Determining whether the sequence ${f_{n}}$ converges uniformly on the set $A.$
$begingroup$
Determining whether the sequence ${f_{n}}$ converges uniformly on the set $A$
begin{equation*}
f_{n}(x) = g(nx),quad g(x)=
begin{cases}
x & text{if }0 leq x < 1/2,\
1-x & text{if } 1/2 leq x leq 1,\
0 & text{if } x > 1.
end{cases}
end{equation*}
I know that the first step is to find the limit of $f_{n}(x),$ which I do not know how, could anyone help me in doing so?
Also the book did not wrote what is the set $A$, can I assume that it is $mathbb{R}$ and start my solution?
calculus sequences-and-series uniform-convergence
edited Jan 21 at 10:00
user549397
1,5101418
asked Jan 21 at 0:35
hopefullyhopefully
247114
$endgroup$
add a comment |
$begingroup$
Determining whether the sequence ${f_{n}}$ converges uniformly on the set $A$
begin{equation*}
f_{n}(x) = g(nx),quad g(x)=
begin{cases}
x & text{if }0 leq x < 1/2,\
1-x & text{if } 1/2 leq x leq 1,\
0 & text{if } x > 1.
end{cases}
end{equation*}
I know that the first step is to find the limit of $f_{n}(x),$ which I do not know how, could anyone help me in doing so?
Also the book did not wrote what is the set $A$, can I assume that it is $mathbb{R}$ and start my solution?
calculus sequences-and-series uniform-convergence
edited Jan 21 at 10:00
user549397
1,5101418
asked Jan 21 at 0:35
hopefullyhopefully
247114
$endgroup$
add a comment |
$begingroup$
Determining whether the sequence ${f_{n}}$ converges uniformly on the set $A$
begin{equation*}
f_{n}(x) = g(nx),quad g(x)=
begin{cases}
x & text{if }0 leq x < 1/2,\
1-x & text{if } 1/2 leq x leq 1,\
0 & text{if } x > 1.
end{cases}
end{equation*}
I know that the first step is to find the limit of $f_{n}(x),$ which I do not know how, could anyone help me in doing so?
Also the book did not wrote what is the set $A$, can I assume that it is $mathbb{R}$ and start my solution?
calculus sequences-and-series uniform-convergence
edited Jan 21 at 10:00
user549397
1,5101418
asked Jan 21 at 0:35
hopefullyhopefully
247114
$endgroup$
Determining whether the sequence ${f_{n}}$ converges uniformly on the set $A$
begin{equation*}
f_{n}(x) = g(nx),quad g(x)=
begin{cases}
x & text{if }0 leq x < 1/2,\
1-x & text{if } 1/2 leq x leq 1,\
0 & text{if } x > 1.
end{cases}
end{equation*}
I know that the first step is to find the limit of $f_{n}(x),$ which I do not know how, could anyone help me in doing so?
Also the book did not wrote what is the set $A$, can I assume that it is $mathbb{R}$ and start my solution?
calculus sequences-and-series uniform-convergence
calculus sequences-and-series uniform-convergence
edited Jan 21 at 10:00
user549397
1,5101418
asked Jan 21 at 0:35
hopefullyhopefully
247114
edited Jan 21 at 10:00
user549397
1,5101418
asked Jan 21 at 0:35
hopefullyhopefully
247114
edited Jan 21 at 10:00
user549397
1,5101418
edited Jan 21 at 10:00
user549397
1,5101418
edited Jan 21 at 10:00
user549397
1,5101418
1,5101418
asked Jan 21 at 0:35
hopefullyhopefully
247114
asked Jan 21 at 0:35
hopefullyhopefully
247114
asked Jan 21 at 0:35
hopefullyhopefully
247114
247114
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$f_n(frac 1 {2n})=g(frac 1 2)=frac 1 2$ for all $n$. This shows that (while $f_nto 0$ pointwise) it does not tend to $0$ uniformly. It converges to $0$ uniformly on any set which is contained in $mathbb R setminus (0,epsilon)$ for some $epsilon >0$. Some details: it is clear that $f_n(x) to 0$ for each $x$. If $f_n to 0$ uniformly then, given $epsilon >0$, there exists an integer $n_0$ such that $|f_n(x)-0| <epsilon$ for all $x$ for all $n geq n_0$. The main point here is the same $n_0$ works for all $x$. Even if you make $x$ dependent on $x$ this inequality must hold as long as $n geq n_0$. To get a contradiction from this inequality you choose appropriate vales of $x$ depending o $n$. This is what I have done. [I get a contradiction when $epsilon <frac 1 2$].
answered Jan 21 at 0:40
Kavi Rama MurthyKavi Rama Murthy
61.9k42262
$endgroup$
$begingroup$
what is the value of this $epsilon$?
