Uniform Continuity of $f(x)=x^3$

Question

1.)Determine whether $f(x)=x^3$ is uniformly continuous on [0,2)

So far, I have $\delta$ = 2 and $\epsilon$ = 8, and plan on using the sandwich theorem with $x^2$ and eventually equating $\delta = \epsilon$.

Certainly, when $\epsilon=8$ you can do it, but you need to find a $\delta>0$ given any $\epsilon>0$, not for one $\epsilon>0$ — Thomas Andrews, Mar 06 '13 at 19:52

score 7 · Accepted Answer · answered Mar 06 '13 at 19:55

7

For given $\epsilon >0$, choose $\delta=\epsilon/12$, then for $x,y\in [0,2),$ such that

$|x-y|<\delta\implies |f(x)-f(y)|=|x^3-y^3|$

$=|x-y||x^2+xy+y^2|<|x-y|(|x|^2+|x||y|+|y|^2)<12|x-y|=12\delta=\epsilon$

Thus, for $\epsilon>0$, $\exists \delta >0$(independent of point where continuity is to be checked), such that $|x-y|<\delta \implies |f(x)-f(y)|<\epsilon$.

Hence, Proved.

answered Mar 06 '13 at 19:55

Aang

14,672

Maybe I'm thinking wrong ,but shouldn't it be $\delta=\epsilon/12$? – Axiom Mar 06 '13 at 19:58
Thanks for pointing out. I edited it. – Aang Mar 06 '13 at 20:00
Why epsilon/12 though? – user65384 Mar 06 '13 at 20:02
Because $x,y<{2}$, so $|x^2+xy+y^2|<{4+4+4}=12$ – Axiom Mar 06 '13 at 20:03
because $(|x|^2+|x||y|+|y|^2)<12$ for $x,y<2$. so to make $|f(x)-f(y)|<\epsilon$, you need to have $|x-y|<\epsilon/12$ – Aang Mar 06 '13 at 20:04

score 4 · Answer 2 · answered Mar 06 '13 at 20:33

4

For a quick way, recall that any continuous function is uniformly continuous on a closed interval $[a,b]$. If $f(x)$ is uniformly continuous on $[a,b]$ then it must be uniformly continuous on $[a,b)$ (with the same $\delta$ for every $\epsilon$).

answered Mar 06 '13 at 20:33

Tom Oldfield

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score 1 · Answer 3 · edited Apr 13 '17 at 12:21

1

A related problem.

Hint: Another approach is the mean value theorem. Here is a related problem.

Added:

Mean Value Theorem: If a function $f$ is continuous on the closed interval [a, b], where a < b, and differentiable on the open interval (a, b), then there exists a point c in (a, b) such that $$ f'(c) = \frac{f(b) - f(a)}{b-a}. $$

In your case, we start as

$$ \frac{f(x)-f(y)}{x-y}=f'(\eta),\,\,\eta\in(x,y) \implies \frac{x^3-y^3}{x-y}=3\eta^2 $$

$$ \implies |{x^3-y^3}|=|3\eta^2||x-y|\leq 12|x-y| < \epsilon $$

$$\implies |x-y| < \frac{\epsilon}{12}=\delta. $$

Now, choosing $ \delta=\frac{\epsilon}{12} $, we have

$$ |x-y|< \delta \implies |x^3-y^3|<\epsilon. $$

edited Apr 13 '17 at 12:21

Community

1

answered Mar 06 '13 at 20:02

Mhenni Benghorbal

47,431

would you care to elaborate some more? – Mar 06 '13 at 20:31
@user1709828: See the added. – Mhenni Benghorbal Mar 06 '13 at 22:38
The implications at the end are a mishmash. To reach a valid proof, at least the last one should be reversed. – Did Mar 09 '13 at 10:52
@Did: Thanks for the comment. It is corrected. – Mhenni Benghorbal Sep 20 '13 at 08:06
@Downvoter: There was a typo and it was corrected. – Mhenni Benghorbal Sep 20 '13 at 08:07
You are welcome. The implications at the end are still a mishmash. – Did Sep 20 '13 at 08:09
For more detailes about $\delta-\epsilon$ approach see my answer here. – Mhenni Benghorbal Sep 20 '13 at 08:23
@Downvoter: There was no a real mistake. In fact it was left it to the OP to work things out. – Mhenni Benghorbal Sep 20 '13 at 08:27
The last part you just added is correct. The middle part is still wrong. – Did Sep 20 '13 at 08:43
@Did: What middle part are you talking about? – Mhenni Benghorbal Sep 20 '13 at 08:47
Always the same for 3 comments now: the part which starts with "In your case" and ends just before "Now". – Did Sep 20 '13 at 08:50
@Did: It is correct. – Mhenni Benghorbal Sep 20 '13 at 09:08
2

Mhenni, Mhenni, Mhenni... Not again? – Did Sep 20 '13 at 09:11

Uniform Continuity of $f(x)=x^3$

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