Mixing
Proving that continuity and the lim sup of a given set being 0 are equivalent
Suppose we have a function defined as follows:
$alpha(f,x_o)=limsup{|f(a)-f(b)|:a,bin (x_o-frac{1}{n},x_o+frac{1}{n})}$
with $f:Rrightarrow R$ and $x_o in R$. I need to prove that $alpha=0$ iff $f$ is continuous at $x_o$.
It seems trivial to show the $alpha=0 Rightarrow$ direction, since $|f(a)-f(b)|geq0$, so if the $lim,sup$ is $0$ we must have $f(a)=f(b)$ everywhere on the interval.
However, I'm really drawing a blank on the other direction. How can I prove that continuity of $f$ implies that $alpha=0$? It seems like I should use the $epsilon-delta$ definition of continuity, since I'm working with $|f(a)-f(b)|$, but I think the $lim,sup$ stuff is throwing me off the scent. Any help would be much appreciated!
real-analysis continuity epsilon-delta limsup-and-liminf
edited Nov 21 '18 at 3:35
Chinnapparaj R
5,2601826
asked Nov 21 '18 at 2:52
notadoctor
947
add a comment |
Suppose we have a function defined as follows:
$alpha(f,x_o)=limsup{|f(a)-f(b)|:a,bin (x_o-frac{1}{n},x_o+frac{1}{n})}$
with $f:Rrightarrow R$ and $x_o in R$. I need to prove that $alpha=0$ iff $f$ is continuous at $x_o$.
It seems trivial to show the $alpha=0 Rightarrow$ direction, since $|f(a)-f(b)|geq0$, so if the $lim,sup$ is $0$ we must have $f(a)=f(b)$ everywhere on the interval.
However, I'm really drawing a blank on the other direction. How can I prove that continuity of $f$ implies that $alpha=0$? It seems like I should use the $epsilon-delta$ definition of continuity, since I'm working with $|f(a)-f(b)|$, but I think the $lim,sup$ stuff is throwing me off the scent. Any help would be much appreciated!
real-analysis continuity epsilon-delta limsup-and-liminf
edited Nov 21 '18 at 3:35
Chinnapparaj R
5,2601826
asked Nov 21 '18 at 2:52
notadoctor
947
Write out the definition of lim sup. It will become very obvious.
– Don Thousand
Nov 21 '18 at 2:59
Is it simply that as $nrightarrow infty$, $(a, b)rightarrow (x_o, x_o)$? But where do I use the fact that $f$ is continuous?
– notadoctor
Nov 21 '18 at 3:07
That's not the definition of lim sup...
– Don Thousand
Nov 21 '18 at 3:08
1
@notadoctor think of the $limsup$ as being two parts: the limit as $nrightarrow infty$, and the $sup$ over all pairs $a,b$ in that interval.
– user25959
Nov 21 '18 at 3:10
I'm still not able to convince myself adequately, I'm afraid. I think I'll have to spend some more time getting familiar with continuity.
– notadoctor
Nov 21 '18 at 4:02
add a comment |
Suppose we have a function defined as follows:
$alpha(f,x_o)=limsup{|f(a)-f(b)|:a,bin (x_o-frac{1}{n},x_o+frac{1}{n})}$
with $f:Rrightarrow R$ and $x_o in R$. I need to prove that $alpha=0$ iff $f$ is continuous at $x_o$.
It seems trivial to show the $alpha=0 Rightarrow$ direction, since $|f(a)-f(b)|geq0$, so if the $lim,sup$ is $0$ we must have $f(a)=f(b)$ everywhere on the interval.
However, I'm really drawing a blank on the other direction. How can I prove that continuity of $f$ implies that $alpha=0$? It seems like I should use the $epsilon-delta$ definition of continuity, since I'm working with $|f(a)-f(b)|$, but I think the $lim,sup$ stuff is throwing me off the scent. Any help would be much appreciated!
real-analysis continuity epsilon-delta limsup-and-liminf
edited Nov 21 '18 at 3:35
Chinnapparaj R
5,2601826
asked Nov 21 '18 at 2:52
notadoctor
947
Suppose we have a function defined as follows:
$alpha(f,x_o)=limsup{|f(a)-f(b)|:a,bin (x_o-frac{1}{n},x_o+frac{1}{n})}$
with $f:Rrightarrow R$ and $x_o in R$. I need to prove that $alpha=0$ iff $f$ is continuous at $x_o$.
