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


*

*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).

*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.
A: 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.
A: (1). If $f:(0,\infty)\to \Bbb R$ is differentiable and $\lim_{x\to \infty}f'(x)=\infty$ then $f$ is not uniformly continuous:  For any $\delta >0$ and any $x>0$,  the MVT implies there exists $y\in (x,x +\delta)$ such that $\frac {f(x+\delta)-f(x)}{\delta}= f'(y).$ Now if $x$ is large enough that $\forall y>x\;(f'(y)>1/\delta)$ then $f(x+\delta)-f(x)=\delta f'(y)>1.$
(2). Given $\delta >0$ take $x>\max (1,1/\delta).$ Then $$(x+\delta)^4-x^4= 4x^3\delta+6x^2\delta^2+4x\delta^3+\delta^4>$$  $$>4x^3\delta=4(x^2)x\delta>4x\delta>4.$$
By (1) or by (2) we have $\sup_{x\in \Bbb R} \{|(x+x')^4-x^4|: |x'-x|<\delta\}> 1$ for every $\delta>0.$ (In fact the $\sup$ is $\infty $.)
Uniform continuity of $f:D\to \Bbb R$ for some (any) domain $D\subset  \Bbb R$ means $$0=\lim_{\delta \to 0^+}\sup \{|f(x')-f(x)|:x,x'\in D\land |x'-x|\leq\delta\}.$$ 
A: I think you should make this a little simpler (and for uniform continuity in general) All you need to do to show $f:X \to Y$ is not uniformly continuous on $X$
(let's suppose there both subsets of $\Bbb R$), then just give me a SINGLE epsilon such that, NO MATTER HOW SMALL delta is chosen, there will be x and y closer than delta for which the difference in function values exceed epsilon. Thus for instance $|(N+\theta)^4- N^4| \ge 4\theta N^3$ so if you choose $x$ really big (with respect to $\delta, x=N+\theta\,\,\,\ \text{and}\,\,\, y = N,$ then if $0 < \delta/2 < \theta < \delta$ you have $|x-y| < \delta$ yet you still have the variable N to play with to make the difference in function values as large as you like (in particular the difference in function values can always be made bigger than 3 regardless of how small $\delta$ is). Nevertheless, I think your proof is an accurate job! 
A: ** Simple Trick:**
Given $a>0$ if we take 
$$x = \frac{1}{a}+2 ~~~and  ~~~y = \frac{1}{a}+2+\frac{a}{2} \color{red}{= x +\frac{a}{2}}
 $$ we have $x,y\in [2,\infty)~~~ and~~~|x-y| =\frac{a}{2}<a $
But since $a\cdot x = 1+ 2a~~~$  and $~~x,y>2\implies x^2+y^2 >8$  we get,
$$|f(x)-f(y)| =(x^2+y^2) |y^2-x^2|\ge  8|y^2-x^2|\\~~~~~~~~~~~~~~~=8| x^2 + 2\frac{a}{2}x +\frac{a^2}{4} - x^2 | \\= 8(ax +\frac{a^2}{4})= 8(1+2a+\frac{a^2}{4} )>8$$
That is $$|f(x)-f(y)| >8$$
Thus $$\color{blue}{\exists \varepsilon_0 =8,\forall~a>0, \exists ~x,y\in \mathbb R: |x-y|< a~~and ~~|f(x)-f(y)|>8}$$
just take $$x = \frac{1}{a}+2 ~~~and  ~~~y = \frac{1}{a}+2+\frac{a}{2} \color{red}{= x +\frac{a}{2}}
 $$
