Compactness of set of indicator functions
$begingroup$
Let $chi_A(x)$ denote an indicator function on $Asubset [0,1]$. Consider the set
$$K={chi_A(x): text{ A is Lebesgue measurable in }[0,1]}.$$
Is this set compact in $L^infty(0,1)$ with respect to weak-* topology? How about weak topology on $L^infty(0,1)$?
Using Banach-Alaoglu, the set $K$ is bounded and closed in $L^infty(0,1)$ so it is compact with respect to weak-* topology. It is closed, because if $chi_{A_1}$ converges to $f$ in $L^infty(0,1)$, then $f$ has to be an indicator function.
fa.functional-analysis real-analysis measure-theory convex-analysis compactness
asked yesterday
Sara WinsletSara Winslet
1438
$endgroup$
add a comment |
$begingroup$
Let $chi_A(x)$ denote an indicator function on $Asubset [0,1]$. Consider the set
$$K={chi_A(x): text{ A is Lebesgue measurable in }[0,1]}.$$
Is this set compact in $L^infty(0,1)$ with respect to weak-* topology? How about weak topology on $L^infty(0,1)$?
Using Banach-Alaoglu, the set $K$ is bounded and closed in $L^infty(0,1)$ so it is compact with respect to weak-* topology. It is closed, because if $chi_{A_1}$ converges to $f$ in $L^infty(0,1)$, then $f$ has to be an indicator function.
fa.functional-analysis real-analysis measure-theory convex-analysis compactness
asked yesterday
Sara WinsletSara Winslet
1438
$endgroup$
add a comment |
$begingroup$
Let $chi_A(x)$ denote an indicator function on $Asubset [0,1]$. Consider the set
$$K={chi_A(x): text{ A is Lebesgue measurable in }[0,1]}.$$
Is this set compact in $L^infty(0,1)$ with respect to weak-* topology? How about weak topology on $L^infty(0,1)$?
Using Banach-Alaoglu, the set $K$ is bounded and closed in $L^infty(0,1)$ so it is compact with respect to weak-* topology. It is closed, because if $chi_{A_1}$ converges to $f$ in $L^infty(0,1)$, then $f$ has to be an indicator function.
fa.functional-analysis real-analysis measure-theory convex-analysis compactness
asked yesterday
Sara WinsletSara Winslet
1438
$endgroup$
Let $chi_A(x)$ denote an indicator function on $Asubset [0,1]$. Consider the set
$$K={chi_A(x): text{ A is Lebesgue measurable in }[0,1]}.$$
Is this set compact in $L^infty(0,1)$ with respect to weak-* topology? How about weak topology on $L^infty(0,1)$?
Using Banach-Alaoglu, the set $K$ is bounded and closed in $L^infty(0,1)$ so it is compact with respect to weak-* topology. It is closed, because if $chi_{A_1}$ converges to $f$ in $L^infty(0,1)$, then $f$ has to be an indicator function.
fa.functional-analysis real-analysis measure-theory convex-analysis compactness
fa.functional-analysis real-analysis measure-theory convex-analysis compactness
asked yesterday
Sara WinsletSara Winslet
1438
asked yesterday
Sara WinsletSara Winslet
1438
asked yesterday
Sara WinsletSara Winslet
1438
asked yesterday
Sara WinsletSara Winslet
1438
asked yesterday
Sara WinsletSara Winslet
1438
1438
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This set is not closed in the weak* topology. Indeed, for each $n$ consider the set $A_n=: [0, frac{1}{2n})cup [frac{2}{2n}, frac{3}{2n}] cupdotscup [frac{2n-2}{2n}, frac{2n-1}{2n})$, i.e. the sum of intervals that covers half of the interval. I claim that the sequence $chi_{A_n}$ converges to the function $frac{1}{2}$ in the weak* topology. Indeed, since the sequence is uniformly bounded, it suffices to check the convergence on a dense subset of $L^1$; we will take the continuous functions. Continuous functions on $[0,1]$ are uniformly continuous, so if we fix a function $f$, then for sufficiently big $n$ it will be almost constant on every interval of the form $[frac{k}{n}, frac{k+1}{n})$, so the integral on $[frac{2k}{2n}, frac{2k+1}{2n})$ will be almost equal to the integral on $[frac{2k+1}{2n}, frac{2k+2}{2n})$, up to $frac{varepsilon}{n}$, where $frac{1}{n}$ is coming from the length of the interval. It follows that the integral on $A_n$ is, up to $varepsilon$, equal to $frac{1}{2} int_{0}^{1} f(x) dx$.
