Results 1 to 6 of 6

Thread: uniformly continuous function

  1. #1
    Junior Member
    Joined
    Mar 2009
    Posts
    29

    uniformly continuous function

    Let $\displaystyle a>1$ and let $\displaystyle f: (1,2] \rightarrow \mathbb{R}$ be defined by $\displaystyle f(x)=\frac{x}{x-1}$. Prove that $\displaystyle f$ is uniformly continuous on the interval $\displaystyle [a,2]$, but $\displaystyle f$ is not uniformly continuous on $\displaystyle (1,2]$.

    I am lost on this problem. Any help is appreciated. Thanks in advance.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,766
    Thanks
    3027
    Quote Originally Posted by eskimo343 View Post
    Let $\displaystyle a>1$ and let $\displaystyle f: (1,2] \rightarrow \mathbb{R}$ be defined by $\displaystyle f(x)=\frac{x}{x-1}$. Prove that $\displaystyle f$ is uniformly continuous on the interval $\displaystyle [a,2]$, but $\displaystyle f$ is not uniformly continuous on $\displaystyle (1,2]$.

    I am lost on this problem. Any help is appreciated. Thanks in advance.
    What do you have to work with? For example there is a theorem that says if a function is continuous on a compact set, then it is uniformly continuous there. If you do not have that theorem, Think about how you would prove continuity: If you want $\displaystyle |f(x)- f(x_0)|= \frac{x}{x-1}- \frac{x_0}{x_0-1}|< \epsilon$, then you want $\displaystyle \frac{x(x_0)-1)- x_0(x-1)}{(x-1)(x_0-1)}|= |\frac{x_0- x}{(x-1)(x_0-1|< \epsilon$. How would you choose $\displaystyle \delta$? Show that if $\displaystyle 1< a< x_0$ you can pick $\displaystyle \delta$ independent of $\displaystyle x_0$ but if $\displaystyle x_0$ can be arbitrarily close to 1, you can't.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2009
    Posts
    29
    Quote Originally Posted by HallsofIvy View Post
    What do you have to work with? For example there is a theorem that says if a function is continuous on a compact set, then it is uniformly continuous there. If you do not have that theorem, Think about how you would prove continuity: If you want $\displaystyle |f(x)- f(x_0)|= \frac{x}{x-1}- \frac{x_0}{x_0-1}|< \epsilon$, then you want $\displaystyle \frac{x(x_0)-1)- x_0(x-1)}{(x-1)(x_0-1)}|= |\frac{x_0- x}{(x-1)(x_0-1|< \epsilon$. How would you choose $\displaystyle \delta$? Show that if $\displaystyle 1< a< x_0$ you can pick $\displaystyle \delta$ independent of $\displaystyle x_0$ but if $\displaystyle x_0$ can be arbitrarily close to 1, you can't.
    HallsofIvy,

    Thanks, I tried to prove [the first part] of this just using the definition of uniform continuity. Here is what I did:
    Let $\displaystyle \delta= (a-1)^2 \cdot \epsilon$. We assume that $\displaystyle |x-y|<\delta$. Then $\displaystyle |\frac{x}{x-1}-\frac{y}{y-1}|$$\displaystyle =\frac{|x(y-1)-y(x-1)|}{|(x-1)(y-1)|} \leq \frac{|x(y-1)-y(x-1)|}{(a-1)^2}$$\displaystyle =\frac{|y-x|}{(a-1)^2}=\frac{|x-y|}{(a-1)^2}
    < \epsilon.$ Does this argument work, just by using the definition of uniform continuity?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10
    If a function has a bounded derivative on some interval then the function is uniformly continuous on that interval. (That is an easy consequence of the mean value theorem: if $\displaystyle |f'(x)|\leqslant M$ then $\displaystyle |f(y)-f(x)| = |f'(c)||y-x|$ for some c between x and y. So if $\displaystyle \delta = \varepsilon/M$ then $\displaystyle |y-x|<\delta\,\Rightarrow\, |f(y)-f(x)| \leqslant M|y-x|<\varepsilon$.)

    The function $\displaystyle f(x) = \frac x{x-1}$ has a bounded derivative on [a,2] if a>1. But as x→1, f'(x) becomes infinite. That means that you can find point x, y close to 1 such that |y-x| is very small but |f(y)f(x)| is very large. So f is not uniformly continuous on (1,2].
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Mar 2009
    Posts
    29
    We have not seen that result yet, so I am trying to prove this just using the definition of uniform continuity. So, I did this:

    Let $\displaystyle \delta= (a-1)^2 \cdot \epsilon$. We assume that $\displaystyle |x-y|<\delta$. Then $\displaystyle |\frac{x}{x-1}-\frac{y}{y-1}|$$\displaystyle =\frac{|x(y-1)-y(x-1)|}{|(x-1)(y-1)|} \leq \frac{|x(y-1)-y(x-1)|}{(a-1)^2}$$\displaystyle =\frac{|y-x|}{(a-1)^2}=\frac{|x-y|}{(a-1)^2}
    < \frac{\delta}{(a-1)^2}= \frac{(a-1)^2 \cdot \epsilon}{(a-1)^2}= \epsilon.$ So, I am really just wondering if this works to prove uniform continuity on $\displaystyle [a,2]$. Does this show uniform continuity on $\displaystyle [a,2]$ just using the definition of uniform continuity?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10
    Quote Originally Posted by eskimo343 View Post
    We have not seen that result yet, so I am trying to prove this just using the definition of uniform continuity. So, I did this:

    Let $\displaystyle \delta= (a-1)^2 \cdot \epsilon$. We assume that $\displaystyle |x-y|<\delta$. Then $\displaystyle |\frac{x}{x-1}-\frac{y}{y-1}|$$\displaystyle =\frac{|x(y-1)-y(x-1)|}{|(x-1)(y-1)|} \leq \frac{|x(y-1)-y(x-1)|}{(a-1)^2}$$\displaystyle =\frac{|y-x|}{(a-1)^2}=\frac{|x-y|}{(a-1)^2}
    < \frac{\delta}{(a-1)^2}= \frac{(a-1)^2 \cdot \epsilon}{(a-1)^2}= \epsilon.$ So, I am really just wondering if this works to prove uniform continuity on $\displaystyle [a,2]$. Does this show uniform continuity on $\displaystyle [a,2]$ just using the definition of uniform continuity?
    Yes, that's quite correct.

    To show that the function is not uniformly continuous on (1,2], you could look at what happens when $\displaystyle x=1+\tfrac1n, \ y = 1+\tfrac1{2n}$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Apr 18th 2011, 08:19 AM
  2. Replies: 3
    Last Post: Apr 18th 2011, 07:24 AM
  3. how to prove this function to be uniformly continuous ?
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Mar 13th 2011, 07:56 AM
  4. Is the following function uniformly continuous?
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Mar 17th 2010, 07:08 PM
  5. Uniformly continuous function
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Mar 7th 2010, 12:07 PM

Search Tags


/mathhelpforum @mathhelpforum