Example of convergence under $|cdot|_2$, but not pointwise.
$begingroup$
Let $(f_n)_n$ be a sequence of continuous functions on $[0,1]$ and $f$ a continuous function on $[0,1]$. Assume that $|f_n-f|_2rightarrow 0$. Does it follow that $f_n(x)rightarrow f(x)$ for some $xin [0,1]$? Give a proof or counterexample. Also, can NOT use Lebesgue here.
$text{Proof}$
Let $f(x)=0$ and $I_{n,k}(x)={chi_{left[frac{k-1}{2^n},frac{k}{2^n}right]}:nin mathbb{N}, k=0,...,2^n}$ where $chi_{left[frac{k-1}{2^n},frac{k}{2^n}right]}$ is defined as
[begin{cases}
f_{n,k}(x) & text{if}space xin left[frac{k-1}{2^n},frac{k}{2^n}right] \
0 & text{if}space xnotin left[frac{k-1}{2^n},frac{k}{2^n}right] \
end{cases}
and $f_{n,k}(x)$ defined as
begin{cases}
(2^{n+1})x + (2-2k) & text{if}space xin left[ frac{k-1}{2^n},frac{2k-1}{2^{n+1}} right] \
-(2^{n+1})x + 2k & text{if}space xin left[ frac{2k-1}{2^{n+1}},frac{k}{2^n} right] \
end{cases}
Each "bump" is a triangle over the interval $left[frac{k-1}{2^n},frac{k}{2^n}right]$ with height $1$, so the area would be $frac{1}{2^{n+1}}$.
Hence, $largeint^{frac{k-1}{2^n}}_{frac{k}{2^n}}f_{n,k}(x)dx=frac{1}{2^{n+1}}$ and, therefore, we have that
$$ underset{nrightarrow infty}{lim}int^{1}_{0}I_{n,k}(x)dx=underset{nrightarrow infty}{lim}sum^{2^n}_{k=1}int^{frac{k-1}{2^n}}_{frac{k}{2^n}}f_{n,k}(x)dx=underset{nrightarrow infty}{lim}sum^{2^n}_{k=1}frac{1}{2^{n+1}}=sum^{infty}_{k=1}0=0 $$
Hence, the sequence converges under $|cdot|_2$, but for any $xin [0,1]$ we can choose $Nin mathbb{N}$ so that
$$ |f_{N,k}(x)|>frac{1}{2} $$
Therefore, $I_{n,k}(x)$ converges no-where to $f(x)$.
Can anyone please check if my reasoning is correct? I know I'm probably not very convincing at the end since I do not know what value to make $N$, but I know that the interval can be divided up enough so that it isn't $0$. I'm also not sure if I have shown convergence under $|cdot|_2$ here either.
real-analysis integration
$endgroup$
add a comment |
$begingroup$
Let $(f_n)_n$ be a sequence of continuous functions on $[0,1]$ and $f$ a continuous function on $[0,1]$. Assume that $|f_n-f|_2rightarrow 0$. Does it follow that $f_n(x)rightarrow f(x)$ for some $xin [0,1]$? Give a proof or counterexample. Also, can NOT use Lebesgue here.
$text{Proof}$
Let $f(x)=0$ and $I_{n,k}(x)={chi_{left[frac{k-1}{2^n},frac{k}{2^n}right]}:nin mathbb{N}, k=0,...,2^n}$ where $chi_{left[frac{k-1}{2^n},frac{k}{2^n}right]}$ is defined as
[begin{cases}
f_{n,k}(x) & text{if}space xin left[frac{k-1}{2^n},frac{k}{2^n}right] \
0 & text{if}space xnotin left[frac{k-1}{2^n},frac{k}{2^n}right] \
end{cases}
and $f_{n,k}(x)$ defined as
begin{cases}
(2^{n+1})x + (2-2k) & text{if}space xin left[ frac{k-1}{2^n},frac{2k-1}{2^{n+1}} right] \
-(2^{n+1})x + 2k & text{if}space xin left[ frac{2k-1}{2^{n+1}},frac{k}{2^n} right] \
end{cases}
Each "bump" is a triangle over the interval $left[frac{k-1}{2^n},frac{k}{2^n}right]$ with height $1$, so the area would be $frac{1}{2^{n+1}}$.
