# Proving or disproving uniform continuity of various real functions

Okay before anyone does anything, I don't want any of you guys to just write out proofs for all of these, that's asking a bit much :P

Maybe just do one, and make it detailed because I really need to see how a problem like this is done. I need a lot of exposure and practice to how to write a proof like this..

also, if anyone can clarify this for me, i'd really appreciate it: I fail to see how $\epsilon$ and $\delta$ relate in the definition of uniform continuity. How does the $\delta$ (unique if the function actually is uniform continuous) actually impose anything on $\epsilon$? this is actually even less clear in "regular" continuity.

• In this kind of excercise, you should use Lagrange's mean value theorem. – blues66 Mar 13 '14 at 2:42
• never heard Langrange's before it -- is it any different from the tried and true MVT? en.wikipedia.org/wiki/Mean_value_theorem – terrible at math Mar 13 '14 at 2:44
• It's that one, cultural differences. – blues66 Mar 13 '14 at 2:45
• either way, I can't use it, my professor told us we can't since we haven't established existence of derivatives yet (differentiation chapter is next) :( – terrible at math Mar 13 '14 at 2:49
• To answer your question on the differences: Uniform continuity is a stronger statement than continuity in the sense that $\delta \equiv \delta(\epsilon)$ for uniform continuity, whereas $\delta \equiv \delta(\epsilon,x)$ for continuity. That is there exists a $\delta>0$ such that the statement holds $\forall x$, rather than for a fixed $x$. That is we find $\delta$ first, then pick $x$; whereas the opposite, fix $x$ and then find $\delta$ is true for continuity. – Chris K Mar 13 '14 at 3:18

We say that $a_n$ and $b_n$ are equivalent if $a_n-b_n \to 0$. If $f$ is uniformly continuous then maps equivalent sequences to equivalent sequences.

1) Consider the sequences $a_n =(1/n)$ and $b_n=(1/2n)$ both are equivalent in $(0,1]$. But $f(a_n)$, $f(b_n)$ are not equivalent. So $f$ is not uniformly continuous in $(0,1]$.

[Alternatively you can show that if $f$ is uniformly continuous then maps Cauchy sequences to Cauchy sequences, and in particular for this case, $(1/n)$ is a Cauchy sequence in $(0,1]$, but $f(1/n)$ is not a Cauchy sequence].

2) You can show that uniformly continuous using $\varepsilon$-$\delta$ definition is logically equivalent to say that $f$ maps equivalent sequences to equivalent sequences, i.e., $f(a_n)-f(b_n)\to 0$ whenever $a_n -b_n \to 0$. Using this second definition:

Given $(a_n), (b_n) \subset [1, \infty)$ and $a_n-b_n \to 0$. We shall show that $f(a_n)-f(b_n) \to 0$. Let $\varepsilon>0$ be arbitrary and choose $n_0>0$ such that $|a_n-b_n|<\varepsilon$ for all $n\ge n_0$. Thus

$$|f(a_n)-f(b_n)|= \bigg|\frac{1}{a_n}-\frac{1}{b_n} \bigg|=\frac{|a_n-b_n|}{a_n b_n}\le |a_n-b_n|< \varepsilon$$

This is possible since $1\le a_n, b_n$. Hence $f(a_n)-f(b_n) \to 0$.

3) Consider the sequences $a_n = n$ and $b_n = n+1/n$ both are equivalent sequences in $[0,\infty)$. But $f(b_n)=n^2+2+1/n^2=f(a_n)+2+1/n^2$, so $f(b_n)-f(a_n)\ge2$. Thus we can conclude that is not uniformly continuous.

4) Since $f$ is continuous on $[0,1]$ then must be uniformly continuous (why?). Let $(0,1) \hookrightarrow [0,1]$. We claim that is uniformly continuous, let $x_n-y_n\to 0$ and $(x_n),(y_n) \subset (0,1)$, so $i(x_n)-i(y_n) = x_n -y_n \to 0$, since the sequences are arbitrary it holds for all equivalent sequences and so $i$ is uniformly continuous.