It seems trivial to show the $alpha=0 Rightarrow$ direction, since $|f(a)-f(b)|geq0$, so if the $lim,sup$ is $0$ we must have $f(a)=f(b)$ everywhere on the interval.
However, I'm really drawing a blank on the other direction. How can I prove that continuity of $f$ implies that $alpha=0$? It seems like I should use the $epsilon-delta$ definition of continuity, since I'm working with $|f(a)-f(b)|$, but I think the $lim,sup$ stuff is throwing me off the scent. Any help would be much appreciated!
real-analysis continuity epsilon-delta limsup-and-liminf
real-analysis continuity epsilon-delta limsup-and-liminf
edited Nov 21 '18 at 3:35
Chinnapparaj R
5,2601826
asked Nov 21 '18 at 2:52
notadoctor
947
edited Nov 21 '18 at 3:35
Chinnapparaj R
5,2601826
asked Nov 21 '18 at 2:52
notadoctor
947
edited Nov 21 '18 at 3:35
Chinnapparaj R
5,2601826
edited Nov 21 '18 at 3:35
Chinnapparaj R
5,2601826
edited Nov 21 '18 at 3:35
Chinnapparaj R
5,2601826
5,2601826
asked Nov 21 '18 at 2:52
notadoctor
947
asked Nov 21 '18 at 2:52
notadoctor
947
asked Nov 21 '18 at 2:52
notadoctor
947
947
Write out the definition of lim sup. It will become very obvious.
– Don Thousand
Nov 21 '18 at 2:59
Is it simply that as $nrightarrow infty$, $(a, b)rightarrow (x_o, x_o)$? But where do I use the fact that $f$ is continuous?
– notadoctor
Nov 21 '18 at 3:07
That's not the definition of lim sup...
– Don Thousand
Nov 21 '18 at 3:08
1
@notadoctor think of the $limsup$ as being two parts: the limit as $nrightarrow infty$, and the $sup$ over all pairs $a,b$ in that interval.
– user25959
Nov 21 '18 at 3:10
I'm still not able to convince myself adequately, I'm afraid. I think I'll have to spend some more time getting familiar with continuity.
– notadoctor
Nov 21 '18 at 4:02
add a comment |
Write out the definition of lim sup. It will become very obvious.
– Don Thousand
Nov 21 '18 at 2:59
Is it simply that as $nrightarrow infty$, $(a, b)rightarrow (x_o, x_o)$? But where do I use the fact that $f$ is continuous?
– notadoctor
Nov 21 '18 at 3:07
That's not the definition of lim sup...
– Don Thousand
Nov 21 '18 at 3:08
1
@notadoctor think of the $limsup$ as being two parts: the limit as $nrightarrow infty$, and the $sup$ over all pairs $a,b$ in that interval.
– user25959
Nov 21 '18 at 3:10
I'm still not able to convince myself adequately, I'm afraid. I think I'll have to spend some more time getting familiar with continuity.
– notadoctor
Nov 21 '18 at 4:02
Write out the definition of lim sup. It will become very obvious.
– Don Thousand
Nov 21 '18 at 2:59
Write out the definition of lim sup. It will become very obvious.
– Don Thousand
Nov 21 '18 at 2:59
Is it simply that as $nrightarrow infty$, $(a, b)rightarrow (x_o, x_o)$? But where do I use the fact that $f$ is continuous?
– notadoctor
Nov 21 '18 at 3:07
Is it simply that as $nrightarrow infty$, $(a, b)rightarrow (x_o, x_o)$? But where do I use the fact that $f$ is continuous?
– notadoctor
Nov 21 '18 at 3:07
That's not the definition of lim sup...
– Don Thousand
Nov 21 '18 at 3:08
That's not the definition of lim sup...
– Don Thousand
Nov 21 '18 at 3:08
1
1
@notadoctor think of the $limsup$ as being two parts: the limit as $nrightarrow infty$, and the $sup$ over all pairs $a,b$ in that interval.
– user25959
Nov 21 '18 at 3:10
@notadoctor think of the $limsup$ as being two parts: the limit as $nrightarrow infty$, and the $sup$ over all pairs $a,b$ in that interval.
– user25959
Nov 21 '18 at 3:10
I'm still not able to convince myself adequately, I'm afraid. I think I'll have to spend some more time getting familiar with continuity.
