Sequence of integral functions convergesConvergence of sequence of functionsProve that there is a subsequence...
Non-trope happy ending?
Is there any evidence that Cleopatra and Caesarion considered fleeing to India to escape the Romans?
How to draw a matrix with arrows in limited space
Has any country ever had 2 former presidents in jail simultaneously?
Why can't the Brexit deadlock in the UK parliament be solved with a plurality vote?
A Trivial Diagnosis
Why is so much work done on numerical verification of the Riemann Hypothesis?
C++ check if statement can be evaluated constexpr
Stack Interview Code methods made from class Node and Smart Pointers
What does "Scientists rise up against statistical significance" mean? (Comment in Nature)
C++ copy constructor called at return
Taxes on Dividends in a Roth IRA
Microchip documentation does not label CAN buss pins on micro controller pinout diagram
Are Captain Marvel's powers affected by Thanos breaking the Tesseract and claiming the stone?
What is the highest possible scrabble score for placing a single tile
Shouldn’t conservatives embrace universal basic income?
Does Doodling or Improvising on the Piano Have Any Benefits?
Which was the first story featuring espers?
How much theory knowledge is actually used while playing?
Does the Linux kernel need a file system to run?
How much of a Devil Fruit must be consumed to gain the power?
How do I fix the group tension caused by my character stealing and possibly killing without provocation?
awk assign to multiple variables at once
Why Shazam when there is already Superman?
Sequence of integral functions converges
Convergence of sequence of functionsProve that there is a subsequence of functions which converges uniformlyHow to prove a sequence of a function converges uniformly?Sequence of differentiable functions that converges pointwise.Converging sequence of integrals for uniformly bounded functionsIs a sequence of decreasing functions in $C^0$ pointwise convergent to $0$ implies the sequence is equicontinuous?Real Analysis - Convergence of a FunctionShowing there is a subsequence that uniformly converges?Show that the sequence of functions converges point-wise but no uniformlyEquicontinuous and uniform norm examples
$begingroup$
$forall ngeq 1$, Let $f_n:[0,1]to mathbb{R},forall xin [0,1],|f_n(x)|leq 1+frac{n}{1+n^2x^2}$. Define $F_n(x)=int_0^xf_n(t) dt$. Show that there is a subsequence of ${F_n}_{ngeq 1}$ that converges pointwise on $[0,1]$.
I have already shown $forall ain (0,1)$, there exists a subsequence of ${F_n}_{ngeq 1}$ that converges uniformly on $[a,1]$. What should I do next?
real-analysis
asked Mar 13 at 6:26
Fluffy SkyeFluffy Skye
31919
$endgroup$
add a comment |
$begingroup$
$forall ngeq 1$, Let $f_n:[0,1]to mathbb{R},forall xin [0,1],|f_n(x)|leq 1+frac{n}{1+n^2x^2}$. Define $F_n(x)=int_0^xf_n(t) dt$. Show that there is a subsequence of ${F_n}_{ngeq 1}$ that converges pointwise on $[0,1]$.
I have already shown $forall ain (0,1)$, there exists a subsequence of ${F_n}_{ngeq 1}$ that converges uniformly on $[a,1]$. What should I do next?
real-analysis
asked Mar 13 at 6:26
Fluffy SkyeFluffy Skye
31919
$endgroup$
$begingroup$
Uniform implies pointwise... no?
$endgroup$
– Sean Roberson
Mar 13 at 6:30
$begingroup$
Yes, but it's on $[a,1]$ not $[0,1]$.
$endgroup$
– Fluffy Skye
Mar 13 at 6:31
add a comment |
$begingroup$
$forall ngeq 1$, Let $f_n:[0,1]to mathbb{R},forall xin [0,1],|f_n(x)|leq 1+frac{n}{1+n^2x^2}$. Define $F_n(x)=int_0^xf_n(t) dt$. Show that there is a subsequence of ${F_n}_{ngeq 1}$ that converges pointwise on $[0,1]$.
I have already shown $forall ain (0,1)$, there exists a subsequence of ${F_n}_{ngeq 1}$ that converges uniformly on $[a,1]$. What should I do next?
real-analysis
asked Mar 13 at 6:26
Fluffy SkyeFluffy Skye
31919
$endgroup$
$forall ngeq 1$, Let $f_n:[0,1]to mathbb{R},forall xin [0,1],|f_n(x)|leq 1+frac{n}{1+n^2x^2}$. Define $F_n(x)=int_0^xf_n(t) dt$. Show that there is a subsequence of ${F_n}_{ngeq 1}$ that converges pointwise on $[0,1]$.