On the other hand, this sequence does not have a convergent subnet in the weak topology. Indeed, if it did, it would have to converge to $frac{1}{2}$. Recall that the weak closure of a convex set is equal to its norm closure. It means that if I take the convex hull of the functions $chi_{A_n}$, then I can find a sequence of elements converging uniformly to $frac{1}{2}$. Note that $chi_{A_n}$ vanishes on the interval $[frac{2n-1}{2n}, 1]$, so any element of the convex hull vanishes on an interval containing $1$. It follows that the distance of any element in the convex hull to $frac{1}{2}$ is at least $frac{1}{2}$, so there can be no convergence.
answered yesterday
Mateusz WasilewskiMateusz Wasilewski
3,1551023
$endgroup$
$begingroup$
In the first sentence, do you mean to say that the set is not closed in the weak (and not weak-*) topology?
$endgroup$
– Willie Wong
yesterday
1
$begingroup$
@WillieWong No, I really wanted to say weak*, to correct the erroneous claim from the OP that this set is compact in the weak*-topology. The second part of the answer was devoted to the weak topology; it was not really necessary, but I thought that an explicit counterexample in this setting would be illuminating. Maybe I will edit the answer later to make it clearer.
$endgroup$
– Mateusz Wasilewski
yesterday
$begingroup$
After re-reading your answer four times, I finally figure out what I misunderstood: I kept thinking of the constant function 1/2 as being just like a characteristic function, but of course it is not in the set K since the OP is not allowing changing the coefficient.
$endgroup$
– Willie Wong
7 hours ago
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: "504"
};
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%2fmathoverflow.net%2fquestions%2f325432%2fcompactness-of-set-of-indicator-functions%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
$begingroup$
This set is not closed in the weak* topology. Indeed, for each $n$ consider the set $A_n=: [0, frac{1}{2n})cup [frac{2}{2n}, frac{3}{2n}] cupdotscup [frac{2n-2}{2n}, frac{2n-1}{2n})$, i.e. the sum of intervals that covers half of the interval. I claim that the sequence $chi_{A_n}$ converges to the function $frac{1}{2}$ in the weak* topology. Indeed, since the sequence is uniformly bounded, it suffices to check the convergence on a dense subset of $L^1$; we will take the continuous functions. Continuous functions on $[0,1]$ are uniformly continuous, so if we fix a function $f$, then for sufficiently big $n$ it will be almost constant on every interval of the form $[frac{k}{n}, frac{k+1}{n})$, so the integral on $[frac{2k}{2n}, frac{2k+1}{2n})$ will be almost equal to the integral on $[frac{2k+1}{2n}, frac{2k+2}{2n})$, up to $frac{varepsilon}{n}$, where $frac{1}{n}$ is coming from the length of the interval. It follows that the integral on $A_n$ is, up to $varepsilon$, equal to $frac{1}{2} int_{0}^{1} f(x) dx$.
