Showing that a curve is not rectifiable if its arc length is not a continuous function
$begingroup$
This is a (translated) proof from a textbook of the fact that arc length of a rectifiable curve is a continuous function.
Let $phi:[T_0,T_1]rightarrowmathbb C$ be a function whose real part and imaginary part are continuous, and $C$ be a curve represented by $phi$. Suppose $C$ is rectifiable. Define $f:[T_0,T_1]rightarrowmathbb R$ by $f(t) = L(C|[T_0,t])$, where $L(C|I)$ is the arc length of $C$ restricted to the interval $I$. To show by contradiction that $f$ is continuous, assume $f$ is not continuous at $t_0in [T_0, T_1]$. Since $f$ is monotonously increasing, either of
$
lim_{trightarrow t_0-0} f(t) < f(t_0)
$ or $lim_{trightarrow t_0+0} f(t) > f(t_0)$ holds. WLOG we may assume the former holds. Here $t_0 > t$. Let $epsilon_0 = f(t_0) - lim_{trightarrow t_0-0}f(t)$. By definition, there exists an infinite number of $t_j < tilde{t_j} < t_{j+1} < tilde {t}_{j+1}quad(j = 1,2,dots)$ s.t. $L(C|[t_j,tilde{t}_j])>epsilon_0/2$. Then we have $L(C) = +infty$. Contradiction.
I can't figure out why the sentence that begin with "By definition" is true. Why are there such $t_j$'s?
EDIT: In the book, $L(C)$ is defined to be the supremum (possibly $+infty$) of $sum_{j=1}^{n}|phi(s_j)-phi(s_{j-1})|$ for any partition $T_0 = s_0 < s_1 < dots < s_n = T_1$. $C$ is rectifiable iff $L(C) <infty$.
calculus real-analysis complex-analysis
asked Mar 25 '12 at 12:42
PteromysPteromys
2,44421644
$endgroup$
add a comment |
$begingroup$
This is a (translated) proof from a textbook of the fact that arc length of a rectifiable curve is a continuous function.
Let $phi:[T_0,T_1]rightarrowmathbb C$ be a function whose real part and imaginary part are continuous, and $C$ be a curve represented by $phi$. Suppose $C$ is rectifiable. Define $f:[T_0,T_1]rightarrowmathbb R$ by $f(t) = L(C|[T_0,t])$, where $L(C|I)$ is the arc length of $C$ restricted to the interval $I$. To show by contradiction that $f$ is continuous, assume $f$ is not continuous at $t_0in [T_0, T_1]$. Since $f$ is monotonously increasing, either of
$
lim_{trightarrow t_0-0} f(t) < f(t_0)
$ or $lim_{trightarrow t_0+0} f(t) > f(t_0)$ holds. WLOG we may assume the former holds. Here $t_0 > t$. Let $epsilon_0 = f(t_0) - lim_{trightarrow t_0-0}f(t)$. By definition, there exists an infinite number of $t_j < tilde{t_j} < t_{j+1} < tilde {t}_{j+1}quad(j = 1,2,dots)$ s.t. $L(C|[t_j,tilde{t}_j])>epsilon_0/2$. Then we have $L(C) = +infty$. Contradiction.
I can't figure out why the sentence that begin with "By definition" is true. Why are there such $t_j$'s?
EDIT: In the book, $L(C)$ is defined to be the supremum (possibly $+infty$) of $sum_{j=1}^{n}|phi(s_j)-phi(s_{j-1})|$ for any partition $T_0 = s_0 < s_1 < dots < s_n = T_1$. $C$ is rectifiable iff $L(C) <infty$.
calculus real-analysis complex-analysis
asked Mar 25 '12 at 12:42
PteromysPteromys
2,44421644
$endgroup$
$begingroup$
this looks strange to me. Could you add the information how rectifiability and arc length are defined in that book?
$endgroup$
– user20266
Mar 25 '12 at 12:51
$begingroup$
I added it to the question statement.
