About the continuity of $x^3$.

Question

This is a question for stefen abott's Understanding Analysis. A 3 part question

1. Show that $f(x)=x^3$ is continuous on $\mathbb R$.(Done)
2. Find sequences $x_n, y_n$ such that $|x_n-y_n|\rightarrow0$ but there exists an $\epsilon_0>0$ such that $|f(x_n)-f(y_n)|>\epsilon_0$.
3. Show f is uniformly continuous on a bounded interval in $\mathbb R$.

I need help with 2nd and 3rd part. Second part is basically proving that $f(x)=x^3$ isn't uniformly continuous. But I can't think of the sequences required to prove the result.

Answer 1

You want $\lvert x_n^3-y_n^3\rvert > \epsilon_0$. Maybe it can be done with $x_n>y_n$, in which case this is $x^3_n>y_n^3+\epsilon_0$. Let's see if $x_n^3=y_n^3+2\epsilon_0$ might somehow work... With $y_n=n$, then $x_n=(n^3-2\epsilon_0)^{1/3}$. Does this satisfy the hypothesis? $x_n-y_n=(n^3-2\epsilon_0)^{1/3}-n$, and one can show that this approaches $0$ as $n\to\infty$.

For the third part, do you know any connections between bounded intervals, compact subsets, and uniform continuity?

Answer 2

Well, let's think this out.

We want $|x_n^3 - y_n^3| > \epsilon$ despite the fact that $|x_n - y_n| < \epsilon$.

Okay $|x_n^3 - y_n^3| = |x_n - y_n||x_n^2 + x_ny_n + y_n^2|$

So we want $|x_n - y_n||x_n^2 + x_ny_n + y_n^2| > \epsilon$ so

$|x_n^2 + x_ny_n + y_n^2|> \frac {\epsilon}{|x_n - y_n|} > 1$ while $|x_n - y_n|< \epsilon$.

Can we do that?

Say, $x_n = n$ and $y_n = n + \frac 1n$. So for any $0 < \epsilon < 1$, if we let $N > \frac 1 \epsilon > 1$ and $n > N$ then $|x_n - y_n| = \frac 1n <\frac 1N < \epsilon$ .

But then $|x_n^3 - y_n^3| = |n^3 - (n^3+3\frac {n^2}n + 3\frac n{n^2} + \frac 1{n^2})| = |3n + \frac 3n + \frac 1{n^2}| > 3N$. .... which is sooo not less than $\epsilon$ it isn't even funny