Mixing
Continuity of homotopy in proof of Hopf's Umlaufsatz
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The standard proof of Hopf's Umlaufsatz proceeds something like
this: We have a unit speed $mathcal{C}^1$ curve $beta:mathbb{R}tomathbb{R}^2$. Furthermore,
$beta$ is a simple loop with period $L$.
We now define $T={(t_1,t_2) in mathbb{R}^2 : 0 leq t_1 leq t_2 leq L }$
and a function $f$ on $T$ as follows:
$$ f(t_1,t_2) =
begin{cases}
beta'(t_1) & t_1=t_2 \
-beta'(0) & (t_1,t_2)=(0,L) \
frac{beta(t_2)-beta(t_1)}{|beta(t_2)-beta(t_1)|} & text{otherwise}
end{cases} $$
It is visually obvious
that $f$ is continuous, but I haven't found a strict proof of this anywhere.
Below is my attempt at a proof, but I think it's ugly and long-winded and it
also isn't complete. My question is how to finish it. I also suspect (or
rather hope) that there is a significantly shorter (or more elegant) proof
that is nevertheless complete and doesn't use hand-waving.
First of all, $f$ is continuous for all $(t_1,t_2)in T$ with $t_1neq t_2$
and $(t_1,t_2)neq(0,L)$ because (as $beta$ is simple) the denominator of
$(beta(t_2)-beta(t_1))/(|beta(t_2)-beta(t_1)|)$ doesn't
vanish and $beta$ and the norm are continuous.
Now for the case $t_1=t_2$. Fix $t_1in[0,L]$. We have
$$ lim_{t_2to t_1} frac{beta(t_2)-beta(t_1)}{t_2-t_1} = beta'(t_1) $$
by definition. This implies
$$ lim_{substack{t_2to t_1\t_2>t_1}} frac{|beta(t_2)-beta(t_1)|}{t_2-t_1} = lim_{substack{t_2to t_1}} left|frac{beta(t_2)-beta(t_1)}{t_2-t_1}right| = |beta'(t_1)| = 1 $$
as $beta$ is a unit speed curve.
Combining these two we get
$$ lim_{substack{t_2to t_1\t_2>t_1}} f(t_1,t_2) = lim_{t_2to t_1} frac{beta(t_2)-beta(t_1)}{t_2-t_1} cdot
lim_{substack{t_2to t_1\t_2>t_1}} frac{t_2-t_1}{|beta(t_2)-beta(t_1)|} = beta'(t_1) $$
This means that for every $varepsilon>0$ we can find a $delta>0$ such that
for all $t_2$ with $|t_2-t_1|<delta$ and $(t_1,t_2)in T$ we have
$|f(t_1,t_2)-beta'(t_1)|<varepsilon$.
As $[0,L]$ is compact, we can even, for a given $varepsilon>0$, find a $delta_1>0$
such that for all $(t_1,t_2)in T$ with $|t_2-t_1|<delta_1$ the inequality
$|f(t_1,t_2)-beta'(t_1)|<varepsilon/2$ holds.
Also, as $beta'$ is continuous and $[0,L]$ is compact, we can find a
$delta_2>0$ such that for all $t_1^ast,t_1in[0,L]$ with $|t_1-t_1^ast|<delta_2$ we
have $|beta'(t_1) - beta'(t_1^ast)|<varepsilon/2$.
Now let $P=(t_1,t_2)in T$ be arbitrary with
$|P-(t_1^ast,t_1^ast)|<min{delta_1/2,delta_2}$. That implies
$|t_1-t_1^ast|<delta_2$ and $|t_2-t_1|<delta_1$. By combining the
previous two inequalities we get
$|f(P)-beta'(t_1^ast)|<varepsilon$. We have thus proved that
$f$ is continuous in $(t_1^ast,t_1^ast)$.
The third case is the point $Q=(0,L)$. As $beta$ is a loop, we have:
$$ lim_{tto L} frac{beta(t)-beta(0)}{L-t} = -lim_{tto L} frac{beta(L)-beta(t)}{L-t} = -beta'(L) = -beta'(0) $$
And as a consequence we have:
$$ lim_{substack{tto L\t<L}} frac{|beta(t)-beta(0)|}{L-t} = lim_{substack{tto L}} left|frac{beta(t)-beta(0)}{L-t}right| = |-beta'(0)| = 1 $$
Combining these two we get, as above:
$$ lim_{substack{tto L\t<L}} f(0,t) = -beta'(0) $$
At this point I'm stuck. I think that we need to show that $f$ is uniformly
continuous around $Q$ and that maybe this is the case because $beta$ has
unit speed. If we can prove that, we can approach $Q$ in a way similar to the
second case: starting from a point near enough we first move - by virtue of
uniform continuity - parallel to the $x$ axis until the first component is
zero. Then we move vertically towards $Q$.
EDIT:
Using Ted Shifrin's advice, I've rewritten the second case of the proof.
However, it seems to me that this depends on the Taylor remainder $R$ being
continuous in both variables (see question mark below) which - if justified in
detail - is not much different from what I did above and still doesn't solve
the third case. Or am I missing something?
As $beta$ is differentiable, we can write, for each $t_1in[0,L]$,
$$ beta(t_2)=beta(t_1)+(t_2-t_1)cdotbeta'(t_1) +(t_2-t_1)cdot R(t_1,t_2-t_1) $$
with $lim_{tto0}R(t_1,t)=mathbf0$.
We thus have
$$ lim_{(t_1,t_2)to(t_1^ast,t_1^ast)} frac{beta(t_2)-beta(t_1)}{t_2-t_1} = lim_{(t_1,t_2)to(t_1^ast,t_1^ast)} (beta'(t_1) + R(t_1,t_2-t_1)) stackrel{color{red}?}{=} beta'(t_1^ast) $$
and
$$ lim_{substack{(t_1,t_2)to(t_1^ast,t_1^ast)\t_2>t_1}} frac{|beta(t_2)-beta(t_1)|}{t_2-t_1} = lim_{substack{(t_1,t_2)to(t_1^ast,t_1^ast)\t_2>t_1}} left|frac{beta(t_2)-beta(t_1)}{t_2-t_1}right| = |beta'(t_1^ast)|=1 $$
which eventually leads to
$$ lim_{substack{(t_1,t_2)to(t_1^ast,t_1^ast)\t_2>t_1}} f(t_1,t_2) = beta'(t_1^ast) $$
as above.
real-analysis differential-geometry continuity
asked Jan 22 at 17:03
FrunobulaxFrunobulax
4,7971636
$endgroup$
add a comment |
$begingroup$
The standard proof of Hopf's Umlaufsatz proceeds something like
this: We have a unit speed $mathcal{C}^1$ curve $beta:mathbb{R}tomathbb{R}^2$. Furthermore,
$beta$ is a simple loop with period $L$.
We now define $T={(t_1,t_2) in mathbb{R}^2 : 0 leq t_1 leq t_2 leq L }$
and a function $f$ on $T$ as follows:
$$ f(t_1,t_2) =
begin{cases}
beta'(t_1) & t_1=t_2 \
-beta'(0) & (t_1,t_2)=(0,L) \
frac{beta(t_2)-beta(t_1)}{|beta(t_2)-beta(t_1)|} & text{otherwise}
end{cases} $$
It is visually obvious
that $f$ is continuous, but I haven't found a strict proof of this anywhere.