$endgroup$
– Pteromys
Mar 26 '12 at 4:23
1
$begingroup$
I don't see the reasoning either, maybe there is some more context that is missing. Anyway, continuity is not difficult to demonstrate by other means. Burago, Burago, and Ivanov do it nicely in a few lines in "A Course in Metric Geometry", p. 35 Proposition 2.3.4 (you can find it online).
$endgroup$
– yasmar
Mar 26 '12 at 18:54
add a comment |
$begingroup$
This is a (translated) proof from a textbook of the fact that arc length of a rectifiable curve is a continuous function.
Let $phi:[T_0,T_1]rightarrowmathbb C$ be a function whose real part and imaginary part are continuous, and $C$ be a curve represented by $phi$. Suppose $C$ is rectifiable. Define $f:[T_0,T_1]rightarrowmathbb R$ by $f(t) = L(C|[T_0,t])$, where $L(C|I)$ is the arc length of $C$ restricted to the interval $I$. To show by contradiction that $f$ is continuous, assume $f$ is not continuous at $t_0in [T_0, T_1]$. Since $f$ is monotonously increasing, either of
$
lim_{trightarrow t_0-0} f(t) < f(t_0)
$ or $lim_{trightarrow t_0+0} f(t) > f(t_0)$ holds. WLOG we may assume the former holds. Here $t_0 > t$. Let $epsilon_0 = f(t_0) - lim_{trightarrow t_0-0}f(t)$. By definition, there exists an infinite number of $t_j < tilde{t_j} < t_{j+1} < tilde {t}_{j+1}quad(j = 1,2,dots)$ s.t. $L(C|[t_j,tilde{t}_j])>epsilon_0/2$. Then we have $L(C) = +infty$. Contradiction.
I can't figure out why the sentence that begin with "By definition" is true. Why are there such $t_j$'s?
EDIT: In the book, $L(C)$ is defined to be the supremum (possibly $+infty$) of $sum_{j=1}^{n}|phi(s_j)-phi(s_{j-1})|$ for any partition $T_0 = s_0 < s_1 < dots < s_n = T_1$. $C$ is rectifiable iff $L(C) <infty$.
calculus real-analysis complex-analysis
asked Mar 25 '12 at 12:42
PteromysPteromys
2,44421644
$endgroup$
This is a (translated) proof from a textbook of the fact that arc length of a rectifiable curve is a continuous function.
Let $phi:[T_0,T_1]rightarrowmathbb C$ be a function whose real part and imaginary part are continuous, and $C$ be a curve represented by $phi$. Suppose $C$ is rectifiable. Define $f:[T_0,T_1]rightarrowmathbb R$ by $f(t) = L(C|[T_0,t])$, where $L(C|I)$ is the arc length of $C$ restricted to the interval $I$. To show by contradiction that $f$ is continuous, assume $f$ is not continuous at $t_0in [T_0, T_1]$. Since $f$ is monotonously increasing, either of
$
lim_{trightarrow t_0-0} f(t) < f(t_0)
$ or $lim_{trightarrow t_0+0} f(t) > f(t_0)$ holds. WLOG we may assume the former holds. Here $t_0 > t$. Let $epsilon_0 = f(t_0) - lim_{trightarrow t_0-0}f(t)$. By definition, there exists an infinite number of $t_j < tilde{t_j} < t_{j+1} < tilde {t}_{j+1}quad(j = 1,2,dots)$ s.t. $L(C|[t_j,tilde{t}_j])>epsilon_0/2$. Then we have $L(C) = +infty$. Contradiction.
I can't figure out why the sentence that begin with "By definition" is true. Why are there such $t_j$'s?
EDIT: In the book, $L(C)$ is defined to be the supremum (possibly $+infty$) of $sum_{j=1}^{n}|phi(s_j)-phi(s_{j-1})|$ for any partition $T_0 = s_0 < s_1 < dots < s_n = T_1$. $C$ is rectifiable iff $L(C) <infty$.
calculus real-analysis complex-analysis
calculus real-analysis complex-analysis
asked Mar 25 '12 at 12:42
PteromysPteromys
2,44421644
asked Mar 25 '12 at 12:42
PteromysPteromys
2,44421644
edited Mar 26 '12 at 4:37
Pteromys
asked Mar 25 '12 at 12:42
PteromysPteromys
2,44421644
asked Mar 25 '12 at 12:42
PteromysPteromys
2,44421644
asked Mar 25 '12 at 12:42
PteromysPteromys
2,44421644
2,44421644
$begingroup$
this looks strange to me. Could you add the information how rectifiability and arc length are defined in that book?
