Is there a fixed point theorem I could use to solve this problem?
Clash Royale CLAN TAG#URR8PPP
let $E = C([0,1]),,,$ $K : E to E, ,,
(Kf)(x) = int_0^1K(x,y)f(y)dy$
also $|K| leq a < 1$
I want to prove that there for $g in E$ there exists a unique $f_g in E$ that satisfies the following equation :
$f_g + Kf_g = g$
which is equivalent to showing that $T : E to E,,,T(f) = g-Kf$ has a fixed point.
with what I have in hands I feel like there must be some theorem I'm missing.
any help will be greatly appreciated !
functional-analysis fixed-point-theorems
asked Dec 26 '18 at 10:38
rapidracimrapidracim
1,4361319
add a comment |
let $E = C([0,1]),,,$ $K : E to E, ,,
(Kf)(x) = int_0^1K(x,y)f(y)dy$
also $|K| leq a < 1$
I want to prove that there for $g in E$ there exists a unique $f_g in E$ that satisfies the following equation :
$f_g + Kf_g = g$
which is equivalent to showing that $T : E to E,,,T(f) = g-Kf$ has a fixed point.
with what I have in hands I feel like there must be some theorem I'm missing.
any help will be greatly appreciated !
functional-analysis fixed-point-theorems
asked Dec 26 '18 at 10:38
rapidracimrapidracim
1,4361319
You have not specified what kind of function $K(x,y)$ is what $|K|$ stands for.
– Kavi Rama Murthy
Dec 26 '18 at 11:53
You need to fix the notation! You're using "$K$" for two different things...
– David C. Ullrich
Dec 26 '18 at 14:30
add a comment |
let $E = C([0,1]),,,$ $K : E to E, ,,
(Kf)(x) = int_0^1K(x,y)f(y)dy$
also $|K| leq a < 1$
I want to prove that there for $g in E$ there exists a unique $f_g in E$ that satisfies the following equation :
$f_g + Kf_g = g$
which is equivalent to showing that $T : E to E,,,T(f) = g-Kf$ has a fixed point.
with what I have in hands I feel like there must be some theorem I'm missing.
any help will be greatly appreciated !
functional-analysis fixed-point-theorems
asked Dec 26 '18 at 10:38
rapidracimrapidracim
1,4361319
let $E = C([0,1]),,,$ $K : E to E, ,,
(Kf)(x) = int_0^1K(x,y)f(y)dy$
also $|K| leq a < 1$
I want to prove that there for $g in E$ there exists a unique $f_g in E$ that satisfies the following equation :
$f_g + Kf_g = g$
which is equivalent to showing that $T : E to E,,,T(f) = g-Kf$ has a fixed point.
with what I have in hands I feel like there must be some theorem I'm missing.
any help will be greatly appreciated !
functional-analysis fixed-point-theorems
functional-analysis fixed-point-theorems
asked Dec 26 '18 at 10:38
rapidracimrapidracim
1,4361319
asked Dec 26 '18 at 10:38
rapidracimrapidracim
1,4361319
asked Dec 26 '18 at 10:38
rapidracimrapidracim
1,4361319
asked Dec 26 '18 at 10:38
rapidracimrapidracim
1,4361319
asked Dec 26 '18 at 10:38
rapidracimrapidracim
1,4361319
1,4361319
You have not specified what kind of function $K(x,y)$ is what $|K|$ stands for.
– Kavi Rama Murthy
Dec 26 '18 at 11:53
You need to fix the notation! You're using "$K$" for two different things...
– David C. Ullrich
Dec 26 '18 at 14:30
add a comment |
You have not specified what kind of function $K(x,y)$ is what $|K|$ stands for.
– Kavi Rama Murthy
Dec 26 '18 at 11:53
You need to fix the notation! You're using "$K$" for two different things...
– David C. Ullrich
Dec 26 '18 at 14:30
You have not specified what kind of function $K(x,y)$ is what $|K|$ stands for.
– Kavi Rama Murthy
Dec 26 '18 at 11:53
You have not specified what kind of function $K(x,y)$ is what $|K|$ stands for.
– Kavi Rama Murthy
Dec 26 '18 at 11:53
You need to fix the notation! You're using "$K$" for two different things...
– David C. Ullrich
Dec 26 '18 at 14:30
You need to fix the notation! You're using "$K$" for two different things...
