Showing function is not Lipschitz continuous












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For an analysis exercise, I had to show that the function $sqrt{1-x^2}$ was uniformly continuous, but not lipschitz continuous on the interval $[-1,1]$. I was able to show it was uniformly continuous, however I keep running into problems showing that it is not lipschitz.



Any help is appreciated.






real-analysis lipschitz-functions







share|cite|improve this question









$endgroup$












  • $begingroup$
    What is the definition of Lipschitz continuity? That is a good starting point
    $endgroup$
    – Zubin Mukerjee
    Jan 14 at 19:58






  • 1




    $begingroup$
    consider the Lipschitz condition near $x=1$, when one of the points is set to $1$.
    $endgroup$
    – Hayk
    Jan 14 at 20:02










  • $begingroup$
    It is that there exists $K>0$ such that for $x,y in [-1,1]$, $|f(x)-f(y)|<K|x-y|$. I know I want to show that such a $K$ cannot exist, but I am lost as to how I can do this.
    $endgroup$
    – Tyler6
    Jan 14 at 20:04
















0












$begingroup$


For an analysis exercise, I had to show that the function $sqrt{1-x^2}$ was uniformly continuous, but not lipschitz continuous on the interval $[-1,1]$. I was able to show it was uniformly continuous, however I keep running into problems showing that it is not lipschitz.



Any help is appreciated.






real-analysis lipschitz-functions







share|cite|improve this question









$endgroup$












  • $begingroup$
    What is the definition of Lipschitz continuity? That is a good starting point
    $endgroup$
    – Zubin Mukerjee
    Jan 14 at 19:58






  • 1




    $begingroup$
    consider the Lipschitz condition near $x=1$, when one of the points is set to $1$.
    $endgroup$
    – Hayk
    Jan 14 at 20:02










  • $begingroup$
    It is that there exists $K>0$ such that for $x,y in [-1,1]$, $|f(x)-f(y)|<K|x-y|$. I know I want to show that such a $K$ cannot exist, but I am lost as to how I can do this.
    $endgroup$
    – Tyler6
    Jan 14 at 20:04














0












0








0


1



$begingroup$


For an analysis exercise, I had to show that the function $sqrt{1-x^2}$ was uniformly continuous, but not lipschitz continuous on the interval $[-1,1]$. I was able to show it was uniformly continuous, however I keep running into problems showing that it is not lipschitz.



Any help is appreciated.






real-analysis lipschitz-functions







share|cite|improve this question









$endgroup$




For an analysis exercise, I had to show that the function $sqrt{1-x^2}$ was uniformly continuous, but not lipschitz continuous on the interval $[-1,1]$. I was able to show it was uniformly continuous, however I keep running into problems showing that it is not lipschitz.



Any help is appreciated.






real-analysis lipschitz-functions




real-analysis lipschitz-functions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 14 at 19:54









Tyler6Tyler6

838413




838413












  • $begingroup$
    What is the definition of Lipschitz continuity? That is a good starting point
    $endgroup$
    – Zubin Mukerjee
    Jan 14 at 19:58






  • 1




    $begingroup$
    consider the Lipschitz condition near $x=1$, when one of the points is set to $1$.
    $endgroup$
    – Hayk
    Jan 14 at 20:02










  • $begingroup$
    It is that there exists $K>0$ such that for $x,y in [-1,1]$, $|f(x)-f(y)|<K|x-y|$. I know I want to show that such a $K$ cannot exist, but I am lost as to how I can do this.
    $endgroup$
    – Tyler6
    Jan 14 at 20:04


















  • $begingroup$
    What is the definition of Lipschitz continuity? That is a good starting point
    $endgroup$
    – Zubin Mukerjee
    Jan 14 at 19:58






  • 1




    $begingroup$
    consider the Lipschitz condition near $x=1$, when one of the points is set to $1$.
    $endgroup$
    – Hayk
    Jan 14 at 20:02










  • $begingroup$
    It is that there exists $K>0$ such that for $x,y in [-1,1]$, $|f(x)-f(y)|<K|x-y|$. I know I want to show that such a $K$ cannot exist, but I am lost as to how I can do this.
    $endgroup$
    – Tyler6
    Jan 14 at 20:04
















$begingroup$
What is the definition of Lipschitz continuity? That is a good starting point
$endgroup$
– Zubin Mukerjee
Jan 14 at 19:58




$begingroup$
What is the definition of Lipschitz continuity? That is a good starting point
$endgroup$
– Zubin Mukerjee
Jan 14 at 19:58




