Show the Cube root of x is uniform continuous on R.

[billy2908]

Homework Statement

Let f(x)=$$x^{1/3}$$ show that it is uniform continuous on the Real metric space.

Homework Equations

By def. of uniform continuity $$\forall\epsilon$$>0 $$\exists\delta>0$$ s.t for $$\forall x,y\in\Re$$ where |x-y|<$$\delta$$ implies |f(x)-f(y)|< $$\epsilon$$

The Attempt at a Solution

I started w/ |$$x^{1/3}$$ -$$y^{1/3}$$|* (|$$x^{2/3}$$ +$$xy^{1/3}$$+$$y^{2/3}$$|/|$$x^{2/3}$$ +$$xy^{1/3}$$+$$y^{2/3}$$|)

=|x-y|/(|$$x^{2/3}$$ +$$xy^{1/3}$$+$$y^{2/3}$$|)

But it doesn't seem to be uniform cont. if I set $$\delta$$=$$\epsilon$$*(|$$x^{2/3}$$ +$$xy^{1/3}$$+$$y^{2/3}$$|)

 

[LCKurtz]

Hint: Consider two cases separately: |x| ≤ 1 and |x| > 1. Then combine the results.

 

[billy2908]

that's exactly how I wanted to prove it. By showing it is uniformly cont. on (-inf, -1) U (1, inf) and given it is uniformly cont. on [-1,1] since it's compact.

But even when I choose x,y say from (1,inf) I still get left off with d=e*(x^2/3 +(xy)^1/3 +y^2/3)

Which doesn't help to show uniform continuity. So far I only am familiar w/ the Lipgarbagez proof for Uniform continuity. i.e.
if |f(x)-f(y)|<M for some real M. then f is unif. cont.

I'm aware there's another type of unif. cont. maybe dealing w/ f(x)=x^1/3 but I have no experience in proving for those.

 

[Dick]

But x^(1/3) is Lipgarbagez on (1,inf) isn't it? It has a bounded derivative, right?

 

[billy2908]

Sorry, can't use the def. of derivative yet at this point. Also even if I can, and the derivative is bounded on (1,inf) is unbounded as x-> 0 since it looks like a verticle line.

 

[Dick]

Sorry, can't use the def. of derivative yet at this point. Also even if I can, and the derivative is bounded on (1,inf) is unbounded as x-> 0 since it looks like a verticle line.

It's hardly 'verticle'. What the minimum value of (x^(2/3)+(xy)^(1/3)+y^(2/3)) for x and y in (1,inf)?

 

[billy2908]

minimum value should be x=1, y=1 in (1,inf)

so (x^(2/3)+(xy)^(1/3)+y^(2/3))= 3.

But I don't know what you are trying to lead me to...

 

[Dick]

minimum value should be x=1, y=1 in (1,inf)

so (x^(2/3)+(xy)^(1/3)+y^(2/3))= 3.

But I don't know what you are trying to lead me to...

I want you to use that to show me that |x^(1/3)-y^(1/3)|<|x-y|/3 for x and y in (1,inf).

 

[billy2908]

I think I got the proof, but want to make sure it's correct.

let z=(x^(2/3)+(xy)^(1/3)+y^(2/3)) then z>3 since x and y are in (1,inf)
=> 1/z <1/3
It actually works on (-inf, -1) as wellso again |x^1/3 - y^1/3| = |x-y|/z < d/3

So we can pick d=3*e.

Also since we know x^1/3 is cont. on [-1,1] which is compact. Then for all e>0 exist a d'>0 that works.

So then I just pick d=min{3e, d'}. Since I already showed it is uniform cont. on |x|>1.

would that suffice as a proof?

 

[Dick]

It looks ok to me.