– David C. Ullrich
Dec 26 '18 at 14:30
add a comment |
2 Answers
2
active
oldest
votes
You can apply the Contraction mapping, a.k.a. Banach's Fixed Point Theorem. Given $f,hin C([0,1])$,
$$
|Tf-Th|leint_0^1|K(x,y)|,|f(y)-h(y)|,dyle|K|,|f-h|<a,|f-h|,
$$
with $0<a<1$.
answered Dec 26 '18 at 11:05
Julián AguirreJulián Aguirre
67.7k24094
add a comment |
This is not a fixed point theorem, but it is well-known that if $T:Eto E$ is a bounded linear operator with $|I-T|<1,$ then $T$ has a bounded inverse
$$
T^-1=sum_k=0^infty (I-T)^k.
$$ In your case, since $|I-(I+K)|<1$, we have
$$
f_g = (I+K)^-1g,quad |f_g|leq |(I+K)^-1||g|.
$$
answered Dec 26 '18 at 12:38
SongSong
6,040318
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can apply the Contraction mapping, a.k.a. Banach's Fixed Point Theorem. Given $f,hin C([0,1])$,
$$
|Tf-Th|leint_0^1|K(x,y)|,|f(y)-h(y)|,dyle|K|,|f-h|<a,|f-h|,
$$
with $0<a<1$.
answered Dec 26 '18 at 11:05
Julián AguirreJulián Aguirre
67.7k24094
add a comment |
You can apply the Contraction mapping, a.k.a. Banach's Fixed Point Theorem. Given $f,hin C([0,1])$,
$$
|Tf-Th|leint_0^1|K(x,y)|,|f(y)-h(y)|,dyle|K|,|f-h|<a,|f-h|,
$$
with $0<a<1$.
answered Dec 26 '18 at 11:05
Julián AguirreJulián Aguirre
67.7k24094
add a comment |
You can apply the Contraction mapping, a.k.a. Banach's Fixed Point Theorem. Given $f,hin C([0,1])$,
$$
|Tf-Th|leint_0^1|K(x,y)|,|f(y)-h(y)|,dyle|K|,|f-h|<a,|f-h|,
$$
with $0<a<1$.
answered Dec 26 '18 at 11:05
Julián AguirreJulián Aguirre
67.7k24094
You can apply the Contraction mapping, a.k.a. Banach's Fixed Point Theorem. Given $f,hin C([0,1])$,
$$
|Tf-Th|leint_0^1|K(x,y)|,|f(y)-h(y)|,dyle|K|,|f-h|<a,|f-h|,
$$
with $0<a<1$.
answered Dec 26 '18 at 11:05
Julián AguirreJulián Aguirre
67.7k24094
answered Dec 26 '18 at 11:05
Julián AguirreJulián Aguirre
67.7k24094
answered Dec 26 '18 at 11:05
Julián AguirreJulián Aguirre
67.7k24094
answered Dec 26 '18 at 11:05
Julián AguirreJulián Aguirre
67.7k24094
67.7k24094
add a comment |
add a comment |
This is not a fixed point theorem, but it is well-known that if $T:Eto E$ is a bounded linear operator with $|I-T|<1,$ then $T$ has a bounded inverse
$$
T^-1=sum_k=0^infty (I-T)^k.
$$ In your case, since $|I-(I+K)|<1$, we have
$$
f_g = (I+K)^-1g,quad |f_g|leq |(I+K)^-1||g|.
$$
answered Dec 26 '18 at 12:38
SongSong
6,040318
add a comment |
This is not a fixed point theorem, but it is well-known that if $T:Eto E$ is a bounded linear operator with $|I-T|<1,$ then $T$ has a bounded inverse
$$
T^-1=sum_k=0^infty (I-T)^k.
$$ In your case, since $|I-(I+K)|<1$, we have
$$
f_g = (I+K)^-1g,quad |f_g|leq |(I+K)^-1||g|.
$$
answered Dec 26 '18 at 12:38
SongSong
6,040318
add a comment |
This is not a fixed point theorem, but it is well-known that if $T:Eto E$ is a bounded linear operator with $|I-T|<1,$ then $T$ has a bounded inverse
$$
T^-1=sum_k=0^infty (I-T)^k.
$$ In your case, since $|I-(I+K)|<1$, we have
$$
f_g = (I+K)^-1g,quad |f_g|leq |(I+K)^-1||g|.
$$
answered Dec 26 '18 at 12:38
SongSong
6,040318
This is not a fixed point theorem, but it is well-known that if $T:Eto E$ is a bounded linear operator with $|I-T|<1,$ then $T$ has a bounded inverse
$$
T^-1=sum_k=0^infty (I-T)^k.
$$ In your case, since $|I-(I+K)|<1$, we have
$$
f_g = (I+K)^-1g,quad |f_g|leq |(I+K)^-1||g|.
$$
answered Dec 26 '18 at 12:38
SongSong
6,040318
answered Dec 26 '18 at 12:38
SongSong
6,040318
answered Dec 26 '18 at 12:38
SongSong
6,040318
answered Dec 26 '18 at 12:38
SongSong
6,040318
6,040318
add a comment |
add a comment |
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You have not specified what kind of function $K(x,y)$ is what $|K|$ stands for.
– Kavi Rama Murthy
Dec 26 '18 at 11:53
You need to fix the notation! You're using "$K$" for two different things...
– David C. Ullrich
Dec 26 '18 at 14:30