1. ## Uniformly Cont. Func

1.) Let $f(x) = \frac{x}{x-1}$. Prove that $f(x)$ is uniformly continuous on the interval [ $1.5,\infty$).

2.) Suppose that $f$ and $g$ are uniformly continuous on some interval $A$. Either prove or disprove $f+g$ is uniformly continous on $A$.

2. This website should help.

3. Originally Posted by fifthrapiers
1.) Let $f(x) = \frac{x}{x-1}$. Prove that $f(x)$ is uniformly continuous on the interval [ $1.5,\infty$).
Theorem: If $f$ is continuous on a closed interval $[a,b]$, then $f$ is uniformly continuous on $[a,b]$

4. Originally Posted by fifthrapiers
2.) Suppose that $f$ and $g$ are uniformly continuous on some interval $A$. Either prove or disprove $f+g$ is uniformly continous on $A$.
go by the definition. there may be a shorter way, but i can't remember now.

Defintion: Let $f$ be a real valued function defined on a set $S \subset \mathbb{R}$. Then $f$ is uniformly continuous on $S$ if

for each $\epsilon > 0$ there exists $\delta > 0$ such that for all $x,y \in S$ and $|x - y|< \delta$ we have $|f(x) - f(y)|< \epsilon$.

I will start you off in the proof.

Proof: Let $f,g$ be uniformly continuous on some interval $I \subset \mathbb{R}$.

Then we have that for each $\epsilon > 0$ there exists $\delta_1 > 0$ such that for all $x,y \in I$ and $|x - y|< \delta_1$ we have $|f(x) - f(y)|< \frac {\epsilon}2$.

A similar thing holds true for $g$ if we choose some $\delta_2 > 0$.

Now let $h(x) = f(x) + g(x)$, fix such an $\epsilon > 0$ and choose $\delta = \mbox{min} \{ \delta_1, \delta_2 \}$.

Then, for all $x,y \in I$ and $|x - y|< \delta$, we have:

$|h(x) - h(y)| = |(f(x) + g(x)) - (f(y) + g(y))| = ...$

finish it off

5. Originally Posted by Jhevon
Theorem: If $f$ is continuous on a closed interval $[a,b]$, then $f$ is uniformly continuous on $[a,b]$
But $[1.5,\infty)$ is not a finite closed interval. How can you use the theorem?

6. Originally Posted by ThePerfectHacker
But $[1.5,\infty)$ is not a finite closed interval. How can you use the theorem?
the theorem didn't say anything about the interval being finite, so i thought it was ok. blame the author

EDIT: ...oh, i see [a,b] implies finite...ok

7. could we use the theorem about the derivative being bounded on the interval?

8. Originally Posted by Jhevon
could we use the theorem about the derivative being bounded on the interval?
That is how I would do it.

9. Originally Posted by ThePerfectHacker
That is how I would do it.
that's what i originally thought of using, but i saw that the first theorem (if applicable) would get the job done a lot faster, with a lot less work.

for the benefit of the class, here's the new theorem we are proposing:

Theorem: Let $f$ be a continuous function on an interval $I$ ( $I$ may be bounded or unbounded, hehe, something my last theorem was missing). Let $I^o$ be the interval obtained by removing from $I$ any endpoints that $I$ may have. If $f$ is differentiable on $I^o$ and if $f'$ is bounded on $I^o$, then $f$ is uniformly continuous on $I$.

10. I'm still very stuck on #1. I hate proving unif. cont.

I attempted #2, and I'd appreciate if you could look at it for me:

Let $\epsilon > 0$ be given. Then, $\frac{\epsilon}{2} > 0$, thus, $\exists R \in \mathbb{N} \ni r \geq R \implies ||f_r - f|| < \frac{\epsilon}{2}$ since $(f_n)$ is unif. convergence to $f$ on $A$. Similarly for g, $\exists P \in \mathbb{N} \ni p \geq P \implies ||g_p - g|| < \frac{\epsilon}{2}$. Let M = max{R,P}. Now, for $n \geq N$, we have:

$||(f_n + g_n) - (f + g)|| = ||(f_n - f) + (g_n - g)|| \leq ||f_n - f|| + ||g_n - g|| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$

Thus, $f + g$ is unif. cont. on A.

11. Originally Posted by fifthrapiers
I'm still very stuck on #1. I hate proving unif. cont.

I attempted #2, and I'd appreciate if you could look at it for me:

Let $\epsilon > 0$ be given. Then, $\frac{\epsilon}{2} > 0$, thus, $\exists R \in \mathbb{N} \ni r \geq R \implies ||f_r - f|| < \frac{\epsilon}{2}$ since $(f_n)$ is unif. convergence to $f$ on $A$. Similarly for g, $\exists P \in \mathbb{N} \ni p \geq P \implies ||g_p - g|| < \frac{\epsilon}{2}$. Let M = max{R,P}. Now, for $n \geq N$, we have:

$||(f_n + g_n) - (f + g)|| = ||(f_n - f) + (g_n - g)|| \leq ||f_n - f|| + ||g_n - g|| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$

Thus, $f + g$ is unif. cont. on A.
Grr, I mis-read it. I proved that they are uniformly CONVERGENT, not uniformly CONTINUOUS!!

