Showing $x^3$ is not uniformly continuous on $\mathbb{R}$

Referring to the original source http://math.stanford.edu/~ksound/Math171S10/Hw8Sol_171.pdf

Prove that $f(x) = x^3$ is not uniformly continuous on $\Bbb R$.

I want to use a more elementary method, no mention of metric spaces. Their selection of $\delta$ confuses me, so I am want to present a more "elementary" method. I am unsure if it is correct.

Here I picked $|a| > \dfrac{\epsilon}{3\delta}$ and $x > a - \delta$

Thus $$\begin{align} |x^3 - a^3| &= |x - a||x^2 + ax + a^2| \\&>|x^2+ax +a^2| \\&> |(a-\delta)^2+a(a - \delta) + a^2|\\&=|3a^2-3a\delta+\delta^2| \\&> 3\delta|a| \end{align}$$

Solution 1:

Their solution doesn't really use any ideas from metric spaces. Here is an elaboration.

Suppose $f(x)=x^3$ were uniformly continuous. Then for any $\epsilon$, there is a $\delta$ that works for all $a$. Let's take $\epsilon=1$; we are now given $\delta$ such that for all $a$ and for all $x$ with $|x-a|<\delta$, we must have $|f(x)-f(a)|<\epsilon$.

Choose $a$ large enough so that $\frac{3\delta a^2}{2}>1$; for example $a=\sqrt{\frac{2}{3\delta}}+1$ works. Now take $x=a+\frac{\delta}{2}$; this satisfies $|x-a|=\frac{\delta}{2}<\delta$. Hence we SHOULD have $|f(x)-f(a)|<\epsilon=1$. Instead we have $|f(x)-f(a)|=|f(a+\frac{\delta}{2})-f(a)|=|(a+\frac{\delta}{2})^3-a^3|=\left|3a^2\frac{\delta}{2}+3a\frac{\delta^2}{4}+\frac{\delta^3}{8}\right|\ge \left|3a^2\frac{\delta}{2}\right|>1$.

This is a contradiction, and hence $f(x)$ is not uniformly continuous.

Solution 2:

Recall that the definition of uniform continuity is

$f:A\to\Bbb R$ is uniformly continuous iff for each $\epsilon>0$ there exists a $\delta >0$ (depending solely on $\epsilon$) such that whenever $x,y\in A$ and $|x-y|<\delta$; it follows $$|f(x)-f(y)|<\epsilon$$

Now, in order to prove a function is not uniformly continuous, we ought to show that

There exists $\epsilon >0$ such that for any $\delta >0$ we can find $x,y\in A$ for which $|x-y|<\delta$ yet $$|f(x)-f(y)|\geq \epsilon$$

This translates almost immediately to

There exist a pair of sequences $\langle x_n\rangle ,\langle y_n\rangle$ in $A$ such that $$\lim\limits_{n\to\infty}(x_n-y_n)=0$$ yet $$\lim\limits_{n\to\infty}(f(x_n)-f(y_n))\neq 0$$

Now, can you find such sequences for $f(x)=x^3$?

HINT Think about $x_n=(n+1)^{1/3}$, $y_n=n^{1/3}$. It shouldn't be surprising the sequences are unbounded, and yet become arbitrarily close. In fact, $x^3$ is uniformly continuous on any bounded interval, so it is necessary the sequences are unbounded. This is in essence why uniform continuity fails here. Note the functional inverse $x^{1/3}$ is uniformly continuous over $\Bbb R$.

OBS The above can be used to show $x^n$ is not uniformly continuous whenever $n> 1$.