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The length of a regular curve is different from its measure? (for rectifiable non absolute continuous path)
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Consider a path $pi : [a,b]to C$ where $C$ is a curve. By definition, then length of the path is $$ell(pi)=sup sum_i=0^n|pi(t_i)-pi(t_i+1)|,$$
where $t_0,...,t_n+1$ is a partition of $[a,b]$, and the sup is taken over all partition.
I know that if $pi$ is rectifiable but not absolutely continuous, then
$$ell(pi)geq int_a^b |pi '(t)|dt= m(C),$$
where $m$ is the Lebesgue measure. So there are case where the measure of a path is in fact not the length of the path ? That looks weird...
real-analysis measure-theory
edited Sep 2 at 13:13
Bernard
112k635102
asked Sep 2 at 13:00
Henri
1176
add a comment |Â
up vote
4
down vote
favorite
Consider a path $pi : [a,b]to C$ where $C$ is a curve. By definition, then length of the path is $$ell(pi)=sup sum_i=0^n|pi(t_i)-pi(t_i+1)|,$$
where $t_0,...,t_n+1$ is a partition of $[a,b]$, and the sup is taken over all partition.
I know that if $pi$ is rectifiable but not absolutely continuous, then
$$ell(pi)geq int_a^b |pi '(t)|dt= m(C),$$
where $m$ is the Lebesgue measure. So there are case where the measure of a path is in fact not the length of the path ? That looks weird...
real-analysis measure-theory
edited Sep 2 at 13:13
Bernard
112k635102
asked Sep 2 at 13:00
Henri
1176
1
That $m(C)$ thing makes no sense. You must mean that $m$ is one-dimensional Lebesgue measure. But if so then there's no such thing as $m(C)$, since $C$ is not a subset of $Bbb R$. You meant to say something about $h_1(C)$, where $h_1$ is one-dimensional Hausdorff measure. With that change it makes sense, but no, $int_a^b||pi'||ne h_1(C)$.
– David C. Ullrich
Sep 2 at 14:31
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Consider a path $pi : [a,b]to C$ where $C$ is a curve. By definition, then length of the path is $$ell(pi)=sup sum_i=0^n|pi(t_i)-pi(t_i+1)|,$$
where $t_0,...,t_n+1$ is a partition of $[a,b]$, and the sup is taken over all partition.
I know that if $pi$ is rectifiable but not absolutely continuous, then
$$ell(pi)geq int_a^b |pi '(t)|dt= m(C),$$
where $m$ is the Lebesgue measure. So there are case where the measure of a path is in fact not the length of the path ? That looks weird...
real-analysis measure-theory
edited Sep 2 at 13:13
Bernard
112k635102
asked Sep 2 at 13:00
Henri
1176
Consider a path $pi : [a,b]to C$ where $C$ is a curve. By definition, then length of the path is $$ell(pi)=sup sum_i=0^n|pi(t_i)-pi(t_i+1)|,$$
where $t_0,...,t_n+1$ is a partition of $[a,b]$, and the sup is taken over all partition.
I know that if $pi$ is rectifiable but not absolutely continuous, then
$$ell(pi)geq int_a^b |pi '(t)|dt= m(C),$$
where $m$ is the Lebesgue measure. So there are case where the measure of a path is in fact not the length of the path ? That looks weird...
real-analysis measure-theory
edited Sep 2 at 13:13
Bernard
112k635102
asked Sep 2 at 13:00
Henri
1176
edited Sep 2 at 13:13
Bernard
112k635102
edited Sep 2 at 13:13
Bernard
112k635102
edited Sep 2 at 13:13
Bernard
112k635102
112k635102
asked Sep 2 at 13:00
Henri
1176
asked Sep 2 at 13:00
Henri
1176
asked Sep 2 at 13:00
Henri
1176
1176
1
That $m(C)$ thing makes no sense. You must mean that $m$ is one-dimensional Lebesgue measure. But if so then there's no such thing as $m(C)$, since $C$ is not a subset of $Bbb R$. You meant to say something about $h_1(C)$, where $h_1$ is one-dimensional Hausdorff measure. With that change it makes sense, but no, $int_a^b||pi'||ne h_1(C)$.
