Continuous functions on unit discs can be extended to whole plane
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Say $B$ and $D$ be closed and open unit discs in 2 dimensional euclidean plane. I have two doubts.
- Given continuous function $g : B to mathbbR$, there is a continuous function $f : mathbbR ^2 to mathbbR$ such that $f=g$ on $B$.
- Given continuous function $u : D to mathbbR$, there is a continuous function $v : mathbbR ^2 to mathbbR$ such that $ v=u $ on $D$.
I think the first one is doable. What I think is if we define $f$ as $ f(re^iz) = r*f(e^iz)$ outside $B$ and $f=g$ inside $B$, we are done. The same can not be done with the second part. Am I on the right track?
general-topology algebraic-topology continuity metric-spaces
add a comment |
Say $B$ and $D$ be closed and open unit discs in 2 dimensional euclidean plane. I have two doubts.
- Given continuous function $g : B to mathbbR$, there is a continuous function $f : mathbbR ^2 to mathbbR$ such that $f=g$ on $B$.
- Given continuous function $u : D to mathbbR$, there is a continuous function $v : mathbbR ^2 to mathbbR$ such that $ v=u $ on $D$.
I think the first one is doable. What I think is if we define $f$ as $ f(re^iz) = r*f(e^iz)$ outside $B$ and $f=g$ inside $B$, we are done. The same can not be done with the second part. Am I on the right track?
general-topology algebraic-topology continuity metric-spaces
For the second part think of a function that goes to $infty$ as you get close to the border of the disk
– Alessandro Codenotti
Dec 25 '18 at 16:51
You're on the right track. You could also simply let $f(re^iz)=f(e^iz)$ for $rgt1$. You could even let $f(re^iz)=f(e^iz)/r$. But whatever you do, you need to prove that your extended function is continuous.
– Barry Cipra
Dec 25 '18 at 16:52
add a comment |
Say $B$ and $D$ be closed and open unit discs in 2 dimensional euclidean plane. I have two doubts.
- Given continuous function $g : B to mathbbR$, there is a continuous function $f : mathbbR ^2 to mathbbR$ such that $f=g$ on $B$.
- Given continuous function $u : D to mathbbR$, there is a continuous function $v : mathbbR ^2 to mathbbR$ such that $ v=u $ on $D$.
I think the first one is doable. What I think is if we define $f$ as $ f(re^iz) = r*f(e^iz)$ outside $B$ and $f=g$ inside $B$, we are done. The same can not be done with the second part. Am I on the right track?
general-topology algebraic-topology continuity metric-spaces
Say $B$ and $D$ be closed and open unit discs in 2 dimensional euclidean plane. I have two doubts.
- Given continuous function $g : B to mathbbR$, there is a continuous function $f : mathbbR ^2 to mathbbR$ such that $f=g$ on $B$.
- Given continuous function $u : D to mathbbR$, there is a continuous function $v : mathbbR ^2 to mathbbR$ such that $ v=u $ on $D$.
I think the first one is doable. What I think is if we define $f$ as $ f(re^iz) = r*f(e^iz)$ outside $B$ and $f=g$ inside $B$, we are done. The same can not be done with the second part. Am I on the right track?
general-topology algebraic-topology continuity metric-spaces
general-topology algebraic-topology continuity metric-spaces
edited Dec 25 '18 at 16:49
mathcounterexamples.net
25.2k21953
25.2k21953
asked Dec 25 '18 at 16:45
ChakSayantan
1426
1426
For the second part think of a function that goes to $infty$ as you get close to the border of the disk
– Alessandro Codenotti
Dec 25 '18 at 16:51
You're on the right track. You could also simply let $f(re^iz)=f(e^iz)$ for $rgt1$. You could even let $f(re^iz)=f(e^iz)/r$. But whatever you do, you need to prove that your extended function is continuous.
– Barry Cipra
Dec 25 '18 at 16:52
add a comment |
For the second part think of a function that goes to $infty$ as you get close to the border of the disk
– Alessandro Codenotti
Dec 25 '18 at 16:51
You're on the right track. You could also simply let $f(re^iz)=f(e^iz)$ for $rgt1$. You could even let $f(re^iz)=f(e^iz)/r$. But whatever you do, you need to prove that your extended function is continuous.
– Barry Cipra
Dec 25 '18 at 16:52
For the second part think of a function that goes to $infty$ as you get close to the border of the disk
– Alessandro Codenotti
Dec 25 '18 at 16:51
For the second part think of a function that goes to $infty$ as you get close to the border of the disk
– Alessandro Codenotti
Dec 25 '18 at 16:51
You're on the right track. You could also simply let $f(re^iz)=f(e^iz)$ for $rgt1$. You could even let $f(re^iz)=f(e^iz)/r$. But whatever you do, you need to prove that your extended function is continuous.
– Barry Cipra
Dec 25 '18 at 16:52
You're on the right track. You could also simply let $f(re^iz)=f(e^iz)$ for $rgt1$. You could even let $f(re^iz)=f(e^iz)/r$. But whatever you do, you need to prove that your extended function is continuous.
