1. Uniform Continuity

$f(x)=1/x$ how would you show this is continuous on $A=[1,\infty)$ but not on $B=(0,\infty)$?

2. The first by exhibiting a function $\varepsilon(\delta)$ valid on that interval, and the second by contradiction.

3. Originally Posted by CrazyCat87
$f(x)=1/x$ how would you show this is continuous on $A=[1,\infty)$ but not on $B=(0,\infty)$?
I assume you mean uniformly continuous

First note that 1/x - 1/y = y-x/xy

For the first part:

For x, y in B, we have x,y>1 so 1/xy < 1

Thus, for all x,y in B lf(x) - f(y)l = l1/x - 1/yl = lx-yl/lxyl < lx-yl

and uniform continuity is not hard to show (let delta = epsilon).

For the second part, you have to show that there exists an epsilon > 0, you can find an x, y small enough such that their difference is greater than epsilon.

4. I'm still having trouble with proving uniform continuity on $B=(0,\infty)$.

How would you show that $g(x)=\frac{1}{x^2}$ is uniformly continuous on $B$?

5. Proving Non-Uniform Continuity

I'd like to see that one also...

6. Originally Posted by CrazyCat87
$f(x)=1/x$ how would you show this is continuous on $A=[1,\infty)$ but not on $B=(0,\infty)$?
To prove that it's uniformly convergent on $[1,\infty)$ merely note that it's differentiable and $\left|\frac{-2}{x^3}\right|\leqslant 2$. Thus, the function is Lipschitz and trivially unfiromly continuous.

7. Originally Posted by CrazyCat87
$f(x)=1/x$ how would you show this is continuous on $A=[1,\infty)$ but not on $B=(0,\infty)$?
To prove that 1/x is not uniformly continuous on (0, $\infty$) we must show that:

There exists ε>0 such that for all δ>0 there exists x,wε(0, $\infty$) such that:

|x-w|<δ and |f(x)-f(w)| $\geq\epsilon$.

Let ε=2.

Take 0<δ<1 .

Choose x= δ and w= δ/4 then we have:

|x-w| = |δ- $\frac{\delta}{4}|= \frac{3\delta}{4}<\delta$

AND

|f(x)-f(w)| = $|\frac{1}{\delta}-\frac{4}{\delta}|=\frac{3}{\delta}\geq 3\geq\epsilon$.

Hence 1/x is not uniformly continuous on (0, $\infty$)

8. as aside note, $f(x)=\frac1x$ is uniformly continuous for each $[\alpha,\infty)$ where $\alpha>0.$

the proof follows easily since given $\epsilon>0$ and by taking $\delta=\alpha^2\epsilon$ we have $\left| \frac{1}{x}-\frac{1}{x_{0}} \right|=\frac{\left| x-x_{0} \right|}{x\cdot x_{0}}\le \frac{1}{\alpha ^{2}}\left| x-x_{0} \right|<\frac{1}{\alpha ^{2}}\cdot \delta =\frac{1}{\alpha ^{2}}\cdot \alpha ^{2}\epsilon =\epsilon.\quad\blacksquare$

9. Originally Posted by Krizalid
as aside note, $f(x)=\frac1x$ is uniformly continuous for each $[\alpha,\infty)$ where $\alpha>0.$

the proof follows easily since given $\epsilon>0$ and by taking $\delta=\alpha^2\epsilon$ we have $\left| \frac{1}{x}-\frac{1}{x_{0}} \right|=\frac{\left| x-x_{0} \right|}{x\cdot x_{0}}\le \frac{1}{\alpha ^{2}}\left| x-x_{0} \right|<\frac{1}{\alpha ^{2}}\cdot \delta =\frac{1}{\alpha ^{2}}\cdot \alpha ^{2}\epsilon =\epsilon.\quad\blacksquare$
Or that $|f'(x)|\leqslant\frac{2}{\alpha^2}$

10. i know, the problem is that bounded derivative implies Lipschitz is not something that many people learn.