– David C. Ullrich
Sep 2 at 14:31
add a comment |Â
1
That $m(C)$ thing makes no sense. You must mean that $m$ is one-dimensional Lebesgue measure. But if so then there's no such thing as $m(C)$, since $C$ is not a subset of $Bbb R$. You meant to say something about $h_1(C)$, where $h_1$ is one-dimensional Hausdorff measure. With that change it makes sense, but no, $int_a^b||pi'||ne h_1(C)$.
– David C. Ullrich
Sep 2 at 14:31
1
1
That $m(C)$ thing makes no sense. You must mean that $m$ is one-dimensional Lebesgue measure. But if so then there's no such thing as $m(C)$, since $C$ is not a subset of $Bbb R$. You meant to say something about $h_1(C)$, where $h_1$ is one-dimensional Hausdorff measure. With that change it makes sense, but no, $int_a^b||pi'||ne h_1(C)$.
– David C. Ullrich
Sep 2 at 14:31
That $m(C)$ thing makes no sense. You must mean that $m$ is one-dimensional Lebesgue measure. But if so then there's no such thing as $m(C)$, since $C$ is not a subset of $Bbb R$. You meant to say something about $h_1(C)$, where $h_1$ is one-dimensional Hausdorff measure. With that change it makes sense, but no, $int_a^b||pi'||ne h_1(C)$.
– David C. Ullrich
Sep 2 at 14:31
add a comment |Â
2 Answers
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An example is the graph $mathcalG(f)$ of the Cantor function $f$.
The Cantor function is a continuous, increasing function from $f:[0,1]to[0,1]$ with $f(0)=0$, $f(1)=1$. Let
$$gamma(t)=(t,f(t)),qquad tin [0,1] $$
A simple argument shows that the length of the graph, i.e. the length of the curve $gamma$ is larger than $sqrt2$.
For any partition $0=t_0<t_1<dots <t_n=1$ we have
beginalign*ell(mathcalG(f))geqsum_i=0^n-1|gamma(t_i+1)-gamma(t_i)|&=sum_i=0^n-1sqrtf(t_i+1)-f(t_i) geq \
text(Cauchy-Schwarz)quad &geq frac1sqrt2left[
sum_i=0^n-1|t_i+1-t_i|+sum_i=0^n-1left|f(t_i+1)-f(t_i)right|right]=\
text(ftextit is increasing)quad &=frac1sqrt2+frac1sqrt2sum_i=0^n-1left(f(t_i+1)-f(t_i)right)
=frac1sqrt2left[1+f(t_n)-f(t_0)right]=\
(f(0)=0,;f(1)=1)quad&=frac1sqrt2left[1+f(1)+f(0)right] =sqrt2
endalign*
This could also be seen as an argument that any rectifiable curve connecting the points $(0,0)$ and $(1,1)$ cannot be shorter than $sqrt2$, i.e. the length of the segment connecting those two points.
With the same computations but using the opposite of Cauchy-Schwarz inequality
beginalign*sum_i=0^n-1|gamma(t_i+1)-gamma(t_i)|&=sum_i=0^n-1sqrtf(t_i+1)-f(t_i) leq \
&leq
sum_i=0^n-1|t_i+1-t_i|+sum_i=0^n-1left|f(t_i+1)-f(t_i)right|=2endalign*
and taking the $sup$ over all partitions of $[0,1]$, we get that $mathcalG(f)$ is, indeed, a rectifiable curve, with $ell(mathcalG(f))leq 2$.
However, we also $f'=0$ a.e. on $[0,1]$, and hence
$$ell (mathcalG(f))geq sqrt2 > 1 =int_0^1dt=int_0^1sqrt^2dt=int_0^1|gamma'(t)|dt$$
answered Sep 2 at 13:55
Lorenzo Quarisa
2,734315
add a comment |Â
up vote
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This is indeed the case. The Cantor function $c$:
- is continuous everywhere but has zero derivative almost everywhere. Hence $displaystyle int_0^1 sqrt1+left(dfracdcdtright)^2 dt$ is equal to $1$.