Below is my attempt at a proof, but I think it's ugly and long-winded and it
also isn't complete. My question is how to finish it. I also suspect (or
rather hope) that there is a significantly shorter (or more elegant) proof
that is nevertheless complete and doesn't use hand-waving.
First of all, $f$ is continuous for all $(t_1,t_2)in T$ with $t_1neq t_2$
and $(t_1,t_2)neq(0,L)$ because (as $beta$ is simple) the denominator of
$(beta(t_2)-beta(t_1))/(|beta(t_2)-beta(t_1)|)$ doesn't
vanish and $beta$ and the norm are continuous.
Now for the case $t_1=t_2$. Fix $t_1in[0,L]$. We have
$$ lim_{t_2to t_1} frac{beta(t_2)-beta(t_1)}{t_2-t_1} = beta'(t_1) $$
by definition. This implies
$$ lim_{substack{t_2to t_1\t_2>t_1}} frac{|beta(t_2)-beta(t_1)|}{t_2-t_1} = lim_{substack{t_2to t_1}} left|frac{beta(t_2)-beta(t_1)}{t_2-t_1}right| = |beta'(t_1)| = 1 $$
as $beta$ is a unit speed curve.
Combining these two we get
$$ lim_{substack{t_2to t_1\t_2>t_1}} f(t_1,t_2) = lim_{t_2to t_1} frac{beta(t_2)-beta(t_1)}{t_2-t_1} cdot
lim_{substack{t_2to t_1\t_2>t_1}} frac{t_2-t_1}{|beta(t_2)-beta(t_1)|} = beta'(t_1) $$
This means that for every $varepsilon>0$ we can find a $delta>0$ such that
for all $t_2$ with $|t_2-t_1|<delta$ and $(t_1,t_2)in T$ we have
$|f(t_1,t_2)-beta'(t_1)|<varepsilon$.
As $[0,L]$ is compact, we can even, for a given $varepsilon>0$, find a $delta_1>0$
such that for all $(t_1,t_2)in T$ with $|t_2-t_1|<delta_1$ the inequality
$|f(t_1,t_2)-beta'(t_1)|<varepsilon/2$ holds.
Also, as $beta'$ is continuous and $[0,L]$ is compact, we can find a
$delta_2>0$ such that for all $t_1^ast,t_1in[0,L]$ with $|t_1-t_1^ast|<delta_2$ we
have $|beta'(t_1) - beta'(t_1^ast)|<varepsilon/2$.
Now let $P=(t_1,t_2)in T$ be arbitrary with
$|P-(t_1^ast,t_1^ast)|<min{delta_1/2,delta_2}$. That implies
$|t_1-t_1^ast|<delta_2$ and $|t_2-t_1|<delta_1$. By combining the
previous two inequalities we get
$|f(P)-beta'(t_1^ast)|<varepsilon$. We have thus proved that
$f$ is continuous in $(t_1^ast,t_1^ast)$.
The third case is the point $Q=(0,L)$. As $beta$ is a loop, we have:
$$ lim_{tto L} frac{beta(t)-beta(0)}{L-t} = -lim_{tto L} frac{beta(L)-beta(t)}{L-t} = -beta'(L) = -beta'(0) $$
And as a consequence we have:
$$ lim_{substack{tto L\t<L}} frac{|beta(t)-beta(0)|}{L-t} = lim_{substack{tto L}} left|frac{beta(t)-beta(0)}{L-t}right| = |-beta'(0)| = 1 $$
Combining these two we get, as above:
$$ lim_{substack{tto L\t<L}} f(0,t) = -beta'(0) $$
At this point I'm stuck. I think that we need to show that $f$ is uniformly
continuous around $Q$ and that maybe this is the case because $beta$ has
unit speed. If we can prove that, we can approach $Q$ in a way similar to the
second case: starting from a point near enough we first move - by virtue of
uniform continuity - parallel to the $x$ axis until the first component is
zero. Then we move vertically towards $Q$.
EDIT:
Using Ted Shifrin's advice, I've rewritten the second case of the proof.
However, it seems to me that this depends on the Taylor remainder $R$ being
continuous in both variables (see question mark below) which - if justified in
detail - is not much different from what I did above and still doesn't solve
the third case. Or am I missing something?
As $beta$ is differentiable, we can write, for each $t_1in[0,L]$,
$$ beta(t_2)=beta(t_1)+(t_2-t_1)cdotbeta'(t_1) +(t_2-t_1)cdot R(t_1,t_2-t_1) $$
with $lim_{tto0}R(t_1,t)=mathbf0$.
We thus have
$$ lim_{(t_1,t_2)to(t_1^ast,t_1^ast)} frac{beta(t_2)-beta(t_1)}{t_2-t_1} = lim_{(t_1,t_2)to(t_1^ast,t_1^ast)} (beta'(t_1) + R(t_1,t_2-t_1)) stackrel{color{red}?}{=} beta'(t_1^ast) $$
and
$$ lim_{substack{(t_1,t_2)to(t_1^ast,t_1^ast)\t_2>t_1}} frac{|beta(t_2)-beta(t_1)|}{t_2-t_1} = lim_{substack{(t_1,t_2)to(t_1^ast,t_1^ast)\t_2>t_1}} left|frac{beta(t_2)-beta(t_1)}{t_2-t_1}right| = |beta'(t_1^ast)|=1 $$
which eventually leads to
$$ lim_{substack{(t_1,t_2)to(t_1^ast,t_1^ast)\t_2>t_1}} f(t_1,t_2) = beta'(t_1^ast) $$
as above.
real-analysis differential-geometry continuity
asked Jan 22 at 17:03
FrunobulaxFrunobulax
4,7971636
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1
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The proof I've usually suggested is to use a Taylor expansion of $beta$ around $t=t_1$.
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– Ted Shifrin
Jan 22 at 17:31
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@TedShifrin Thanks for your help, but if I understand you correctly, this doesn't solve my problem. I've added an alternative proof based on your suggestion to my question but I think I'm still stuck.
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– Frunobulax
Jan 23 at 14:27
1
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I admit I hadn't read your original post as carefully as I should have. For the main case you added based on my comment, you should use the Taylor expansion at $t=t_1^*$, fixed. I also don't know why you're saying in your original exposition that you have continuity at $P$; it should say at $(t_1^*,t_1^*)$? If you replace the interval $t_2in (L-delta,L]$ with the interval $t_2in (-delta,0]$ the previous diagonal argument should work just fine (if you pay attention to the sign).
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– Ted Shifrin
Jan 23 at 21:13
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@TedShifrin I've now added a proof as an answer which I think should work. I also tried the Taylor expansion with $t=t_1^ast$ fixed, but couldn't make it work. (And, yes, "$P$" was a typo which I've fixed.) Thanks again for your help.
