Mixing
$f in L^1, g in L^1$ s.t. $f cdot g not in L^1$
-1
$begingroup$
For what functions for instance stands: $f in L^1, g in L^1$ but $f cdot g$ are not in $L^1$? Thank you
real-analysis functions
edited Jan 18 at 12:18
Andrews
6741318
asked Jan 18 at 9:10
Lukáš AltmanLukáš Altman
11
$endgroup$
add a comment |
-1
$begingroup$
For what functions for instance stands: $f in L^1, g in L^1$ but $f cdot g$ are not in $L^1$? Thank you
real-analysis functions
edited Jan 18 at 12:18
Andrews
6741318
asked Jan 18 at 9:10
Lukáš AltmanLukáš Altman
11
$endgroup$
$begingroup$
The 'convolution' tag is confusing because if it the operation is convolution, then $f ast g in L^1$. Are you asking for multiplication or convolution?
$endgroup$
– twnly
Jan 18 at 9:22
$begingroup$
I am asking for multiplication. It is connected with definition of convolution
$endgroup$
– Lukáš Altman
Jan 18 at 9:24
add a comment |
-1
-1
-1
$begingroup$
For what functions for instance stands: $f in L^1, g in L^1$ but $f cdot g$ are not in $L^1$? Thank you
real-analysis functions
edited Jan 18 at 12:18
Andrews
6741318
asked Jan 18 at 9:10
Lukáš AltmanLukáš Altman
11
$endgroup$
For what functions for instance stands: $f in L^1, g in L^1$ but $f cdot g$ are not in $L^1$? Thank you
real-analysis functions
real-analysis functions
edited Jan 18 at 12:18
Andrews
6741318
asked Jan 18 at 9:10
Lukáš AltmanLukáš Altman
11
edited Jan 18 at 12:18
Andrews
6741318
asked Jan 18 at 9:10
Lukáš AltmanLukáš Altman
11
edited Jan 18 at 12:18
Andrews
6741318
edited Jan 18 at 12:18
Andrews
6741318
edited Jan 18 at 12:18
Andrews
6741318
6741318
asked Jan 18 at 9:10
Lukáš AltmanLukáš Altman
11
asked Jan 18 at 9:10
Lukáš AltmanLukáš Altman
11
asked Jan 18 at 9:10
Lukáš AltmanLukáš Altman
11
11
$begingroup$
The 'convolution' tag is confusing because if it the operation is convolution, then $f ast g in L^1$. Are you asking for multiplication or convolution?
$endgroup$
– twnly
Jan 18 at 9:22
$begingroup$
I am asking for multiplication. It is connected with definition of convolution
$endgroup$
– Lukáš Altman
Jan 18 at 9:24
add a comment |
$begingroup$
The 'convolution' tag is confusing because if it the operation is convolution, then $f ast g in L^1$. Are you asking for multiplication or convolution?
$endgroup$
– twnly
Jan 18 at 9:22
$begingroup$
I am asking for multiplication. It is connected with definition of convolution
$endgroup$
– Lukáš Altman
Jan 18 at 9:24
$begingroup$
The 'convolution' tag is confusing because if it the operation is convolution, then $f ast g in L^1$. Are you asking for multiplication or convolution?
$endgroup$
– twnly
Jan 18 at 9:22
$begingroup$
The 'convolution' tag is confusing because if it the operation is convolution, then $f ast g in L^1$. Are you asking for multiplication or convolution?
$endgroup$
– twnly
Jan 18 at 9:22
$begingroup$
I am asking for multiplication. It is connected with definition of convolution
$endgroup$
– Lukáš Altman
Jan 18 at 9:24
$begingroup$
I am asking for multiplication. It is connected with definition of convolution
$endgroup$
– Lukáš Altman
Jan 18 at 9:24
add a comment |
1 Answer
1
active
oldest
votes
2
$begingroup$
$f(x) = g(x) = frac{1}{sqrt{x}}$ on $[0,1]$
answered Jan 18 at 9:12
KlausKlaus
2,0099
$endgroup$
$begingroup$
Thank you. For $g(x)f(x) in L^1$ so $ int_0^1 1/x dx = [ln|x|]_0^1 $ how can I set $0$ to x?
$endgroup$
– Lukáš Altman
Jan 18 at 9:20
$begingroup$
You can't. $limlimits_{x to 0} ln|x| = -infty$ and that's why $f cdot g$ is not in $L^1$, which is what you asked for.
$endgroup$
– Klaus
Jan 18 at 9:39
$begingroup$
Thank you very much
$endgroup$
– Lukáš Altman
Jan 18 at 10:58
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
});
}
});
draft saved
draft discarded
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%2f3078009%2ff-in-l1-g-in-l1-s-t-f-cdot-g-not-in-l1%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
2
$begingroup$
$f(x) = g(x) = frac{1}{sqrt{x}}$ on $[0,1]$
answered Jan 18 at 9:12
KlausKlaus
2,0099
$endgroup$
$begingroup$
Thank you. For $g(x)f(x) in L^1$ so $ int_0^1 1/x dx = [ln|x|]_0^1 $ how can I set $0$ to x?
