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I'm trying to use continuity to show that $f(x)=x^3$ is uniformly continuous on $[0,1]$ but not $[0,\infty)$.

I've tried setting up an epsilon-delta proof, but I'm struggling a little:

By definition of uniform continuity, we know that $\forall \epsilon >0, \exists \delta >0$ such that

$|x-y|<\delta \Rightarrow |f(x) - f(y)| < \epsilon$.

And so, forcing

$\delta = \min\{1, \frac{\epsilon}{p_x}\}$ where $p_x = (x^2+xy+y^2)$

And so, we havve that

$|(x)^3 - (y)^3|=|(x-y)(x^2+xy+y^2)| < |\delta (x^2 + xy+y^2)| < \epsilon$

I'm not sure if this is the correct way to go about proving it, or if I landed myself into a circular argument. Furthermore, intuitively I'm guessing we only have uniform continuity on $[0,1]$ but not [0,$\infty)$ because our $p_x$ would get too large?

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  • $\begingroup$ This might help you : math.stackexchange.com/questions/428528/… $\endgroup$ – user399481 Dec 19 '16 at 1:36
  • $\begingroup$ Note that in general, ANY continuous function on a compact set is uniformly continuous. $\endgroup$ – Alex R. Dec 19 '16 at 1:36
  • $\begingroup$ @AlexR. Ah yes. And we have compactness because we have a closed set (by the Heine Borel theorem)? $\endgroup$ – Nikitau Dec 19 '16 at 1:45
  • $\begingroup$ @Nikitau $[0,1]$ is closed and bounded and hence compact by Heine Borel. $\endgroup$ – Henricus V. Dec 19 '16 at 2:42
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Note that in the definition of uniform continuity, given $\varepsilon > 0$, you need to provide a $\delta > 0$ that works for all $x,y \in [0,1]$. In particular, you cannot have a $\delta$ that depends on $x$. This is in contrast to proving that $f$ is continuous at a specific $x$ where the $\delta$ you provide can depend on $x$.

In your case, we have

$$ |x^3 - y^3| = |(x - y)(x^2 + xy + y^2)| \leq |x - y||x^2 + xy + y^2|. $$

Now, if $x,y \in [0,1]$, we have $|x^2 + xy + y^2| \leq 3$ so we can deduce that

$$ |x^3 - y^3| \leq 3|x - y|. $$

Hence, given $\varepsilon > 0$, we can take $\delta = \frac{\varepsilon}{3}$ and then if $|x - y| < \delta$ then $|x^3 - y^3| \leq 3|x - y| < 3\delta = \varepsilon$.

Note that the same argument would work if you wanted to prove uniform continuity on $[0,L]$ with the constant $3$ replaced by a different constant $C_L$ which bounds $|x^2 + xy + y^2|$ on $[0,L]$ (for example, $C_L$ can be $3L^2$).

Expressed in this way, we see that your basic intuition is correct. As $L$ gets larger, our constant $C_L$ also gets larger and hence our $\delta = \frac{\varepsilon}{C_L}$ gets smaller and smaller. In the limit, this should lead us to suspect (but this is not a formal proof!) that uniform continuity will fail.

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  • $\begingroup$ I'm a little confused on the part where you say that "this is in contrast to proving that f is continuous at a specific x". In an example provided in class, the professor wanted to show $f(x) = \frac{x^3}{1+x^2}$ was continuous on $\mathbb{R}$. She then fixed $x_0 \in \mathbb{R}$, and the delta included a function $P(x_0)$. This is might be a fairly silly question, but I'm confused why we were allowed do to that for the whole of $\mathbb{R}$ but not an interval [0,1]? $\endgroup$ – Nikitau Dec 19 '16 at 1:49
  • $\begingroup$ @Nikitau: You are allowed to do it on an interval or on the whole of $\mathbb{R}$, but this only proves that your function is continuous, not uniformly continuous. Being uniformly continuous is a stronger property. If you look at the definition, this is expressed precisely by the fact that to show uniform continuity over some domain $D$, you need to provide a $\delta$ that works for all points in $D$ simultaneously. This is in contrast to proving that $f$ is continuous on $D$ where you can fix a point $x_0 \in D$ and then provide a $\delta = \delta(x_0)$ that depends on $x_0$. $\endgroup$ – levap Dec 19 '16 at 1:52
  • $\begingroup$ Oh thanks for the clarification! So if I was just proving continuity, the original post would have been OK? Or would having 'y' in my delta pose a problem? (Sorry for all the questions! I have an exam coming up and I just want to make sure I'm understanding all the nuances). $\endgroup$ – Nikitau Dec 19 '16 at 1:56
  • $\begingroup$ @Nikitau: That is a problem. Even if you are only interested in regular continuity at a point $x_0$, you cannot have your $\delta$ depend on the argument of the function, only on $x_0$ (and $\varepsilon$). $\endgroup$ – levap Dec 19 '16 at 1:58
  • $\begingroup$ Ah, that makes sense. Thank you for pointing that out :) $\endgroup$ – Nikitau Dec 19 '16 at 2:04
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To show that $x^3$ fails to be uniformly continuous on $[0,\infty)$, we take $\epsilon=\frac{3}{2}$. Then, for all $\delta>0$, and for $x=\frac{1}{\sqrt\delta}$ and $y=\frac{1}{\sqrt \delta}+\frac{\delta}{2}$ we have $|x-y|<\delta$ and

$$\begin{align}|x^3-y^3|&=\left|\left(\frac{1}{\sqrt\delta}+\frac{\delta}{2}\right)^3-\left(\frac{1}{\sqrt\delta}\right)^3\right|\\\\ &\ge \frac32\\\\ &=\epsilon \end{align}$$

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  • $\begingroup$ I can see why this works, but how does one know what values to set for $\epsilon$ and x,y? $\endgroup$ – Nikitau Dec 19 '16 at 2:38
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    $\begingroup$ You need only find one such $\epsilon>0$ such that for any $\delta>0$ there are points $x$ and $y$ such that $|x-y|<\delta$ and $|x^3-y^3|\ge \epsilon$. Certainly, any $\epsilon$ smaller than $3/2$ would have worked also. $\endgroup$ – Mark Viola Dec 19 '16 at 2:40

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