Is there a fixed point theorem I could use to solve this problem?
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 1 hour ago
rapidracim
1,3961319
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 1 hour ago
rapidracim
1,3961319
You have not specified what kind of function $K(x,y)$ is what $|K|$ stands for.
– Kavi Rama Murthy
11 mins ago
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 1 hour ago
rapidracim
1,3961319
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 1 hour ago
rapidracim
1,3961319
asked 1 hour ago
rapidracim
1,3961319
asked 1 hour ago
rapidracim
1,3961319
asked 1 hour ago
rapidracim
1,3961319
asked 1 hour ago
rapidracim
1,3961319
1,3961319
You have not specified what kind of function $K(x,y)$ is what $|K|$ stands for.
– Kavi Rama Murthy
11 mins ago
add a comment |
You have not specified what kind of function $K(x,y)$ is what $|K|$ stands for.
– Kavi Rama Murthy
11 mins ago
You have not specified what kind of function $K(x,y)$ is what $|K|$ stands for.
– Kavi Rama Murthy
11 mins ago
You have not specified what kind of function $K(x,y)$ is what $|K|$ stands for.
– Kavi Rama Murthy
11 mins ago
add a comment |
1 Answer
1
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 59 mins ago
Julián Aguirre
67.4k24094
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
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 59 mins ago
Julián Aguirre
67.4k24094
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 59 mins ago
Julián Aguirre
67.4k24094
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 59 mins ago
Julián Aguirre
67.4k24094
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 59 mins ago
Julián Aguirre
67.4k24094
answered 59 mins ago
Julián Aguirre
67.4k24094
answered 59 mins ago
Julián Aguirre
67.4k24094
answered 59 mins ago
Julián Aguirre
67.4k24094
67.4k24094
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
11 mins ago