Well, I know how to prove f is unif. continuous but not f + g...

So proving f is unif. continuous...:

Let (f_n) be the sequence of functions where $f_n : A \rightarrow \mathbb{R}$ are unif. cont. on A and that [tex](f_n) converges unif. to f on A. Let $\epsilon > 0$ be given. Then, $\exists$ a natural number $N \in \mathbb{N} \ni \forall n \geq N, |f_{n}(x) - f(x)| < \frac{\epsilon}{3} \forall x \in A$ by unif. conv. of $f_n$ to f on A. Particularly, $|f_{N}(x) - f(x)| < \frac{\epsilon}{3} \forall x \in A$.

As $f_N$ is unif. cont on A, $\exists \delta > 0 \ni |f_{N}(x) - f_{N}(y)| < \frac{\epsilon}{3}$ whenever $x,y \in A$ satisfy $|x-y| < \delta$. Then we have for this $\delta$, suppose $x,y \in A$ satisfy $|x - y| < \delta$ and consider:

\begin{eqnarray}
|f(x) - f(y)| &=& |f(x) - [f_{N}(x) - f_{N}(y)] - [f_{N}(y) - f_{N}(y)] - f(y)|\\
&=& |[f(x) - f_{N}(x)] + [f_{N}(x) - f_{N}(y)] + [f_{N}(y) - f(y)]|\\
&\leq& |f(x) - f_{N}(x)| + |f_{N}(x) - f_{N}(y)| + |f_{N}(y) - f(y)|\\
&<& \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon \end{eqnarray}

Thus, f is unif. cont. on A.

(NOTE: I get LaTeX syntax error on the last part (for the eqnarray stuff, not sure why) so I left it without [tex] so you can see the general jist. I think it's easy to read just in latex format.

But, I have no idea how to show f + g will be unif cont...

12. Originally Posted by fifthrapiers
Grr, I mis-read it. I proved that they are uniformly CONVERGENT, not uniformly CONTINUOUS!!

Well, I know how to prove f is unif. continuous but not f + g...

So proving f is unif. continuous...:

Let (f_n) be the sequence of functions where $f_n : A \rightarrow \mathbb{R}$ are unif. cont. on A and that [tex](f_n) converges unif. to f on A. Let $\epsilon > 0$ be given. Then, $\exists$ a natural number $N \in \mathbb{N} \ni \forall n \geq N, |f_{n}(x) - f(x)| < \frac{\epsilon}{3} \forall x \in A$ by unif. conv. of $f_n$ to f on A. Particularly, $|f_{N}(x) - f(x)| < \frac{\epsilon}{3} \forall x \in A$.

As $f_N$ is unif. cont on A, $\exists \delta > 0 \ni |f_{N}(x) - f_{N}(y)| < \frac{\epsilon}{3}$ whenever $x,y \in A$ satisfy $|x-y| < \delta$. Then we have for this $\delta$, suppose $x,y \in A$ satisfy $|x - y| < \delta$ and consider:

\begin{eqnarray}
|f(x) - f(y)| &=& |f(x) - [f_{N}(x) - f_{N}(y)] - [f_{N}(y) - f_{N}(y)] - f(y)|\\
&=& |[f(x) - f_{N}(x)] + [f_{N}(x) - f_{N}(y)] + [f_{N}(y) - f(y)]|\\
&\leq& |f(x) - f_{N}(x)| + |f_{N}(x) - f_{N}(y)| + |f_{N}(y) - f(y)|\\
&<& \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon \end{eqnarray}

Thus, f is unif. cont. on A.

(NOTE: I get LaTeX syntax error on the last part (for the eqnarray stuff, not sure why) so I left it without [tex] so you can see the general jist. I think it's easy to read just in latex format.

But, I have no idea how to show f + g will be unif cont...
firstly, you do not have to prove (nor do you have enough information to do so) that f is uniformly continuous, that was given. secondly, we are not dealing with a sequence of functions here. were that the case, we would be concerned with uniform convergence, not uniform continuity.

your proof should start EXACTLY as mine did (you can neglect to define the new function as h(x) if you wish), all you have to do is complete my proof

13. Originally Posted by fifthrapiers
I'm still very stuck on #1. I hate proving unif. cont.
all you have to show is that the derivative of the function exists and is bounded on (1.5, oo). then apply the theorem i mentioned in post #9