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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 $\delta = \epsilon$.

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  • $\begingroup$ Certainly, when $\epsilon=8$ you can do it, but you need to find a $\delta>0$ given any $\epsilon>0$, not for one $\epsilon>0$ $\endgroup$ – Thomas Andrews Mar 6 '13 at 19:52
  • $\begingroup$ 1. do you now a theorem which gives a connection between the limits of the function at the boundary of the interval and uniform continuity? 2. you shouldn't choose $\epsilon=8$. this should be arbitrary. 3. Do you know Heine's theorem? $\endgroup$ – Quickbeam2k1 Mar 6 '13 at 19:52
  • $\begingroup$ 1. If $f$ is uniformly continuous on a set $A$ and $B\subseteq A$, then $f$ is uniformly continuous on $B$. 2. If $f$ is continuous on a bounded close interval $A$ (e.g. $A=[0,2]$...) then it is uniformly continuous on $I$. 3. Can you find a suitable $B\subseteq A$...? $\endgroup$ – AndreasT Mar 6 '13 at 19:57
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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.

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  • $\begingroup$ Maybe I'm thinking wrong ,but shouldn't it be $\delta=\epsilon/12$? $\endgroup$ – Axiom Mar 6 '13 at 19:58
  • $\begingroup$ Thanks for pointing out. I edited it. $\endgroup$ – Aang Mar 6 '13 at 20:00
  • $\begingroup$ Why epsilon/12 though? $\endgroup$ – user65384 Mar 6 '13 at 20:02
  • $\begingroup$ Because $x,y<{2}$, so $|x^2+xy+y^2|<{4+4+4}=12$ $\endgroup$ – Axiom Mar 6 '13 at 20:03
  • $\begingroup$ because $(|x|^2+|x||y|+|y|^2)<12$ for $x,y<2$. so to make $|f(x)-f(y)|<\epsilon$, you need to have $|x-y|<\epsilon/12$ $\endgroup$ – Aang Mar 6 '13 at 20:04
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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 $\epsilon$).

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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

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

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  • $\begingroup$ would you care to elaborate some more? $\endgroup$ – user45099 Mar 6 '13 at 20:31
  • $\begingroup$ @user1709828: See the added. $\endgroup$ – Mhenni Benghorbal Mar 6 '13 at 22:38
  • $\begingroup$ The implications at the end are a mishmash. To reach a valid proof, at least the last one should be reversed. $\endgroup$ – Did Mar 9 '13 at 10:52
  • $\begingroup$ @Did: Thanks for the comment. It is corrected. $\endgroup$ – Mhenni Benghorbal Sep 20 '13 at 8:06
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    $\begingroup$ Mhenni, Mhenni, Mhenni... Not again? $\endgroup$ – Did Sep 20 '13 at 9:11

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