$endgroup$
– Frunobulax
Feb 3 at 15:41
add a comment |
$begingroup$
The standard proof of Hopf's Umlaufsatz proceeds something like
this: We have a unit speed $mathcal{C}^1$ curve $beta:mathbb{R}tomathbb{R}^2$. Furthermore,
$beta$ is a simple loop with period $L$.
We now define $T={(t_1,t_2) in mathbb{R}^2 : 0 leq t_1 leq t_2 leq L }$
and a function $f$ on $T$ as follows:
$$ f(t_1,t_2) =
begin{cases}
beta'(t_1) & t_1=t_2 \
-beta'(0) & (t_1,t_2)=(0,L) \
frac{beta(t_2)-beta(t_1)}{|beta(t_2)-beta(t_1)|} & text{otherwise}
end{cases} $$
It is visually obvious
that $f$ is continuous, but I haven't found a strict proof of this anywhere.
Below is my attempt at a proof, but I think it's ugly and long-winded and it
also isn't complete. My question is how to finish it. I also suspect (or
rather hope) that there is a significantly shorter (or more elegant) proof
that is nevertheless complete and doesn't use hand-waving.
First of all, $f$ is continuous for all $(t_1,t_2)in T$ with $t_1neq t_2$
and $(t_1,t_2)neq(0,L)$ because (as $beta$ is simple) the denominator of
$(beta(t_2)-beta(t_1))/(|beta(t_2)-beta(t_1)|)$ doesn't
vanish and $beta$ and the norm are continuous.
Now for the case $t_1=t_2$. Fix $t_1in[0,L]$. We have
$$ lim_{t_2to t_1} frac{beta(t_2)-beta(t_1)}{t_2-t_1} = beta'(t_1) $$
by definition. This implies
$$ lim_{substack{t_2to t_1\t_2>t_1}} frac{|beta(t_2)-beta(t_1)|}{t_2-t_1} = lim_{substack{t_2to t_1}} left|frac{beta(t_2)-beta(t_1)}{t_2-t_1}right| = |beta'(t_1)| = 1 $$
as $beta$ is a unit speed curve.
Combining these two we get
$$ lim_{substack{t_2to t_1\t_2>t_1}} f(t_1,t_2) = lim_{t_2to t_1} frac{beta(t_2)-beta(t_1)}{t_2-t_1} cdot
lim_{substack{t_2to t_1\t_2>t_1}} frac{t_2-t_1}{|beta(t_2)-beta(t_1)|} = beta'(t_1) $$
This means that for every $varepsilon>0$ we can find a $delta>0$ such that
for all $t_2$ with $|t_2-t_1|<delta$ and $(t_1,t_2)in T$ we have
$|f(t_1,t_2)-beta'(t_1)|<varepsilon$.
As $[0,L]$ is compact, we can even, for a given $varepsilon>0$, find a $delta_1>0$
such that for all $(t_1,t_2)in T$ with $|t_2-t_1|<delta_1$ the inequality
$|f(t_1,t_2)-beta'(t_1)|<varepsilon/2$ holds.
Also, as $beta'$ is continuous and $[0,L]$ is compact, we can find a
$delta_2>0$ such that for all $t_1^ast,t_1in[0,L]$ with $|t_1-t_1^ast|<delta_2$ we
have $|beta'(t_1) - beta'(t_1^ast)|<varepsilon/2$.
Now let $P=(t_1,t_2)in T$ be arbitrary with
$|P-(t_1^ast,t_1^ast)|<min{delta_1/2,delta_2}$. That implies
$|t_1-t_1^ast|<delta_2$ and $|t_2-t_1|<delta_1$. By combining the
previous two inequalities we get
$|f(P)-beta'(t_1^ast)|<varepsilon$. We have thus proved that
$f$ is continuous in $(t_1^ast,t_1^ast)$.
The third case is the point $Q=(0,L)$. As $beta$ is a loop, we have:
$$ lim_{tto L} frac{beta(t)-beta(0)}{L-t} = -lim_{tto L} frac{beta(L)-beta(t)}{L-t} = -beta'(L) = -beta'(0) $$
And as a consequence we have:
$$ lim_{substack{tto L\t<L}} frac{|beta(t)-beta(0)|}{L-t} = lim_{substack{tto L}} left|frac{beta(t)-beta(0)}{L-t}right| = |-beta'(0)| = 1 $$
Combining these two we get, as above:
$$ lim_{substack{tto L\t<L}} f(0,t) = -beta'(0) $$
At this point I'm stuck. I think that we need to show that $f$ is uniformly
continuous around $Q$ and that maybe this is the case because $beta$ has
unit speed. If we can prove that, we can approach $Q$ in a way similar to the
second case: starting from a point near enough we first move - by virtue of
uniform continuity - parallel to the $x$ axis until the first component is
zero. Then we move vertically towards $Q$.
EDIT:
Using Ted Shifrin's advice, I've rewritten the second case of the proof.
However, it seems to me that this depends on the Taylor remainder $R$ being
continuous in both variables (see question mark below) which - if justified in
detail - is not much different from what I did above and still doesn't solve
the third case. Or am I missing something?
As $beta$ is differentiable, we can write, for each $t_1in[0,L]$,
$$ beta(t_2)=beta(t_1)+(t_2-t_1)cdotbeta'(t_1) +(t_2-t_1)cdot R(t_1,t_2-t_1) $$
with $lim_{tto0}R(t_1,t)=mathbf0$.
We thus have
$$ lim_{(t_1,t_2)to(t_1^ast,t_1^ast)} frac{beta(t_2)-beta(t_1)}{t_2-t_1} = lim_{(t_1,t_2)to(t_1^ast,t_1^ast)} (beta'(t_1) + R(t_1,t_2-t_1)) stackrel{color{red}?}{=} beta'(t_1^ast) $$
and
$$ lim_{substack{(t_1,t_2)to(t_1^ast,t_1^ast)\t_2>t_1}} frac{|beta(t_2)-beta(t_1)|}{t_2-t_1} = lim_{substack{(t_1,t_2)to(t_1^ast,t_1^ast)\t_2>t_1}} left|frac{beta(t_2)-beta(t_1)}{t_2-t_1}right| = |beta'(t_1^ast)|=1 $$
which eventually leads to
$$ lim_{substack{(t_1,t_2)to(t_1^ast,t_1^ast)\t_2>t_1}} f(t_1,t_2) = beta'(t_1^ast) $$
as above.
real-analysis differential-geometry continuity
asked Jan 22 at 17:03
FrunobulaxFrunobulax
4,7971636
$endgroup$
The standard proof of Hopf's Umlaufsatz proceeds something like
this: We have a unit speed $mathcal{C}^1$ curve $beta:mathbb{R}tomathbb{R}^2$. Furthermore,
$beta$ is a simple loop with period $L$.
We now define $T={(t_1,t_2) in mathbb{R}^2 : 0 leq t_1 leq t_2 leq L }$
and a function $f$ on $T$ as follows:
$$ f(t_1,t_2) =
begin{cases}
beta'(t_1) & t_1=t_2 \
-beta'(0) & (t_1,t_2)=(0,L) \
frac{beta(t_2)-beta(t_1)}{|beta(t_2)-beta(t_1)|} & text{otherwise}
end{cases} $$
It is visually obvious
that $f$ is continuous, but I haven't found a strict proof of this anywhere.
