+a+b}.$$ Observe that $1-frac{1}{1+a+b}leq 1-frac{1}{1+a}$(why?). Since $1-frac{1}{1+b}geq 0$, the inequality follows directly.






share|cite|improve this answer











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    1












    $begingroup$

    Note also that $f(x) ge 0$ for $x ge 0$.
    $$d(x,z) = f(|x-z|) le f(|x-y|+|y-z|) le f(|x-y|) + f(|y-z|) = d(x,y) + d(y,z).$$
    The first inequality is due to $f$ being increasing, and the secoond inequality is the subadditivity property.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

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      3












      $begingroup$

      Note that $f(x)=frac{x}{1+x}=1-frac{1}{1+x}$. For $a,bgeq 0 $,
      $$f(a+b)=1-frac{1}{1+a+b}.$$ Observe that $1-frac{1}{1+a+b}leq 1-frac{1}{1+a}$(why?). Since $1-frac{1}{1+b}geq 0$, the inequality follows directly.






      share|cite|improve this answer











      $endgroup$


















        3












        $begingroup$

        Note that $f(x)=frac{x}{1+x}=1-frac{1}{1+x}$. For $a,bgeq 0 $,
        $$f(a+b)=1-frac{1}{1+a+b}.$$ Observe that $1-frac{1}{1+a+b}leq 1-frac{1}{1+a}$(why?). Since $1-frac{1}{1+b}geq 0$, the inequality follows directly.






        share|cite|improve this answer











        $endgroup$
















          3












          3








          3





          $begingroup$

          Note that $f(x)=frac{x}{1+x}=1-frac{1}{1+x}$. For $a,bgeq 0 $,
          $$f(a+b)=1-frac{1}{1+a+b}.$$ Observe that $1-frac{1}{1+a+b}leq 1-frac{1}{1+a}$(why?). Since $1-frac{1}{1+b}geq 0$, the inequality follows directly.






          share|cite|improve this answer











          $endgroup$



          Note that $f(x)=frac{x}{1+x}=1-frac{1}{1+x}$. For $a,bgeq 0 $,
          $$f(a+b)=1-frac{1}{1+a+b}.$$ Observe that $1-frac{1}{1+a+b}leq 1-frac{1}{1+a}$(why?). Since $1-frac{1}{1+b}geq 0$, the inequality follows directly.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 9 at 6:56

























          answered Jan 9 at 2:24









          Thomas ShelbyThomas Shelby

          2,745421




          2,745421























              1












              $begingroup$

              Note also that $f(x) ge 0$ for $x ge 0$.
              $$d(x,z) = f(|x-z|) le f(|x-y|+|y-z|) le f(|x-y|) + f(|y-z|) = d(x,y) + d(y,z).$$
              The first inequality is due to $f$ being increasing, and the secoond inequality is the subadditivity property.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Note also that $f(x) ge 0$ for $x ge 0$.
                $$d(x,z) = f(|x-z|) le f(|x-y|+|y-z|) le f(|x-y|) + f(|y-z|) = d(x,y) + d(y,z).$$
                The first inequality is due to $f$ being increasing, and the secoond inequality is the subadditivity property.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Note also that $f(x) ge 0$ for $x ge 0$.
                  $$d(x,z) = f(|x-z|) le f(|x-y|+|y-z|) le f(|x-y|) + f(|y-z|) = d(x,y) + d(y,z).$$
                  The first inequality is due to $f$ being increasing, and the secoond inequality is the subadditivity property.






                  share|cite|improve this answer









                  $endgroup$



                  Note also that $f(x) ge 0$ for $x ge 0$.
                  $$d(x,z) = f(|x-z|) le f(|x-y|+|y-z|) le f(|x-y|) + f(|y-z|) = d(x,y) + d(y,z).$$
                  The first inequality is due to $f$ being increasing, and the secoond inequality is the subadditivity property.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 9 at 2:17









                  angryavianangryavian

                  40.6k23380




                  40.6k23380






























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