1. ## Uniformly continous

I have f: (0,1] -> R defined by f(x)= xsin(1/x) where 0 < x =< 1.

I have to show that it is uniformly continuous on it's domain.

I know the definition of uniform continuity is

f is uniformly continuous on (0,1] if for all E>0 there exists a D>0 such that d(x,y)<D => d(f(x), f(y)) <E

I know what the graph of this function looks like, and I can see that it is continuous on (0,1]

I'm not sure what x and y to pick to prove the continuity. I think I should use something to do with (0,1] being a compact set? Thanks

2. Originally Posted by EricaMae
I have f: (0,1] -> R defined by f(x)= xsin(1/x) where 0 < x =< 1.

I have to show that it is uniformly continuous on it's domain.

I know the definition of uniform continuity is

f is uniformly continuous on (0,1] if for all E>0 there exists a D>0 such that d(x,y)<D => d(f(x), f(y)) <E

I know what the graph of this function looks like, and I can see that it is continuous on (0,1]

I'm not sure what x and y to pick to prove the continuity. I think I should use something to do with (0,1] being a compact set (This is not compact, it is not closed)? Thanks
$\left|x\sin\left(\frac{1}{x}\right)-x'\sin\left(\frac{1}{x'}\right)\right|\leqslant|x-x'|$

So take $\delta=\varepsilon$

3. Okay, that makes sense, but does that show that it is continuous on it's domain?

4. Originally Posted by EricaMae
Okay, that makes sense, but does that show that it is continuous on it's domain?
No, it shows that it is uniformly continous on its domain which is what you asked for. The concepts are equivalent only on compact metric spaces. $(0,1]$ is not compact because by the Heine-Borel theorem we must have that in Euclidean n-space that a compact set is bounded and CLOSED. This set is the former but not the latter since 0 is a limit point of $(0,1]$ but not a point of $(0,1]$.

5. Yes, sorry I meant to say does that show that it's uniformly continuous on it's domain, not just continuous. Thanks very much for your help!

6. Originally Posted by Mathstud28
$\left|x\sin\left(\frac{1}{x}\right)-x'\sin\left(\frac{1}{x'}\right)\right|\leqslant|x-x'|$
I don't think this is true... (the derivative of $x\mapsto x\sin\frac{1}{x}$ gets arbitrarily large on neighbourhoods of $0$)

On the other hand, you can say: the function $f:x\mapsto f(x)=x\sin\frac{1}{x}$ is continuous on $(0,1]$, and $f(x)\to_{x\to 0^+} 0$ (because $|f(x)|\leq x$), so that $f$ can be extended to a continuous function on $[0,1]$ by letting $f(0)=0$. Now, we have a continuous function on the compact subset $[0,1]$, so it is uniformly continuous on $[0,1]$ by Heine's theorem. As a consequence, $f$ is uniformly continuous on $(0,1]$ (this is weaker).

7. Originally Posted by Laurent
I don't think this is true... (the derivative of $x\mapsto x\sin\frac{1}{x}$ gets arbitrarily large on neighbourhoods of $0$)

On the other hand, you can say: the function $f:x\mapsto f(x)=x\sin\frac{1}{x}$ is continuous on $(0,1]$, and $f(x)\to_{x\to 0^+} 0$ (because $|f(x)|\leq x$), so that $f$ can be extended to a continuous function on $[0,1]$ by letting $f(0)=0$. Now, we have a continuous function on the compact subset $[0,1]$, so it is uniformly continuous on $[0,1]$ by Heine's theorem. As a consequence, $f$ is uniformly continuous on $(0,1]$ (this is weaker).
Your analysis of the inequality seems correct but all of my calculations have shown that it is in fact true ...maybe I screwed up. But anyways I think you covered this with (this is weaker) but I actually thought of doing this except I thought that by extending $f$ from $f0,1]\mapsto\mathbb{R}" alt="f0,1]\mapsto\mathbb{R}" /> to $f:[0,1]\mapsto\mathbb{R}$ we are changing the problem because as you stated we would have to make the stipulation that $f(0)=0$ which means that the function you have shown is uniformly continous is not neccessarly the function that was asked to be shown was uniformly continous. Actually, can't we infact state that this function is not uniformly continous on this interval because $f'$ is unbounded?

8. Originally Posted by Mathstud28
$\left|x\sin\left(\frac{1}{x}\right)-x'\sin\left(\frac{1}{x'}\right)\right|\leqslant|x-x'|$

So take $\delta=\varepsilon$
Except its not true, just put $x=1,\ x'=0.4$

CB

9. Originally Posted by CaptainBlack
Except its not true, just put $x=1,\ x'=0.4$
CB
Thank you.

10. Okay, now I'm really confused. The question says to show that f is uniformly continuous on it's domain of (0,1]. with f(x)=xsin(1/x). I would assume that it is uniformly continuous on it's domain just because I dont see the teacher asking a question that isn't true. Then again, who knows.

11. Originally Posted by EricaMae
Okay, now I'm really confused. The question says to show that f is uniformly continuous on it's domain of (0,1]. with f(x)=xsin(1/x). I would assume that it is uniformly continuous on it's domain just because I dont see the teacher asking a question that isn't true. Then again, who knows.
The proof I gave is correct (and the remark about the derivative was an argument proving that MathStud28's inequality couldn't be correct: $f'(x)$ is a limit of ratios $\frac{f(x)-f(x')}{x-x'}$, so since the derivative is not bounded, the ratios aren't either).

I said that if you let $f(0)=0$, the function is now continuous on $[0,1]$, and hence uniformly continuous on $[0,1]$ (if you want to be quite rigorous, you can call the new function $\widetilde{f}$).

Now, look at your definition: if a function is uniformly continuous on a set, it is uniformly continuous on any subset. This is straightforward: what is true for every $x,y$ is true for the $x,y$ in a subset... Hence $f$ (or $\widetilde{f}$) is uniformly continuous on $(0,1]$.

A final remark: the fact that the derivatives are not bounded is not an obstruction. Take any continuous function on $[0,1]$. Then it is uniformly continuous. And it can be non-differentiable at all. You don't have to have $|f(x')-f(x)|\leq C |x'-x|$ (i.e. a Lipschitz function) ; for instance (and this is still strong) $|f(x)-f(x')|\leq C\sqrt{|x-x'|}$ (Hölder function of parameter $\frac{1}{2}$) would be enough to imply uniformity.