Below is my attempt at a proof, but I think it's ugly and long-winded and it
also isn't complete. My question is how to finish it. I also suspect (or
rather hope) that there is a significantly shorter (or more elegant) proof
that is nevertheless complete and doesn't use hand-waving.
First of all, $f$ is continuous for all $(t_1,t_2)in T$ with $t_1neq t_2$
and $(t_1,t_2)neq(0,L)$ because (as $beta$ is simple) the denominator of
$(beta(t_2)-beta(t_1))/(|beta(t_2)-beta(t_1)|)$ doesn't
vanish and $beta$ and the norm are continuous.
Now for the case $t_1=t_2$. Fix $t_1in[0,L]$. We have
$$ lim_{t_2to t_1} frac{beta(t_2)-beta(t_1)}{t_2-t_1} = beta'(t_1) $$
by definition. This implies
$$ lim_{substack{t_2to t_1\t_2>t_1}} frac{|beta(t_2)-beta(t_1)|}{t_2-t_1} = lim_{substack{t_2to t_1}} left|frac{beta(t_2)-beta(t_1)}{t_2-t_1}right| = |beta'(t_1)| = 1 $$
as $beta$ is a unit speed curve.
Combining these two we get
$$ lim_{substack{t_2to t_1\t_2>t_1}} f(t_1,t_2) = lim_{t_2to t_1} frac{beta(t_2)-beta(t_1)}{t_2-t_1} cdot
lim_{substack{t_2to t_1\t_2>t_1}} frac{t_2-t_1}{|beta(t_2)-beta(t_1)|} = beta'(t_1) $$
This means that for every $varepsilon>0$ we can find a $delta>0$ such that
for all $t_2$ with $|t_2-t_1|<delta$ and $(t_1,t_2)in T$ we have
$|f(t_1,t_2)-beta'(t_1)|<varepsilon$.
As $[0,L]$ is compact, we can even, for a given $varepsilon>0$, find a $delta_1>0$
such that for all $(t_1,t_2)in T$ with $|t_2-t_1|<delta_1$ the inequality
$|f(t_1,t_2)-beta'(t_1)|<varepsilon/2$ holds.
Also, as $beta'$ is continuous and $[0,L]$ is compact, we can find a
$delta_2>0$ such that for all $t_1^ast,t_1in[0,L]$ with $|t_1-t_1^ast|<delta_2$ we
have $|beta'(t_1) - beta'(t_1^ast)|<varepsilon/2$.
Now let $P=(t_1,t_2)in T$ be arbitrary with
$|P-(t_1^ast,t_1^ast)|<min{delta_1/2,delta_2}$. That implies
$|t_1-t_1^ast|<delta_2$ and $|t_2-t_1|<delta_1$. By combining the
previous two inequalities we get
$|f(P)-beta'(t_1^ast)|<varepsilon$. We have thus proved that
$f$ is continuous in $(t_1^ast,t_1^ast)$.
The third case is the point $Q=(0,L)$. As $beta$ is a loop, we have:
$$ lim_{tto L} frac{beta(t)-beta(0)}{L-t} = -lim_{tto L} frac{beta(L)-beta(t)}{L-t} = -beta'(L) = -beta'(0) $$
And as a consequence we have:
$$ lim_{substack{tto L\t<L}} frac{|beta(t)-beta(0)|}{L-t} = lim_{substack{tto L}} left|frac{beta(t)-beta(0)}{L-t}right| = |-beta'(0)| = 1 $$
Combining these two we get, as above:
$$ lim_{substack{tto L\t<L}} f(0,t) = -beta'(0) $$
At this point I'm stuck. I think that we need to show that $f$ is uniformly
continuous around $Q$ and that maybe this is the case because $beta$ has
unit speed. If we can prove that, we can approach $Q$ in a way similar to the
second case: starting from a point near enough we first move - by virtue of
uniform continuity - parallel to the $x$ axis until the first component is
zero. Then we move vertically towards $Q$.
EDIT:
Using Ted Shifrin's advice, I've rewritten the second case of the proof.
However, it seems to me that this depends on the Taylor remainder $R$ being
continuous in both variables (see question mark below) which - if justified in
detail - is not much different from what I did above and still doesn't solve
the third case. Or am I missing something?
As $beta$ is differentiable, we can write, for each $t_1in[0,L]$,
$$ beta(t_2)=beta(t_1)+(t_2-t_1)cdotbeta'(t_1) +(t_2-t_1)cdot R(t_1,t_2-t_1) $$
with $lim_{tto0}R(t_1,t)=mathbf0$.
We thus have
$$ lim_{(t_1,t_2)to(t_1^ast,t_1^ast)} frac{beta(t_2)-beta(t_1)}{t_2-t_1} = lim_{(t_1,t_2)to(t_1^ast,t_1^ast)} (beta'(t_1) + R(t_1,t_2-t_1)) stackrel{color{red}?}{=} beta'(t_1^ast) $$
and
$$ lim_{substack{(t_1,t_2)to(t_1^ast,t_1^ast)\t_2>t_1}} frac{|beta(t_2)-beta(t_1)|}{t_2-t_1} = lim_{substack{(t_1,t_2)to(t_1^ast,t_1^ast)\t_2>t_1}} left|frac{beta(t_2)-beta(t_1)}{t_2-t_1}right| = |beta'(t_1^ast)|=1 $$
which eventually leads to
$$ lim_{substack{(t_1,t_2)to(t_1^ast,t_1^ast)\t_2>t_1}} f(t_1,t_2) = beta'(t_1^ast) $$
as above.
real-analysis differential-geometry continuity
real-analysis differential-geometry continuity
asked Jan 22 at 17:03
FrunobulaxFrunobulax
4,7971636
asked Jan 22 at 17:03
FrunobulaxFrunobulax
4,7971636
edited Feb 3 at 11:50
Frunobulax
asked Jan 22 at 17:03
FrunobulaxFrunobulax
4,7971636
asked Jan 22 at 17:03
FrunobulaxFrunobulax
4,7971636
asked Jan 22 at 17:03
FrunobulaxFrunobulax
4,7971636
4,7971636
1
$begingroup$
The proof I've usually suggested is to use a Taylor expansion of $beta$ around $t=t_1$.
$endgroup$
– Ted Shifrin
Jan 22 at 17:31
$begingroup$
@TedShifrin Thanks for your help, but if I understand you correctly, this doesn't solve my problem. I've added an alternative proof based on your suggestion to my question but I think I'm still stuck.
$endgroup$
– Frunobulax
Jan 23 at 14:27
1
$begingroup$
I admit I hadn't read your original post as carefully as I should have. For the main case you added based on my comment, you should use the Taylor expansion at $t=t_1^*$, fixed. I also don't know why you're saying in your original exposition that you have continuity at $P$; it should say at $(t_1^*,t_1^*)$? If you replace the interval $t_2in (L-delta,L]$ with the interval $t_2in (-delta,0]$ the previous diagonal argument should work just fine (if you pay attention to the sign).