On the other hand, this sequence does not have a convergent subnet in the weak topology. Indeed, if it did, it would have to converge to $frac{1}{2}$. Recall that the weak closure of a convex set is equal to its norm closure. It means that if I take the convex hull of the functions $chi_{A_n}$, then I can find a sequence of elements converging uniformly to $frac{1}{2}$. Note that $chi_{A_n}$ vanishes on the interval $[frac{2n-1}{2n}, 1]$, so any element of the convex hull vanishes on an interval containing $1$. It follows that the distance of any element in the convex hull to $frac{1}{2}$ is at least $frac{1}{2}$, so there can be no convergence.
answered yesterday
Mateusz WasilewskiMateusz Wasilewski
3,1551023
$endgroup$
$begingroup$
In the first sentence, do you mean to say that the set is not closed in the weak (and not weak-*) topology?
$endgroup$
– Willie Wong
yesterday
1
$begingroup$
@WillieWong No, I really wanted to say weak*, to correct the erroneous claim from the OP that this set is compact in the weak*-topology. The second part of the answer was devoted to the weak topology; it was not really necessary, but I thought that an explicit counterexample in this setting would be illuminating. Maybe I will edit the answer later to make it clearer.
$endgroup$
– Mateusz Wasilewski
yesterday
$begingroup$
After re-reading your answer four times, I finally figure out what I misunderstood: I kept thinking of the constant function 1/2 as being just like a characteristic function, but of course it is not in the set K since the OP is not allowing changing the coefficient.
$endgroup$
– Willie Wong
7 hours ago
add a comment |
$begingroup$
This set is not closed in the weak* topology. Indeed, for each $n$ consider the set $A_n=: [0, frac{1}{2n})cup [frac{2}{2n}, frac{3}{2n}] cupdotscup [frac{2n-2}{2n}, frac{2n-1}{2n})$, i.e. the sum of intervals that covers half of the interval. I claim that the sequence $chi_{A_n}$ converges to the function $frac{1}{2}$ in the weak* topology. Indeed, since the sequence is uniformly bounded, it suffices to check the convergence on a dense subset of $L^1$; we will take the continuous functions. Continuous functions on $[0,1]$ are uniformly continuous, so if we fix a function $f$, then for sufficiently big $n$ it will be almost constant on every interval of the form $[frac{k}{n}, frac{k+1}{n})$, so the integral on $[frac{2k}{2n}, frac{2k+1}{2n})$ will be almost equal to the integral on $[frac{2k+1}{2n}, frac{2k+2}{2n})$, up to $frac{varepsilon}{n}$, where $frac{1}{n}$ is coming from the length of the interval. It follows that the integral on $A_n$ is, up to $varepsilon$, equal to $frac{1}{2} int_{0}^{1} f(x) dx$.
On the other hand, this sequence does not have a convergent subnet in the weak topology. Indeed, if it did, it would have to converge to $frac{1}{2}$. Recall that the weak closure of a convex set is equal to its norm closure. It means that if I take the convex hull of the functions $chi_{A_n}$, then I can find a sequence of elements converging uniformly to $frac{1}{2}$. Note that $chi_{A_n}$ vanishes on the interval $[frac{2n-1}{2n}, 1]$, so any element of the convex hull vanishes on an interval containing $1$. It follows that the distance of any element in the convex hull to $frac{1}{2}$ is at least $frac{1}{2}$, so there can be no convergence.
answered yesterday
Mateusz WasilewskiMateusz Wasilewski
3,1551023
$endgroup$
$begingroup$
In the first sentence, do you mean to say that the set is not closed in the weak (and not weak-*) topology?
$endgroup$
– Willie Wong
yesterday
1
$begingroup$
@WillieWong No, I really wanted to say weak*, to correct the erroneous claim from the OP that this set is compact in the weak*-topology. The second part of the answer was devoted to the weak topology; it was not really necessary, but I thought that an explicit counterexample in this setting would be illuminating. Maybe I will edit the answer later to make it clearer.
$endgroup$
– Mateusz Wasilewski
yesterday
$begingroup$
After re-reading your answer four times, I finally figure out what I misunderstood: I kept thinking of the constant function 1/2 as being just like a characteristic function, but of course it is not in the set K since the OP is not allowing changing the coefficient.