- However its arc length is greater or equal than $2$. See Arc length of the Cantor function.
answered Sep 2 at 13:51
mathcounterexamples.net
25.6k21754
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
An example is the graph $mathcalG(f)$ of the Cantor function $f$.
The Cantor function is a continuous, increasing function from $f:[0,1]to[0,1]$ with $f(0)=0$, $f(1)=1$. Let
$$gamma(t)=(t,f(t)),qquad tin [0,1] $$
A simple argument shows that the length of the graph, i.e. the length of the curve $gamma$ is larger than $sqrt2$.
For any partition $0=t_0<t_1<dots <t_n=1$ we have
beginalign*ell(mathcalG(f))geqsum_i=0^n-1|gamma(t_i+1)-gamma(t_i)|&=sum_i=0^n-1sqrtf(t_i+1)-f(t_i) geq \
text(Cauchy-Schwarz)quad &geq frac1sqrt2left[
sum_i=0^n-1|t_i+1-t_i|+sum_i=0^n-1left|f(t_i+1)-f(t_i)right|right]=\
text(ftextit is increasing)quad &=frac1sqrt2+frac1sqrt2sum_i=0^n-1left(f(t_i+1)-f(t_i)right)
=frac1sqrt2left[1+f(t_n)-f(t_0)right]=\
(f(0)=0,;f(1)=1)quad&=frac1sqrt2left[1+f(1)+f(0)right] =sqrt2
endalign*
This could also be seen as an argument that any rectifiable curve connecting the points $(0,0)$ and $(1,1)$ cannot be shorter than $sqrt2$, i.e. the length of the segment connecting those two points.
With the same computations but using the opposite of Cauchy-Schwarz inequality
beginalign*sum_i=0^n-1|gamma(t_i+1)-gamma(t_i)|&=sum_i=0^n-1sqrtf(t_i+1)-f(t_i) leq \
&leq
sum_i=0^n-1|t_i+1-t_i|+sum_i=0^n-1left|f(t_i+1)-f(t_i)right|=2endalign*
and taking the $sup$ over all partitions of $[0,1]$, we get that $mathcalG(f)$ is, indeed, a rectifiable curve, with $ell(mathcalG(f))leq 2$.
However, we also $f'=0$ a.e. on $[0,1]$, and hence
$$ell (mathcalG(f))geq sqrt2 > 1 =int_0^1dt=int_0^1sqrt^2dt=int_0^1|gamma'(t)|dt$$
answered Sep 2 at 13:55
Lorenzo Quarisa
2,734315
add a comment |Â
up vote
2
down vote
An example is the graph $mathcalG(f)$ of the Cantor function $f$.
The Cantor function is a continuous, increasing function from $f:[0,1]to[0,1]$ with $f(0)=0$, $f(1)=1$. Let
$$gamma(t)=(t,f(t)),qquad tin [0,1] $$
A simple argument shows that the length of the graph, i.e. the length of the curve $gamma$ is larger than $sqrt2$.
For any partition $0=t_0<t_1<dots <t_n=1$ we have
beginalign*ell(mathcalG(f))geqsum_i=0^n-1|gamma(t_i+1)-gamma(t_i)|&=sum_i=0^n-1sqrtf(t_i+1)-f(t_i) geq \
text(Cauchy-Schwarz)quad &geq frac1sqrt2left[
sum_i=0^n-1|t_i+1-t_i|+sum_i=0^n-1left|f(t_i+1)-f(t_i)right|right]=\
text(ftextit is increasing)quad &=frac1sqrt2+frac1sqrt2sum_i=0^n-1left(f(t_i+1)-f(t_i)right)
=frac1sqrt2left[1+f(t_n)-f(t_0)right]=\
(f(0)=0,;f(1)=1)quad&=frac1sqrt2left[1+f(1)+f(0)right] =sqrt2
endalign*
This could also be seen as an argument that any rectifiable curve connecting the points $(0,0)$ and $(1,1)$ cannot be shorter than $sqrt2$, i.e. the length of the segment connecting those two points.