1




1




$begingroup$
consider the Lipschitz condition near $x=1$, when one of the points is set to $1$.
$endgroup$
– Hayk
Jan 14 at 20:02




$begingroup$
consider the Lipschitz condition near $x=1$, when one of the points is set to $1$.
$endgroup$
– Hayk
Jan 14 at 20:02












$begingroup$
It is that there exists $K>0$ such that for $x,y in [-1,1]$, $|f(x)-f(y)|<K|x-y|$. I know I want to show that such a $K$ cannot exist, but I am lost as to how I can do this.
$endgroup$
– Tyler6
Jan 14 at 20:04




$begingroup$
It is that there exists $K>0$ such that for $x,y in [-1,1]$, $|f(x)-f(y)|<K|x-y|$. I know I want to show that such a $K$ cannot exist, but I am lost as to how I can do this.
$endgroup$
– Tyler6
Jan 14 at 20:04










1 Answer
1






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$begingroup$

Following the definition of Lipschitz condition for $f$, we need to show that for some constant $K>0 $ one has
$$
|f(x) - f(y)| leq K |x- y| text{ for all } x,y in [-1,1].
$$
Now take $y = 1$, then $f(y) = 0$ and the above becomes
$$
tag{1} sqrt{1 - x^2} leq K |x - 1|, text{ for all } x in [-1,1].
$$
However,
$$
limlimits_{xto 1-}frac{sqrt{1 - x^2}}{1-x} = limlimits_{xto 1-}frac{sqrt{1 + x}}{sqrt{1 - x}} = + infty,
$$
hence no $K$ satisfies $(1)$.






share|cite|improve this answer









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    $begingroup$

    Following the definition of Lipschitz condition for $f$, we need to show that for some constant $K>0 $ one has
    $$
    |f(x) - f(y)| leq K |x- y| text{ for all } x,y in [-1,1].
    $$
    Now take $y = 1$, then $f(y) = 0$ and the above becomes
    $$
    tag{1} sqrt{1 - x^2} leq K |x - 1|, text{ for all } x in [-1,1].
    $$
    However,
    $$
    limlimits_{xto 1-}frac{sqrt{1 - x^2}}{1-x} = limlimits_{xto 1-}frac{sqrt{1 + x}}{sqrt{1 - x}} = + infty,
    $$
    hence no $K$ satisfies $(1)$.






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      Following the definition of Lipschitz condition for $f$, we need to show that for some constant $K>0 $ one has
      $$
      |f(x) - f(y)| leq K |x- y| text{ for all } x,y in [-1,1].
      $$
      Now take $y = 1$, then $f(y) = 0$ and the above becomes
      $$
      tag{1} sqrt{1 - x^2} leq K |x - 1|, text{ for all } x in [-1,1].
      $$
      However,
      $$
      limlimits_{xto 1-}frac{sqrt{1 - x^2}}{1-x} = limlimits_{xto 1-}frac{sqrt{1 + x}}{sqrt{1 - x}} = + infty,
      $$
      hence no $K$ satisfies $(1)$.






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        Following the definition of Lipschitz condition for $f$, we need to show that for some constant $K>0 $ one has
        $$
        |f(x) - f(y)| leq K |x- y| text{ for all } x,y in [-1,1].
        $$
        Now take $y = 1$, then $f(y) = 0$ and the above becomes
        $$
        tag{1} sqrt{1 - x^2} leq K |x - 1|, text{ for all } x in [-1,1].
        $$
        However,
        $$
        limlimits_{xto 1-}frac{sqrt{1 - x^2}}{1-x} = limlimits_{xto 1-}frac{sqrt{1 + x}}{sqrt{1 - x}} = + infty,
        $$
        hence no $K$ satisfies $(1)$.






        share|cite|improve this answer









        $endgroup$



        Following the definition of Lipschitz condition for $f$, we need to show that for some constant $K>0 $ one has
        $$
        |f(x) - f(y)| leq K |x- y| text{ for all } x,y in [-1,1].
        $$
        Now take $y = 1$, then $f(y) = 0$ and the above becomes
        $$
        tag{1} sqrt{1 - x^2} leq K |x - 1|, text{ for all } x in [-1,1].
        $$
        However,
        $$
        limlimits_{xto 1-}frac{sqrt{1 - x^2}}{1-x} = limlimits_{xto 1-}frac{sqrt{1 + x}}{sqrt{1 - x}} = + infty,
        $$
        hence no $K$ satisfies $(1)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 14 at 20:15









        HaykHayk

        2,6271214




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