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

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 6 '13 at 19:52
• 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? – Quickbeam2k1 Mar 6 '13 at 19:52
• 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 $B\subseteq A$...? – AndreasT Mar 6 '13 at 19:57

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.

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

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 $\epsilon$).

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

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

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

• would you care to elaborate some more? – user45099 Mar 6 '13 at 20:31
• @user1709828: See the added. – Mhenni Benghorbal Mar 6 '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 9 '13 at 10:52
• @Did: Thanks for the comment. It is corrected. – Mhenni Benghorbal Sep 20 '13 at 8:06
• Mhenni, Mhenni, Mhenni... Not again? – Did Sep 20 '13 at 9:11