I have already shown $forall ain (0,1)$, there exists a subsequence of ${F_n}_{ngeq 1}$ that converges uniformly on $[a,1]$. What should I do next?
real-analysis
real-analysis
asked Mar 13 at 6:26
Fluffy SkyeFluffy Skye
31919
asked Mar 13 at 6:26
Fluffy SkyeFluffy Skye
31919
asked Mar 13 at 6:26
Fluffy SkyeFluffy Skye
31919
asked Mar 13 at 6:26
Fluffy SkyeFluffy Skye
31919
asked Mar 13 at 6:26
Fluffy SkyeFluffy Skye
31919
31919
$begingroup$
Uniform implies pointwise... no?
$endgroup$
– Sean Roberson
Mar 13 at 6:30
$begingroup$
Yes, but it's on $[a,1]$ not $[0,1]$.
$endgroup$
– Fluffy Skye
Mar 13 at 6:31
add a comment |
$begingroup$
Uniform implies pointwise... no?
$endgroup$
– Sean Roberson
Mar 13 at 6:30
$begingroup$
Yes, but it's on $[a,1]$ not $[0,1]$.
$endgroup$
– Fluffy Skye
Mar 13 at 6:31
$begingroup$
Uniform implies pointwise... no?
$endgroup$
– Sean Roberson
Mar 13 at 6:30
$begingroup$
Uniform implies pointwise... no?
$endgroup$
– Sean Roberson
Mar 13 at 6:30
$begingroup$
Yes, but it's on $[a,1]$ not $[0,1]$.
$endgroup$
– Fluffy Skye
Mar 13 at 6:31
$begingroup$
Yes, but it's on $[a,1]$ not $[0,1]$.
$endgroup$
– Fluffy Skye
Mar 13 at 6:31
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Use a diagonal argument. For each $n$ you have a subsequence that converges uniformly on $(frac 1 n ,1)$ and so you can construct a diagonal subsequence which converges at every point. [Note that $F_n(0)=0$].
answered Mar 13 at 6:35
Kavi Rama MurthyKavi Rama Murthy
68.9k53169
$endgroup$
1
$begingroup$
Umm, so I first pick a subsequence that converges uniformly on $[1/2,1]$, and pick from it a subsequence that converges uniformly on $[1/3,1]$ and so on. Then pick the diagonal element of all these sequences?
$endgroup$
– Fluffy Skye
Mar 13 at 6:37
$begingroup$
@FluffySkye Exactly. You got the idea.
$endgroup$
– Kavi Rama Murthy
Mar 13 at 6:38
$begingroup$
But I don't understand how we end up with poinwise convergence though. Is it because $forall xin (0,1]$, far far along down our diagonal sequence it will converge, but 0 is not ok? I don't quite get it.
$endgroup$
– Fluffy Skye
Mar 13 at 6:41
1
$begingroup$
A given $x>0$ lies in $(a,1)$ for some $a in (0,1)$ so we get pointwise convergence but you don't get uniform convergence because no single $a$ works for all $x$.
$endgroup$
– Kavi Rama Murthy
Mar 13 at 6:44
$begingroup$
So it's like our diagonal sequence uniformly converges on $[a,1],forall ain (0,1)$, just can't reach $[0,1]$ right? And that is because we only consider the supreme distance between $F_n$ and it's poinwise limit $F$ on $[a,1]$ but never $[0,1]$, so we can't find an $n_epsilon$ such that the suprem distance on $[0,1]$ is less than $epsilon$
$endgroup$
– Fluffy Skye
Mar 13 at 6:51
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%2f3146172%2fsequence-of-integral-functions-converges%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
$begingroup$
Use a diagonal argument. For each $n$ you have a subsequence that converges uniformly on $(frac 1 n ,1)$ and so you can construct a diagonal subsequence which converges at every point. [Note that $F_n(0)=0$].
answered Mar 13 at 6:35
Kavi Rama MurthyKavi Rama Murthy
68.9k53169
$endgroup$
1
$begingroup$
Umm, so I first pick a subsequence that converges uniformly on $[1/2,1]$, and pick from it a subsequence that converges uniformly on $[1/3,1]$ and so on. Then pick the diagonal element of all these sequences?
$endgroup$
– Fluffy Skye
Mar 13 at 6:37
$begingroup$
@FluffySkye Exactly. You got the idea.
$endgroup$
– Kavi Rama Murthy
Mar 13 at 6:38
$begingroup$
But I don't understand how we end up with poinwise convergence though. Is it because $forall xin (0,1]$, far far along down our diagonal sequence it will converge, but 0 is not ok? I don't quite get it.
$endgroup$
– Fluffy Skye
Mar 13 at 6:41
1
$begingroup$
A given $x>0$ lies in $(a,1)$ for some $a in (0,1)$ so we get pointwise convergence but you don't get uniform convergence because no single $a$ works for all $x$.