$endgroup$
– Ted Shifrin
Jan 23 at 21:13
$begingroup$
@TedShifrin I've now added a proof as an answer which I think should work. I also tried the Taylor expansion with $t=t_1^ast$ fixed, but couldn't make it work. (And, yes, "$P$" was a typo which I've fixed.) Thanks again for your help.
$endgroup$
– Frunobulax
Feb 3 at 15:41
add a comment |
1
$begingroup$
The proof I've usually suggested is to use a Taylor expansion of $beta$ around $t=t_1$.
$endgroup$
– Ted Shifrin
Jan 22 at 17:31
$begingroup$
@TedShifrin Thanks for your help, but if I understand you correctly, this doesn't solve my problem. I've added an alternative proof based on your suggestion to my question but I think I'm still stuck.
$endgroup$
– Frunobulax
Jan 23 at 14:27
1
$begingroup$
I admit I hadn't read your original post as carefully as I should have. For the main case you added based on my comment, you should use the Taylor expansion at $t=t_1^*$, fixed. I also don't know why you're saying in your original exposition that you have continuity at $P$; it should say at $(t_1^*,t_1^*)$? If you replace the interval $t_2in (L-delta,L]$ with the interval $t_2in (-delta,0]$ the previous diagonal argument should work just fine (if you pay attention to the sign).
$endgroup$
– Ted Shifrin
Jan 23 at 21:13
$begingroup$
@TedShifrin I've now added a proof as an answer which I think should work. I also tried the Taylor expansion with $t=t_1^ast$ fixed, but couldn't make it work. (And, yes, "$P$" was a typo which I've fixed.) Thanks again for your help.
$endgroup$
– Frunobulax
Feb 3 at 15:41
1
1
$begingroup$
The proof I've usually suggested is to use a Taylor expansion of $beta$ around $t=t_1$.
$endgroup$
– Ted Shifrin
Jan 22 at 17:31
$begingroup$
The proof I've usually suggested is to use a Taylor expansion of $beta$ around $t=t_1$.
$endgroup$
– Ted Shifrin
Jan 22 at 17:31
$begingroup$
@TedShifrin Thanks for your help, but if I understand you correctly, this doesn't solve my problem. I've added an alternative proof based on your suggestion to my question but I think I'm still stuck.
$endgroup$
– Frunobulax
Jan 23 at 14:27
$begingroup$
@TedShifrin Thanks for your help, but if I understand you correctly, this doesn't solve my problem. I've added an alternative proof based on your suggestion to my question but I think I'm still stuck.
$endgroup$
– Frunobulax
Jan 23 at 14:27
1
1
$begingroup$
I admit I hadn't read your original post as carefully as I should have. For the main case you added based on my comment, you should use the Taylor expansion at $t=t_1^*$, fixed. I also don't know why you're saying in your original exposition that you have continuity at $P$; it should say at $(t_1^*,t_1^*)$? If you replace the interval $t_2in (L-delta,L]$ with the interval $t_2in (-delta,0]$ the previous diagonal argument should work just fine (if you pay attention to the sign).
$endgroup$
– Ted Shifrin
Jan 23 at 21:13
$begingroup$
I admit I hadn't read your original post as carefully as I should have. For the main case you added based on my comment, you should use the Taylor expansion at $t=t_1^*$, fixed. I also don't know why you're saying in your original exposition that you have continuity at $P$; it should say at $(t_1^*,t_1^*)$? If you replace the interval $t_2in (L-delta,L]$ with the interval $t_2in (-delta,0]$ the previous diagonal argument should work just fine (if you pay attention to the sign).
$endgroup$
– Ted Shifrin
Jan 23 at 21:13
$begingroup$
@TedShifrin I've now added a proof as an answer which I think should work. I also tried the Taylor expansion with $t=t_1^ast$ fixed, but couldn't make it work. (And, yes, "$P$" was a typo which I've fixed.) Thanks again for your help.
$endgroup$
– Frunobulax
Feb 3 at 15:41
$begingroup$
@TedShifrin I've now added a proof as an answer which I think should work. I also tried the Taylor expansion with $t=t_1^ast$ fixed, but couldn't make it work. (And, yes, "$P$" was a typo which I've fixed.) Thanks again for your help.
$endgroup$
– Frunobulax
Feb 3 at 15:41
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For the sake of completeness, I'll put the whole proof here. The "trick" is to enlarge the domain of $f$ so that both special cases are essentially the same except for the sign.
We have a unit speed $mathcal{C}^1$ curve $beta$. Furthermore,
$beta$ is a simple loop with period $L$.
We now define $T={(t_1,t_2) in mathbb{R}^2 : 0 leq t_1 leq L text{ and } t_1 leq t_2 leq t_1+L }$
and a function $f$ on $T$ as follows:
$$ f(t_1,t_2) =
begin{cases}
beta'(t_1) & t_2=t_1 \
-beta'(t_1) & t_2=t_1+L \
frac{beta(t_2)-beta(t_1)}{|beta(t_2)-beta(t_1)|} & text{otherwise}
end{cases} $$
We want to prove that $f$ is continuous on $T$.
First of all, $f$ is continuous for all $(t_1,t_2)in T$ with
$t_2notin{t_1,t_1+L}$ (the "otherwise" case in the definition) because (as
$beta$ is simple and has period $L$) the denominator of
$(beta(t_2)-beta(t_1))/(|beta(t_2)-beta(t_1)|)$ doesn't
vanish and $beta$ and the norm are continuous.
Now for the case $t_2=t_1$. Fix $t_1in[0,L]$. We have
$$ lim_{t_2to t_1} frac{beta(t_2)-beta(t_1)}{t_2-t_1} = beta'(t_1) $$
by definition. This implies
$$ lim_{substack{t_2to t_1\t_2>t_1}} frac{|beta(t_2)-beta(t_1)|}{t_2-t_1} = lim_{substack{t_2to t_1}} left|frac{beta(t_2)-beta(t_1)}{t_2-t_1}right| = |beta'(t_1)| = 1 $$
as $beta$ is a unit speed curve.
Combining these two we get
$$ lim_{substack{t_2to t_1\t_2>t_1}} f(t_1,t_2) = lim_{t_2to t_1} frac{beta(t_2)-beta(t_1)}{t_2-t_1} cdot
lim_{substack{t_2to t_1\t_2>t_1}} frac{t_2-t_1}{|beta(t_2)-beta(t_1)|} = beta'(t_1) $$
This means that for every $varepsilon>0$ we can find a $delta>0$ such that
for all $t_2$ with $|t_2-t_1|<delta$ and $(t_1,t_2)in T$ we have
$|f(t_1,t_2)-beta'(t_1)|<varepsilon$.
As $[0,L]$ is compact, we can even, for a given $varepsilon>0$, find a $delta_1>0$
such that for textit{all} $(t_1,t_2)in T$ with $|t_2-t_1|<delta_1$ the inequality
$|f(t_1,t_2)-beta'(t_1)|<varepsilon/2$ holds.
Also, as $beta'$ is continuous and $[0,L]$ is compact, we can find a
$delta_2>0$ such that for all $t_1^ast,t_1in[0,L]$ with $|t_1-t_1^ast|<delta_2$ we
have $|beta'(t_1) - beta'(t_1^ast)|<varepsilon/2$.