– Barry Cipra
Dec 25 '18 at 16:52
add a comment |
4 Answers
4
active
oldest
votes
Is indeed true.
Is false. Consider $u : (x,y) mapsto frac11-sqrtx^2+y^2$. $u$ cannot be extended to a continuous function on $mathbb R^2$ as it is unbounded on $D$.
add a comment |
As for the first one, you are correct. If $B$ is closed, then $f:Bto mathbbR$ being continuous determines the boundary behavior of $f|_partial D$. Indeed, radially extending as you propose should work fine.
As for the second part, your doubts are well-founded. While the function may be tame on the interior of the domain, it may not extend continuously to the boundary. Take for instance
$$ f(x)=sinleft(frac1x-1right)$$
defined on $[0,1)$. It is quite tame, away from $x=1$, but has an oscillating discontinuity at $1$. Now, define
$$g(x,y)=sinleft(frac1sqrtx^2+y^2-1right)$$
on $D$. This function cannot be extended continuously to any point in $partial D$, let alone to $mathbbR^2$.
add a comment |
Your approach is good. A concrete example for the second part is $f(z)=frac1z$ which tends to $infty$ as $z$ tends to 1 and thus cannot be extended to the unit disk, let alone the whole plane
1
f is a map from R² to R. So f(z) can't be what you said. But f(z) = 1/(|z|-1) would work. Thanks.
– ChakSayantan
Dec 25 '18 at 16:59
Good point @ChakSayantan I will update my answer
– Sorin Tirc
Dec 25 '18 at 17:00
add a comment |
Though it may be a bit advanced, the Tietze extension theorem covers the case of $B$. Your method works just as well, but this would sort of be an assurance that it's possible.
For the open disc, it's easy to find an unbounded continuous function which obviously can't be extended.
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
Is indeed true.
Is false. Consider $u : (x,y) mapsto frac11-sqrtx^2+y^2$. $u$ cannot be extended to a continuous function on $mathbb R^2$ as it is unbounded on $D$.
add a comment |
Is indeed true.
Is false. Consider $u : (x,y) mapsto frac11-sqrtx^2+y^2$. $u$ cannot be extended to a continuous function on $mathbb R^2$ as it is unbounded on $D$.
add a comment |
Is indeed true.
Is false. Consider $u : (x,y) mapsto frac11-sqrtx^2+y^2$. $u$ cannot be extended to a continuous function on $mathbb R^2$ as it is unbounded on $D$.
Is indeed true.
Is false. Consider $u : (x,y) mapsto frac11-sqrtx^2+y^2$. $u$ cannot be extended to a continuous function on $mathbb R^2$ as it is unbounded on $D$.
answered Dec 25 '18 at 16:52
mathcounterexamples.net
25.2k21953
25.2k21953
add a comment |
add a comment |
As for the first one, you are correct. If $B$ is closed, then $f:Bto mathbbR$ being continuous determines the boundary behavior of $f|_partial D$. Indeed, radially extending as you propose should work fine.
As for the second part, your doubts are well-founded. While the function may be tame on the interior of the domain, it may not extend continuously to the boundary. Take for instance
$$ f(x)=sinleft(frac1x-1right)$$
defined on $[0,1)$. It is quite tame, away from $x=1$, but has an oscillating discontinuity at $1$. Now, define
$$g(x,y)=sinleft(frac1sqrtx^2+y^2-1right)$$
on $D$. This function cannot be extended continuously to any point in $partial D$, let alone to $mathbbR^2$.
add a comment |
As for the first one, you are correct. If $B$ is closed, then $f:Bto mathbbR$ being continuous determines the boundary behavior of $f|_partial D$. Indeed, radially extending as you propose should work fine.
As for the second part, your doubts are well-founded. While the function may be tame on the interior of the domain, it may not extend continuously to the boundary. Take for instance
$$ f(x)=sinleft(frac1x-1right)$$
defined on $[0,1)$. It is quite tame, away from $x=1$, but has an oscillating discontinuity at $1$. Now, define
$$g(x,y)=sinleft(frac1sqrtx^2+y^2-1right)$$
on $D$. This function cannot be extended continuously to any point in $partial D$, let alone to $mathbbR^2$.
add a comment |
As for the first one, you are correct. If $B$ is closed, then $f:Bto mathbbR$ being continuous determines the boundary behavior of $f|_partial D$. Indeed, radially extending as you propose should work fine.
As for the second part, your doubts are well-founded. While the function may be tame on the interior of the domain, it may not extend continuously to the boundary. Take for instance
$$ f(x)=sinleft(frac1x-1right)$$
defined on $[0,1)$. It is quite tame, away from $x=1$, but has an oscillating discontinuity at $1$. Now, define
$$g(x,y)=sinleft(frac1sqrtx^2+y^2-1right)$$
on $D$. This function cannot be extended continuously to any point in $partial D$, let alone to $mathbbR^2$.