With the same computations but using the opposite of Cauchy-Schwarz inequality
beginalign*sum_i=0^n-1|gamma(t_i+1)-gamma(t_i)|&=sum_i=0^n-1sqrtf(t_i+1)-f(t_i) leq \
&leq
sum_i=0^n-1|t_i+1-t_i|+sum_i=0^n-1left|f(t_i+1)-f(t_i)right|=2endalign*
and taking the $sup$ over all partitions of $[0,1]$, we get that $mathcalG(f)$ is, indeed, a rectifiable curve, with $ell(mathcalG(f))leq 2$.
However, we also $f'=0$ a.e. on $[0,1]$, and hence
$$ell (mathcalG(f))geq sqrt2 > 1 =int_0^1dt=int_0^1sqrt^2dt=int_0^1|gamma'(t)|dt$$
answered Sep 2 at 13:55
Lorenzo Quarisa
2,734315
add a comment |Â
up vote
2
down vote
up vote
2
down vote
An example is the graph $mathcalG(f)$ of the Cantor function $f$.
The Cantor function is a continuous, increasing function from $f:[0,1]to[0,1]$ with $f(0)=0$, $f(1)=1$. Let
$$gamma(t)=(t,f(t)),qquad tin [0,1] $$
A simple argument shows that the length of the graph, i.e. the length of the curve $gamma$ is larger than $sqrt2$.
For any partition $0=t_0<t_1<dots <t_n=1$ we have
beginalign*ell(mathcalG(f))geqsum_i=0^n-1|gamma(t_i+1)-gamma(t_i)|&=sum_i=0^n-1sqrtf(t_i+1)-f(t_i) geq \
text(Cauchy-Schwarz)quad &geq frac1sqrt2left[
sum_i=0^n-1|t_i+1-t_i|+sum_i=0^n-1left|f(t_i+1)-f(t_i)right|right]=\
text(ftextit is increasing)quad &=frac1sqrt2+frac1sqrt2sum_i=0^n-1left(f(t_i+1)-f(t_i)right)
=frac1sqrt2left[1+f(t_n)-f(t_0)right]=\
(f(0)=0,;f(1)=1)quad&=frac1sqrt2left[1+f(1)+f(0)right] =sqrt2
endalign*
This could also be seen as an argument that any rectifiable curve connecting the points $(0,0)$ and $(1,1)$ cannot be shorter than $sqrt2$, i.e. the length of the segment connecting those two points.
With the same computations but using the opposite of Cauchy-Schwarz inequality
beginalign*sum_i=0^n-1|gamma(t_i+1)-gamma(t_i)|&=sum_i=0^n-1sqrtf(t_i+1)-f(t_i) leq \
&leq
sum_i=0^n-1|t_i+1-t_i|+sum_i=0^n-1left|f(t_i+1)-f(t_i)right|=2endalign*
and taking the $sup$ over all partitions of $[0,1]$, we get that $mathcalG(f)$ is, indeed, a rectifiable curve, with $ell(mathcalG(f))leq 2$.
However, we also $f'=0$ a.e. on $[0,1]$, and hence
$$ell (mathcalG(f))geq sqrt2 > 1 =int_0^1dt=int_0^1sqrt^2dt=int_0^1|gamma'(t)|dt$$
answered Sep 2 at 13:55
Lorenzo Quarisa
2,734315
An example is the graph $mathcalG(f)$ of the Cantor function $f$.
The Cantor function is a continuous, increasing function from $f:[0,1]to[0,1]$ with $f(0)=0$, $f(1)=1$. Let
$$gamma(t)=(t,f(t)),qquad tin [0,1] $$
A simple argument shows that the length of the graph, i.e. the length of the curve $gamma$ is larger than $sqrt2$.
For any partition $0=t_0<t_1<dots <t_n=1$ we have
beginalign*ell(mathcalG(f))geqsum_i=0^n-1|gamma(t_i+1)-gamma(t_i)|&=sum_i=0^n-1sqrtf(t_i+1)-f(t_i) geq \
text(Cauchy-Schwarz)quad &geq frac1sqrt2left[
sum_i=0^n-1|t_i+1-t_i|+sum_i=0^n-1left|f(t_i+1)-f(t_i)right|right]=\
text(ftextit is increasing)quad &=frac1sqrt2+frac1sqrt2sum_i=0^n-1left(f(t_i+1)-f(t_i)right)
=frac1sqrt2left[1+f(t_n)-f(t_0)right]=\
(f(0)=0,;f(1)=1)quad&=frac1sqrt2left[1+f(1)+f(0)right] =sqrt2
endalign*
This could also be seen as an argument that any rectifiable curve connecting the points $(0,0)$ and $(1,1)$ cannot be shorter than $sqrt2$, i.e. the length of the segment connecting those two points.