$endgroup$
– Kavi Rama Murthy
Mar 13 at 6:44
$begingroup$
So it's like our diagonal sequence uniformly converges on $[a,1],forall ain (0,1)$, just can't reach $[0,1]$ right? And that is because we only consider the supreme distance between $F_n$ and it's poinwise limit $F$ on $[a,1]$ but never $[0,1]$, so we can't find an $n_epsilon$ such that the suprem distance on $[0,1]$ is less than $epsilon$
$endgroup$
– Fluffy Skye
Mar 13 at 6:51
add a comment |
$begingroup$
Use a diagonal argument. For each $n$ you have a subsequence that converges uniformly on $(frac 1 n ,1)$ and so you can construct a diagonal subsequence which converges at every point. [Note that $F_n(0)=0$].
answered Mar 13 at 6:35
Kavi Rama MurthyKavi Rama Murthy
68.9k53169
$endgroup$
1
$begingroup$
Umm, so I first pick a subsequence that converges uniformly on $[1/2,1]$, and pick from it a subsequence that converges uniformly on $[1/3,1]$ and so on. Then pick the diagonal element of all these sequences?
$endgroup$
– Fluffy Skye
Mar 13 at 6:37
$begingroup$
@FluffySkye Exactly. You got the idea.
$endgroup$
– Kavi Rama Murthy
Mar 13 at 6:38
$begingroup$
But I don't understand how we end up with poinwise convergence though. Is it because $forall xin (0,1]$, far far along down our diagonal sequence it will converge, but 0 is not ok? I don't quite get it.
$endgroup$
– Fluffy Skye
Mar 13 at 6:41
1
$begingroup$
A given $x>0$ lies in $(a,1)$ for some $a in (0,1)$ so we get pointwise convergence but you don't get uniform convergence because no single $a$ works for all $x$.
$endgroup$
– Kavi Rama Murthy
Mar 13 at 6:44
$begingroup$
So it's like our diagonal sequence uniformly converges on $[a,1],forall ain (0,1)$, just can't reach $[0,1]$ right? And that is because we only consider the supreme distance between $F_n$ and it's poinwise limit $F$ on $[a,1]$ but never $[0,1]$, so we can't find an $n_epsilon$ such that the suprem distance on $[0,1]$ is less than $epsilon$
$endgroup$
– Fluffy Skye
Mar 13 at 6:51
add a comment |
$begingroup$
Use a diagonal argument. For each $n$ you have a subsequence that converges uniformly on $(frac 1 n ,1)$ and so you can construct a diagonal subsequence which converges at every point. [Note that $F_n(0)=0$].
answered Mar 13 at 6:35
Kavi Rama MurthyKavi Rama Murthy
68.9k53169
$endgroup$
Use a diagonal argument. For each $n$ you have a subsequence that converges uniformly on $(frac 1 n ,1)$ and so you can construct a diagonal subsequence which converges at every point. [Note that $F_n(0)=0$].
answered Mar 13 at 6:35
Kavi Rama MurthyKavi Rama Murthy
68.9k53169
answered Mar 13 at 6:35
Kavi Rama MurthyKavi Rama Murthy
68.9k53169
answered Mar 13 at 6:35
Kavi Rama MurthyKavi Rama Murthy
68.9k53169
answered Mar 13 at 6:35
Kavi Rama MurthyKavi Rama Murthy
68.9k53169
68.9k53169
1
$begingroup$
Umm, so I first pick a subsequence that converges uniformly on $[1/2,1]$, and pick from it a subsequence that converges uniformly on $[1/3,1]$ and so on. Then pick the diagonal element of all these sequences?
$endgroup$
– Fluffy Skye
Mar 13 at 6:37
$begingroup$
@FluffySkye Exactly. You got the idea.
$endgroup$
– Kavi Rama Murthy
Mar 13 at 6:38
$begingroup$
But I don't understand how we end up with poinwise convergence though. Is it because $forall xin (0,1]$, far far along down our diagonal sequence it will converge, but 0 is not ok? I don't quite get it.
$endgroup$
– Fluffy Skye
Mar 13 at 6:41
1
$begingroup$
A given $x>0$ lies in $(a,1)$ for some $a in (0,1)$ so we get pointwise convergence but you don't get uniform convergence because no single $a$ works for all $x$.
$endgroup$
– Kavi Rama Murthy
Mar 13 at 6:44
$begingroup$
So it's like our diagonal sequence uniformly converges on $[a,1],forall ain (0,1)$, just can't reach $[0,1]$ right? And that is because we only consider the supreme distance between $F_n$ and it's poinwise limit $F$ on $[a,1]$ but never $[0,1]$, so we can't find an $n_epsilon$ such that the suprem distance on $[0,1]$ is less than $epsilon$
$endgroup$
– Fluffy Skye
Mar 13 at 6:51
add a comment |
1
$begingroup$
Umm, so I first pick a subsequence that converges uniformly on $[1/2,1]$, and pick from it a subsequence that converges uniformly on $[1/3,1]$ and so on. Then pick the diagonal element of all these sequences?
