Application of DCT, integrability on $[0,infty)$
$begingroup$
Let $f$ be integrable on $[0,infty$).
Show that lim$_{n rightarrow infty}$ $int_0^{infty}$ $f(x) frac{x}{n+x}$ $dx$ = $0$.
Ok so I want to say I can take the limit inside, but this is a consequence of DCT or MCT, so to apply DCT, this is what I have come up with so far:
define $f_n$ = $f(x)$ $cdot$ $frac{x}{n+x}$ (my intuition from previous measure theory problems is that I need this step)
then {$f_n}$ is a sequence of measurable functions,
but are the $f_n$ bounded above $f(x)$ ?
*Note by assumption we have $int_0^{infty}$ $f(x)$ $dx < infty$.
Using all of this am I able to somehow deduce that
lim$_{n rightarrow infty}$ $int_0^{infty}$ $f(x) frac{x}{n+x}$ $dx$ = $int_0^{infty}$ $f(x)$ lim$_{n rightarrow infty}$ $frac{x}{n+x}$ $dx$ = $int_0^{infty}$ $f(x)$ $dx$ $int_0^{infty}$ lim$_{n rightarrow infty}$ $frac{x}{n+x}$ $dx$
I am VERY hesitant on my above two equalities, specially the last one, where am I wrong or need input? Thanks in advance and sorry if I am missing something obvious.
real-analysis measure-theory lebesgue-integral lebesgue-measure
asked Dec 23 '18 at 22:14
Hossien SahebjameHossien Sahebjame
979
$endgroup$
add a comment |
$begingroup$
Let $f$ be integrable on $[0,infty$).
Show that lim$_{n rightarrow infty}$ $int_0^{infty}$ $f(x) frac{x}{n+x}$ $dx$ = $0$.
Ok so I want to say I can take the limit inside, but this is a consequence of DCT or MCT, so to apply DCT, this is what I have come up with so far:
define $f_n$ = $f(x)$ $cdot$ $frac{x}{n+x}$ (my intuition from previous measure theory problems is that I need this step)
then {$f_n}$ is a sequence of measurable functions,
but are the $f_n$ bounded above $f(x)$ ?
*Note by assumption we have $int_0^{infty}$ $f(x)$ $dx < infty$.
Using all of this am I able to somehow deduce that
lim$_{n rightarrow infty}$ $int_0^{infty}$ $f(x) frac{x}{n+x}$ $dx$ = $int_0^{infty}$ $f(x)$ lim$_{n rightarrow infty}$ $frac{x}{n+x}$ $dx$ = $int_0^{infty}$ $f(x)$ $dx$ $int_0^{infty}$ lim$_{n rightarrow infty}$ $frac{x}{n+x}$ $dx$
I am VERY hesitant on my above two equalities, specially the last one, where am I wrong or need input? Thanks in advance and sorry if I am missing something obvious.
real-analysis measure-theory lebesgue-integral lebesgue-measure
asked Dec 23 '18 at 22:14
Hossien SahebjameHossien Sahebjame
979
$endgroup$
$begingroup$
Are you really asking if $frac{x}{n+x} le 1$ for $x > 0$ and $n ge 1$?
$endgroup$
– mathworker21
Dec 23 '18 at 22:17
$begingroup$
omg duh, $x > 0$. I didn't consider that, thats why I said forgive me If im asking something obvious, Im learning, I don't learn fast. SORRY FOR BEING STUPID.
$endgroup$
– Hossien Sahebjame
Dec 23 '18 at 22:27
$begingroup$
happens to all of us
$endgroup$
– mathworker21
Dec 23 '18 at 22:30
$begingroup$
lol thanks, again, Im new to measure theory, im really trying to get a lot better so feel free to comment if you ever see me ask noob questions.
$endgroup$
– Hossien Sahebjame
Dec 23 '18 at 22:32
add a comment |
$begingroup$
Let $f$ be integrable on $[0,infty$).
Show that lim$_{n rightarrow infty}$ $int_0^{infty}$ $f(x) frac{x}{n+x}$ $dx$ = $0$.
Ok so I want to say I can take the limit inside, but this is a consequence of DCT or MCT, so to apply DCT, this is what I have come up with so far:
define $f_n$ = $f(x)$ $cdot$ $frac{x}{n+x}$ (my intuition from previous measure theory problems is that I need this step)
then {$f_n}$ is a sequence of measurable functions,
but are the $f_n$ bounded above $f(x)$ ?
*Note by assumption we have $int_0^{infty}$ $f(x)$ $dx < infty$.
Using all of this am I able to somehow deduce that
lim$_{n rightarrow infty}$ $int_0^{infty}$ $f(x) frac{x}{n+x}$ $dx$ = $int_0^{infty}$ $f(x)$ lim$_{n rightarrow infty}$ $frac{x}{n+x}$ $dx$ = $int_0^{infty}$ $f(x)$ $dx$ $int_0^{infty}$ lim$_{n rightarrow infty}$ $frac{x}{n+x}$ $dx$
I am VERY hesitant on my above two equalities, specially the last one, where am I wrong or need input? Thanks in advance and sorry if I am missing something obvious.
real-analysis measure-theory lebesgue-integral lebesgue-measure
asked Dec 23 '18 at 22:14
Hossien SahebjameHossien Sahebjame
979
$endgroup$
Let $f$ be integrable on $[0,infty$).
