1
$\begingroup$

enter image description here

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.

$\endgroup$
  • $\begingroup$ In this kind of excercise, you should use Lagrange's mean value theorem. $\endgroup$ – blues66 Mar 13 '14 at 2:42
  • 1
    $\begingroup$ never heard Langrange's before it -- is it any different from the tried and true MVT? en.wikipedia.org/wiki/Mean_value_theorem $\endgroup$ – terrible at math Mar 13 '14 at 2:44
  • $\begingroup$ It's that one, cultural differences. $\endgroup$ – blues66 Mar 13 '14 at 2:45
  • $\begingroup$ 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) :( $\endgroup$ – terrible at math Mar 13 '14 at 2:49
  • $\begingroup$ 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. $\endgroup$ – Chris K Mar 13 '14 at 3:18
3
$\begingroup$

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 $1\le a_n, b_n$.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – terrible at math Mar 13 '14 at 3:22
  • $\begingroup$ 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. $\endgroup$ – terrible at math Mar 13 '14 at 3:23
  • $\begingroup$ @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 $a_n -b_n \to 0$. $\endgroup$ – Jose Antonio Mar 13 '14 at 3:55
  • $\begingroup$ @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) $\endgroup$ – Jose Antonio Mar 13 '14 at 4:18
  • $\begingroup$ @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". $\endgroup$ – Jose Antonio Mar 13 '14 at 6:11
1
$\begingroup$

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 $y\ge1$.

$\endgroup$
  • $\begingroup$ interesting, so zero is the only thing that screws up the uniform continuity of this function. $\endgroup$ – terrible at math Mar 13 '14 at 3:17
  • $\begingroup$ can you explain how you went from $\mid \frac{1}{x} - \frac{1}{y}\mid$ to $\frac{\mid x-y \mid}{xy}$ ? $\endgroup$ – terrible at math Mar 13 '14 at 3:39
  • $\begingroup$ $\mid \frac{1}{x} - \frac{1}{y}\mid\Rightarrow\mid \frac{y-x}{xy}\mid\Rightarrow\frac{\mid x-y \mid}{|x||y|}$ $\endgroup$ – blues66 Mar 13 '14 at 4:20
  • $\begingroup$ The main thing that screws up the uniform continuity, is the not bounded derivate. $\endgroup$ – blues66 Mar 13 '14 at 4:21
0
$\begingroup$

My answer is:

i) No.

ii) Yes.

iii) No.

iv) Yes.

v) Yes.

$\endgroup$
  • $\begingroup$ 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 ? $\endgroup$ – terrible at math Mar 13 '14 at 3:48
  • $\begingroup$ @terribleatmath: let me double check. $\endgroup$ – DeepSea Mar 13 '14 at 3:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.