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 9 hours ago
rapidracim
1,4091319
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 9 hours ago
rapidracim
1,4091319
You have not specified what kind of function $K(x,y)$ is what $|K|$ stands for.
– Kavi Rama Murthy
8 hours ago
You need to fix the notation! You're using "$K$" for two different things...
– David C. Ullrich
5 hours 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 9 hours ago
rapidracim
1,4091319
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 9 hours ago
rapidracim
1,4091319
asked 9 hours ago
rapidracim
1,4091319
asked 9 hours ago
rapidracim
1,4091319
asked 9 hours ago
rapidracim
1,4091319
asked 9 hours ago
rapidracim
1,4091319
1,4091319
You have not specified what kind of function $K(x,y)$ is what $|K|$ stands for.
– Kavi Rama Murthy
8 hours ago
You need to fix the notation! You're using "$K$" for two different things...
– David C. Ullrich
5 hours ago
add a comment |
You have not specified what kind of function $K(x,y)$ is what $|K|$ stands for.
– Kavi Rama Murthy
8 hours ago
You need to fix the notation! You're using "$K$" for two different things...
– David C. Ullrich
5 hours ago
You have not specified what kind of function $K(x,y)$ is what $|K|$ stands for.
– Kavi Rama Murthy
8 hours ago
You have not specified what kind of function $K(x,y)$ is what $|K|$ stands for.
– Kavi Rama Murthy
8 hours ago
You need to fix the notation! You're using "$K$" for two different things...
– David C. Ullrich
5 hours ago
You need to fix the notation! You're using "$K$" for two different things...
– David C. Ullrich
5 hours ago
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 8 hours ago
Julián Aguirre
67.5k24094
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
$$
f_g = (I+K)^{-1}g,quad |f_g|leq |(I+K)^{-1}||g|.
$$
answered 7 hours ago
Song
4,225316
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3052823%2fis-there-a-fixed-point-theorem-i-could-use-to-solve-this-problem%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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 8 hours ago
Julián Aguirre
67.5k24094
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 8 hours ago
Julián Aguirre
67.5k24094
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 8 hours ago
Julián Aguirre
67.5k24094
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 8 hours ago
Julián Aguirre
67.5k24094
answered 8 hours ago
Julián Aguirre
67.5k24094
answered 8 hours ago
Julián Aguirre
67.5k24094
answered 8 hours ago
Julián Aguirre
67.5k24094
67.5k24094
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
$$
f_g = (I+K)^{-1}g,quad |f_g|leq |(I+K)^{-1}||g|.
$$
answered 7 hours ago
Song
4,225316
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
$$
f_g = (I+K)^{-1}g,quad |f_g|leq |(I+K)^{-1}||g|.
$$
answered 7 hours ago
Song
4,225316
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
$$
f_g = (I+K)^{-1}g,quad |f_g|leq |(I+K)^{-1}||g|.
$$
answered 7 hours ago
Song
4,225316
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
$$
f_g = (I+K)^{-1}g,quad |f_g|leq |(I+K)^{-1}||g|.
$$
answered 7 hours ago
Song
4,225316
answered 7 hours ago
Song
4,225316
answered 7 hours ago
Song
4,225316
answered 7 hours ago
Song
4,225316
4,225316
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3052823%2fis-there-a-fixed-point-theorem-i-could-use-to-solve-this-problem%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
You have not specified what kind of function $K(x,y)$ is what $|K|$ stands for.
– Kavi Rama Murthy
8 hours ago
You need to fix the notation! You're using "$K$" for two different things...
– David C. Ullrich
5 hours ago