Now let $P=(t_1,t_2)in T$ be arbitrary with
$|P-(t_1^ast,t_1^ast)|<min{delta_1/2,delta_2}$. That implies
$|t_1-t_1^ast|<delta_2$ and $|t_2-t_1|<delta_1$. By combining the
previous two inequalities we get
$|f(P)-beta'(t_1^ast)|<varepsilon$. We have thus proved that
$f$ is continuous in $(t_1^ast,t_1^ast)$.
The third case is $t_2=t_1+L$. This is almost identical to the second case.
He we have
$$ lim_{t_2to t_1+L} frac{beta(t_2)-beta(t_1+L)}{(t_1+L)-t_2} = -beta'(t_1+L) = -beta'(t_1) $$
because $beta$ has period $L$. And thus:
$$ lim_{substack{t_2to t_1+L\t_2<t_1+L}} frac{|beta(t_2)-beta(t_1+L)|}{(t_1+L)-t_2} = lim_{substack{t_2to t_1+L}} left|frac{beta(t_2)-beta(t_1+L)}{(t_1+L)-t_2}right| = 1 $$
When we combine these two, the only difference is the sign:
$$ lim_{substack{t_2to t_1+L\t_2<t_1+L}} f(t_1+L,t_2) = -beta'(t_1) $$
We can now proceed like above to show that $f$ is continuous for all points of
the form $(t_1^ast,t_1^ast+L)$ and thus in particular at $(0,L)$.
answered Feb 3 at 15:38
FrunobulaxFrunobulax
4,7971636
$endgroup$
add a comment |
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$begingroup$
For the sake of completeness, I'll put the whole proof here. The "trick" is to enlarge the domain of $f$ so that both special cases are essentially the same except for the sign.
We have a unit speed $mathcal{C}^1$ curve $beta$. Furthermore,
$beta$ is a simple loop with period $L$.
We now define $T={(t_1,t_2) in mathbb{R}^2 : 0 leq t_1 leq L text{ and } t_1 leq t_2 leq t_1+L }$
and a function $f$ on $T$ as follows:
$$ f(t_1,t_2) =
begin{cases}
beta'(t_1) & t_2=t_1 \
-beta'(t_1) & t_2=t_1+L \
frac{beta(t_2)-beta(t_1)}{|beta(t_2)-beta(t_1)|} & text{otherwise}
end{cases} $$
We want to prove that $f$ is continuous on $T$.
First of all, $f$ is continuous for all $(t_1,t_2)in T$ with
$t_2notin{t_1,t_1+L}$ (the "otherwise" case in the definition) because (as
$beta$ is simple and has period $L$) the denominator of
$(beta(t_2)-beta(t_1))/(|beta(t_2)-beta(t_1)|)$ doesn't
vanish and $beta$ and the norm are continuous.
Now for the case $t_2=t_1$. Fix $t_1in[0,L]$. We have
$$ lim_{t_2to t_1} frac{beta(t_2)-beta(t_1)}{t_2-t_1} = beta'(t_1) $$
by definition. This implies
$$ lim_{substack{t_2to t_1\t_2>t_1}} frac{|beta(t_2)-beta(t_1)|}{t_2-t_1} = lim_{substack{t_2to t_1}} left|frac{beta(t_2)-beta(t_1)}{t_2-t_1}right| = |beta'(t_1)| = 1 $$
as $beta$ is a unit speed curve.
Combining these two we get
$$ lim_{substack{t_2to t_1\t_2>t_1}} f(t_1,t_2) = lim_{t_2to t_1} frac{beta(t_2)-beta(t_1)}{t_2-t_1} cdot
lim_{substack{t_2to t_1\t_2>t_1}} frac{t_2-t_1}{|beta(t_2)-beta(t_1)|} = beta'(t_1) $$
This means that for every $varepsilon>0$ we can find a $delta>0$ such that
for all $t_2$ with $|t_2-t_1|<delta$ and $(t_1,t_2)in T$ we have
$|f(t_1,t_2)-beta'(t_1)|<varepsilon$.
As $[0,L]$ is compact, we can even, for a given $varepsilon>0$, find a $delta_1>0$
such that for textit{all} $(t_1,t_2)in T$ with $|t_2-t_1|<delta_1$ the inequality
$|f(t_1,t_2)-beta'(t_1)|<varepsilon/2$ holds.
Also, as $beta'$ is continuous and $[0,L]$ is compact, we can find a
$delta_2>0$ such that for all $t_1^ast,t_1in[0,L]$ with $|t_1-t_1^ast|<delta_2$ we
have $|beta'(t_1) - beta'(t_1^ast)|<varepsilon/2$.
Now let $P=(t_1,t_2)in T$ be arbitrary with
$|P-(t_1^ast,t_1^ast)|<min{delta_1/2,delta_2}$. That implies
$|t_1-t_1^ast|<delta_2$ and $|t_2-t_1|<delta_1$. By combining the
previous two inequalities we get
$|f(P)-beta'(t_1^ast)|<varepsilon$. We have thus proved that
$f$ is continuous in $(t_1^ast,t_1^ast)$.
The third case is $t_2=t_1+L$. This is almost identical to the second case.
He we have
$$ lim_{t_2to t_1+L} frac{beta(t_2)-beta(t_1+L)}{(t_1+L)-t_2} = -beta'(t_1+L) = -beta'(t_1) $$
because $beta$ has period $L$. And thus:
$$ lim_{substack{t_2to t_1+L\t_2<t_1+L}} frac{|beta(t_2)-beta(t_1+L)|}{(t_1+L)-t_2} = lim_{substack{t_2to t_1+L}} left|frac{beta(t_2)-beta(t_1+L)}{(t_1+L)-t_2}right| = 1 $$
When we combine these two, the only difference is the sign:
$$ lim_{substack{t_2to t_1+L\t_2<t_1+L}} f(t_1+L,t_2) = -beta'(t_1) $$
We can now proceed like above to show that $f$ is continuous for all points of
the form $(t_1^ast,t_1^ast+L)$ and thus in particular at $(0,L)$.
answered Feb 3 at 15:38
FrunobulaxFrunobulax
4,7971636
$endgroup$
add a comment |
$begingroup$
For the sake of completeness, I'll put the whole proof here. The "trick" is to enlarge the domain of $f$ so that both special cases are essentially the same except for the sign.
We have a unit speed $mathcal{C}^1$ curve $beta$. Furthermore,
$beta$ is a simple loop with period $L$.
We now define $T={(t_1,t_2) in mathbb{R}^2 : 0 leq t_1 leq L text{ and } t_1 leq t_2 leq t_1+L }$
and a function $f$ on $T$ as follows:
$$ f(t_1,t_2) =
begin{cases}
beta'(t_1) & t_2=t_1 \
-beta'(t_1) & t_2=t_1+L \
frac{beta(t_2)-beta(t_1)}{|beta(t_2)-beta(t_1)|} & text{otherwise}
end{cases} $$
We want to prove that $f$ is continuous on $T$.