We claim that $f\circ i$ is uniformly continuous. We shall show that maps equivalent sequences to equivalent sequences. Since $i(x_n)-i(y_n)\to 0$ and $f$ is uniformly continuous then $f(i(x_n))-f(i(y_n))=(f\circ i)(x_n)-(f \circ i)(y_n) \to 0$.

Hence $f\circ i$ is uniformly continuous. But $f\circ i=f \restriction _{(0,1)}$ which is just the square root function define on $(0,1)$.

5) Suppose that $(a_n),(b_n) \subset [1,\infty)$ and $a_n-b_n \to 0$. We shall show that $a_n^{1/2}-b_n^{1/2} \to 0$. Given $\varepsilon>0$, choose $n_0$ such that $|a_n-b_n|< \varepsilon$ for all $n\ge n_0$. Thus

$$|f(a_n)-f(b_n)|=|\sqrt{a_n}-\sqrt{b_n}|= \bigg|\frac{a_n-b_n}{\sqrt{a_n}+\sqrt{b_n}} \bigg|=\frac{|a_n-b_n|}{\sqrt{a_n}+\sqrt{b_n}}\le \frac{|a_n-b_n|}{2}< \varepsilon$$

Since $1\le a_n, b_n$.

• for number 1, that's actually a really interesting argument, thank you. Seems like thinking about what it does to cauchy sequences makes reasoning about it a lot easier. – terrible at math Mar 13 '14 at 3:22
• just for the record, is it not sufficient to consider just 1 cauchy sequence and check that it maps to a non-cauchy sequence? Like you said, $(1/n)$ is cauchy on the domain but $f(1/n)$ = n is not. – terrible at math Mar 13 '14 at 3:23
• @terribleatmath: Uniformly continuous using $\varepsilon$-$\delta$ definition is logically equivalent to say that $f$ maps equivalent sequences to equivalent sequences, i.e., $f(a_n)-f(b_n)\to 0$ whenever $a_n -b_n \to 0$. – Jose Antonio Mar 13 '14 at 3:55
• @terribleatmath: If f is uniformly continuous then maps Cauchy sequences to Cauchy sequences, is a necessary condition. So, if were the case in which does not map Cauchy seq. to Cauchy then is not uniformly continuous (by contraposition) – Jose Antonio Mar 13 '14 at 4:18
• @terribleatmath: I've already finished all the exercises. Hopefully this would help you. To twist a little bit the usual argument I've used the "sequences" definition of uniformly continuous function which I've always found more "natural". – Jose Antonio Mar 13 '14 at 6:11

As you're lacking a positive answer, I'll do the second. We have $$|f(x)-f(y)|=|\frac{1}{x}-\frac{1}{y}|=\frac{|x-y|}{xy}\le|x-y|$$ as you have $x\ge1$ and $y\ge1$.

• interesting, so zero is the only thing that screws up the uniform continuity of this function. – terrible at math Mar 13 '14 at 3:17
• can you explain how you went from $\mid \frac{1}{x} - \frac{1}{y}\mid$ to $\frac{\mid x-y \mid}{xy}$ ? – terrible at math Mar 13 '14 at 3:39
• $\mid \frac{1}{x} - \frac{1}{y}\mid\Rightarrow\mid \frac{y-x}{xy}\mid\Rightarrow\frac{\mid x-y \mid}{|x||y|}$ – blues66 Mar 13 '14 at 4:20
• The main thing that screws up the uniform continuity, is the not bounded derivate. – blues66 Mar 13 '14 at 4:21

• are you sure iv) is a yes? I'm getting this: since the domain is $(0,1)$, for $0<x<1$ note that $\sqrt(x) > x$ so $|f(x) - f(y)| = |\sqrt(x) - \sqrt(y)| \geq |x-y|$ thus it can't be uniform cont ? – terrible at math Mar 13 '14 at 3:48