– notadoctor
Nov 21 '18 at 4:02
I'm still not able to convince myself adequately, I'm afraid. I think I'll have to spend some more time getting familiar with continuity.
– notadoctor
Nov 21 '18 at 4:02
add a comment |
1 Answer
1
active
oldest
votes
Let $epsilon >0$. If $n$ is sufficiently large then (by continuity) $|f(a)-f(b)| leq |f(a)-f(x_0)|+|f(x_0)-f(b)|<2epsilon$ whenever $a,bin (x_0-frac 1 n, x_0+frac 1 n)$. Take sup over all such $a,b$ and then let $n to infty$. You will get $alpha (f,x_0) leq 2epsilon$. Since this is true for all $epsilon >0$ we must have $alpha (f,x_0)=0$.
answered Nov 21 '18 at 5:12
Kavi Rama Murthy
51k31854
Thank you! This helps a lot. However, I am unsure where your inequality $alpha (f,x_0) leq 2epsilon$ comes from. If the sup is $frac{2}{n}$, doesn't $n rightarrow infty$ just give me $0$ right away from that?
– notadoctor
Nov 21 '18 at 6:44
$|f(x_0)-f(a)| <epsilon$ for $|a-x_0| <delta$. If $frac 1 n <delta$ we get $|f(x_0)-f(a)| <epsilon$. Similarly, $|f(x_0)-f(b)| <epsilon$.
– Kavi Rama Murthy
Nov 21 '18 at 7:42
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%2f3007192%2fproving-that-continuity-and-the-lim-sup-of-a-given-set-being-0-are-equivalent%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
Let $epsilon >0$. If $n$ is sufficiently large then (by continuity) $|f(a)-f(b)| leq |f(a)-f(x_0)|+|f(x_0)-f(b)|<2epsilon$ whenever $a,bin (x_0-frac 1 n, x_0+frac 1 n)$. Take sup over all such $a,b$ and then let $n to infty$. You will get $alpha (f,x_0) leq 2epsilon$. Since this is true for all $epsilon >0$ we must have $alpha (f,x_0)=0$.
answered Nov 21 '18 at 5:12
Kavi Rama Murthy
51k31854
Thank you! This helps a lot. However, I am unsure where your inequality $alpha (f,x_0) leq 2epsilon$ comes from. If the sup is $frac{2}{n}$, doesn't $n rightarrow infty$ just give me $0$ right away from that?
– notadoctor
Nov 21 '18 at 6:44
$|f(x_0)-f(a)| <epsilon$ for $|a-x_0| <delta$. If $frac 1 n <delta$ we get $|f(x_0)-f(a)| <epsilon$. Similarly, $|f(x_0)-f(b)| <epsilon$.
– Kavi Rama Murthy
Nov 21 '18 at 7:42
add a comment |
Let $epsilon >0$. If $n$ is sufficiently large then (by continuity) $|f(a)-f(b)| leq |f(a)-f(x_0)|+|f(x_0)-f(b)|<2epsilon$ whenever $a,bin (x_0-frac 1 n, x_0+frac 1 n)$. Take sup over all such $a,b$ and then let $n to infty$. You will get $alpha (f,x_0) leq 2epsilon$. Since this is true for all $epsilon >0$ we must have $alpha (f,x_0)=0$.
answered Nov 21 '18 at 5:12
Kavi Rama Murthy
51k31854
Thank you! This helps a lot. However, I am unsure where your inequality $alpha (f,x_0) leq 2epsilon$ comes from. If the sup is $frac{2}{n}$, doesn't $n rightarrow infty$ just give me $0$ right away from that?
– notadoctor
Nov 21 '18 at 6:44
$|f(x_0)-f(a)| <epsilon$ for $|a-x_0| <delta$. If $frac 1 n <delta$ we get $|f(x_0)-f(a)| <epsilon$. Similarly, $|f(x_0)-f(b)| <epsilon$.
