Mixing
How to show that the set of functions $E={f(0)=f(1)=0}$ is closed in $C[0, 1]$ with respect to the $infty$...
$begingroup$
To do it I need to show that all limit points of $E$ are inside $C[0, 1]$. How can it be showed?
One option is to show that addition of $E$ is an open set.
The second is to build some sequence which has a limit in $E$, I think.
But how can it be done correctly?
The first steps of my solution using the second option, so we need to prove that $E$ contains all its limit points:
1) Using the theorem: $f_0(x)$ is a limit point of the set $E subset C[0, 1] iff exists {f_n} subset E: f_n to f_0$ as $n to infty$, and $f_n neq f_0 forall n in mathbb N $.
2) Now we need to show that $f_0 in E$. And here I've stopped.
functional-analysis
edited Jan 14 at 21:19
postmortes
2,04121119
asked Dec 26 '18 at 3:31
shpindlershpindler
172
$endgroup$
add a comment |
$begingroup$
To do it I need to show that all limit points of $E$ are inside $C[0, 1]$. How can it be showed?
One option is to show that addition of $E$ is an open set.
The second is to build some sequence which has a limit in $E$, I think.
But how can it be done correctly?
The first steps of my solution using the second option, so we need to prove that $E$ contains all its limit points:
1) Using the theorem: $f_0(x)$ is a limit point of the set $E subset C[0, 1] iff exists {f_n} subset E: f_n to f_0$ as $n to infty$, and $f_n neq f_0 forall n in mathbb N $.
2) Now we need to show that $f_0 in E$. And here I've stopped.
functional-analysis
edited Jan 14 at 21:19
postmortes
2,04121119
asked Dec 26 '18 at 3:31
shpindlershpindler
172
$endgroup$
add a comment |
$begingroup$
To do it I need to show that all limit points of $E$ are inside $C[0, 1]$. How can it be showed?
One option is to show that addition of $E$ is an open set.
The second is to build some sequence which has a limit in $E$, I think.
But how can it be done correctly?
The first steps of my solution using the second option, so we need to prove that $E$ contains all its limit points:
1) Using the theorem: $f_0(x)$ is a limit point of the set $E subset C[0, 1] iff exists {f_n} subset E: f_n to f_0$ as $n to infty$, and $f_n neq f_0 forall n in mathbb N $.
2) Now we need to show that $f_0 in E$. And here I've stopped.
functional-analysis
edited Jan 14 at 21:19
postmortes
2,04121119
asked Dec 26 '18 at 3:31
shpindlershpindler
172
$endgroup$
To do it I need to show that all limit points of $E$ are inside $C[0, 1]$. How can it be showed?
One option is to show that addition of $E$ is an open set.
The second is to build some sequence which has a limit in $E$, I think.
But how can it be done correctly?
The first steps of my solution using the second option, so we need to prove that $E$ contains all its limit points:
1) Using the theorem: $f_0(x)$ is a limit point of the set $E subset C[0, 1] iff exists {f_n} subset E: f_n to f_0$ as $n to infty$, and $f_n neq f_0 forall n in mathbb N $.
2) Now we need to show that $f_0 in E$. And here I've stopped.
functional-analysis
functional-analysis
edited Jan 14 at 21:19
postmortes
2,04121119
asked Dec 26 '18 at 3:31
shpindlershpindler
172
edited Jan 14 at 21:19
postmortes
2,04121119
asked Dec 26 '18 at 3:31
shpindlershpindler
172
edited Jan 14 at 21:19
postmortes
2,04121119
edited Jan 14 at 21:19
postmortes
2,04121119
edited Jan 14 at 21:19
postmortes
2,04121119
2,04121119
asked Dec 26 '18 at 3:31
shpindlershpindler
172
asked Dec 26 '18 at 3:31
shpindlershpindler
172
asked Dec 26 '18 at 3:31
shpindlershpindler
172
172
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You have a sequence $(f_n)_n$ of continuous functions on $[0,1]$ such that $f_n(0) =0$ and $f_n(1)=0$. You know that $f_n to f_0$, where $f_0 in C([0,1])$ and the symbol $to$ means convergence in supermum norm.
You have to prove that $f_0 (0) =f_0(1) =0$. This follows from the fact that $f_n to f_0$. Can you explain why?
answered Dec 26 '18 at 12:26
HermioneHermione
19619
$endgroup$
$begingroup$
I would add for completeness that $to$ means converges in supremum norm.
$endgroup$
– Jonas Lenz
Dec 26 '18 at 12:29
$begingroup$
Your are right, now it is more clear. Thanks!