As for the first one, you are correct. If $B$ is closed, then $f:Bto mathbbR$ being continuous determines the boundary behavior of $f|_partial D$. Indeed, radially extending as you propose should work fine.
As for the second part, your doubts are well-founded. While the function may be tame on the interior of the domain, it may not extend continuously to the boundary. Take for instance
$$ f(x)=sinleft(frac1x-1right)$$
defined on $[0,1)$. It is quite tame, away from $x=1$, but has an oscillating discontinuity at $1$. Now, define
$$g(x,y)=sinleft(frac1sqrtx^2+y^2-1right)$$
on $D$. This function cannot be extended continuously to any point in $partial D$, let alone to $mathbbR^2$.
answered Dec 25 '18 at 16:52
Antonios-Alexandros Robotis
9,56241640
9,56241640
add a comment |
add a comment |
Your approach is good. A concrete example for the second part is $f(z)=frac1z$ which tends to $infty$ as $z$ tends to 1 and thus cannot be extended to the unit disk, let alone the whole plane
1
f is a map from R² to R. So f(z) can't be what you said. But f(z) = 1/(|z|-1) would work. Thanks.
– ChakSayantan
Dec 25 '18 at 16:59
Good point @ChakSayantan I will update my answer
– Sorin Tirc
Dec 25 '18 at 17:00
add a comment |
Your approach is good. A concrete example for the second part is $f(z)=frac1z$ which tends to $infty$ as $z$ tends to 1 and thus cannot be extended to the unit disk, let alone the whole plane
1
f is a map from R² to R. So f(z) can't be what you said. But f(z) = 1/(|z|-1) would work. Thanks.
– ChakSayantan
Dec 25 '18 at 16:59
Good point @ChakSayantan I will update my answer
– Sorin Tirc
Dec 25 '18 at 17:00
add a comment |
Your approach is good. A concrete example for the second part is $f(z)=frac1z$ which tends to $infty$ as $z$ tends to 1 and thus cannot be extended to the unit disk, let alone the whole plane
Your approach is good. A concrete example for the second part is $f(z)=frac1z$ which tends to $infty$ as $z$ tends to 1 and thus cannot be extended to the unit disk, let alone the whole plane
edited Dec 25 '18 at 17:00
answered Dec 25 '18 at 16:50
Sorin Tirc
1,520113
1,520113
1
f is a map from R² to R. So f(z) can't be what you said. But f(z) = 1/(|z|-1) would work. Thanks.
– ChakSayantan
Dec 25 '18 at 16:59
Good point @ChakSayantan I will update my answer
– Sorin Tirc
Dec 25 '18 at 17:00
add a comment |
1
f is a map from R² to R. So f(z) can't be what you said. But f(z) = 1/(|z|-1) would work. Thanks.
– ChakSayantan
Dec 25 '18 at 16:59
Good point @ChakSayantan I will update my answer
– Sorin Tirc
Dec 25 '18 at 17:00
1
1
f is a map from R² to R. So f(z) can't be what you said. But f(z) = 1/(|z|-1) would work. Thanks.
– ChakSayantan
Dec 25 '18 at 16:59
f is a map from R² to R. So f(z) can't be what you said. But f(z) = 1/(|z|-1) would work. Thanks.
– ChakSayantan
Dec 25 '18 at 16:59
Good point @ChakSayantan I will update my answer
– Sorin Tirc
Dec 25 '18 at 17:00
Good point @ChakSayantan I will update my answer
– Sorin Tirc
Dec 25 '18 at 17:00
add a comment |
Though it may be a bit advanced, the Tietze extension theorem covers the case of $B$. Your method works just as well, but this would sort of be an assurance that it's possible.
For the open disc, it's easy to find an unbounded continuous function which obviously can't be extended.
add a comment |
Though it may be a bit advanced, the Tietze extension theorem covers the case of $B$. Your method works just as well, but this would sort of be an assurance that it's possible.
For the open disc, it's easy to find an unbounded continuous function which obviously can't be extended.
add a comment |
Though it may be a bit advanced, the Tietze extension theorem covers the case of $B$. Your method works just as well, but this would sort of be an assurance that it's possible.
For the open disc, it's easy to find an unbounded continuous function which obviously can't be extended.
Though it may be a bit advanced, the Tietze extension theorem covers the case of $B$. Your method works just as well, but this would sort of be an assurance that it's possible.
For the open disc, it's easy to find an unbounded continuous function which obviously can't be extended.
answered Dec 25 '18 at 16:51
Matt Samuel
37.5k63565
37.5k63565
add a comment |
add a comment |
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For the second part think of a function that goes to $infty$ as you get close to the border of the disk
– Alessandro Codenotti
Dec 25 '18 at 16:51
You're on the right track. You could also simply let $f(re^iz)=f(e^iz)$ for $rgt1$. You could even let $f(re^iz)=f(e^iz)/r$. But whatever you do, you need to prove that your extended function is continuous.
– Barry Cipra
Dec 25 '18 at 16:52