$endgroup$
– Lukáš Altman
Jan 18 at 9:20
$begingroup$
You can't. $limlimits_{x to 0} ln|x| = -infty$ and that's why $f cdot g$ is not in $L^1$, which is what you asked for.
$endgroup$
– Klaus
Jan 18 at 9:39
$begingroup$
Thank you very much
$endgroup$
– Lukáš Altman
Jan 18 at 10:58
add a comment |
2
$begingroup$
$f(x) = g(x) = frac{1}{sqrt{x}}$ on $[0,1]$
answered Jan 18 at 9:12
KlausKlaus
2,0099
$endgroup$
$begingroup$
Thank you. For $g(x)f(x) in L^1$ so $ int_0^1 1/x dx = [ln|x|]_0^1 $ how can I set $0$ to x?
$endgroup$
– Lukáš Altman
Jan 18 at 9:20
$begingroup$
You can't. $limlimits_{x to 0} ln|x| = -infty$ and that's why $f cdot g$ is not in $L^1$, which is what you asked for.
$endgroup$
– Klaus
Jan 18 at 9:39
$begingroup$
Thank you very much
$endgroup$
– Lukáš Altman
Jan 18 at 10:58
add a comment |
2
2
2
$begingroup$
$f(x) = g(x) = frac{1}{sqrt{x}}$ on $[0,1]$
answered Jan 18 at 9:12
KlausKlaus
2,0099
$endgroup$
$f(x) = g(x) = frac{1}{sqrt{x}}$ on $[0,1]$
answered Jan 18 at 9:12
KlausKlaus
2,0099
answered Jan 18 at 9:12
KlausKlaus
2,0099
answered Jan 18 at 9:12
KlausKlaus
2,0099
answered Jan 18 at 9:12
KlausKlaus
2,0099
2,0099
$begingroup$
Thank you. For $g(x)f(x) in L^1$ so $ int_0^1 1/x dx = [ln|x|]_0^1 $ how can I set $0$ to x?
$endgroup$
– Lukáš Altman
Jan 18 at 9:20
$begingroup$
You can't. $limlimits_{x to 0} ln|x| = -infty$ and that's why $f cdot g$ is not in $L^1$, which is what you asked for.
$endgroup$
– Klaus
Jan 18 at 9:39
$begingroup$
Thank you very much
$endgroup$
– Lukáš Altman
Jan 18 at 10:58
add a comment |
$begingroup$
Thank you. For $g(x)f(x) in L^1$ so $ int_0^1 1/x dx = [ln|x|]_0^1 $ how can I set $0$ to x?
$endgroup$
– Lukáš Altman
Jan 18 at 9:20
$begingroup$
You can't. $limlimits_{x to 0} ln|x| = -infty$ and that's why $f cdot g$ is not in $L^1$, which is what you asked for.
$endgroup$
– Klaus
Jan 18 at 9:39
$begingroup$
Thank you very much
$endgroup$
– Lukáš Altman
Jan 18 at 10:58
$begingroup$
Thank you. For $g(x)f(x) in L^1$ so $ int_0^1 1/x dx = [ln|x|]_0^1 $ how can I set $0$ to x?
$endgroup$
– Lukáš Altman
Jan 18 at 9:20
$begingroup$
Thank you. For $g(x)f(x) in L^1$ so $ int_0^1 1/x dx = [ln|x|]_0^1 $ how can I set $0$ to x?
$endgroup$
– Lukáš Altman
Jan 18 at 9:20
$begingroup$
You can't. $limlimits_{x to 0} ln|x| = -infty$ and that's why $f cdot g$ is not in $L^1$, which is what you asked for.
$endgroup$
– Klaus
Jan 18 at 9:39
$begingroup$
You can't. $limlimits_{x to 0} ln|x| = -infty$ and that's why $f cdot g$ is not in $L^1$, which is what you asked for.
$endgroup$
– Klaus
Jan 18 at 9:39
$begingroup$
Thank you very much
$endgroup$
– Lukáš Altman
Jan 18 at 10:58
$begingroup$
Thank you very much
$endgroup$
– Lukáš Altman
Jan 18 at 10:58
add a comment |
draft saved
draft discarded
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.
draft saved
draft discarded
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%2f3078009%2ff-in-l1-g-in-l1-s-t-f-cdot-g-not-in-l1%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
$begingroup$
The 'convolution' tag is confusing because if it the operation is convolution, then $f ast g in L^1$. Are you asking for multiplication or convolution?
$endgroup$
– twnly
Jan 18 at 9:22
$begingroup$
I am asking for multiplication. It is connected with definition of convolution
$endgroup$
– Lukáš Altman
Jan 18 at 9:24