$endgroup$
– Hermione
Dec 26 '18 at 12:34
$begingroup$
@Hermione Am I right that since $max_{x in [0, 1]}|f_n(x) - f_0(x)| < epsilon$ $forall epsilon > 0$ $Rightarrow$ if x = 0 then $|f_0(0)| < epsilon$ $forall epsilon > 0$ $Rightarrow$ $f_0(0) = 0$. Analogically for x = 1. Therefore $f_0(0) = f_0(1) = 0$ $Rightarrow$ $f_0 in E$. So E is a closed set.
$endgroup$
– shpindler
Dec 26 '18 at 15:25
$begingroup$
Yes @shpindler, you are right
$endgroup$
– Hermione
Dec 26 '18 at 17:05
$begingroup$
@Hermione, thanks!
$endgroup$
– shpindler
Dec 26 '18 at 17:24
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%2f3052617%2fhow-to-show-that-the-set-of-functions-e-f0-f1-0-is-closed-in-c0-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You have a sequence $(f_n)_n$ of continuous functions on $[0,1]$ such that $f_n(0) =0$ and $f_n(1)=0$. You know that $f_n to f_0$, where $f_0 in C([0,1])$ and the symbol $to$ means convergence in supermum norm.
You have to prove that $f_0 (0) =f_0(1) =0$. This follows from the fact that $f_n to f_0$. Can you explain why?
answered Dec 26 '18 at 12:26
HermioneHermione
19619
$endgroup$
$begingroup$
I would add for completeness that $to$ means converges in supremum norm.
$endgroup$
– Jonas Lenz
Dec 26 '18 at 12:29
$begingroup$
Your are right, now it is more clear. Thanks!
$endgroup$
– Hermione
Dec 26 '18 at 12:34
$begingroup$
@Hermione Am I right that since $max_{x in [0, 1]}|f_n(x) - f_0(x)| < epsilon$ $forall epsilon > 0$ $Rightarrow$ if x = 0 then $|f_0(0)| < epsilon$ $forall epsilon > 0$ $Rightarrow$ $f_0(0) = 0$. Analogically for x = 1. Therefore $f_0(0) = f_0(1) = 0$ $Rightarrow$ $f_0 in E$. So E is a closed set.
$endgroup$
– shpindler
Dec 26 '18 at 15:25
$begingroup$
Yes @shpindler, you are right
$endgroup$
– Hermione
Dec 26 '18 at 17:05
$begingroup$
@Hermione, thanks!
$endgroup$
– shpindler
Dec 26 '18 at 17:24
add a comment |
$begingroup$
You have a sequence $(f_n)_n$ of continuous functions on $[0,1]$ such that $f_n(0) =0$ and $f_n(1)=0$. You know that $f_n to f_0$, where $f_0 in C([0,1])$ and the symbol $to$ means convergence in supermum norm.
You have to prove that $f_0 (0) =f_0(1) =0$. This follows from the fact that $f_n to f_0$. Can you explain why?
answered Dec 26 '18 at 12:26
HermioneHermione
19619
$endgroup$
$begingroup$
I would add for completeness that $to$ means converges in supremum norm.
$endgroup$
– Jonas Lenz
Dec 26 '18 at 12:29
$begingroup$
Your are right, now it is more clear. Thanks!
$endgroup$
– Hermione
Dec 26 '18 at 12:34
$begingroup$
@Hermione Am I right that since $max_{x in [0, 1]}|f_n(x) - f_0(x)| < epsilon$ $forall epsilon > 0$ $Rightarrow$ if x = 0 then $|f_0(0)| < epsilon$ $forall epsilon > 0$ $Rightarrow$ $f_0(0) = 0$. Analogically for x = 1. Therefore $f_0(0) = f_0(1) = 0$ $Rightarrow$ $f_0 in E$. So E is a closed set.
$endgroup$
– shpindler
Dec 26 '18 at 15:25
$begingroup$
Yes @shpindler, you are right
$endgroup$
– Hermione
Dec 26 '18 at 17:05
$begingroup$
@Hermione, thanks!
$endgroup$
– shpindler
Dec 26 '18 at 17:24
add a comment |
$begingroup$
You have a sequence $(f_n)_n$ of continuous functions on $[0,1]$ such that $f_n(0) =0$ and $f_n(1)=0$. You know that $f_n to f_0$, where $f_0 in C([0,1])$ and the symbol $to$ means convergence in supermum norm.
You have to prove that $f_0 (0) =f_0(1) =0$. This follows from the fact that $f_n to f_0$. Can you explain why?
answered Dec 26 '18 at 12:26
HermioneHermione
19619
$endgroup$
You have a sequence $(f_n)_n$ of continuous functions on $[0,1]$ such that $f_n(0) =0$ and $f_n(1)=0$. You know that $f_n to f_0$, where $f_0 in C([0,1])$ and the symbol $to$ means convergence in supermum norm.