Hence, $largeint^{frac{k-1}{2^n}}_{frac{k}{2^n}}f_{n,k}(x)dx=frac{1}{2^{n+1}}$ and, therefore, we have that
$$ underset{nrightarrow infty}{lim}int^{1}_{0}I_{n,k}(x)dx=underset{nrightarrow infty}{lim}sum^{2^n}_{k=1}int^{frac{k-1}{2^n}}_{frac{k}{2^n}}f_{n,k}(x)dx=underset{nrightarrow infty}{lim}sum^{2^n}_{k=1}frac{1}{2^{n+1}}=sum^{infty}_{k=1}0=0 $$
Hence, the sequence converges under $|cdot|_2$, but for any $xin [0,1]$ we can choose $Nin mathbb{N}$ so that
$$ |f_{N,k}(x)|>frac{1}{2} $$
Therefore, $I_{n,k}(x)$ converges no-where to $f(x)$.
Can anyone please check if my reasoning is correct? I know I'm probably not very convincing at the end since I do not know what value to make $N$, but I know that the interval can be divided up enough so that it isn't $0$. I'm also not sure if I have shown convergence under $|cdot|_2$ here either.
real-analysis integration
$endgroup$
$begingroup$
In your statement of your question, when you say "for some $x in [0, 1]$", do you mean "for all $x in [0, 1]$"?
$endgroup$
– Theo Bendit
Oct 1 '18 at 3:46
$begingroup$
It is for some $xin [0,1]$.
$endgroup$
– Joe Man Analysis
Oct 1 '18 at 3:52
1
$begingroup$
In your last line you have $lim_{nto infty}A(n)=0$ where $A(n)=sum_{k=1}^{2^n}frac {1}{2^{n+1}}$ but this is $false$ because $A(n)=1-2^{-(n+1)}$. Furthermore what you are interested in is whether $int_0^1 (sum_{k=1}^nI_k)^2dx $ goes to $0.$
$endgroup$
– DanielWainfleet
Oct 1 '18 at 5:57
1
$begingroup$
BTW you can use text {....} between dollar-signs or between double-dollar signs, around material (like the word "if") that should be un-Latexed... E.g .text { if }. You can also use , and ; and quad to insert various sizes of extra space. E.g. in a displayed line I like to write ; text { if }; for the word "if"
$endgroup$
– DanielWainfleet
Oct 1 '18 at 6:06
add a comment |
$begingroup$
Let $(f_n)_n$ be a sequence of continuous functions on $[0,1]$ and $f$ a continuous function on $[0,1]$. Assume that $|f_n-f|_2rightarrow 0$. Does it follow that $f_n(x)rightarrow f(x)$ for some $xin [0,1]$? Give a proof or counterexample. Also, can NOT use Lebesgue here.
$text{Proof}$
Let $f(x)=0$ and $I_{n,k}(x)={chi_{left[frac{k-1}{2^n},frac{k}{2^n}right]}:nin mathbb{N}, k=0,...,2^n}$ where $chi_{left[frac{k-1}{2^n},frac{k}{2^n}right]}$ is defined as
[begin{cases}
f_{n,k}(x) & text{if}space xin left[frac{k-1}{2^n},frac{k}{2^n}right] \
0 & text{if}space xnotin left[frac{k-1}{2^n},frac{k}{2^n}right] \
end{cases}
and $f_{n,k}(x)$ defined as
begin{cases}
(2^{n+1})x + (2-2k) & text{if}space xin left[ frac{k-1}{2^n},frac{2k-1}{2^{n+1}} right] \
-(2^{n+1})x + 2k & text{if}space xin left[ frac{2k-1}{2^{n+1}},frac{k}{2^n} right] \
end{cases}
Each "bump" is a triangle over the interval $left[frac{k-1}{2^n},frac{k}{2^n}right]$ with height $1$, so the area would be $frac{1}{2^{n+1}}$.
Hence, $largeint^{frac{k-1}{2^n}}_{frac{k}{2^n}}f_{n,k}(x)dx=frac{1}{2^{n+1}}$ and, therefore, we have that
$$ underset{nrightarrow infty}{lim}int^{1}_{0}I_{n,k}(x)dx=underset{nrightarrow infty}{lim}sum^{2^n}_{k=1}int^{frac{k-1}{2^n}}_{frac{k}{2^n}}f_{n,k}(x)dx=underset{nrightarrow infty}{lim}sum^{2^n}_{k=1}frac{1}{2^{n+1}}=sum^{infty}_{k=1}0=0 $$
Hence, the sequence converges under $|cdot|_2$, but for any $xin [0,1]$ we can choose $Nin mathbb{N}$ so that
$$ |f_{N,k}(x)|>frac{1}{2} $$
Therefore, $I_{n,k}(x)$ converges no-where to $f(x)$.
