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:

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

Take $x=\frac{\delta}{2}+\frac{1}{\delta}$ and $y=\frac{1}{\delta}$ then we have that $|x-y|=|\frac{\delta}{2}+\frac{1}{\delta}-\frac{1}{\delta}|=|\frac{\delta}{2}|<\delta$ 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 $\epsilon < \frac{3}{4}$ 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

• So, is this an exam question? – user21436 Apr 22 '12 at 12:17
• No, I'm just practicing for my exam where questions like this (not this one though) will come it. This is from one of the past papers that are for revision – hmmmm Apr 22 '12 at 12:29
• Just trying to ensure we are not taken for a ride. Hope you don't mind. :) – user21436 Apr 22 '12 at 12:33
• @KannappanSampath no its fine- it annoys me when I see people posting assessment questions on forums :) – hmmmm Apr 22 '12 at 12:36

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 $|x-y|<\delta$ 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 $|f(x)-f(y)|<\varepsilon$ 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 $$|x^4-y^4| < \epsilon$$ $$|(x-y)(x+y)(x^2+y^2)| < \epsilon$$ $$|x-y|\cdot|x+y|\cdot|x^2+y^2| < \epsilon$$ $$|x-y| < \frac{\epsilon}{|x+y|\cdot|x^2+y^2|}$$

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

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

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.

• Thanks for the reply, I think that I use that we are considering all of $\mathbb{R}$ when i choose $x=\delta+\frac{1}{\delta}$ and $y=\frac{1}{\delta}$ as these would not be valid for small $\delta$ in bounded interval? – hmmmm Apr 22 '12 at 13:38
• @hmmmm Ok, I misunderstood what you were saying there. If the calculations are alright, then your proof is fine. – dtldarek Apr 22 '12 at 13:44
• Is it correct to fix $\delta>0$ and choosing $y = x + \delta$ specifically? Since the distance between x and y will be exactly $\delta$, and we want $|x-y|<\delta$. – DrHAL May 2 '16 at 8:08
• @dtldarek Is it possible that one method will fail to show and other will succeed. For example, for $f(x)=1/x$ on $(0,\infty)$, first method leads the limit to 0? – chandresh Feb 3 at 10:18
• @chandresh Function $f(x)=1/x$ is uniformly continuous on $(1, \infty)$, or even on $(a, \infty)$ for any fixed $a > 0$. On the other hand, $f$ is very steep near zero, and so the limit you want to calculate is $$\lim_{x \to 0^+}\left|\frac{1}{x}-\frac{1}{x+\delta}\right|.$$ – dtldarek Feb 5 at 8:04

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 $0<\varepsilon<3$ 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.

• Hey sorry about that, I edited it-hopefully it's right now? – hmmmm Apr 22 '12 at 13:39
• I also edited now my calculation which still differs abit from your new one. Could you show the steps of how you got this answer for $|f(x)-f(y)|$? – T. Eskin Apr 22 '12 at 13:47
• yeah I messed that up quite a bit sorry (I had the wrong power and the wrong delta's) – hmmmm Apr 22 '12 at 13:50
• It should be $|(\frac{\delta}{2}+\frac{1}{\delta})^4-\frac{1}{\delta^4}|$ which would give $|\frac{\delta^4}{16}+\frac{\delta^2}{2}+\frac{2}{\delta}+\frac{3}{2}|$ I think I could conclude a similar thing from here? – hmmmm Apr 22 '12 at 13:52
• Except that is the last $-\frac{1}{\delta^{4}}$ missing from there? Otherwise it looks close to mine. – T. Eskin Apr 22 '12 at 13:57

(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 $$0=\lim_{\delta \to 0^+}\sup \{|f(x')-f(x)|:x,x'\in D\land |x'-x|\leq\delta\}.$$

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!

** 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 $$x = \frac{1}{a}+2 ~~~and ~~~y = \frac{1}{a}+2+\frac{a}{2} \color{red}{= x +\frac{a}{2}}$$