uniformly continuous function

**eskimo343** · March 3rd 2009, 01:38 PM

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.

**HallsofIvy** · March 3rd 2009, 01:57 PM

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.

**eskimo343** · March 4th 2009, 08:25 PM

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 $|\frac{x}{x-1}-\frac{y}{y-1}|$ $=\frac{|x(y-1)-y(x-1)|}{|(x-1)(y-1)|} \leq \frac{|x(y-1)-y(x-1)|}{(a-1)^2}$ $=\frac{|y-x|}{(a-1)^2}=\frac{|x-y|}{(a-1)^2}<br /> < \epsilon.$ Does this argument work, just by using the definition of uniform continuity?

**Opalg** · March 5th 2009, 12:31 AM

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].

**eskimo343** · March 5th 2009, 09:58 AM

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 $|\frac{x}{x-1}-\frac{y}{y-1}|$ $=\frac{|x(y-1)-y(x-1)|}{|(x-1)(y-1)|} \leq \frac{|x(y-1)-y(x-1)|}{(a-1)^2}$ $=\frac{|y-x|}{(a-1)^2}=\frac{|x-y|}{(a-1)^2}<br /> < \frac{\delta}{(a-1)^2}= \frac{(a-1)^2 \cdot \epsilon}{(a-1)^2}= \epsilon.$ 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?

**Opalg** · March 6th 2009, 12:10 AM

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 $|\frac{x}{x-1}-\frac{y}{y-1}|$ $=\frac{|x(y-1)-y(x-1)|}{|(x-1)(y-1)|} \leq \frac{|x(y-1)-y(x-1)|}{(a-1)^2}$ $=\frac{|y-x|}{(a-1)^2}=\frac{|x-y|}{(a-1)^2}<br /> < \frac{\delta}{(a-1)^2}= \frac{(a-1)^2 \cdot \epsilon}{(a-1)^2}= \epsilon.$ 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 $x=1+\tfrac1n, \ y = 1+\tfrac1{2n}$ .

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