Can anyone please check if my reasoning is correct? I know I'm probably not very convincing at the end since I do not know what value to make $N$, but I know that the interval can be divided up enough so that it isn't $0$. I'm also not sure if I have shown convergence under $|cdot|_2$ here either.
real-analysis integration
$endgroup$
Let $(f_n)_n$ be a sequence of continuous functions on $[0,1]$ and $f$ a continuous function on $[0,1]$. Assume that $|f_n-f|_2rightarrow 0$. Does it follow that $f_n(x)rightarrow f(x)$ for some $xin [0,1]$? Give a proof or counterexample. Also, can NOT use Lebesgue here.
$text{Proof}$
Let $f(x)=0$ and $I_{n,k}(x)={chi_{left[frac{k-1}{2^n},frac{k}{2^n}right]}:nin mathbb{N}, k=0,...,2^n}$ where $chi_{left[frac{k-1}{2^n},frac{k}{2^n}right]}$ is defined as
[begin{cases}
f_{n,k}(x) & text{if}space xin left[frac{k-1}{2^n},frac{k}{2^n}right] \
0 & text{if}space xnotin left[frac{k-1}{2^n},frac{k}{2^n}right] \
end{cases}
and $f_{n,k}(x)$ defined as
begin{cases}
(2^{n+1})x + (2-2k) & text{if}space xin left[ frac{k-1}{2^n},frac{2k-1}{2^{n+1}} right] \
-(2^{n+1})x + 2k & text{if}space xin left[ frac{2k-1}{2^{n+1}},frac{k}{2^n} right] \
end{cases}
Each "bump" is a triangle over the interval $left[frac{k-1}{2^n},frac{k}{2^n}right]$ with height $1$, so the area would be $frac{1}{2^{n+1}}$.
Hence, $largeint^{frac{k-1}{2^n}}_{frac{k}{2^n}}f_{n,k}(x)dx=frac{1}{2^{n+1}}$ and, therefore, we have that
$$ underset{nrightarrow infty}{lim}int^{1}_{0}I_{n,k}(x)dx=underset{nrightarrow infty}{lim}sum^{2^n}_{k=1}int^{frac{k-1}{2^n}}_{frac{k}{2^n}}f_{n,k}(x)dx=underset{nrightarrow infty}{lim}sum^{2^n}_{k=1}frac{1}{2^{n+1}}=sum^{infty}_{k=1}0=0 $$
Hence, the sequence converges under $|cdot|_2$, but for any $xin [0,1]$ we can choose $Nin mathbb{N}$ so that
$$ |f_{N,k}(x)|>frac{1}{2} $$
Therefore, $I_{n,k}(x)$ converges no-where to $f(x)$.
Can anyone please check if my reasoning is correct? I know I'm probably not very convincing at the end since I do not know what value to make $N$, but I know that the interval can be divided up enough so that it isn't $0$. I'm also not sure if I have shown convergence under $|cdot|_2$ here either.
real-analysis integration
real-analysis integration
edited Dec 9 '18 at 9:16
Gaby Alfonso
697315
697315
asked Oct 1 '18 at 3:39
Joe Man AnalysisJoe Man Analysis
33419
33419
$begingroup$
In your statement of your question, when you say "for some $x in [0, 1]$", do you mean "for all $x in [0, 1]$"?
$endgroup$
– Theo Bendit
Oct 1 '18 at 3:46
$begingroup$
It is for some $xin [0,1]$.
$endgroup$
– Joe Man Analysis
Oct 1 '18 at 3:52
1
$begingroup$
In your last line you have $lim_{nto infty}A(n)=0$ where $A(n)=sum_{k=1}^{2^n}frac {1}{2^{n+1}}$ but this is $false$ because $A(n)=1-2^{-(n+1)}$. Furthermore what you are interested in is whether $int_0^1 (sum_{k=1}^nI_k)^2dx $ goes to $0.$
$endgroup$
– DanielWainfleet
Oct 1 '18 at 5:57
1
$begingroup$
BTW you can use text {....} between dollar-signs or between double-dollar signs, around material (like the word "if") that should be un-Latexed... E.g .text { if }. You can also use , and ; and quad to insert various sizes of extra space. E.g. in a displayed line I like to write ; text { if }; for the word "if"
$endgroup$
– DanielWainfleet
Oct 1 '18 at 6:06
add a comment |
$begingroup$
In your statement of your question, when you say "for some $x in [0, 1]$", do you mean "for all $x in [0, 1]$"?
$endgroup$
– Theo Bendit
Oct 1 '18 at 3:46
$begingroup$
It is for some $xin [0,1]$.