$endgroup$
– Willie Wong
7 hours ago
add a comment |
$begingroup$
This set is not closed in the weak* topology. Indeed, for each $n$ consider the set $A_n=: [0, frac{1}{2n})cup [frac{2}{2n}, frac{3}{2n}] cupdotscup [frac{2n-2}{2n}, frac{2n-1}{2n})$, i.e. the sum of intervals that covers half of the interval. I claim that the sequence $chi_{A_n}$ converges to the function $frac{1}{2}$ in the weak* topology. Indeed, since the sequence is uniformly bounded, it suffices to check the convergence on a dense subset of $L^1$; we will take the continuous functions. Continuous functions on $[0,1]$ are uniformly continuous, so if we fix a function $f$, then for sufficiently big $n$ it will be almost constant on every interval of the form $[frac{k}{n}, frac{k+1}{n})$, so the integral on $[frac{2k}{2n}, frac{2k+1}{2n})$ will be almost equal to the integral on $[frac{2k+1}{2n}, frac{2k+2}{2n})$, up to $frac{varepsilon}{n}$, where $frac{1}{n}$ is coming from the length of the interval. It follows that the integral on $A_n$ is, up to $varepsilon$, equal to $frac{1}{2} int_{0}^{1} f(x) dx$.
On the other hand, this sequence does not have a convergent subnet in the weak topology. Indeed, if it did, it would have to converge to $frac{1}{2}$. Recall that the weak closure of a convex set is equal to its norm closure. It means that if I take the convex hull of the functions $chi_{A_n}$, then I can find a sequence of elements converging uniformly to $frac{1}{2}$. Note that $chi_{A_n}$ vanishes on the interval $[frac{2n-1}{2n}, 1]$, so any element of the convex hull vanishes on an interval containing $1$. It follows that the distance of any element in the convex hull to $frac{1}{2}$ is at least $frac{1}{2}$, so there can be no convergence.
answered yesterday
Mateusz WasilewskiMateusz Wasilewski
3,1551023
$endgroup$
This set is not closed in the weak* topology. Indeed, for each $n$ consider the set $A_n=: [0, frac{1}{2n})cup [frac{2}{2n}, frac{3}{2n}] cupdotscup [frac{2n-2}{2n}, frac{2n-1}{2n})$, i.e. the sum of intervals that covers half of the interval. I claim that the sequence $chi_{A_n}$ converges to the function $frac{1}{2}$ in the weak* topology. Indeed, since the sequence is uniformly bounded, it suffices to check the convergence on a dense subset of $L^1$; we will take the continuous functions. Continuous functions on $[0,1]$ are uniformly continuous, so if we fix a function $f$, then for sufficiently big $n$ it will be almost constant on every interval of the form $[frac{k}{n}, frac{k+1}{n})$, so the integral on $[frac{2k}{2n}, frac{2k+1}{2n})$ will be almost equal to the integral on $[frac{2k+1}{2n}, frac{2k+2}{2n})$, up to $frac{varepsilon}{n}$, where $frac{1}{n}$ is coming from the length of the interval. It follows that the integral on $A_n$ is, up to $varepsilon$, equal to $frac{1}{2} int_{0}^{1} f(x) dx$.
On the other hand, this sequence does not have a convergent subnet in the weak topology. Indeed, if it did, it would have to converge to $frac{1}{2}$. Recall that the weak closure of a convex set is equal to its norm closure. It means that if I take the convex hull of the functions $chi_{A_n}$, then I can find a sequence of elements converging uniformly to $frac{1}{2}$. Note that $chi_{A_n}$ vanishes on the interval $[frac{2n-1}{2n}, 1]$, so any element of the convex hull vanishes on an interval containing $1$. It follows that the distance of any element in the convex hull to $frac{1}{2}$ is at least $frac{1}{2}$, so there can be no convergence.
answered yesterday
Mateusz WasilewskiMateusz Wasilewski
3,1551023
answered yesterday
Mateusz WasilewskiMateusz Wasilewski
3,1551023
answered yesterday
Mateusz WasilewskiMateusz Wasilewski
3,1551023
answered yesterday
Mateusz WasilewskiMateusz Wasilewski
3,1551023
3,1551023
$begingroup$
In the first sentence, do you mean to say that the set is not closed in the weak (and not weak-*) topology?
