$lim_{|h|rightarrow 0} ||f(x+h)-f(x)||_p=0$ for $p =infty$
$begingroup$
For $fin L^p(X)$ where $pin [1,infty)$. we have
$$lim_{|h|rightarrow 0} ||f(x+h)-f(x)||_p=0$$
I use the fact that continuous function with compact support is dense in $L^p$ to prove the case for $p<infty$. If I consider the case $p=infty$, the method that I used fails here, but how do I find an example that fails at $p=infty$?
edit:
If I consider $f(x)=chi_{[0,x]}$, then $||f(x+h)-f(x)||_infty=1$, therefore
$$lim_{|h|rightarrow 0} ||f(x+h)-f(x)||_p= lim_{|h|rightarrow 0} 1 =1$$
Is this correct?
real-analysis functional-analysis
asked Jan 7 at 12:44
ReinherdReinherd
764
$endgroup$
|
show 2 more comments
$begingroup$
For $fin L^p(X)$ where $pin [1,infty)$. we have
$$lim_{|h|rightarrow 0} ||f(x+h)-f(x)||_p=0$$
I use the fact that continuous function with compact support is dense in $L^p$ to prove the case for $p<infty$. If I consider the case $p=infty$, the method that I used fails here, but how do I find an example that fails at $p=infty$?
edit:
If I consider $f(x)=chi_{[0,x]}$, then $||f(x+h)-f(x)||_infty=1$, therefore
$$lim_{|h|rightarrow 0} ||f(x+h)-f(x)||_p= lim_{|h|rightarrow 0} 1 =1$$
Is this correct?
real-analysis functional-analysis
asked Jan 7 at 12:44
ReinherdReinherd
764
$endgroup$
2
$begingroup$
For example, consider the characteristic function of an interval.
$endgroup$
– Rigel
Jan 7 at 12:52
$begingroup$
@Rigel thank you. If I consider $f(x)=chi_{[0,x]}$, then the norm still go to $0$. How do I construct it?
$endgroup$
– Reinherd
Jan 7 at 13:09
$begingroup$
@Reinherd Why do you think that the norm goes to zero?
$endgroup$
– saz
Jan 7 at 13:21
$begingroup$
@saz I think I might be wrong about taking the limit directly to make $h=0$. Actually , the norm is always 1 before taking the limit. So it is still 1 even though I take the limit, is it?
$endgroup$
– Reinherd
Jan 7 at 13:28
1
$begingroup$
Saying $f(x)=chi_{[0,x]}$ doesn't make much sense - that would say that $f:Bbb Rto L^infty$, not $fin L^infty$. As already suggested, try $f$ equal to the characteristic function of an interval. Like $f=chi_{[0,1]}$.
$endgroup$
– David C. Ullrich
Jan 7 at 14:35
|
show 2 more comments
$begingroup$
For $fin L^p(X)$ where $pin [1,infty)$. we have
$$lim_{|h|rightarrow 0} ||f(x+h)-f(x)||_p=0$$
I use the fact that continuous function with compact support is dense in $L^p$ to prove the case for $p<infty$. If I consider the case $p=infty$, the method that I used fails here, but how do I find an example that fails at $p=infty$?
edit:
If I consider $f(x)=chi_{[0,x]}$, then $||f(x+h)-f(x)||_infty=1$, therefore
$$lim_{|h|rightarrow 0} ||f(x+h)-f(x)||_p= lim_{|h|rightarrow 0} 1 =1$$
Is this correct?
real-analysis functional-analysis
asked Jan 7 at 12:44
ReinherdReinherd
764
$endgroup$
For $fin L^p(X)$ where $pin [1,infty)$. we have
$$lim_{|h|rightarrow 0} ||f(x+h)-f(x)||_p=0$$
I use the fact that continuous function with compact support is dense in $L^p$ to prove the case for $p<infty$. If I consider the case $p=infty$, the method that I used fails here, but how do I find an example that fails at $p=infty$?
edit:
If I consider $f(x)=chi_{[0,x]}$, then $||f(x+h)-f(x)||_infty=1$, therefore
$$lim_{|h|rightarrow 0} ||f(x+h)-f(x)||_p= lim_{|h|rightarrow 0} 1 =1$$
Is this correct?
real-analysis functional-analysis
real-analysis functional-analysis
asked Jan 7 at 12:44
ReinherdReinherd
764
asked Jan 7 at 12:44
ReinherdReinherd
764
edited Jan 7 at 13:33
Reinherd
asked Jan 7 at 12:44
ReinherdReinherd
764
asked Jan 7 at 12:44
ReinherdReinherd
764
asked Jan 7 at 12:44
ReinherdReinherd
764
764
2
$begingroup$
For example, consider the characteristic function of an interval.
$endgroup$
– Rigel
Jan 7 at 12:52
$begingroup$
@Rigel thank you. If I consider $f(x)=chi_{[0,x]}$, then the norm still go to $0$. How do I construct it?
$endgroup$
– Reinherd
Jan 7 at 13:09
$begingroup$
@Reinherd Why do you think that the norm goes to zero?