$endgroup$
– hopefully
Jan 21 at 0:50
$begingroup$
what is the intuition behind this solution?
$endgroup$
– hopefully
Jan 21 at 0:51
$begingroup$
what about considering the case $x = 1/3n$, will the solution remain the same? If so why?
$endgroup$
– hopefully
Jan 21 at 0:52
1
$begingroup$
@hopefully If $x>0$ the $nx>1$ for $n$ sufficiently large. This implies $f_n(x)=0$ for $n$ sufficiently large, so $f_n(x)to 0$.
$endgroup$
– Kavi Rama Murthy
Jan 21 at 23:14
1
$begingroup$
@hopefully Yes, you are right.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 0:12
|
show 8 more comments
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%2f3081335%2fdetermining-whether-the-sequence-f-n-converges-uniformly-on-the-set-a%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
$begingroup$
$f_n(frac 1 {2n})=g(frac 1 2)=frac 1 2$ for all $n$. This shows that (while $f_nto 0$ pointwise) it does not tend to $0$ uniformly. It converges to $0$ uniformly on any set which is contained in $mathbb R setminus (0,epsilon)$ for some $epsilon >0$. Some details: it is clear that $f_n(x) to 0$ for each $x$. If $f_n to 0$ uniformly then, given $epsilon >0$, there exists an integer $n_0$ such that $|f_n(x)-0| <epsilon$ for all $x$ for all $n geq n_0$. The main point here is the same $n_0$ works for all $x$. Even if you make $x$ dependent on $x$ this inequality must hold as long as $n geq n_0$. To get a contradiction from this inequality you choose appropriate vales of $x$ depending o $n$. This is what I have done. [I get a contradiction when $epsilon <frac 1 2$].
answered Jan 21 at 0:40
Kavi Rama MurthyKavi Rama Murthy
61.9k42262
$endgroup$
$begingroup$
what is the value of this $epsilon$?
$endgroup$
– hopefully
Jan 21 at 0:50
$begingroup$
what is the intuition behind this solution?
$endgroup$
– hopefully
Jan 21 at 0:51
$begingroup$
what about considering the case $x = 1/3n$, will the solution remain the same? If so why?
$endgroup$
– hopefully
Jan 21 at 0:52
1
$begingroup$
@hopefully If $x>0$ the $nx>1$ for $n$ sufficiently large. This implies $f_n(x)=0$ for $n$ sufficiently large, so $f_n(x)to 0$.
$endgroup$
– Kavi Rama Murthy
Jan 21 at 23:14
1
$begingroup$
@hopefully Yes, you are right.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 0:12
|
show 8 more comments
$begingroup$
$f_n(frac 1 {2n})=g(frac 1 2)=frac 1 2$ for all $n$. This shows that (while $f_nto 0$ pointwise) it does not tend to $0$ uniformly. It converges to $0$ uniformly on any set which is contained in $mathbb R setminus (0,epsilon)$ for some $epsilon >0$. Some details: it is clear that $f_n(x) to 0$ for each $x$. If $f_n to 0$ uniformly then, given $epsilon >0$, there exists an integer $n_0$ such that $|f_n(x)-0| <epsilon$ for all $x$ for all $n geq n_0$. The main point here is the same $n_0$ works for all $x$. Even if you make $x$ dependent on $x$ this inequality must hold as long as $n geq n_0$. To get a contradiction from this inequality you choose appropriate vales of $x$ depending o $n$. This is what I have done. [I get a contradiction when $epsilon <frac 1 2$].
answered Jan 21 at 0:40
Kavi Rama MurthyKavi Rama Murthy
61.9k42262
$endgroup$
$begingroup$
what is the value of this $epsilon$?
$endgroup$
– hopefully
Jan 21 at 0:50
$begingroup$
what is the intuition behind this solution?
$endgroup$
– hopefully
Jan 21 at 0:51
$begingroup$
what about considering the case $x = 1/3n$, will the solution remain the same? If so why?
$endgroup$
– hopefully
Jan 21 at 0:52
1
$begingroup$
@hopefully If $x>0$ the $nx>1$ for $n$ sufficiently large. This implies $f_n(x)=0$ for $n$ sufficiently large, so $f_n(x)to 0$.