$endgroup$
– Willie Wong
yesterday
1
$begingroup$
@WillieWong No, I really wanted to say weak*, to correct the erroneous claim from the OP that this set is compact in the weak*-topology. The second part of the answer was devoted to the weak topology; it was not really necessary, but I thought that an explicit counterexample in this setting would be illuminating. Maybe I will edit the answer later to make it clearer.
$endgroup$
– Mateusz Wasilewski
yesterday
$begingroup$
After re-reading your answer four times, I finally figure out what I misunderstood: I kept thinking of the constant function 1/2 as being just like a characteristic function, but of course it is not in the set K since the OP is not allowing changing the coefficient.
$endgroup$
– Willie Wong
7 hours ago
add a comment |
$begingroup$
In the first sentence, do you mean to say that the set is not closed in the weak (and not weak-*) topology?
$endgroup$
– Willie Wong
yesterday
1
$begingroup$
@WillieWong No, I really wanted to say weak*, to correct the erroneous claim from the OP that this set is compact in the weak*-topology. The second part of the answer was devoted to the weak topology; it was not really necessary, but I thought that an explicit counterexample in this setting would be illuminating. Maybe I will edit the answer later to make it clearer.
$endgroup$
– Mateusz Wasilewski
yesterday
$begingroup$
After re-reading your answer four times, I finally figure out what I misunderstood: I kept thinking of the constant function 1/2 as being just like a characteristic function, but of course it is not in the set K since the OP is not allowing changing the coefficient.
$endgroup$
– Willie Wong
7 hours ago
$begingroup$
In the first sentence, do you mean to say that the set is not closed in the weak (and not weak-*) topology?
$endgroup$
– Willie Wong
yesterday
$begingroup$
In the first sentence, do you mean to say that the set is not closed in the weak (and not weak-*) topology?
$endgroup$
– Willie Wong
yesterday
1
1
$begingroup$
@WillieWong No, I really wanted to say weak*, to correct the erroneous claim from the OP that this set is compact in the weak*-topology. The second part of the answer was devoted to the weak topology; it was not really necessary, but I thought that an explicit counterexample in this setting would be illuminating. Maybe I will edit the answer later to make it clearer.
$endgroup$
– Mateusz Wasilewski
yesterday
$begingroup$
@WillieWong No, I really wanted to say weak*, to correct the erroneous claim from the OP that this set is compact in the weak*-topology. The second part of the answer was devoted to the weak topology; it was not really necessary, but I thought that an explicit counterexample in this setting would be illuminating. Maybe I will edit the answer later to make it clearer.
$endgroup$
– Mateusz Wasilewski
yesterday
$begingroup$
After re-reading your answer four times, I finally figure out what I misunderstood: I kept thinking of the constant function 1/2 as being just like a characteristic function, but of course it is not in the set K since the OP is not allowing changing the coefficient.
$endgroup$
– Willie Wong
7 hours ago
$begingroup$
After re-reading your answer four times, I finally figure out what I misunderstood: I kept thinking of the constant function 1/2 as being just like a characteristic function, but of course it is not in the set K since the OP is not allowing changing the coefficient.
$endgroup$
– Willie Wong
7 hours ago
add a comment |
Thanks for contributing an answer to MathOverflow!
- 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%2fmathoverflow.net%2fquestions%2f325432%2fcompactness-of-set-of-indicator-functions%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