$endgroup$
– saz
Jan 7 at 13:21
$begingroup$
@saz I think I might be wrong about taking the limit directly to make $h=0$. Actually , the norm is always 1 before taking the limit. So it is still 1 even though I take the limit, is it?
$endgroup$
– Reinherd
Jan 7 at 13:28
1
$begingroup$
Saying $f(x)=chi_{[0,x]}$ doesn't make much sense - that would say that $f:Bbb Rto L^infty$, not $fin L^infty$. As already suggested, try $f$ equal to the characteristic function of an interval. Like $f=chi_{[0,1]}$.
$endgroup$
– David C. Ullrich
Jan 7 at 14:35
|
show 2 more comments
2
$begingroup$
For example, consider the characteristic function of an interval.
$endgroup$
– Rigel
Jan 7 at 12:52
$begingroup$
@Rigel thank you. If I consider $f(x)=chi_{[0,x]}$, then the norm still go to $0$. How do I construct it?
$endgroup$
– Reinherd
Jan 7 at 13:09
$begingroup$
@Reinherd Why do you think that the norm goes to zero?
$endgroup$
– saz
Jan 7 at 13:21
$begingroup$
@saz I think I might be wrong about taking the limit directly to make $h=0$. Actually , the norm is always 1 before taking the limit. So it is still 1 even though I take the limit, is it?
$endgroup$
– Reinherd
Jan 7 at 13:28
1
$begingroup$
Saying $f(x)=chi_{[0,x]}$ doesn't make much sense - that would say that $f:Bbb Rto L^infty$, not $fin L^infty$. As already suggested, try $f$ equal to the characteristic function of an interval. Like $f=chi_{[0,1]}$.
$endgroup$
– David C. Ullrich
Jan 7 at 14:35
2
2
$begingroup$
For example, consider the characteristic function of an interval.
$endgroup$
– Rigel
Jan 7 at 12:52
$begingroup$
For example, consider the characteristic function of an interval.
$endgroup$
– Rigel
Jan 7 at 12:52
$begingroup$
@Rigel thank you. If I consider $f(x)=chi_{[0,x]}$, then the norm still go to $0$. How do I construct it?
$endgroup$
– Reinherd
Jan 7 at 13:09
$begingroup$
@Rigel thank you. If I consider $f(x)=chi_{[0,x]}$, then the norm still go to $0$. How do I construct it?
$endgroup$
– Reinherd
Jan 7 at 13:09
$begingroup$
@Reinherd Why do you think that the norm goes to zero?
$endgroup$
– saz
Jan 7 at 13:21
$begingroup$
@Reinherd Why do you think that the norm goes to zero?
$endgroup$
– saz
Jan 7 at 13:21
$begingroup$
@saz I think I might be wrong about taking the limit directly to make $h=0$. Actually , the norm is always 1 before taking the limit. So it is still 1 even though I take the limit, is it?
$endgroup$
– Reinherd
Jan 7 at 13:28
$begingroup$
@saz I think I might be wrong about taking the limit directly to make $h=0$. Actually , the norm is always 1 before taking the limit. So it is still 1 even though I take the limit, is it?
$endgroup$
– Reinherd
Jan 7 at 13:28
1
1
$begingroup$
Saying $f(x)=chi_{[0,x]}$ doesn't make much sense - that would say that $f:Bbb Rto L^infty$, not $fin L^infty$. As already suggested, try $f$ equal to the characteristic function of an interval. Like $f=chi_{[0,1]}$.
$endgroup$
– David C. Ullrich
Jan 7 at 14:35
$begingroup$
Saying $f(x)=chi_{[0,x]}$ doesn't make much sense - that would say that $f:Bbb Rto L^infty$, not $fin L^infty$. As already suggested, try $f$ equal to the characteristic function of an interval. Like $f=chi_{[0,1]}$.
$endgroup$
– David C. Ullrich
Jan 7 at 14:35
|
show 2 more comments
0
active
oldest
votes
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%2f3064968%2flim-h-rightarrow-0-fxh-fx-p-0-for-p-infty%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3064968%2flim-h-rightarrow-0-fxh-fx-p-0-for-p-infty%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
2
$begingroup$
For example, consider the characteristic function of an interval.
$endgroup$
– Rigel
Jan 7 at 12:52
$begingroup$
@Rigel thank you. If I consider $f(x)=chi_{[0,x]}$, then the norm still go to $0$. How do I construct it?
$endgroup$
– Reinherd
Jan 7 at 13:09
$begingroup$
@Reinherd Why do you think that the norm goes to zero?
$endgroup$
– saz
Jan 7 at 13:21
$begingroup$
@saz I think I might be wrong about taking the limit directly to make $h=0$. Actually , the norm is always 1 before taking the limit. So it is still 1 even though I take the limit, is it?
$endgroup$
– Reinherd
Jan 7 at 13:28
1
$begingroup$
Saying $f(x)=chi_{[0,x]}$ doesn't make much sense - that would say that $f:Bbb Rto L^infty$, not $fin L^infty$. As already suggested, try $f$ equal to the characteristic function of an interval. Like $f=chi_{[0,1]}$.
$endgroup$
– David C. Ullrich
Jan 7 at 14:35