Results 1 to 2 of 2

Math Help - a function that has an asymptote is uniformly continuous

  1. #1
    Member
    Joined
    Feb 2011
    Posts
    147
    Thanks
    3

    a function that has an asymptote is uniformly continuous

    Let f:[0,\infty]\longrightarrow \mathbb{R} be a continuous function and \lim_{x\rightarrow\infty}(f(x)-(ax+b))=0. Show that f is uniformly continuous.

    I'd like to use this charactetization of uniform continuity. A function f is uniformly continuous iff for all sequences x_n,\,y_n if |x_n-y_n|\rightarrow 0, then also |f(x_n)-f(y_n)|\rightarrow 0.

    Suppose |x_n-y_n|\rightarrow 0.
    If x_n is a bounded sequence, then y_n must be bounded too, and so if |x_n-y_n|\rightarrow 0, then also |f(x_n)-f(y_n)|\rightarrow 0, because f is continuous and therefore uniformly continuous on a compact set. If, on the other hand, x_n\rightarrow\infty, then also y_n\rightarrow\infty and we have

    |f(x_n)-f(y_n)|\leq |f(x_n)-(ax_n+b)|+|f(y_n)-(ay_n+b)|+|(ax_n+b)-(ay_n+b)|\rightarrow 0

    But what if x_n is neither bounded nor convergent to infinity?
    Last edited by ymar; June 28th 2011 at 12:01 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Feb 2008
    Posts
    410

    Re: a function that has an asymptote is uniformly continuous

    Quote Originally Posted by ymar View Post
    Let f:[0,\infty]\longrightarrow \mathbb{R} be a continuous function and \lim_{x\rightarrow\infty}(f(x)-(ax+b))=0. Show that f is uniformly continuous.

    I'd like to use this charactetization of uniform continuity. A function f is uniformly continuous iff for all sequences x_n,\,y_n if |x_n-y_n|\rightarrow 0, then also |f(x_n)-f(y_n)|\rightarrow 0.

    Suppose |x_n-y_n|\rightarrow 0.
    If x_n is a bounded sequence, then y_n must be bounded too, and so if |x_n-y_n|\rightarrow 0, then also |f(x_n)-f(y_n)|\rightarrow 0, because f is continuous and therefore uniformly continuous on a compact set. If, on the other hand, x_n\rightarrow\infty, then also y_n\rightarrow\infty and we have

    |f(x_n)-f(y_n)|\leq |f(x_n)-(ax_n+b)|+|f(y_n)-(ay_n+b)|+|(ax_n+b)-(ay_n+b)|\rightarrow 0

    But what if x_n is neither bounded nor convergent to infinity?
    Let g=f-(ax+b) and \epsilon>0. Then there is M>0 such that if x,y\geq M then |g(x)|,|g(y)|\leq\epsilon/3. Notice also that f is uniformly continuous on [0,M+\delta_1] (by Heine-Cantor), where \delta_1=\epsilon/3|a|. So there is \delta_2>0 such that if |x-y|<\delta_2 for x,y\in[0,M+\delta_1] then |f(x)-f(y)|<\epsilon. Now let \delta=\min\{\delta_1,\delta_2\}, and suppose |x-y|<\delta. If x,y\geq M then |f(x)-f(y)|<\epsilon (details omitted). Otherwise M>x or M>y, giving us x,y\in[0,M+\delta_1] and hence |f(x)-f(y)|<\epsilon. So in both cases we have |x-y|<\delta implying |f(x)-f(y)|<\epsilon.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: April 18th 2011, 09:19 AM
  2. Replies: 3
    Last Post: April 18th 2011, 08:24 AM
  3. Is the following function uniformly continuous?
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: March 17th 2010, 08:08 PM
  4. Uniformly continuous function
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: March 7th 2010, 01:07 PM
  5. uniformly continuous function
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: March 6th 2009, 01:10 AM

Search Tags


/mathhelpforum @mathhelpforum