$endgroup$
– Fluffy Skye
Mar 13 at 6:37
$begingroup$
@FluffySkye Exactly. You got the idea.
$endgroup$
– Kavi Rama Murthy
Mar 13 at 6:38
$begingroup$
But I don't understand how we end up with poinwise convergence though. Is it because $forall xin (0,1]$, far far along down our diagonal sequence it will converge, but 0 is not ok? I don't quite get it.
$endgroup$
– Fluffy Skye
Mar 13 at 6:41
1
$begingroup$
A given $x>0$ lies in $(a,1)$ for some $a in (0,1)$ so we get pointwise convergence but you don't get uniform convergence because no single $a$ works for all $x$.
$endgroup$
– Kavi Rama Murthy
Mar 13 at 6:44
$begingroup$
So it's like our diagonal sequence uniformly converges on $[a,1],forall ain (0,1)$, just can't reach $[0,1]$ right? And that is because we only consider the supreme distance between $F_n$ and it's poinwise limit $F$ on $[a,1]$ but never $[0,1]$, so we can't find an $n_epsilon$ such that the suprem distance on $[0,1]$ is less than $epsilon$
$endgroup$
– Fluffy Skye
Mar 13 at 6:51
1
1
$begingroup$
Umm, so I first pick a subsequence that converges uniformly on $[1/2,1]$, and pick from it a subsequence that converges uniformly on $[1/3,1]$ and so on. Then pick the diagonal element of all these sequences?
$endgroup$
– Fluffy Skye
Mar 13 at 6:37
$begingroup$
Umm, so I first pick a subsequence that converges uniformly on $[1/2,1]$, and pick from it a subsequence that converges uniformly on $[1/3,1]$ and so on. Then pick the diagonal element of all these sequences?
$endgroup$
– Fluffy Skye
Mar 13 at 6:37
$begingroup$
@FluffySkye Exactly. You got the idea.
$endgroup$
– Kavi Rama Murthy
Mar 13 at 6:38
$begingroup$
@FluffySkye Exactly. You got the idea.
$endgroup$
– Kavi Rama Murthy
Mar 13 at 6:38
$begingroup$
But I don't understand how we end up with poinwise convergence though. Is it because $forall xin (0,1]$, far far along down our diagonal sequence it will converge, but 0 is not ok? I don't quite get it.
$endgroup$
– Fluffy Skye
Mar 13 at 6:41
$begingroup$
But I don't understand how we end up with poinwise convergence though. Is it because $forall xin (0,1]$, far far along down our diagonal sequence it will converge, but 0 is not ok? I don't quite get it.
$endgroup$
– Fluffy Skye
Mar 13 at 6:41
1
1
$begingroup$
A given $x>0$ lies in $(a,1)$ for some $a in (0,1)$ so we get pointwise convergence but you don't get uniform convergence because no single $a$ works for all $x$.
$endgroup$
– Kavi Rama Murthy
Mar 13 at 6:44
$begingroup$
A given $x>0$ lies in $(a,1)$ for some $a in (0,1)$ so we get pointwise convergence but you don't get uniform convergence because no single $a$ works for all $x$.
$endgroup$
– Kavi Rama Murthy
Mar 13 at 6:44
$begingroup$
So it's like our diagonal sequence uniformly converges on $[a,1],forall ain (0,1)$, just can't reach $[0,1]$ right? And that is because we only consider the supreme distance between $F_n$ and it's poinwise limit $F$ on $[a,1]$ but never $[0,1]$, so we can't find an $n_epsilon$ such that the suprem distance on $[0,1]$ is less than $epsilon$
$endgroup$
– Fluffy Skye
Mar 13 at 6:51
$begingroup$
So it's like our diagonal sequence uniformly converges on $[a,1],forall ain (0,1)$, just can't reach $[0,1]$ right? And that is because we only consider the supreme distance between $F_n$ and it's poinwise limit $F$ on $[a,1]$ but never $[0,1]$, so we can't find an $n_epsilon$ such that the suprem distance on $[0,1]$ is less than $epsilon$
$endgroup$
– Fluffy Skye
Mar 13 at 6:51
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%2f3146172%2fsequence-of-integral-functions-converges%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
$begingroup$
Uniform implies pointwise... no?
$endgroup$
– Sean Roberson
Mar 13 at 6:30
$begingroup$
Yes, but it's on $[a,1]$ not $[0,1]$.
$endgroup$
– Fluffy Skye
Mar 13 at 6:31