# Math Help - Continuity Proof

1. ## Continuity Proof

Question:

Show that the function: $f(x):=\frac{1}{x^2}$ is uniformly continous on $A:=[1,\infty)$, but that it is not uniformly continuous on $B := (0,\infty)$.

My thoughts:

So far I've gotten to $|f(x)-f(u)|\leq|x-u||x+u|$ when in A.

2. Originally Posted by Luxury
Question:

Show that the function: $f(x):=\frac{1}{x^2}$ is uniformly continous on $A:=[1,\infty)$, but that it is not uniformly continuous on $B := (0,\infty)$.

My thoughts:

So far I've gotten to $|f(x)-f(u)|\leq|x-u||x+u|$ when in A.
You may want to browse through the forum because I could swear I saw it recently

EDIT: http://www.mathhelpforum.com/math-he...ontinuity.html OK not exactly the same, but close enough/

3. I actually did see that, and I already knew how to do that proof. I'm still lost on this one thought. It might be just because I've been staring at this stuff for too long.

4. Originally Posted by Luxury
Question:

Show that the function: $f(x):=\frac{1}{x^2}$ is uniformly continous on $A:=[1,\infty)$, but that it is not uniformly continuous on $B := (0,\infty)$.

My thoughts:

So far I've gotten to $|f(x)-f(u)|\leq|x-u||x+u|$ when in A.
Here's one way to show its not uniformly continuous on $B := (0,\infty)$

Suppose f(x) is uniformly continuous on $B := (0,\infty)$. Fix e > 0 Thus, there is a d such that:

lx-yl < d --> l f(x) - f(y) l < e

Choose an n so large such that n > max{1/ $\sqrt{d}$, 7e/3}.

Let x = 1/ $\sqrt{n}$, y = x + d/2

Clearly, lx-yl < d

But, lf(x) - f(y)l = l n - 1/(1/ $\sqrt{n}$+d/2)^2 l = l n + 4n/(4+n*d^2+2d* $\sqrt{n}$) l = n* l 1 - 4/(4+n*d^2+2d* $\sqrt{n}$)l

but we said n > 1/ $\sqrt{d}$, thus

1 - 4/(4+n*d^2+2d* $\sqrt{n}$) > 3/7 and
n*l(1 - 4/(4+n*d^2+2d* $\sqrt{n}$)l > n*3/7 > e

Note, there are likely mistakes here, I rushed it a bit.

5. Also note that a unif. cont. function preserves Cauchy sequences. But, $\frac{1}{\sqrt{n}}\to 0$ but $f\left(\frac{1}{\sqrt{n}}\right)=n$ is not Cauchy.

6. Originally Posted by Drexel28
Also note that a unif. cont. function preserves Cauchy sequences. But, $\frac{1}{\sqrt{n}}\to 0$ but $f\left(\frac{1}{\sqrt{n}}\right)=n$ is not Cauchy.
Interesting, never knew that.

7. South, correct me if I'm wrong, but I think you did it for f(x) = 1/x not f(x)=1/x^2.

8. Originally Posted by Luxury
South, correct me if I'm wrong, but I think you did it for f(x) = 1/x not f(x)=1/x^2.
Correct

Fixed (I hope)