$endgroup$
– Joe Man Analysis
Oct 1 '18 at 3:52
1
$begingroup$
In your last line you have $lim_{nto infty}A(n)=0$ where $A(n)=sum_{k=1}^{2^n}frac {1}{2^{n+1}}$ but this is $false$ because $A(n)=1-2^{-(n+1)}$. Furthermore what you are interested in is whether $int_0^1 (sum_{k=1}^nI_k)^2dx $ goes to $0.$
$endgroup$
– DanielWainfleet
Oct 1 '18 at 5:57
1
$begingroup$
BTW you can use text {....} between dollar-signs or between double-dollar signs, around material (like the word "if") that should be un-Latexed... E.g .text { if }. You can also use , and ; and quad to insert various sizes of extra space. E.g. in a displayed line I like to write ; text { if }; for the word "if"
$endgroup$
– DanielWainfleet
Oct 1 '18 at 6:06
$begingroup$
In your statement of your question, when you say "for some $x in [0, 1]$", do you mean "for all $x in [0, 1]$"?
$endgroup$
– Theo Bendit
Oct 1 '18 at 3:46
$begingroup$
In your statement of your question, when you say "for some $x in [0, 1]$", do you mean "for all $x in [0, 1]$"?
$endgroup$
– Theo Bendit
Oct 1 '18 at 3:46
$begingroup$
It is for some $xin [0,1]$.
$endgroup$
– Joe Man Analysis
Oct 1 '18 at 3:52
$begingroup$
It is for some $xin [0,1]$.
$endgroup$
– Joe Man Analysis
Oct 1 '18 at 3:52
1
1
$begingroup$
In your last line you have $lim_{nto infty}A(n)=0$ where $A(n)=sum_{k=1}^{2^n}frac {1}{2^{n+1}}$ but this is $false$ because $A(n)=1-2^{-(n+1)}$. Furthermore what you are interested in is whether $int_0^1 (sum_{k=1}^nI_k)^2dx $ goes to $0.$
$endgroup$
– DanielWainfleet
Oct 1 '18 at 5:57
$begingroup$
In your last line you have $lim_{nto infty}A(n)=0$ where $A(n)=sum_{k=1}^{2^n}frac {1}{2^{n+1}}$ but this is $false$ because $A(n)=1-2^{-(n+1)}$. Furthermore what you are interested in is whether $int_0^1 (sum_{k=1}^nI_k)^2dx $ goes to $0.$
$endgroup$
– DanielWainfleet
Oct 1 '18 at 5:57
1
1
$begingroup$
BTW you can use text {....} between dollar-signs or between double-dollar signs, around material (like the word "if") that should be un-Latexed... E.g .text { if }. You can also use , and ; and quad to insert various sizes of extra space. E.g. in a displayed line I like to write ; text { if }; for the word "if"
$endgroup$
– DanielWainfleet
Oct 1 '18 at 6:06
$begingroup$
BTW you can use text {....} between dollar-signs or between double-dollar signs, around material (like the word "if") that should be un-Latexed... E.g .text { if }. You can also use , and ; and quad to insert various sizes of extra space. E.g. in a displayed line I like to write ; text { if }; for the word "if"
$endgroup$
– DanielWainfleet
Oct 1 '18 at 6:06
add a comment |
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%2f2937538%2fexample-of-convergence-under-cdot-2-but-not-pointwise%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%2f2937538%2fexample-of-convergence-under-cdot-2-but-not-pointwise%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
$begingroup$
In your statement of your question, when you say "for some $x in [0, 1]$", do you mean "for all $x in [0, 1]$"?
$endgroup$
– Theo Bendit
Oct 1 '18 at 3:46
$begingroup$
It is for some $xin [0,1]$.
$endgroup$
– Joe Man Analysis
Oct 1 '18 at 3:52
1
$begingroup$
In your last line you have $lim_{nto infty}A(n)=0$ where $A(n)=sum_{k=1}^{2^n}frac {1}{2^{n+1}}$ but this is $false$ because $A(n)=1-2^{-(n+1)}$. Furthermore what you are interested in is whether $int_0^1 (sum_{k=1}^nI_k)^2dx $ goes to $0.$
$endgroup$
– DanielWainfleet
Oct 1 '18 at 5:57
1
$begingroup$
BTW you can use text {....} between dollar-signs or between double-dollar signs, around material (like the word "if") that should be un-Latexed... E.g .text { if }. You can also use , and ; and quad to insert various sizes of extra space. E.g. in a displayed line I like to write ; text { if }; for the word "if"
$endgroup$
– DanielWainfleet
Oct 1 '18 at 6:06