$endgroup$
– user20266
Mar 25 '12 at 12:51
$begingroup$
I added it to the question statement.
$endgroup$
– Pteromys
Mar 26 '12 at 4:23
1
$begingroup$
I don't see the reasoning either, maybe there is some more context that is missing. Anyway, continuity is not difficult to demonstrate by other means. Burago, Burago, and Ivanov do it nicely in a few lines in "A Course in Metric Geometry", p. 35 Proposition 2.3.4 (you can find it online).
$endgroup$
– yasmar
Mar 26 '12 at 18:54
add a comment |
$begingroup$
this looks strange to me. Could you add the information how rectifiability and arc length are defined in that book?
$endgroup$
– user20266
Mar 25 '12 at 12:51
$begingroup$
I added it to the question statement.
$endgroup$
– Pteromys
Mar 26 '12 at 4:23
1
$begingroup$
I don't see the reasoning either, maybe there is some more context that is missing. Anyway, continuity is not difficult to demonstrate by other means. Burago, Burago, and Ivanov do it nicely in a few lines in "A Course in Metric Geometry", p. 35 Proposition 2.3.4 (you can find it online).
$endgroup$
– yasmar
Mar 26 '12 at 18:54
$begingroup$
this looks strange to me. Could you add the information how rectifiability and arc length are defined in that book?
$endgroup$
– user20266
Mar 25 '12 at 12:51
$begingroup$
this looks strange to me. Could you add the information how rectifiability and arc length are defined in that book?
$endgroup$
– user20266
Mar 25 '12 at 12:51
$begingroup$
I added it to the question statement.
$endgroup$
– Pteromys
Mar 26 '12 at 4:23
$begingroup$
I added it to the question statement.
$endgroup$
– Pteromys
Mar 26 '12 at 4:23
1
1
$begingroup$
I don't see the reasoning either, maybe there is some more context that is missing. Anyway, continuity is not difficult to demonstrate by other means. Burago, Burago, and Ivanov do it nicely in a few lines in "A Course in Metric Geometry", p. 35 Proposition 2.3.4 (you can find it online).
$endgroup$
– yasmar
Mar 26 '12 at 18:54
$begingroup$
I don't see the reasoning either, maybe there is some more context that is missing. Anyway, continuity is not difficult to demonstrate by other means. Burago, Burago, and Ivanov do it nicely in a few lines in "A Course in Metric Geometry", p. 35 Proposition 2.3.4 (you can find it online).
$endgroup$
– yasmar
Mar 26 '12 at 18:54
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I am not sure this is what the author intended, but the claim that holds "by definition" can be shown as follows:
Take any $t_1<t_0$. Then $L(C|[t_1,t_0])ge epsilon_0$, so there exists a partition $t_1=s_0 < cdots < s_n=t_0$ such that $$sum_{j=1}^n |phi(s_j)-phi(s_{j-1})|> 3 epsilon_0/4.$$ We may assume $|phi(s_n)-phi(s_{n-1})|< epsilon_0/4$, since $phi$ is continuous.
(Otherwise there exists $hat s_n in (s_{n-1},t_0)$ with $|t_0-hat s_n|<1/4 epsilon_0$. Take $hat s_j = s_j$ for $j=1,ldots n-1$, and $hat s_{n+1}=t_0$. Then $$sum_{j=1}^{n+1} |phi(hat s_j)-phi(hat s_{j-1})| ge sum_{j=1}^n |phi(s_j)-phi(s_{j-1})|> 3 epsilon_0/4$$ by triangle inequality.)
Hence $L(C|[t_1,s_{n-1}])>epsilon_0/2$. Set $t_2=s_{n-1}$.