With the same computations but using the opposite of Cauchy-Schwarz inequality
beginalign*sum_i=0^n-1|gamma(t_i+1)-gamma(t_i)|&=sum_i=0^n-1sqrtf(t_i+1)-f(t_i) leq \
&leq
sum_i=0^n-1|t_i+1-t_i|+sum_i=0^n-1left|f(t_i+1)-f(t_i)right|=2endalign*
and taking the $sup$ over all partitions of $[0,1]$, we get that $mathcalG(f)$ is, indeed, a rectifiable curve, with $ell(mathcalG(f))leq 2$.
However, we also $f'=0$ a.e. on $[0,1]$, and hence
$$ell (mathcalG(f))geq sqrt2 > 1 =int_0^1dt=int_0^1sqrt^2dt=int_0^1|gamma'(t)|dt$$
answered Sep 2 at 13:55
Lorenzo Quarisa
2,734315
edited Sep 2 at 14:05
answered Sep 2 at 13:55
Lorenzo Quarisa
2,734315
answered Sep 2 at 13:55
Lorenzo Quarisa
2,734315
answered Sep 2 at 13:55
Lorenzo Quarisa
2,734315
2,734315
add a comment |Â
add a comment |Â
up vote
2
down vote
This is indeed the case. The Cantor function $c$:
- is continuous everywhere but has zero derivative almost everywhere. Hence $displaystyle int_0^1 sqrt1+left(dfracdcdtright)^2 dt$ is equal to $1$.
- However its arc length is greater or equal than $2$. See Arc length of the Cantor function.
answered Sep 2 at 13:51
mathcounterexamples.net
25.6k21754
add a comment |Â
up vote
2
down vote
This is indeed the case. The Cantor function $c$:
- is continuous everywhere but has zero derivative almost everywhere. Hence $displaystyle int_0^1 sqrt1+left(dfracdcdtright)^2 dt$ is equal to $1$.
- However its arc length is greater or equal than $2$. See Arc length of the Cantor function.
answered Sep 2 at 13:51
mathcounterexamples.net
25.6k21754
add a comment |Â
up vote
2
down vote
up vote
2
down vote
This is indeed the case. The Cantor function $c$:
- is continuous everywhere but has zero derivative almost everywhere. Hence $displaystyle int_0^1 sqrt1+left(dfracdcdtright)^2 dt$ is equal to $1$.
- However its arc length is greater or equal than $2$. See Arc length of the Cantor function.
answered Sep 2 at 13:51
mathcounterexamples.net
25.6k21754
This is indeed the case. The Cantor function $c$:
- is continuous everywhere but has zero derivative almost everywhere. Hence $displaystyle int_0^1 sqrt1+left(dfracdcdtright)^2 dt$ is equal to $1$.
- However its arc length is greater or equal than $2$. See Arc length of the Cantor function.
answered Sep 2 at 13:51
mathcounterexamples.net
25.6k21754
edited Sep 2 at 14:17
answered Sep 2 at 13:51
mathcounterexamples.net
25.6k21754
answered Sep 2 at 13:51
mathcounterexamples.net
25.6k21754
answered Sep 2 at 13:51
mathcounterexamples.net
25.6k21754
25.6k21754
add a comment |Â
add a comment |Â
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1
That $m(C)$ thing makes no sense. You must mean that $m$ is one-dimensional Lebesgue measure. But if so then there's no such thing as $m(C)$, since $C$ is not a subset of $Bbb R$. You meant to say something about $h_1(C)$, where $h_1$ is one-dimensional Hausdorff measure. With that change it makes sense, but no, $int_a^b||pi'||ne h_1(C)$.
– David C. Ullrich
Sep 2 at 14:31