$f$ maps $z$ in $\mathbb C$ to $z^4$ in $\mathbb C$. How do I show that this function is not uniformly continuous?

  • $\begingroup$ @Potato I know that bounded derivative implies UC but since I need to show non-UC, I'm not sure how the derivative helps...? $\endgroup$ – Ryan Mar 28 '13 at 1:54
  • $\begingroup$ Consider claim (b) here: math.stackexchange.com/questions/118665/… $\endgroup$ – Potato Mar 28 '13 at 1:57
  • $\begingroup$ No? Is that what you are really trying to ask? I think you made a typo. $\endgroup$ – Potato Mar 28 '13 at 2:22
  • $\begingroup$ @Potato Hold on! The converse is true also?! I.e. UC implies bounded derivative?! (typo corrected and original comment deleted) $\endgroup$ – Ryan Mar 28 '13 at 5:07
  • $\begingroup$ No. Take $\sqrt x$ on $(0,\infty)$, for example. $\endgroup$ – Potato Mar 28 '13 at 5:17

Here is one way $x^4-y^4 = (y+x)(x^2+y^2)(x-y)$ Hence $|x^4-y^4| = |(y+x)(x^2+y^2)||(x-y)|$.

So, regardless of how small $|x-y|$ is (as long as it is not zero, of course), you can choose $x$ so that $|(y+x)(x^2+y^2)|$ is as large as you want. Hence it cannot be uniformly continuous.

Here is another way. The function is smooth, hence differentiable. We have $\lim_{h \to 0} \frac{(z+h)^4-z^4}{h} = 4z^3$, or more relevant to this discussion, $\lim_{h \to 0} \frac{|(z+h)^4-z^4|}{|h|} = 4|z|^3$. Hence at any $z\neq 0$, there exists some $\delta>0$ such that $\frac{|(z+h)^4-z^4|}{|h|} \ge 2|z|^3$ as long as $|h|< \delta$. Since we can choose $|z|$ as large as we want, the function cannot be uniformly continuous.

  • $\begingroup$ I love your first answer; it has boosted my confidence to use algebraic manipulations to prove non-UC. The example to show non-UC given in my textbook is so complicated and specific that it was futile trying to riff off it. $\endgroup$ – Ryan Mar 28 '13 at 1:50

This function is not uniformly continuous on $\mathbb{R}$. Can it be uniformly continuous on $\mathbb{C}$, which contains $\mathbb{R}$?

  • $\begingroup$ No. But a function can be UC on $\mathbb C$ but not UC on $\mathbb R$, right? (just checking) $\endgroup$ – Ryan Mar 28 '13 at 1:48
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    $\begingroup$ Uniform Continuity is a strong condition about nearby points being nearby in the image. When you add more nearby points, it just makes it more difficult to fulfill that condition. So if you fail that condition on $\mathbb{R}$, you must fail on $\mathbb{C}$. However, if you fail on $\mathbb{C}$, then maybe, when you remove points and restrict to $\mathbb{R}$, you do have uniform continuity. $\endgroup$ – Elchanan Solomon Mar 28 '13 at 1:50

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