We still have $L(C|[t_2,t_0])geepsilon_0$, so we can start over to get $t_3, t_4,ldots$ with $L(C|[t_j,t_{j+1}]) > epsilon_0/2$ by induction.
(Not sure why we need $tilde t_j$, but if we want to, we can take any $tilde t_j in (t_j,t_0)$ and then proceed in step 2 with $tilde t_j$ instead of $t_j$.)
answered Dec 1 '18 at 16:36
hifehife
1938
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f124244%2fshowing-that-a-curve-is-not-rectifiable-if-its-arc-length-is-not-a-continuous-fu%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I am not sure this is what the author intended, but the claim that holds "by definition" can be shown as follows:
Take any $t_1<t_0$. Then $L(C|[t_1,t_0])ge epsilon_0$, so there exists a partition $t_1=s_0 < cdots < s_n=t_0$ such that $$sum_{j=1}^n |phi(s_j)-phi(s_{j-1})|> 3 epsilon_0/4.$$ We may assume $|phi(s_n)-phi(s_{n-1})|< epsilon_0/4$, since $phi$ is continuous.
(Otherwise there exists $hat s_n in (s_{n-1},t_0)$ with $|t_0-hat s_n|<1/4 epsilon_0$. Take $hat s_j = s_j$ for $j=1,ldots n-1$, and $hat s_{n+1}=t_0$. Then $$sum_{j=1}^{n+1} |phi(hat s_j)-phi(hat s_{j-1})| ge sum_{j=1}^n |phi(s_j)-phi(s_{j-1})|> 3 epsilon_0/4$$ by triangle inequality.)
Hence $L(C|[t_1,s_{n-1}])>epsilon_0/2$. Set $t_2=s_{n-1}$.
We still have $L(C|[t_2,t_0])geepsilon_0$, so we can start over to get $t_3, t_4,ldots$ with $L(C|[t_j,t_{j+1}]) > epsilon_0/2$ by induction.
(Not sure why we need $tilde t_j$, but if we want to, we can take any $tilde t_j in (t_j,t_0)$ and then proceed in step 2 with $tilde t_j$ instead of $t_j$.)
answered Dec 1 '18 at 16:36
hifehife
1938
$endgroup$
add a comment |
$begingroup$
I am not sure this is what the author intended, but the claim that holds "by definition" can be shown as follows:
Take any $t_1<t_0$. Then $L(C|[t_1,t_0])ge epsilon_0$, so there exists a partition $t_1=s_0 < cdots < s_n=t_0$ such that $$sum_{j=1}^n |phi(s_j)-phi(s_{j-1})|> 3 epsilon_0/4.$$ We may assume $|phi(s_n)-phi(s_{n-1})|< epsilon_0/4$, since $phi$ is continuous.
(Otherwise there exists $hat s_n in (s_{n-1},t_0)$ with $|t_0-hat s_n|<1/4 epsilon_0$. Take $hat s_j = s_j$ for $j=1,ldots n-1$, and $hat s_{n+1}=t_0$. Then $$sum_{j=1}^{n+1} |phi(hat s_j)-phi(hat s_{j-1})| ge sum_{j=1}^n |phi(s_j)-phi(s_{j-1})|> 3 epsilon_0/4$$ by triangle inequality.)
Hence $L(C|[t_1,s_{n-1}])>epsilon_0/2$. Set $t_2=s_{n-1}$.
We still have $L(C|[t_2,t_0])geepsilon_0$, so we can start over to get $t_3, t_4,ldots$ with $L(C|[t_j,t_{j+1}]) > epsilon_0/2$ by induction.
(Not sure why we need $tilde t_j$, but if we want to, we can take any $tilde t_j in (t_j,t_0)$ and then proceed in step 2 with $tilde t_j$ instead of $t_j$.)
answered Dec 1 '18 at 16:36
hifehife
1938
$endgroup$
add a comment |
$begingroup$
I am not sure this is what the author intended, but the claim that holds "by definition" can be shown as follows:
Take any $t_1<t_0$. Then $L(C|[t_1,t_0])ge epsilon_0$, so there exists a partition $t_1=s_0 < cdots < s_n=t_0$ such that $$sum_{j=1}^n |phi(s_j)-phi(s_{j-1})|> 3 epsilon_0/4.$$ We may assume $|phi(s_n)-phi(s_{n-1})|< epsilon_0/4$, since $phi$ is continuous.
