Results 1 to 7 of 7

Math Help - Absolute Value Function

  1. #1
    Member
    Joined
    Jan 2010
    Posts
    80

    Absolute Value Function

    How would you show that the absolute value function f(x) := |x| is continuous at every point c in \mathbb{R}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by CrazyCat87 View Post
    How would you show that the absolute value function f(x) := |x| is continuous at every point c in \mathbb{R}
    You've got to be kidding me. It's Lipschitz with Lipschitz constant one. |f(x)-f(y)||\leqslant |x-y|. So, in fact it's uniformly continuous on \mathbb{R}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,004
    Thanks
    1660
    Well, if he is asking this, he probably doesn't know about "Lipschitz"- or "uniformly continuous"!

    CrazyCat87, The definition of f(x) continuous at x= a is that f(a) exist, \lim_{x\to a} f(x) exist, and \lim_{x\to a} f(x)= f(a). Typically, showing that the two are equal requires showing they exist so that is enough.

    Here, if x> 0, f(x)= |x|= x. If a> 0, we can take \delta< a/2 so that as long as |x-a|<\delta, x> 0 and |f(x)- f(a)|= |x- a|< \epsilon will be true as long as \delta is less than the smaller of a/2 and \epsilon.

    If x< 0, f(x)= |x|= -x. If a< 0, we can take \delta< -a/2 so that as long as |x-a|< \delta, x< 0 and |f(x)- f(a)|= |-x-(-a)|= |x-a|< \epsilon will be true as long as \delta is less than the smaller of -a/2 and \epsilon.

    Finally, if a= 0, |f(x)- f(a)|= | |x|- 0|= |x|. We can make that less than \epsilon by taking \delta= \epsilon.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jan 2010
    Posts
    80
    Thank you HallsofIvy
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jan 2010
    Posts
    80
    Quote Originally Posted by HallsofIvy View Post
    Well, if he is asking this, he probably doesn't know about "Lipschitz"- or "uniformly continuous"!

    CrazyCat87, The definition of f(x) continuous at x= a is that f(a) exist, \lim_{x\to a} f(x) exist, and \lim_{x\to a} f(x)= f(a). Typically, showing that the two are equal requires showing they exist so that is enough.

    Here, if x> 0, f(x)= |x|= x. If a> 0, we can take \delta< a/2 so that as long as |x-a|<\delta, x> 0 and |f(x)- f(a)|= |x- a|< \epsilon will be true as long as \delta is less than the smaller of a/2 and \epsilon.

    If x< 0, f(x)= |x|= -x. If a< 0, we can take \delta< -a/2 so that as long as |x-a|< \delta, x< 0 and |f(x)- f(a)|= |-x-(-a)|= |x-a|< \epsilon will be true as long as \delta is less than the smaller of -a/2 and \epsilon.

    Finally, if a= 0, |f(x)- f(a)|= | |x|- 0|= |x|. We can make that less than \epsilon by taking \delta= \epsilon.
    Why do you take \delta<a/2
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Jan 2010
    Posts
    80
    Here's what I did, let me know if it's missing anything.

    If x>0 then f(x) = |x|=x. As seen in some example, if c\in\mathbb{R}, then the limit of f(x) as x goes to c equals c. Since f(c) = c, f(x) is continuous at every point c.

    Similar approach for x<0....
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,803
    Thanks
    1692
    Awards
    1
    Quote Originally Posted by CrazyCat87 View Post
    If x>0 then f(x) = |x|=x. As seen in some example, if c\in\mathbb{R}, then the limit of f(x) as x goes to c equals c. Since f(c) = c, f(x) is continuous at every point c.
    Similar approach for x<0....
    It is so easy. Learn the most useful inequality.
    \left( {\forall x~\&~y} \right) it is true that \left| {\left| x \right| - \left| y \right|} \right| \leqslant \left| {x - y} \right|.
    Once we realize that, then we let \delta  = \varepsilon .
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: December 1st 2011, 09:56 PM
  2. Replies: 1
    Last Post: July 2nd 2011, 12:35 PM
  3. absolute function
    Posted in the Algebra Forum
    Replies: 2
    Last Post: April 4th 2010, 12:20 PM
  4. Replies: 2
    Last Post: November 8th 2009, 01:52 PM
  5. integration of an absolute value function??
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 27th 2008, 05:52 PM

Search Tags


/mathhelpforum @mathhelpforum