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Show that a mapping is a contraction in $ mathbbR^2 $ [closed]
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Consider $ mathbbR^2 $ with the usual Euclidean distance $ d$. I have to show that the equation $ v = M(v)$ where
$$ M(v) = (1,1) + beta(x,y), quad beta in [0,1) $$
for all $ v=(x,y)$, has a unique fixed point $v^star$, by using the contraction mapping theorem (see here). Clearly $ mathbbR^2 $ is complete and $M:mathbbR^2 to mathbbR^2$. But how can I show that $M$ is a contraction?
Using the definition of contraction, I have to show that:
$$ lVert M(v) - M(v^prime) rVert leq beta lVert v - v^prime rVert $$
for some $beta in [0,1) $, and for all $v,v^prime in mathbbR^2$.
contraction-operator
asked Jul 23 at 22:35
Alessandro
142
closed as off-topic by Clayton, ThomasGrubb, Isaac Browne, Shailesh, Claude Leibovici Jul 24 at 8:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Clayton, ThomasGrubb, Isaac Browne, Shailesh, Claude Leibovici
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up vote
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Consider $ mathbbR^2 $ with the usual Euclidean distance $ d$. I have to show that the equation $ v = M(v)$ where
$$ M(v) = (1,1) + beta(x,y), quad beta in [0,1) $$
for all $ v=(x,y)$, has a unique fixed point $v^star$, by using the contraction mapping theorem (see here). Clearly $ mathbbR^2 $ is complete and $M:mathbbR^2 to mathbbR^2$. But how can I show that $M$ is a contraction?
Using the definition of contraction, I have to show that:
$$ lVert M(v) - M(v^prime) rVert leq beta lVert v - v^prime rVert $$
for some $beta in [0,1) $, and for all $v,v^prime in mathbbR^2$.
contraction-operator
asked Jul 23 at 22:35
Alessandro
142
closed as off-topic by Clayton, ThomasGrubb, Isaac Browne, Shailesh, Claude Leibovici Jul 24 at 8:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Clayton, ThomasGrubb, Isaac Browne, Shailesh, Claude Leibovici
2
Have you tried anything at all yet? (Hint: using the definition of a contraction gets you pretty far.....)
– ThomasGrubb
Jul 23 at 23:00
1
@Thomas I computed the fixed point, $ v^star = ( dfrac11-beta, dfrac11-beta) $ :)
– Alessandro
Jul 23 at 23:23
@Alessandro: Nicely done! You've shown that a fixed point (if it exists) has the form $leftlanglefrac11-beta,frac11-betarightrangle.$ Now, you are supposed to use the CMT to prove that a fixed point exists. You've taken an end-around approach, which (while valid) will not get you full credit on your assignment.
– Cameron Buie
Jul 23 at 23:34
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up vote
1
down vote
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up vote
1
down vote
favorite
Consider $ mathbbR^2 $ with the usual Euclidean distance $ d$. I have to show that the equation $ v = M(v)$ where
$$ M(v) = (1,1) + beta(x,y), quad beta in [0,1) $$
for all $ v=(x,y)$, has a unique fixed point $v^star$, by using the contraction mapping theorem (see here). Clearly $ mathbbR^2 $ is complete and $M:mathbbR^2 to mathbbR^2$. But how can I show that $M$ is a contraction?
Using the definition of contraction, I have to show that:
$$ lVert M(v) - M(v^prime) rVert leq beta lVert v - v^prime rVert $$
for some $beta in [0,1) $, and for all $v,v^prime in mathbbR^2$.
contraction-operator
asked Jul 23 at 22:35
Alessandro
142
Consider $ mathbbR^2 $ with the usual Euclidean distance $ d$. I have to show that the equation $ v = M(v)$ where
$$ M(v) = (1,1) + beta(x,y), quad beta in [0,1) $$
for all $ v=(x,y)$, has a unique fixed point $v^star$, by using the contraction mapping theorem (see here). Clearly $ mathbbR^2 $ is complete and $M:mathbbR^2 to mathbbR^2$. But how can I show that $M$ is a contraction?
Using the definition of contraction, I have to show that:
$$ lVert M(v) - M(v^prime) rVert leq beta lVert v - v^prime rVert $$
for some $beta in [0,1) $, and for all $v,v^prime in mathbbR^2$.
contraction-operator
asked Jul 23 at 22:35
Alessandro
142
edited Jul 24 at 15:40
asked Jul 23 at 22:35
Alessandro
142
asked Jul 23 at 22:35
Alessandro
142
asked Jul 23 at 22:35
Alessandro
142
142
closed as off-topic by Clayton, ThomasGrubb, Isaac Browne, Shailesh, Claude Leibovici Jul 24 at 8:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Clayton, ThomasGrubb, Isaac Browne, Shailesh, Claude Leibovici
closed as off-topic by Clayton, ThomasGrubb, Isaac Browne, Shailesh, Claude Leibovici Jul 24 at 8:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Clayton, ThomasGrubb, Isaac Browne, Shailesh, Claude Leibovici
2
Have you tried anything at all yet? (Hint: using the definition of a contraction gets you pretty far.....)
– ThomasGrubb
Jul 23 at 23:00
1
@Thomas I computed the fixed point, $ v^star = ( dfrac11-beta, dfrac11-beta) $ :)
– Alessandro
Jul 23 at 23:23
@Alessandro: Nicely done! You've shown that a fixed point (if it exists) has the form $leftlanglefrac11-beta,frac11-betarightrangle.$ Now, you are supposed to use the CMT to prove that a fixed point exists. You've taken an end-around approach, which (while valid) will not get you full credit on your assignment.
