Approximating a path in the closed disk by a path in the open disk
$begingroup$
Let $gamma: [0,1]to overline{mathbb{D}}$ be a path in the closed unit disk.
For $epsilon>0$, define the following approximation $gamma_{epsilon}$, which lies in the open unit disk:
$$
gamma_{epsilon}(t)= begin{cases}
gamma(t) &text{if } |gamma(t)|leq 1-epsilon\
(1-epsilon)frac{gamma(t)}{|gamma(t)|} &text{otherwise}.
end{cases}
$$
In other words, for the parts of $gamma$ that are outside the $(1-epsilon)$ disk, we project it onto the $(1-epsilon)$-circle.
Suppose $gamma$ is rectifiable.
Is it true that the arclength of $gamma_{epsilon}$ converges to the arclength of $gamma$?
real-analysis complex-analysis
asked Dec 20 '18 at 10:10
MananoManano
3369
$endgroup$
add a comment |
$begingroup$
Let $gamma: [0,1]to overline{mathbb{D}}$ be a path in the closed unit disk.
For $epsilon>0$, define the following approximation $gamma_{epsilon}$, which lies in the open unit disk:
$$
gamma_{epsilon}(t)= begin{cases}
gamma(t) &text{if } |gamma(t)|leq 1-epsilon\
(1-epsilon)frac{gamma(t)}{|gamma(t)|} &text{otherwise}.
end{cases}
$$
In other words, for the parts of $gamma$ that are outside the $(1-epsilon)$ disk, we project it onto the $(1-epsilon)$-circle.
Suppose $gamma$ is rectifiable.
Is it true that the arclength of $gamma_{epsilon}$ converges to the arclength of $gamma$?
real-analysis complex-analysis
asked Dec 20 '18 at 10:10
MananoManano
3369
$endgroup$
add a comment |
$begingroup$
Let $gamma: [0,1]to overline{mathbb{D}}$ be a path in the closed unit disk.
For $epsilon>0$, define the following approximation $gamma_{epsilon}$, which lies in the open unit disk:
$$
gamma_{epsilon}(t)= begin{cases}
gamma(t) &text{if } |gamma(t)|leq 1-epsilon\
(1-epsilon)frac{gamma(t)}{|gamma(t)|} &text{otherwise}.
end{cases}
$$
In other words, for the parts of $gamma$ that are outside the $(1-epsilon)$ disk, we project it onto the $(1-epsilon)$-circle.
Suppose $gamma$ is rectifiable.
Is it true that the arclength of $gamma_{epsilon}$ converges to the arclength of $gamma$?
real-analysis complex-analysis
asked Dec 20 '18 at 10:10
MananoManano
3369
$endgroup$
Let $gamma: [0,1]to overline{mathbb{D}}$ be a path in the closed unit disk.
For $epsilon>0$, define the following approximation $gamma_{epsilon}$, which lies in the open unit disk:
$$
gamma_{epsilon}(t)= begin{cases}
gamma(t) &text{if } |gamma(t)|leq 1-epsilon\
(1-epsilon)frac{gamma(t)}{|gamma(t)|} &text{otherwise}.
end{cases}
$$
In other words, for the parts of $gamma$ that are outside the $(1-epsilon)$ disk, we project it onto the $(1-epsilon)$-circle.
Suppose $gamma$ is rectifiable.
Is it true that the arclength of $gamma_{epsilon}$ converges to the arclength of $gamma$?
real-analysis complex-analysis
real-analysis complex-analysis
asked Dec 20 '18 at 10:10
MananoManano
3369
asked Dec 20 '18 at 10:10
MananoManano
3369
edited Dec 21 '18 at 23:07
Manano
asked Dec 20 '18 at 10:10
MananoManano
3369
asked Dec 20 '18 at 10:10
MananoManano
3369
asked Dec 20 '18 at 10:10
MananoManano
3369
3369
add a comment |
add a comment |
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