Showing $f(x)=x^4$ is not uniformly continuous

I am looking at uniform continuity (for my exam) at the moment and I'm fine with showing that a function is uniformly continuous but I'm having a bit more trouble showing that it is not uniformly continuous, for example:

show that $x^4$ is not uniformly continuous on $\mathbb{R}$, so my solution would be something like:

Assume that it is uniformly continuous then:

$$\forall\epsilon\geq0\exists\delta>0:\forall{x,y}\in\mathbb{R}\ \mbox{if}\ |x-y|&lt\delta \mbox{then} |x^4-y^4|&lt\epsilon$$

Take $x=\frac{\delta}{2}+\frac{1}{\delta}$ and $y=\frac{1}{\delta}$ then we have that $|x-y|=|\frac{\delta}{2}+\frac{1}{\delta}-\frac{1}{\delta}|=|\frac{\delta}{2}|&lt\delta$ however $$|f(x)-f(y)|=|\frac{\delta^3}{8}+3\frac{\delta}{4}+\frac{3}{2\delta}|$$

Now if $\delta\leq 1$ then $|f(x)-f(y)|>\frac{3}{4}$ and if $\delta\geq 1$ then $|f(x)-f(y)|>\frac{3}{4}$ so there exists not $\delta$ for $\epsilon &lt \frac{3}{4}$ and we have a contradiction.

So I was wondering if this was ok (I think it's fine) but also if this was the general way to go about showing that some function is not uniformly continuous? Or if there was any other ways of doing this that are not from the definition?

Thanks very much for any help

Solution 1:

To show that it is not uniformly continuous on the whole line, there are two usual (and similar) ways to do it:

1. Show that for every $\delta > 0$ there exist $x$ and $y$ such that $|x-y|&lt\delta$ and $|f(x)-f(y)|$ is greater than some positive constant (usually this is even arbitrarily large).
2. Fix the $\varepsilon$ and show that for $|f(x)-f(y)|&lt\varepsilon$ we need $\delta = 0$.

First way:

Fix $\delta > 0$, set $y = x+\delta$ and check $$\lim_{x\to\infty}|x^4 - (x+\delta)^4| = \lim_{x\to\infty} 4x^3\delta + o(x^3) = +\infty.$$

Second way:

Fix $\epsilon > 0$, thus $$|x^4-y^4| &lt \epsilon $$ $$|(x-y)(x+y)(x^2+y^2)| &lt \epsilon $$ $$|x-y|\cdot|x+y|\cdot|x^2+y^2| &lt \epsilon $$ $$|x-y| &lt \frac{\epsilon}{|x+y|\cdot|x^2+y^2|} $$

but this describes a necessary condition, so $\delta$ has to be at least as small as the right side, i.e.

$$|x-y| &lt \delta \leq \frac{\epsilon}{|x+y|\cdot|x^2+y^2|} $$

so if either of $x$ or $y$ tends to infinity then $\delta$ tends to $0$.

Hope that helps ;-)

Edit: after explanation and calculation fixes, I don't disagree with your proof.

Solution 2:

I will comment on your solution after writing another approach. For any $x,y\in\mathbb{R}$ we have: \begin{align*} |x^{4}-y^{4}|=|(x^{2}-y^{2})(x^{2}+y^{2})|=|(x-y)(x+y)(x^{2}+y^{2})|=|x-y|\cdot |x+y|\cdot |x^{2}+y^{2}| \end{align*}

So what you can see is that even if you take arbitrarily close $x$ and $y$, you can grow the distance of $x^{4}$ and $y^{4}$ as much as you want by taking them far enough away from zero. You can easily conclude from here that the function is not uniformly continuous by a contraposition for example.

Alright, then to your solution. If the calculations would be correct, then it would be fine. You could assume at first that such $\delta>0$ exists for $0&lt\varepsilon&lt3$ and conclude with a contradiction. However, I got a bit different calculations than you. Using the above equation we see that: \begin{align*} |f(\frac{\delta}{2}+\frac{1}{\delta})-f(\frac{1}{\delta})|&=|(\frac{\delta}{2}+\frac{1}{\delta})^{4}-\frac{1}{\delta^{4}}|=|\frac{\delta}{2}(\frac{\delta}{2}+\frac{2}{\delta})((\frac{\delta}{2}+\frac{1}{\delta})^{2}+\frac{1}{\delta^{2}})| \\ &= |(\frac{\delta^{2}}{4}+1)(\frac{\delta^{2}}{4}+2\cdot \frac{\delta}{2}\cdot \frac{1}{\delta}+\frac{1}{\delta^{2}}+\frac{1}{\delta^{2}})| \\ &=|(\frac{\delta^{2}}{4}+1)(\frac{\delta^{2}}{4}+1+\frac{2}{\delta^{2}})| \\ &= |\frac{\delta^{4}}{16}+\frac{\delta^{2}}{4}+\frac{1}{2}+\frac{\delta^{2}}{4}+1+\frac{2}{\delta^{2}}| \\ &= |\frac{\delta^{4}}{16}+\frac{\delta^{2}}{2}+\frac{2}{\delta^{2}}+\frac{3}{2}|\\ &= \frac{\delta^{4}}{16}+\frac{\delta^{2}}{2}+\frac{2}{\delta^{2}}+\frac{3}{2} \end{align*} If you're able to find a lower bound for this (which is quite easy) as you did previously, then by choosing an epsilon smaller than that fixed number you may conclude as you did in your original post by contradiction.