Prove $f(x) = x^2sinleft(frac{1}{x}right)$ is Lipschitz (no use of derivative)












1












$begingroup$



Prove that $f:mathbb{R}to mathbb{R}$ such that
$$ f(x) = left{
begin{array}{c l}
x^2, sinleft(frac{1}{x}right) & ,quad xneq 0\
0 & ,quad x=0
end{array} right.$$
is Lipschitz (without use of derivatives).




Attempt. I am aware (Lipschitz-continuous $f(x)=x^2cdot sinleft(frac{1}{x}right)$)
that: $$|f(x)-f(y)|leq 3|x-y| ~~~forall~x,~yin mathbb{R},$$
but I am looking for a proof, without use of derivatives.
I tried: for $x,yneq 0$:



begin{eqnarray}
x^2sinfrac{1}{x} - y^2 sinfrac{1}{y}
&=& (x^2-y^2)sinfrac{1}{x} + y^2left ( sinfrac{1}{x} - sinfrac{1}{y} right ),nonumber
end{eqnarray}
so: $$left | x^2sinfrac{1}{x} - y^2 sinfrac{1}{y} right |
leq |x^2 - y^2| + y^2 left | sinfrac{1}{x} - sinfrac{1}{y} right |.$$
Since:
$$left | sinfrac{1}{x} - sinfrac{1}{y} right | leq left| frac{1}{x} - frac{1}{y}right |= frac{|x - y|}{xy},$$
we get:
$$left | x^2 sinfrac{1}{x} - y^2 sinfrac{1}{y} right |
leq left(x+y+frac{y}{x}right)|x-y|.$$
Unfortunatelly , the quantity $x+y+frac{y}{x}$ grows to
$+infty$, either for large $x$, or for $xapprox 0.$



Thanks in advance for the help.






real-analysis analysis lipschitz-functions







share|cite|improve this question









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    1












    $begingroup$



    Prove that $f:mathbb{R}to mathbb{R}$ such that
    $$ f(x) = left{
    begin{array}{c l}
    x^2, sinleft(frac{1}{x}right) & ,quad xneq 0\
    0 & ,quad x=0
    end{array} right.$$
    is Lipschitz (without use of derivatives).




    Attempt. I am aware (Lipschitz-continuous $f(x)=x^2cdot sinleft(frac{1}{x}right)$)
    that: $$|f(x)-f(y)|leq 3|x-y| ~~~forall~x,~yin mathbb{R},$$
    but I am looking for a proof, without use of derivatives.
    I tried: for $x,yneq 0$:



    begin{eqnarray}
    x^2sinfrac{1}{x} - y^2 sinfrac{1}{y}
    &=& (x^2-y^2)sinfrac{1}{x} + y^2left ( sinfrac{1}{x} - sinfrac{1}{y} right ),nonumber
    end{eqnarray}
    so: $$left | x^2sinfrac{1}{x} - y^2 sinfrac{1}{y} right |
    leq |x^2 - y^2| + y^2 left | sinfrac{1}{x} - sinfrac{1}{y} right |.$$
    Since:
    $$left | sinfrac{1}{x} - sinfrac{1}{y} right | leq left| frac{1}{x} - frac{1}{y}right |= frac{|x - y|}{xy},$$
    we get:
    $$left | x^2 sinfrac{1}{x} - y^2 sinfrac{1}{y} right |
    leq left(x+y+frac{y}{x}right)|x-y|.$$
    Unfortunatelly , the quantity $x+y+frac{y}{x}$ grows to
    $+infty$, either for large $x$, or for $xapprox 0.$



    Thanks in advance for the help.






    real-analysis analysis lipschitz-functions







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$



      Prove that $f:mathbb{R}to mathbb{R}$ such that
      $$ f(x) = left{
      begin{array}{c l}
      x^2, sinleft(frac{1}{x}right) & ,quad xneq 0\
      0 & ,quad x=0
      end{array} right.$$
      is Lipschitz (without use of derivatives).




