Uniform Continuity for 1/x

• November 4th 2010, 01:29 AM
AKTilted
Uniform Continuity for 1/x
I'm trying to prove the following statement:

Prove that the function f(x)=1/x is continuous on (0,1] but it is not uniformly continuous on this interval.

Thanks a lot.
• November 4th 2010, 06:20 AM
Tinyboss
For the "not uniformly continuous" part: do you know how to formally negate a logical proposition, i.e. exchange "for-all's" and "there-exists's" and reverse (in)equalities? Try that with the definition of uniform continuity.
• November 4th 2010, 07:04 AM
Drexel28
Quote:

Originally Posted by AKTilted
I'm trying to prove the following statement:

Prove that the function f(x)=1/x is continuous on (0,1] but it is not uniformly continuous on this interval.

Thanks a lot.

I actually discuss this in my post here. In essence, (we'll speak less generally now) if $f:((0,1]\to\mathbb{R}$ were uniformly continuous, then we could extend it to some uniformly continuous map $\tilde{f}:[0,1]\to\mathbb{R}$. But, this is just nonsense since we'd have to have that $f(0)=\lim f\left(\frac{1}{n}\right)=\lim n$
• November 4th 2010, 07:08 AM
Also sprach Zarathustra
Quote:

Originally Posted by AKTilted
I'm trying to prove the following statement:

Prove that the function f(x)=1/x is continuous on (0,1] but it is not uniformly continuous on this interval.

Thanks a lot.

The negation of uniform continuity:

For some $\epsilon > 0$ and for each $\delta > 0$
there exist $x_0, x_1$ such that $|x_0 - x_1| < \delta$ but $|f(x_0) - f(x_1)| \geq \epsilon$.

Assume that there is exist $\delta <\frac{1}{n}$ $n\in \left \{\mathbb{N} / 0}{ \right \}$.

Now let $x_0,x_1 \in\mathbb{R}^+ x_0=x_1+\frac{\delta }{2} and x_1=\frac{\delta }{2}$

$\left | f(x_0)-f(x_1) \right |=\left | \frac{1}{x_0}-\frac{1}{x_1} \right |=\left |\frac{ x_0-x_1}{x_0x_1} \right |
$

$=\frac{\frac{2}{n}}{x_0(x_0+\frac{\delta }{2})}\frac{\delta }{2}=\frac{\frac{2}{n}}{\delta^2}\frac{\delta }{2}=\frac{\frac{1}{n}}{\delta }>1$
• November 4th 2010, 07:15 AM
Also sprach Zarathustra
Another much better solution:

Choose $x_n=\frac{1}{n}$and $y_n=\frac{1}{n+1}$for any $n\in\mathbb{N}$.

$\left | x_n-y_n \right |<\frac{1}{n}$, but $\left | f(x_n)-f(y_n) \right |=1
$
.