# uniformly continuous function

• March 3rd 2009, 02:38 PM
eskimo343
uniformly continuous function
Let $a>1$ and let $f: (1,2] \rightarrow \mathbb{R}$ be defined by $f(x)=\frac{x}{x-1}$. Prove that $f$ is uniformly continuous on the interval $[a,2]$, but $f$ is not uniformly continuous on $(1,2]$.

I am lost on this problem. Any help is appreciated. Thanks in advance.
• March 3rd 2009, 02:57 PM
HallsofIvy
Quote:

Originally Posted by eskimo343
Let $a>1$ and let $f: (1,2] \rightarrow \mathbb{R}$ be defined by $f(x)=\frac{x}{x-1}$. Prove that $f$ is uniformly continuous on the interval $[a,2]$, but $f$ is not uniformly continuous on $(1,2]$.

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 $|f(x)- f(x_0)|= \frac{x}{x-1}- \frac{x_0}{x_0-1}|< \epsilon$, then you want $\frac{x(x_0)-1)- x_0(x-1)}{(x-1)(x_0-1)}|= |\frac{x_0- x}{(x-1)(x_0-1|< \epsilon$. How would you choose $\delta$? Show that if $1< a< x_0$ you can pick $\delta$ independent of $x_0$ but if $x_0$ can be arbitrarily close to 1, you can't.
• March 4th 2009, 09:25 PM
eskimo343
Quote:

Originally Posted by HallsofIvy
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 $|f(x)- f(x_0)|= \frac{x}{x-1}- \frac{x_0}{x_0-1}|< \epsilon$, then you want $\frac{x(x_0)-1)- x_0(x-1)}{(x-1)(x_0-1)}|= |\frac{x_0- x}{(x-1)(x_0-1|< \epsilon$. How would you choose $\delta$? Show that if $1< a< x_0$ you can pick $\delta$ independent of $x_0$ but if $x_0$ 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 $\delta= (a-1)^2 \cdot \epsilon$. We assume that $|x-y|<\delta$. Then $|\frac{x}{x-1}-\frac{y}{y-1}|$ $=\frac{|x(y-1)-y(x-1)|}{|(x-1)(y-1)|} \leq \frac{|x(y-1)-y(x-1)|}{(a-1)^2}$ $=\frac{|y-x|}{(a-1)^2}=\frac{|x-y|}{(a-1)^2}
< \epsilon.$
Does this argument work, just by using the definition of uniform continuity?
• March 5th 2009, 01:31 AM
Opalg
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 $|f'(x)|\leqslant M$ then $|f(y)-f(x)| = |f'(c)||y-x|$ for some c between x and y. So if $\delta = \varepsilon/M$ then $|y-x|<\delta\,\Rightarrow\, |f(y)-f(x)| \leqslant M|y-x|<\varepsilon$.)

The function $f(x) = \frac x{x-1}$ 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].
• March 5th 2009, 10:58 AM
eskimo343
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 $\delta= (a-1)^2 \cdot \epsilon$. We assume that $|x-y|<\delta$. Then $|\frac{x}{x-1}-\frac{y}{y-1}|$ $=\frac{|x(y-1)-y(x-1)|}{|(x-1)(y-1)|} \leq \frac{|x(y-1)-y(x-1)|}{(a-1)^2}$ $=\frac{|y-x|}{(a-1)^2}=\frac{|x-y|}{(a-1)^2}
< \frac{\delta}{(a-1)^2}= \frac{(a-1)^2 \cdot \epsilon}{(a-1)^2}= \epsilon.$
So, I am really just wondering if this works to prove uniform continuity on $[a,2]$. Does this show uniform continuity on $[a,2]$ just using the definition of uniform continuity?
• March 6th 2009, 01:10 AM
Opalg
Quote:

Originally Posted by eskimo343
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 $\delta= (a-1)^2 \cdot \epsilon$. We assume that $|x-y|<\delta$. Then $|\frac{x}{x-1}-\frac{y}{y-1}|$ $=\frac{|x(y-1)-y(x-1)|}{|(x-1)(y-1)|} \leq \frac{|x(y-1)-y(x-1)|}{(a-1)^2}$ $=\frac{|y-x|}{(a-1)^2}=\frac{|x-y|}{(a-1)^2}
< \frac{\delta}{(a-1)^2}= \frac{(a-1)^2 \cdot \epsilon}{(a-1)^2}= \epsilon.$
So, I am really just wondering if this works to prove uniform continuity on $[a,2]$. Does this show uniform continuity on $[a,2]$ 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 $x=1+\tfrac1n, \ y = 1+\tfrac1{2n}$.