– Kavi Rama Murthy
Nov 21 '18 at 7:42
add a comment |
Let $epsilon >0$. If $n$ is sufficiently large then (by continuity) $|f(a)-f(b)| leq |f(a)-f(x_0)|+|f(x_0)-f(b)|<2epsilon$ whenever $a,bin (x_0-frac 1 n, x_0+frac 1 n)$. Take sup over all such $a,b$ and then let $n to infty$. You will get $alpha (f,x_0) leq 2epsilon$. Since this is true for all $epsilon >0$ we must have $alpha (f,x_0)=0$.
answered Nov 21 '18 at 5:12
Kavi Rama Murthy
51k31854
Let $epsilon >0$. If $n$ is sufficiently large then (by continuity) $|f(a)-f(b)| leq |f(a)-f(x_0)|+|f(x_0)-f(b)|<2epsilon$ whenever $a,bin (x_0-frac 1 n, x_0+frac 1 n)$. Take sup over all such $a,b$ and then let $n to infty$. You will get $alpha (f,x_0) leq 2epsilon$. Since this is true for all $epsilon >0$ we must have $alpha (f,x_0)=0$.
answered Nov 21 '18 at 5:12
Kavi Rama Murthy
51k31854
answered Nov 21 '18 at 5:12
Kavi Rama Murthy
51k31854
answered Nov 21 '18 at 5:12
Kavi Rama Murthy
51k31854
answered Nov 21 '18 at 5:12
Kavi Rama Murthy
51k31854
51k31854
Thank you! This helps a lot. However, I am unsure where your inequality $alpha (f,x_0) leq 2epsilon$ comes from. If the sup is $frac{2}{n}$, doesn't $n rightarrow infty$ just give me $0$ right away from that?
– notadoctor
Nov 21 '18 at 6:44
$|f(x_0)-f(a)| <epsilon$ for $|a-x_0| <delta$. If $frac 1 n <delta$ we get $|f(x_0)-f(a)| <epsilon$. Similarly, $|f(x_0)-f(b)| <epsilon$.
– Kavi Rama Murthy
Nov 21 '18 at 7:42
add a comment |
Thank you! This helps a lot. However, I am unsure where your inequality $alpha (f,x_0) leq 2epsilon$ comes from. If the sup is $frac{2}{n}$, doesn't $n rightarrow infty$ just give me $0$ right away from that?
– notadoctor
Nov 21 '18 at 6:44
$|f(x_0)-f(a)| <epsilon$ for $|a-x_0| <delta$. If $frac 1 n <delta$ we get $|f(x_0)-f(a)| <epsilon$. Similarly, $|f(x_0)-f(b)| <epsilon$.
– Kavi Rama Murthy
Nov 21 '18 at 7:42
Thank you! This helps a lot. However, I am unsure where your inequality $alpha (f,x_0) leq 2epsilon$ comes from. If the sup is $frac{2}{n}$, doesn't $n rightarrow infty$ just give me $0$ right away from that?
– notadoctor
Nov 21 '18 at 6:44
Thank you! This helps a lot. However, I am unsure where your inequality $alpha (f,x_0) leq 2epsilon$ comes from. If the sup is $frac{2}{n}$, doesn't $n rightarrow infty$ just give me $0$ right away from that?
– notadoctor
Nov 21 '18 at 6:44
$|f(x_0)-f(a)| <epsilon$ for $|a-x_0| <delta$. If $frac 1 n <delta$ we get $|f(x_0)-f(a)| <epsilon$. Similarly, $|f(x_0)-f(b)| <epsilon$.
– Kavi Rama Murthy
Nov 21 '18 at 7:42
$|f(x_0)-f(a)| <epsilon$ for $|a-x_0| <delta$. If $frac 1 n <delta$ we get $|f(x_0)-f(a)| <epsilon$. Similarly, $|f(x_0)-f(b)| <epsilon$.
– Kavi Rama Murthy
Nov 21 '18 at 7:42
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3007192%2fproving-that-continuity-and-the-lim-sup-of-a-given-set-being-0-are-equivalent%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
Write out the definition of lim sup. It will become very obvious.
– Don Thousand
Nov 21 '18 at 2:59
Is it simply that as $nrightarrow infty$, $(a, b)rightarrow (x_o, x_o)$? But where do I use the fact that $f$ is continuous?
– notadoctor
Nov 21 '18 at 3:07
That's not the definition of lim sup...
– Don Thousand
Nov 21 '18 at 3:08
1
@notadoctor think of the $limsup$ as being two parts: the limit as $nrightarrow infty$, and the $sup$ over all pairs $a,b$ in that interval.
– user25959
Nov 21 '18 at 3:10
I'm still not able to convince myself adequately, I'm afraid. I think I'll have to spend some more time getting familiar with continuity.
– notadoctor
Nov 21 '18 at 4:02