(Otherwise there exists $hat s_n in (s_{n-1},t_0)$ with $|t_0-hat s_n|<1/4 epsilon_0$. Take $hat s_j = s_j$ for $j=1,ldots n-1$, and $hat s_{n+1}=t_0$. Then $$sum_{j=1}^{n+1} |phi(hat s_j)-phi(hat s_{j-1})| ge sum_{j=1}^n |phi(s_j)-phi(s_{j-1})|> 3 epsilon_0/4$$ by triangle inequality.)
Hence $L(C|[t_1,s_{n-1}])>epsilon_0/2$. Set $t_2=s_{n-1}$.
We still have $L(C|[t_2,t_0])geepsilon_0$, so we can start over to get $t_3, t_4,ldots$ with $L(C|[t_j,t_{j+1}]) > epsilon_0/2$ by induction.
(Not sure why we need $tilde t_j$, but if we want to, we can take any $tilde t_j in (t_j,t_0)$ and then proceed in step 2 with $tilde t_j$ instead of $t_j$.)
answered Dec 1 '18 at 16:36
hifehife
1938
$endgroup$
I am not sure this is what the author intended, but the claim that holds "by definition" can be shown as follows:
Take any $t_1<t_0$. Then $L(C|[t_1,t_0])ge epsilon_0$, so there exists a partition $t_1=s_0 < cdots < s_n=t_0$ such that $$sum_{j=1}^n |phi(s_j)-phi(s_{j-1})|> 3 epsilon_0/4.$$ We may assume $|phi(s_n)-phi(s_{n-1})|< epsilon_0/4$, since $phi$ is continuous.
(Otherwise there exists $hat s_n in (s_{n-1},t_0)$ with $|t_0-hat s_n|<1/4 epsilon_0$. Take $hat s_j = s_j$ for $j=1,ldots n-1$, and $hat s_{n+1}=t_0$. Then $$sum_{j=1}^{n+1} |phi(hat s_j)-phi(hat s_{j-1})| ge sum_{j=1}^n |phi(s_j)-phi(s_{j-1})|> 3 epsilon_0/4$$ by triangle inequality.)
Hence $L(C|[t_1,s_{n-1}])>epsilon_0/2$. Set $t_2=s_{n-1}$.
We still have $L(C|[t_2,t_0])geepsilon_0$, so we can start over to get $t_3, t_4,ldots$ with $L(C|[t_j,t_{j+1}]) > epsilon_0/2$ by induction.
(Not sure why we need $tilde t_j$, but if we want to, we can take any $tilde t_j in (t_j,t_0)$ and then proceed in step 2 with $tilde t_j$ instead of $t_j$.)
answered Dec 1 '18 at 16:36
hifehife
1938
edited Jan 10 at 11:20
answered Dec 1 '18 at 16:36
hifehife
1938
answered Dec 1 '18 at 16:36
hifehife
1938
answered Dec 1 '18 at 16:36
hifehife
1938
1938
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f124244%2fshowing-that-a-curve-is-not-rectifiable-if-its-arc-length-is-not-a-continuous-fu%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
this looks strange to me. Could you add the information how rectifiability and arc length are defined in that book?
$endgroup$
– user20266
Mar 25 '12 at 12:51
$begingroup$
I added it to the question statement.
$endgroup$
– Pteromys
Mar 26 '12 at 4:23
1
$begingroup$
I don't see the reasoning either, maybe there is some more context that is missing. Anyway, continuity is not difficult to demonstrate by other means. Burago, Burago, and Ivanov do it nicely in a few lines in "A Course in Metric Geometry", p. 35 Proposition 2.3.4 (you can find it online).
$endgroup$
– yasmar
Mar 26 '12 at 18:54