Show that lim$_{n rightarrow infty}$ $int_0^{infty}$ $f(x) frac{x}{n+x}$ $dx$ = $0$.
Ok so I want to say I can take the limit inside, but this is a consequence of DCT or MCT, so to apply DCT, this is what I have come up with so far:
define $f_n$ = $f(x)$ $cdot$ $frac{x}{n+x}$ (my intuition from previous measure theory problems is that I need this step)
then {$f_n}$ is a sequence of measurable functions,
but are the $f_n$ bounded above $f(x)$ ?
*Note by assumption we have $int_0^{infty}$ $f(x)$ $dx < infty$.
Using all of this am I able to somehow deduce that
lim$_{n rightarrow infty}$ $int_0^{infty}$ $f(x) frac{x}{n+x}$ $dx$ = $int_0^{infty}$ $f(x)$ lim$_{n rightarrow infty}$ $frac{x}{n+x}$ $dx$ = $int_0^{infty}$ $f(x)$ $dx$ $int_0^{infty}$ lim$_{n rightarrow infty}$ $frac{x}{n+x}$ $dx$
I am VERY hesitant on my above two equalities, specially the last one, where am I wrong or need input? Thanks in advance and sorry if I am missing something obvious.
real-analysis measure-theory lebesgue-integral lebesgue-measure
real-analysis measure-theory lebesgue-integral lebesgue-measure
asked Dec 23 '18 at 22:14
Hossien SahebjameHossien Sahebjame
979
asked Dec 23 '18 at 22:14
Hossien SahebjameHossien Sahebjame
979
asked Dec 23 '18 at 22:14
Hossien SahebjameHossien Sahebjame
979
asked Dec 23 '18 at 22:14
Hossien SahebjameHossien Sahebjame
979
asked Dec 23 '18 at 22:14
Hossien SahebjameHossien Sahebjame
979
979
$begingroup$
Are you really asking if $frac{x}{n+x} le 1$ for $x > 0$ and $n ge 1$?
$endgroup$
– mathworker21
Dec 23 '18 at 22:17
$begingroup$
omg duh, $x > 0$. I didn't consider that, thats why I said forgive me If im asking something obvious, Im learning, I don't learn fast. SORRY FOR BEING STUPID.
$endgroup$
– Hossien Sahebjame
Dec 23 '18 at 22:27
$begingroup$
happens to all of us
$endgroup$
– mathworker21
Dec 23 '18 at 22:30
$begingroup$
lol thanks, again, Im new to measure theory, im really trying to get a lot better so feel free to comment if you ever see me ask noob questions.
$endgroup$
– Hossien Sahebjame
Dec 23 '18 at 22:32
add a comment |
$begingroup$
Are you really asking if $frac{x}{n+x} le 1$ for $x > 0$ and $n ge 1$?
$endgroup$
– mathworker21
Dec 23 '18 at 22:17
$begingroup$
omg duh, $x > 0$. I didn't consider that, thats why I said forgive me If im asking something obvious, Im learning, I don't learn fast. SORRY FOR BEING STUPID.
$endgroup$
– Hossien Sahebjame
Dec 23 '18 at 22:27
$begingroup$
happens to all of us
$endgroup$
– mathworker21
Dec 23 '18 at 22:30
$begingroup$
lol thanks, again, Im new to measure theory, im really trying to get a lot better so feel free to comment if you ever see me ask noob questions.
$endgroup$
– Hossien Sahebjame
Dec 23 '18 at 22:32
$begingroup$
Are you really asking if $frac{x}{n+x} le 1$ for $x > 0$ and $n ge 1$?
$endgroup$
– mathworker21
Dec 23 '18 at 22:17
$begingroup$
Are you really asking if $frac{x}{n+x} le 1$ for $x > 0$ and $n ge 1$?
$endgroup$
– mathworker21
Dec 23 '18 at 22:17
$begingroup$
omg duh, $x > 0$. I didn't consider that, thats why I said forgive me If im asking something obvious, Im learning, I don't learn fast. SORRY FOR BEING STUPID.
$endgroup$
– Hossien Sahebjame
Dec 23 '18 at 22:27
$begingroup$
omg duh, $x > 0$. I didn't consider that, thats why I said forgive me If im asking something obvious, Im learning, I don't learn fast. SORRY FOR BEING STUPID.
$endgroup$
– Hossien Sahebjame
Dec 23 '18 at 22:27
$begingroup$
happens to all of us
$endgroup$
– mathworker21
Dec 23 '18 at 22:30
$begingroup$
happens to all of us
$endgroup$
– mathworker21
Dec 23 '18 at 22:30
$begingroup$
lol thanks, again, Im new to measure theory, im really trying to get a lot better so feel free to comment if you ever see me ask noob questions.