First of all, $f$ is continuous for all $(t_1,t_2)in T$ with
$t_2notin{t_1,t_1+L}$ (the "otherwise" case in the definition) because (as
$beta$ is simple and has period $L$) the denominator of
$(beta(t_2)-beta(t_1))/(|beta(t_2)-beta(t_1)|)$ doesn't
vanish and $beta$ and the norm are continuous.
Now for the case $t_2=t_1$. Fix $t_1in[0,L]$. We have
$$ lim_{t_2to t_1} frac{beta(t_2)-beta(t_1)}{t_2-t_1} = beta'(t_1) $$
by definition. This implies
$$ lim_{substack{t_2to t_1\t_2>t_1}} frac{|beta(t_2)-beta(t_1)|}{t_2-t_1} = lim_{substack{t_2to t_1}} left|frac{beta(t_2)-beta(t_1)}{t_2-t_1}right| = |beta'(t_1)| = 1 $$
as $beta$ is a unit speed curve.
Combining these two we get
$$ lim_{substack{t_2to t_1\t_2>t_1}} f(t_1,t_2) = lim_{t_2to t_1} frac{beta(t_2)-beta(t_1)}{t_2-t_1} cdot
lim_{substack{t_2to t_1\t_2>t_1}} frac{t_2-t_1}{|beta(t_2)-beta(t_1)|} = beta'(t_1) $$
This means that for every $varepsilon>0$ we can find a $delta>0$ such that
for all $t_2$ with $|t_2-t_1|<delta$ and $(t_1,t_2)in T$ we have
$|f(t_1,t_2)-beta'(t_1)|<varepsilon$.
As $[0,L]$ is compact, we can even, for a given $varepsilon>0$, find a $delta_1>0$
such that for textit{all} $(t_1,t_2)in T$ with $|t_2-t_1|<delta_1$ the inequality
$|f(t_1,t_2)-beta'(t_1)|<varepsilon/2$ holds.
Also, as $beta'$ is continuous and $[0,L]$ is compact, we can find a
$delta_2>0$ such that for all $t_1^ast,t_1in[0,L]$ with $|t_1-t_1^ast|<delta_2$ we
have $|beta'(t_1) - beta'(t_1^ast)|<varepsilon/2$.
Now let $P=(t_1,t_2)in T$ be arbitrary with
$|P-(t_1^ast,t_1^ast)|<min{delta_1/2,delta_2}$. That implies
$|t_1-t_1^ast|<delta_2$ and $|t_2-t_1|<delta_1$. By combining the
previous two inequalities we get
$|f(P)-beta'(t_1^ast)|<varepsilon$. We have thus proved that
$f$ is continuous in $(t_1^ast,t_1^ast)$.
The third case is $t_2=t_1+L$. This is almost identical to the second case.
He we have
$$ lim_{t_2to t_1+L} frac{beta(t_2)-beta(t_1+L)}{(t_1+L)-t_2} = -beta'(t_1+L) = -beta'(t_1) $$
because $beta$ has period $L$. And thus:
$$ lim_{substack{t_2to t_1+L\t_2<t_1+L}} frac{|beta(t_2)-beta(t_1+L)|}{(t_1+L)-t_2} = lim_{substack{t_2to t_1+L}} left|frac{beta(t_2)-beta(t_1+L)}{(t_1+L)-t_2}right| = 1 $$
When we combine these two, the only difference is the sign:
$$ lim_{substack{t_2to t_1+L\t_2<t_1+L}} f(t_1+L,t_2) = -beta'(t_1) $$
We can now proceed like above to show that $f$ is continuous for all points of
the form $(t_1^ast,t_1^ast+L)$ and thus in particular at $(0,L)$.
answered Feb 3 at 15:38
FrunobulaxFrunobulax
4,7971636
$endgroup$
add a comment |
$begingroup$
For the sake of completeness, I'll put the whole proof here. The "trick" is to enlarge the domain of $f$ so that both special cases are essentially the same except for the sign.
We have a unit speed $mathcal{C}^1$ curve $beta$. Furthermore,
$beta$ is a simple loop with period $L$.
We now define $T={(t_1,t_2) in mathbb{R}^2 : 0 leq t_1 leq L text{ and } t_1 leq t_2 leq t_1+L }$
and a function $f$ on $T$ as follows:
$$ f(t_1,t_2) =
begin{cases}
beta'(t_1) & t_2=t_1 \
-beta'(t_1) & t_2=t_1+L \
frac{beta(t_2)-beta(t_1)}{|beta(t_2)-beta(t_1)|} & text{otherwise}
end{cases} $$
We want to prove that $f$ is continuous on $T$.
First of all, $f$ is continuous for all $(t_1,t_2)in T$ with
$t_2notin{t_1,t_1+L}$ (the "otherwise" case in the definition) because (as
$beta$ is simple and has period $L$) the denominator of
$(beta(t_2)-beta(t_1))/(|beta(t_2)-beta(t_1)|)$ doesn't
vanish and $beta$ and the norm are continuous.
Now for the case $t_2=t_1$. Fix $t_1in[0,L]$. We have
$$ lim_{t_2to t_1} frac{beta(t_2)-beta(t_1)}{t_2-t_1} = beta'(t_1) $$
by definition. This implies
$$ lim_{substack{t_2to t_1\t_2>t_1}} frac{|beta(t_2)-beta(t_1)|}{t_2-t_1} = lim_{substack{t_2to t_1}} left|frac{beta(t_2)-beta(t_1)}{t_2-t_1}right| = |beta'(t_1)| = 1 $$
as $beta$ is a unit speed curve.
Combining these two we get
$$ lim_{substack{t_2to t_1\t_2>t_1}} f(t_1,t_2) = lim_{t_2to t_1} frac{beta(t_2)-beta(t_1)}{t_2-t_1} cdot
lim_{substack{t_2to t_1\t_2>t_1}} frac{t_2-t_1}{|beta(t_2)-beta(t_1)|} = beta'(t_1) $$
This means that for every $varepsilon>0$ we can find a $delta>0$ such that
for all $t_2$ with $|t_2-t_1|<delta$ and $(t_1,t_2)in T$ we have
$|f(t_1,t_2)-beta'(t_1)|<varepsilon$.
As $[0,L]$ is compact, we can even, for a given $varepsilon>0$, find a $delta_1>0$
such that for textit{all} $(t_1,t_2)in T$ with $|t_2-t_1|<delta_1$ the inequality
$|f(t_1,t_2)-beta'(t_1)|<varepsilon/2$ holds.
Also, as $beta'$ is continuous and $[0,L]$ is compact, we can find a
$delta_2>0$ such that for all $t_1^ast,t_1in[0,L]$ with $|t_1-t_1^ast|<delta_2$ we
have $|beta'(t_1) - beta'(t_1^ast)|<varepsilon/2$.
Now let $P=(t_1,t_2)in T$ be arbitrary with
$|P-(t_1^ast,t_1^ast)|<min{delta_1/2,delta_2}$. That implies
$|t_1-t_1^ast|<delta_2$ and $|t_2-t_1|<delta_1$. By combining the
previous two inequalities we get
$|f(P)-beta'(t_1^ast)|<varepsilon$. We have thus proved that
$f$ is continuous in $(t_1^ast,t_1^ast)$.
The third case is $t_2=t_1+L$. This is almost identical to the second case.