      Attempt. I am aware (Lipschitz-continuous $f(x)=x^2cdot sinleft(frac{1}{x}right)$)
      that: $$|f(x)-f(y)|leq 3|x-y| ~~~forall~x,~yin mathbb{R},$$
      but I am looking for a proof, without use of derivatives.
      I tried: for $x,yneq 0$:



      begin{eqnarray}
      x^2sinfrac{1}{x} - y^2 sinfrac{1}{y}
      &=& (x^2-y^2)sinfrac{1}{x} + y^2left ( sinfrac{1}{x} - sinfrac{1}{y} right ),nonumber
      end{eqnarray}
      so: $$left | x^2sinfrac{1}{x} - y^2 sinfrac{1}{y} right |
      leq |x^2 - y^2| + y^2 left | sinfrac{1}{x} - sinfrac{1}{y} right |.$$
      Since:
      $$left | sinfrac{1}{x} - sinfrac{1}{y} right | leq left| frac{1}{x} - frac{1}{y}right |= frac{|x - y|}{xy},$$
      we get:
      $$left | x^2 sinfrac{1}{x} - y^2 sinfrac{1}{y} right |
      leq left(x+y+frac{y}{x}right)|x-y|.$$
      Unfortunatelly , the quantity $x+y+frac{y}{x}$ grows to
      $+infty$, either for large $x$, or for $xapprox 0.$



      Thanks in advance for the help.






      real-analysis analysis lipschitz-functions







      share|cite|improve this question









      $endgroup$





      Prove that $f:mathbb{R}to mathbb{R}$ such that
      $$ f(x) = left{
      begin{array}{c l}
      x^2, sinleft(frac{1}{x}right) & ,quad xneq 0\
      0 & ,quad x=0
      end{array} right.$$
      is Lipschitz (without use of derivatives).




      Attempt. I am aware (Lipschitz-continuous $f(x)=x^2cdot sinleft(frac{1}{x}right)$)
      that: $$|f(x)-f(y)|leq 3|x-y| ~~~forall~x,~yin mathbb{R},$$
      but I am looking for a proof, without use of derivatives.
      I tried: for $x,yneq 0$:



      begin{eqnarray}
      x^2sinfrac{1}{x} - y^2 sinfrac{1}{y}
      &=& (x^2-y^2)sinfrac{1}{x} + y^2left ( sinfrac{1}{x} - sinfrac{1}{y} right ),nonumber
      end{eqnarray}
      so: $$left | x^2sinfrac{1}{x} - y^2 sinfrac{1}{y} right |
      leq |x^2 - y^2| + y^2 left | sinfrac{1}{x} - sinfrac{1}{y} right |.$$
      Since:
      $$left | sinfrac{1}{x} - sinfrac{1}{y} right | leq left| frac{1}{x} - frac{1}{y}right |= frac{|x - y|}{xy},$$
      we get:
      $$left | x^2 sinfrac{1}{x} - y^2 sinfrac{1}{y} right |
      leq left(x+y+frac{y}{x}right)|x-y|.$$
      Unfortunatelly , the quantity $x+y+frac{y}{x}$ grows to
      $+infty$, either for large $x$, or for $xapprox 0.$



      Thanks in advance for the help.






      real-analysis analysis lipschitz-functions




      real-analysis analysis lipschitz-functions






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      asked Dec 16 '18 at 20:50









      Nikolaos SkoutNikolaos Skout

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      2,322618






















          1 Answer
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          $begingroup$

          Writing
          $$
          f(x) - f(y) = (x^2-y^2)sinfrac{1}{x} + y^2left ( sinfrac{1}{x} - sinfrac{1}{y} right )
          $$
          is a good start. The terms on the right-hand side can be estimated better if we assume that $0 < |y| le |x|$:
          $$
          left| (x^2- y^2) sin frac 1x right| le |x-y| |x+y|| frac 1x| le 2|x-y|
          $$
          and
          $$
          left|y^2left ( sinfrac{1}{x} - sinfrac{1}{y} right )right| le |y^2|
          left| frac 1x - frac 1y right| = left|frac yxright||x-y| le |x-y|
          $$
          and therefore
          $$
          |f(x)-f(y)|leq 3|x-y| , .
          $$



          For $0 < |x| le |y|$ repeat the same calculation with $x$ and $y$ exchanged, or use the symmetry of $f$.



          Finally, for $x ne 0 = y$
          $$
          |f(x) - f(0)| = left |x^2 sin frac 1x right| le |x| = |x-0| , .
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Great use of inequalities. Thank you!
            $endgroup$
            – Nikolaos Skout
            Dec 16 '18 at 22:38











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          $begingroup$

          Writing
          $$
          f(x) - f(y) = (x^2-y^2)sinfrac{1}{x} + y^2left ( sinfrac{1}{x} - sinfrac{1}{y} right )
          $$
          is a good start. The terms on the right-hand side can be estimated better if we assume that $0 < |y| le |x|$:
          $$
          left| (x^2- y^2) sin frac 1x right| le |x-y| |x+y|| frac 1x| le 2|x-y|
          $$
          and
          $$
          left|y^2left ( sinfrac{1}{x} - sinfrac{1}{y} right )right| le |y^2|
          left| frac 1x - frac 1y right| = left|frac yxright||x-y| le |x-y|
          $$
          and therefore
          $$
          |f(x)-f(y)|leq 3|x-y| , .
          $$



          For $0 < |x| le |y|$ repeat the same calculation with $x$ and $y$ exchanged, or use the symmetry of $f$.