– Cameron Buie
Jul 23 at 23:34
add a comment |Â
2
Have you tried anything at all yet? (Hint: using the definition of a contraction gets you pretty far.....)
– ThomasGrubb
Jul 23 at 23:00
1
@Thomas I computed the fixed point, $ v^star = ( dfrac11-beta, dfrac11-beta) $ :)
– Alessandro
Jul 23 at 23:23
@Alessandro: Nicely done! You've shown that a fixed point (if it exists) has the form $leftlanglefrac11-beta,frac11-betarightrangle.$ Now, you are supposed to use the CMT to prove that a fixed point exists. You've taken an end-around approach, which (while valid) will not get you full credit on your assignment.
– Cameron Buie
Jul 23 at 23:34
2
2
Have you tried anything at all yet? (Hint: using the definition of a contraction gets you pretty far.....)
– ThomasGrubb
Jul 23 at 23:00
Have you tried anything at all yet? (Hint: using the definition of a contraction gets you pretty far.....)
– ThomasGrubb
Jul 23 at 23:00
1
1
@Thomas I computed the fixed point, $ v^star = ( dfrac11-beta, dfrac11-beta) $ :)
– Alessandro
Jul 23 at 23:23
@Thomas I computed the fixed point, $ v^star = ( dfrac11-beta, dfrac11-beta) $ :)
– Alessandro
Jul 23 at 23:23
@Alessandro: Nicely done! You've shown that a fixed point (if it exists) has the form $leftlanglefrac11-beta,frac11-betarightrangle.$ Now, you are supposed to use the CMT to prove that a fixed point exists. You've taken an end-around approach, which (while valid) will not get you full credit on your assignment.
– Cameron Buie
Jul 23 at 23:34
@Alessandro: Nicely done! You've shown that a fixed point (if it exists) has the form $leftlanglefrac11-beta,frac11-betarightrangle.$ Now, you are supposed to use the CMT to prove that a fixed point exists. You've taken an end-around approach, which (while valid) will not get you full credit on your assignment.
– Cameron Buie
Jul 23 at 23:34
add a comment |Â
1 Answer
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It was pretty easy :)
Let $ lVert cdot rVert $ denote the Euclidean distance between two vectors. Then
$$ lVert M(v) - M(v^prime) rVert = lVert beta (v_1,v_2)- beta(v^prime_1,v^prime_2) rVert = beta lVert v - v^prime rVert $$
This shows that $M$ is a contraction with modulus $ beta $. Therefore I can apply the CMT, as suggested above, and this buys me the fixed point.
Thanks @Thomas for the advice!
answered Jul 24 at 0:47
Alessandro
142
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
It was pretty easy :)
Let $ lVert cdot rVert $ denote the Euclidean distance between two vectors. Then
$$ lVert M(v) - M(v^prime) rVert = lVert beta (v_1,v_2)- beta(v^prime_1,v^prime_2) rVert = beta lVert v - v^prime rVert $$
This shows that $M$ is a contraction with modulus $ beta $. Therefore I can apply the CMT, as suggested above, and this buys me the fixed point.
Thanks @Thomas for the advice!
answered Jul 24 at 0:47
Alessandro
142
add a comment |Â
up vote
0
down vote
It was pretty easy :)
Let $ lVert cdot rVert $ denote the Euclidean distance between two vectors. Then
$$ lVert M(v) - M(v^prime) rVert = lVert beta (v_1,v_2)- beta(v^prime_1,v^prime_2) rVert = beta lVert v - v^prime rVert $$
This shows that $M$ is a contraction with modulus $ beta $. Therefore I can apply the CMT, as suggested above, and this buys me the fixed point.
Thanks @Thomas for the advice!
answered Jul 24 at 0:47
Alessandro
142
add a comment |Â
up vote
0
down vote
up vote
0
down vote
It was pretty easy :)
Let $ lVert cdot rVert $ denote the Euclidean distance between two vectors. Then
$$ lVert M(v) - M(v^prime) rVert = lVert beta (v_1,v_2)- beta(v^prime_1,v^prime_2) rVert = beta lVert v - v^prime rVert $$
This shows that $M$ is a contraction with modulus $ beta $. Therefore I can apply the CMT, as suggested above, and this buys me the fixed point.
Thanks @Thomas for the advice!
answered Jul 24 at 0:47
Alessandro
142
It was pretty easy :)
Let $ lVert cdot rVert $ denote the Euclidean distance between two vectors. Then
$$ lVert M(v) - M(v^prime) rVert = lVert beta (v_1,v_2)- beta(v^prime_1,v^prime_2) rVert = beta lVert v - v^prime rVert $$
This shows that $M$ is a contraction with modulus $ beta $. Therefore I can apply the CMT, as suggested above, and this buys me the fixed point.
Thanks @Thomas for the advice!
answered Jul 24 at 0:47
Alessandro
142
answered Jul 24 at 0:47
Alessandro
142
answered Jul 24 at 0:47
Alessandro
142
answered Jul 24 at 0:47
Alessandro
142
142
add a comment |Â
add a comment |Â
2
Have you tried anything at all yet? (Hint: using the definition of a contraction gets you pretty far.....)
– ThomasGrubb
Jul 23 at 23:00
1
@Thomas I computed the fixed point, $ v^star = ( dfrac11-beta, dfrac11-beta) $ :)
– Alessandro
Jul 23 at 23:23
@Alessandro: Nicely done! You've shown that a fixed point (if it exists) has the form $leftlanglefrac11-beta,frac11-betarightrangle.$ Now, you are supposed to use the CMT to prove that a fixed point exists. You've taken an end-around approach, which (while valid) will not get you full credit on your assignment.
– Cameron Buie
Jul 23 at 23:34