He we have
$$ lim_{t_2to t_1+L} frac{beta(t_2)-beta(t_1+L)}{(t_1+L)-t_2} = -beta'(t_1+L) = -beta'(t_1) $$
because $beta$ has period $L$. And thus:
$$ lim_{substack{t_2to t_1+L\t_2<t_1+L}} frac{|beta(t_2)-beta(t_1+L)|}{(t_1+L)-t_2} = lim_{substack{t_2to t_1+L}} left|frac{beta(t_2)-beta(t_1+L)}{(t_1+L)-t_2}right| = 1 $$
When we combine these two, the only difference is the sign:
$$ lim_{substack{t_2to t_1+L\t_2<t_1+L}} f(t_1+L,t_2) = -beta'(t_1) $$
We can now proceed like above to show that $f$ is continuous for all points of
the form $(t_1^ast,t_1^ast+L)$ and thus in particular at $(0,L)$.
answered Feb 3 at 15:38
FrunobulaxFrunobulax
4,7971636
$endgroup$
For the sake of completeness, I'll put the whole proof here. The "trick" is to enlarge the domain of $f$ so that both special cases are essentially the same except for the sign.
We have a unit speed $mathcal{C}^1$ curve $beta$. Furthermore,
$beta$ is a simple loop with period $L$.
We now define $T={(t_1,t_2) in mathbb{R}^2 : 0 leq t_1 leq L text{ and } t_1 leq t_2 leq t_1+L }$
and a function $f$ on $T$ as follows:
$$ f(t_1,t_2) =
begin{cases}
beta'(t_1) & t_2=t_1 \
-beta'(t_1) & t_2=t_1+L \
frac{beta(t_2)-beta(t_1)}{|beta(t_2)-beta(t_1)|} & text{otherwise}
end{cases} $$
We want to prove that $f$ is continuous on $T$.
First of all, $f$ is continuous for all $(t_1,t_2)in T$ with
$t_2notin{t_1,t_1+L}$ (the "otherwise" case in the definition) because (as
$beta$ is simple and has period $L$) the denominator of
$(beta(t_2)-beta(t_1))/(|beta(t_2)-beta(t_1)|)$ doesn't
vanish and $beta$ and the norm are continuous.
Now for the case $t_2=t_1$. Fix $t_1in[0,L]$. We have
$$ lim_{t_2to t_1} frac{beta(t_2)-beta(t_1)}{t_2-t_1} = beta'(t_1) $$
by definition. This implies
$$ lim_{substack{t_2to t_1\t_2>t_1}} frac{|beta(t_2)-beta(t_1)|}{t_2-t_1} = lim_{substack{t_2to t_1}} left|frac{beta(t_2)-beta(t_1)}{t_2-t_1}right| = |beta'(t_1)| = 1 $$
as $beta$ is a unit speed curve.
Combining these two we get
$$ lim_{substack{t_2to t_1\t_2>t_1}} f(t_1,t_2) = lim_{t_2to t_1} frac{beta(t_2)-beta(t_1)}{t_2-t_1} cdot
lim_{substack{t_2to t_1\t_2>t_1}} frac{t_2-t_1}{|beta(t_2)-beta(t_1)|} = beta'(t_1) $$
This means that for every $varepsilon>0$ we can find a $delta>0$ such that
for all $t_2$ with $|t_2-t_1|<delta$ and $(t_1,t_2)in T$ we have
$|f(t_1,t_2)-beta'(t_1)|<varepsilon$.
As $[0,L]$ is compact, we can even, for a given $varepsilon>0$, find a $delta_1>0$
such that for textit{all} $(t_1,t_2)in T$ with $|t_2-t_1|<delta_1$ the inequality
$|f(t_1,t_2)-beta'(t_1)|<varepsilon/2$ holds.
Also, as $beta'$ is continuous and $[0,L]$ is compact, we can find a
$delta_2>0$ such that for all $t_1^ast,t_1in[0,L]$ with $|t_1-t_1^ast|<delta_2$ we
have $|beta'(t_1) - beta'(t_1^ast)|<varepsilon/2$.
Now let $P=(t_1,t_2)in T$ be arbitrary with
$|P-(t_1^ast,t_1^ast)|<min{delta_1/2,delta_2}$. That implies
$|t_1-t_1^ast|<delta_2$ and $|t_2-t_1|<delta_1$. By combining the
previous two inequalities we get
$|f(P)-beta'(t_1^ast)|<varepsilon$. We have thus proved that
$f$ is continuous in $(t_1^ast,t_1^ast)$.
The third case is $t_2=t_1+L$. This is almost identical to the second case.
He we have
$$ lim_{t_2to t_1+L} frac{beta(t_2)-beta(t_1+L)}{(t_1+L)-t_2} = -beta'(t_1+L) = -beta'(t_1) $$
because $beta$ has period $L$. And thus:
$$ lim_{substack{t_2to t_1+L\t_2<t_1+L}} frac{|beta(t_2)-beta(t_1+L)|}{(t_1+L)-t_2} = lim_{substack{t_2to t_1+L}} left|frac{beta(t_2)-beta(t_1+L)}{(t_1+L)-t_2}right| = 1 $$
When we combine these two, the only difference is the sign:
$$ lim_{substack{t_2to t_1+L\t_2<t_1+L}} f(t_1+L,t_2) = -beta'(t_1) $$
We can now proceed like above to show that $f$ is continuous for all points of
the form $(t_1^ast,t_1^ast+L)$ and thus in particular at $(0,L)$.
answered Feb 3 at 15:38
FrunobulaxFrunobulax
4,7971636
answered Feb 3 at 15:38
FrunobulaxFrunobulax
4,7971636
answered Feb 3 at 15:38
FrunobulaxFrunobulax
4,7971636
answered Feb 3 at 15:38
FrunobulaxFrunobulax
4,7971636
4,7971636
add a comment |
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1
$begingroup$
The proof I've usually suggested is to use a Taylor expansion of $beta$ around $t=t_1$.
$endgroup$
– Ted Shifrin
Jan 22 at 17:31
$begingroup$
@TedShifrin Thanks for your help, but if I understand you correctly, this doesn't solve my problem. I've added an alternative proof based on your suggestion to my question but I think I'm still stuck.
$endgroup$
– Frunobulax
Jan 23 at 14:27
1
$begingroup$
I admit I hadn't read your original post as carefully as I should have. For the main case you added based on my comment, you should use the Taylor expansion at $t=t_1^*$, fixed. I also don't know why you're saying in your original exposition that you have continuity at $P$; it should say at $(t_1^*,t_1^*)$? If you replace the interval $t_2in (L-delta,L]$ with the interval $t_2in (-delta,0]$ the previous diagonal argument should work just fine (if you pay attention to the sign).
$endgroup$
– Ted Shifrin
Jan 23 at 21:13
$begingroup$
@TedShifrin I've now added a proof as an answer which I think should work. I also tried the Taylor expansion with $t=t_1^ast$ fixed, but couldn't make it work. (And, yes, "$P$" was a typo which I've fixed.) Thanks again for your help.
$endgroup$
– Frunobulax
Feb 3 at 15:41