You have to prove that $f_0 (0) =f_0(1) =0$. This follows from the fact that $f_n to f_0$. Can you explain why?
answered Dec 26 '18 at 12:26
HermioneHermione
19619
edited Dec 26 '18 at 12:33
answered Dec 26 '18 at 12:26
HermioneHermione
19619
answered Dec 26 '18 at 12:26
HermioneHermione
19619
answered Dec 26 '18 at 12:26
HermioneHermione
19619
19619
$begingroup$
I would add for completeness that $to$ means converges in supremum norm.
$endgroup$
– Jonas Lenz
Dec 26 '18 at 12:29
$begingroup$
Your are right, now it is more clear. Thanks!
$endgroup$
– Hermione
Dec 26 '18 at 12:34
$begingroup$
@Hermione Am I right that since $max_{x in [0, 1]}|f_n(x) - f_0(x)| < epsilon$ $forall epsilon > 0$ $Rightarrow$ if x = 0 then $|f_0(0)| < epsilon$ $forall epsilon > 0$ $Rightarrow$ $f_0(0) = 0$. Analogically for x = 1. Therefore $f_0(0) = f_0(1) = 0$ $Rightarrow$ $f_0 in E$. So E is a closed set.
$endgroup$
– shpindler
Dec 26 '18 at 15:25
$begingroup$
Yes @shpindler, you are right
$endgroup$
– Hermione
Dec 26 '18 at 17:05
$begingroup$
@Hermione, thanks!
$endgroup$
– shpindler
Dec 26 '18 at 17:24
add a comment |
$begingroup$
I would add for completeness that $to$ means converges in supremum norm.
$endgroup$
– Jonas Lenz
Dec 26 '18 at 12:29
$begingroup$
Your are right, now it is more clear. Thanks!
$endgroup$
– Hermione
Dec 26 '18 at 12:34
$begingroup$
@Hermione Am I right that since $max_{x in [0, 1]}|f_n(x) - f_0(x)| < epsilon$ $forall epsilon > 0$ $Rightarrow$ if x = 0 then $|f_0(0)| < epsilon$ $forall epsilon > 0$ $Rightarrow$ $f_0(0) = 0$. Analogically for x = 1. Therefore $f_0(0) = f_0(1) = 0$ $Rightarrow$ $f_0 in E$. So E is a closed set.
$endgroup$
– shpindler
Dec 26 '18 at 15:25
$begingroup$
Yes @shpindler, you are right
$endgroup$
– Hermione
Dec 26 '18 at 17:05
$begingroup$
@Hermione, thanks!
$endgroup$
– shpindler
Dec 26 '18 at 17:24
$begingroup$
I would add for completeness that $to$ means converges in supremum norm.
$endgroup$
– Jonas Lenz
Dec 26 '18 at 12:29
$begingroup$
I would add for completeness that $to$ means converges in supremum norm.
$endgroup$
– Jonas Lenz
Dec 26 '18 at 12:29
$begingroup$
Your are right, now it is more clear. Thanks!
$endgroup$
– Hermione
Dec 26 '18 at 12:34
$begingroup$
Your are right, now it is more clear. Thanks!
$endgroup$
– Hermione
Dec 26 '18 at 12:34
$begingroup$
@Hermione Am I right that since $max_{x in [0, 1]}|f_n(x) - f_0(x)| < epsilon$ $forall epsilon > 0$ $Rightarrow$ if x = 0 then $|f_0(0)| < epsilon$ $forall epsilon > 0$ $Rightarrow$ $f_0(0) = 0$. Analogically for x = 1. Therefore $f_0(0) = f_0(1) = 0$ $Rightarrow$ $f_0 in E$. So E is a closed set.
$endgroup$
– shpindler
Dec 26 '18 at 15:25
$begingroup$
@Hermione Am I right that since $max_{x in [0, 1]}|f_n(x) - f_0(x)| < epsilon$ $forall epsilon > 0$ $Rightarrow$ if x = 0 then $|f_0(0)| < epsilon$ $forall epsilon > 0$ $Rightarrow$ $f_0(0) = 0$. Analogically for x = 1. Therefore $f_0(0) = f_0(1) = 0$ $Rightarrow$ $f_0 in E$. So E is a closed set.
$endgroup$
– shpindler
Dec 26 '18 at 15:25
$begingroup$
Yes @shpindler, you are right
$endgroup$
– Hermione
Dec 26 '18 at 17:05
$begingroup$
Yes @shpindler, you are right
$endgroup$
– Hermione
Dec 26 '18 at 17:05
$begingroup$
@Hermione, thanks!
$endgroup$
– shpindler
Dec 26 '18 at 17:24
$begingroup$
@Hermione, thanks!
$endgroup$
– shpindler
Dec 26 '18 at 17:24
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.
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%2f3052617%2fhow-to-show-that-the-set-of-functions-e-f0-f1-0-is-closed-in-c0-1%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