$endgroup$
– Hossien Sahebjame
Dec 23 '18 at 22:32
$begingroup$
lol thanks, again, Im new to measure theory, im really trying to get a lot better so feel free to comment if you ever see me ask noob questions.
$endgroup$
– Hossien Sahebjame
Dec 23 '18 at 22:32
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I will do this from scratch. First, integrability reads $int |f(x)|<infty$. Now, set $f_n(x)= f(x)frac{x}{n+x}$, and observe that, $|f_n(x)|leq |f(x)|$ point wise, as $frac{x}{n+x}leq 1$ holds always, whenever $xgeq 0$, and $ngeq 1$.
Now, also note that, for any fixed $x$, $lim_{ntoinfty}f_n(x)=0$, point wise. Hence, by DCT, $lim_n int f_n(x); dx=int lim_{ntoinfty}f_n(x); dx = 0$, as $lim_n f_n(x)=0$.
answered Dec 23 '18 at 22:19
AaronAaron
1,902415
$endgroup$
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%2f3050745%2fapplication-of-dct-integrability-on-0-infty%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
$begingroup$
I will do this from scratch. First, integrability reads $int |f(x)|<infty$. Now, set $f_n(x)= f(x)frac{x}{n+x}$, and observe that, $|f_n(x)|leq |f(x)|$ point wise, as $frac{x}{n+x}leq 1$ holds always, whenever $xgeq 0$, and $ngeq 1$.
Now, also note that, for any fixed $x$, $lim_{ntoinfty}f_n(x)=0$, point wise. Hence, by DCT, $lim_n int f_n(x); dx=int lim_{ntoinfty}f_n(x); dx = 0$, as $lim_n f_n(x)=0$.
answered Dec 23 '18 at 22:19
AaronAaron
1,902415
$endgroup$
add a comment |
$begingroup$
I will do this from scratch. First, integrability reads $int |f(x)|<infty$. Now, set $f_n(x)= f(x)frac{x}{n+x}$, and observe that, $|f_n(x)|leq |f(x)|$ point wise, as $frac{x}{n+x}leq 1$ holds always, whenever $xgeq 0$, and $ngeq 1$.
Now, also note that, for any fixed $x$, $lim_{ntoinfty}f_n(x)=0$, point wise. Hence, by DCT, $lim_n int f_n(x); dx=int lim_{ntoinfty}f_n(x); dx = 0$, as $lim_n f_n(x)=0$.
answered Dec 23 '18 at 22:19
AaronAaron
1,902415
$endgroup$
add a comment |
$begingroup$
I will do this from scratch. First, integrability reads $int |f(x)|<infty$. Now, set $f_n(x)= f(x)frac{x}{n+x}$, and observe that, $|f_n(x)|leq |f(x)|$ point wise, as $frac{x}{n+x}leq 1$ holds always, whenever $xgeq 0$, and $ngeq 1$.
Now, also note that, for any fixed $x$, $lim_{ntoinfty}f_n(x)=0$, point wise. Hence, by DCT, $lim_n int f_n(x); dx=int lim_{ntoinfty}f_n(x); dx = 0$, as $lim_n f_n(x)=0$.
answered Dec 23 '18 at 22:19
AaronAaron
1,902415
$endgroup$
I will do this from scratch. First, integrability reads $int |f(x)|<infty$. Now, set $f_n(x)= f(x)frac{x}{n+x}$, and observe that, $|f_n(x)|leq |f(x)|$ point wise, as $frac{x}{n+x}leq 1$ holds always, whenever $xgeq 0$, and $ngeq 1$.
Now, also note that, for any fixed $x$, $lim_{ntoinfty}f_n(x)=0$, point wise. Hence, by DCT, $lim_n int f_n(x); dx=int lim_{ntoinfty}f_n(x); dx = 0$, as $lim_n f_n(x)=0$.
answered Dec 23 '18 at 22:19
AaronAaron
1,902415
answered Dec 23 '18 at 22:19
AaronAaron
1,902415
answered Dec 23 '18 at 22:19
AaronAaron
1,902415
answered Dec 23 '18 at 22:19
AaronAaron
1,902415
1,902415
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.
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%2f3050745%2fapplication-of-dct-integrability-on-0-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
$begingroup$
Are you really asking if $frac{x}{n+x} le 1$ for $x > 0$ and $n ge 1$?
$endgroup$
– mathworker21
Dec 23 '18 at 22:17
$begingroup$
omg duh, $x > 0$. I didn't consider that, thats why I said forgive me If im asking something obvious, Im learning, I don't learn fast. SORRY FOR BEING STUPID.
$endgroup$
– Hossien Sahebjame
Dec 23 '18 at 22:27
$begingroup$
happens to all of us
$endgroup$
– mathworker21
Dec 23 '18 at 22:30
$begingroup$
lol thanks, again, Im new to measure theory, im really trying to get a lot better so feel free to comment if you ever see me ask noob questions.
$endgroup$
– Hossien Sahebjame
Dec 23 '18 at 22:32