          Finally, for $x ne 0 = y$
          $$
          |f(x) - f(0)| = left |x^2 sin frac 1x right| le |x| = |x-0| , .
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Great use of inequalities. Thank you!
            $endgroup$
            – Nikolaos Skout
            Dec 16 '18 at 22:38
















          2












          $begingroup$

          Writing
          $$
          f(x) - f(y) = (x^2-y^2)sinfrac{1}{x} + y^2left ( sinfrac{1}{x} - sinfrac{1}{y} right )
          $$
          is a good start. The terms on the right-hand side can be estimated better if we assume that $0 < |y| le |x|$:
          $$
          left| (x^2- y^2) sin frac 1x right| le |x-y| |x+y|| frac 1x| le 2|x-y|
          $$
          and
          $$
          left|y^2left ( sinfrac{1}{x} - sinfrac{1}{y} right )right| le |y^2|
          left| frac 1x - frac 1y right| = left|frac yxright||x-y| le |x-y|
          $$
          and therefore
          $$
          |f(x)-f(y)|leq 3|x-y| , .
          $$



          For $0 < |x| le |y|$ repeat the same calculation with $x$ and $y$ exchanged, or use the symmetry of $f$.



          Finally, for $x ne 0 = y$
          $$
          |f(x) - f(0)| = left |x^2 sin frac 1x right| le |x| = |x-0| , .
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Great use of inequalities. Thank you!
            $endgroup$
            – Nikolaos Skout
            Dec 16 '18 at 22:38














          2












          2








          2





          $begingroup$

          Writing
          $$
          f(x) - f(y) = (x^2-y^2)sinfrac{1}{x} + y^2left ( sinfrac{1}{x} - sinfrac{1}{y} right )
          $$
          is a good start. The terms on the right-hand side can be estimated better if we assume that $0 < |y| le |x|$:
          $$
          left| (x^2- y^2) sin frac 1x right| le |x-y| |x+y|| frac 1x| le 2|x-y|
          $$
          and
          $$
          left|y^2left ( sinfrac{1}{x} - sinfrac{1}{y} right )right| le |y^2|
          left| frac 1x - frac 1y right| = left|frac yxright||x-y| le |x-y|
          $$
          and therefore
          $$
          |f(x)-f(y)|leq 3|x-y| , .
          $$



          For $0 < |x| le |y|$ repeat the same calculation with $x$ and $y$ exchanged, or use the symmetry of $f$.



          Finally, for $x ne 0 = y$
          $$
          |f(x) - f(0)| = left |x^2 sin frac 1x right| le |x| = |x-0| , .
          $$






          share|cite|improve this answer











          $endgroup$



          Writing
          $$
          f(x) - f(y) = (x^2-y^2)sinfrac{1}{x} + y^2left ( sinfrac{1}{x} - sinfrac{1}{y} right )
          $$
          is a good start. The terms on the right-hand side can be estimated better if we assume that $0 < |y| le |x|$:
          $$
          left| (x^2- y^2) sin frac 1x right| le |x-y| |x+y|| frac 1x| le 2|x-y|
          $$
          and
          $$
          left|y^2left ( sinfrac{1}{x} - sinfrac{1}{y} right )right| le |y^2|
          left| frac 1x - frac 1y right| = left|frac yxright||x-y| le |x-y|
          $$
          and therefore
          $$
          |f(x)-f(y)|leq 3|x-y| , .
          $$



          For $0 < |x| le |y|$ repeat the same calculation with $x$ and $y$ exchanged, or use the symmetry of $f$.



          Finally, for $x ne 0 = y$
          $$
          |f(x) - f(0)| = left |x^2 sin frac 1x right| le |x| = |x-0| , .
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 16 '18 at 21:26

























          answered Dec 16 '18 at 21:15









          Martin RMartin R

          28.9k33458




          28.9k33458












          • $begingroup$
            Great use of inequalities. Thank you!
            $endgroup$
            – Nikolaos Skout
            Dec 16 '18 at 22:38


















          • $begingroup$
            Great use of inequalities. Thank you!
            $endgroup$
            – Nikolaos Skout
            Dec 16 '18 at 22:38
















          $begingroup$
          Great use of inequalities. Thank you!
          $endgroup$
          – Nikolaos Skout
          Dec 16 '18 at 22:38




          $begingroup$
          Great use of inequalities. Thank you!
          $endgroup$
          – Nikolaos Skout
          Dec 16 '18 at 22:38


















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