$endgroup$
– Kavi Rama Murthy
Jan 21 at 23:14
1
$begingroup$
@hopefully Yes, you are right.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 0:12
|
show 8 more comments
$begingroup$
$f_n(frac 1 {2n})=g(frac 1 2)=frac 1 2$ for all $n$. This shows that (while $f_nto 0$ pointwise) it does not tend to $0$ uniformly. It converges to $0$ uniformly on any set which is contained in $mathbb R setminus (0,epsilon)$ for some $epsilon >0$. Some details: it is clear that $f_n(x) to 0$ for each $x$. If $f_n to 0$ uniformly then, given $epsilon >0$, there exists an integer $n_0$ such that $|f_n(x)-0| <epsilon$ for all $x$ for all $n geq n_0$. The main point here is the same $n_0$ works for all $x$. Even if you make $x$ dependent on $x$ this inequality must hold as long as $n geq n_0$. To get a contradiction from this inequality you choose appropriate vales of $x$ depending o $n$. This is what I have done. [I get a contradiction when $epsilon <frac 1 2$].
answered Jan 21 at 0:40
Kavi Rama MurthyKavi Rama Murthy
61.9k42262
$endgroup$
$f_n(frac 1 {2n})=g(frac 1 2)=frac 1 2$ for all $n$. This shows that (while $f_nto 0$ pointwise) it does not tend to $0$ uniformly. It converges to $0$ uniformly on any set which is contained in $mathbb R setminus (0,epsilon)$ for some $epsilon >0$. Some details: it is clear that $f_n(x) to 0$ for each $x$. If $f_n to 0$ uniformly then, given $epsilon >0$, there exists an integer $n_0$ such that $|f_n(x)-0| <epsilon$ for all $x$ for all $n geq n_0$. The main point here is the same $n_0$ works for all $x$. Even if you make $x$ dependent on $x$ this inequality must hold as long as $n geq n_0$. To get a contradiction from this inequality you choose appropriate vales of $x$ depending o $n$. This is what I have done. [I get a contradiction when $epsilon <frac 1 2$].
answered Jan 21 at 0:40
Kavi Rama MurthyKavi Rama Murthy
61.9k42262
edited Jan 21 at 9:50
answered Jan 21 at 0:40
Kavi Rama MurthyKavi Rama Murthy
61.9k42262
answered Jan 21 at 0:40
Kavi Rama MurthyKavi Rama Murthy
61.9k42262
answered Jan 21 at 0:40
Kavi Rama MurthyKavi Rama Murthy
61.9k42262
61.9k42262
$begingroup$
what is the value of this $epsilon$?
$endgroup$
– hopefully
Jan 21 at 0:50
$begingroup$
what is the intuition behind this solution?
$endgroup$
– hopefully
Jan 21 at 0:51
$begingroup$
what about considering the case $x = 1/3n$, will the solution remain the same? If so why?
$endgroup$
– hopefully
Jan 21 at 0:52
1
$begingroup$
@hopefully If $x>0$ the $nx>1$ for $n$ sufficiently large. This implies $f_n(x)=0$ for $n$ sufficiently large, so $f_n(x)to 0$.
$endgroup$
– Kavi Rama Murthy
Jan 21 at 23:14
1
$begingroup$
@hopefully Yes, you are right.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 0:12
|
show 8 more comments
$begingroup$
what is the value of this $epsilon$?
$endgroup$
– hopefully
Jan 21 at 0:50
$begingroup$
what is the intuition behind this solution?
$endgroup$
– hopefully
Jan 21 at 0:51
$begingroup$
what about considering the case $x = 1/3n$, will the solution remain the same? If so why?
$endgroup$
– hopefully
Jan 21 at 0:52
1
$begingroup$
@hopefully If $x>0$ the $nx>1$ for $n$ sufficiently large. This implies $f_n(x)=0$ for $n$ sufficiently large, so $f_n(x)to 0$.
$endgroup$
– Kavi Rama Murthy
Jan 21 at 23:14
1
$begingroup$
@hopefully Yes, you are right.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 0:12
$begingroup$
what is the value of this $epsilon$?
$endgroup$
– hopefully
Jan 21 at 0:50
$begingroup$
what is the value of this $epsilon$?
$endgroup$
– hopefully
Jan 21 at 0:50
$begingroup$
what is the intuition behind this solution?
$endgroup$
– hopefully
Jan 21 at 0:51
$begingroup$
what is the intuition behind this solution?
$endgroup$
– hopefully
Jan 21 at 0:51
$begingroup$
what about considering the case $x = 1/3n$, will the solution remain the same? If so why?
$endgroup$
– hopefully
Jan 21 at 0:52
$begingroup$
what about considering the case $x = 1/3n$, will the solution remain the same? If so why?
$endgroup$
– hopefully
Jan 21 at 0:52
1
1
$begingroup$
@hopefully If $x>0$ the $nx>1$ for $n$ sufficiently large. This implies $f_n(x)=0$ for $n$ sufficiently large, so $f_n(x)to 0$.
$endgroup$
– Kavi Rama Murthy
Jan 21 at 23:14
$begingroup$
@hopefully If $x>0$ the $nx>1$ for $n$ sufficiently large. This implies $f_n(x)=0$ for $n$ sufficiently large, so $f_n(x)to 0$.
$endgroup$
– Kavi Rama Murthy
Jan 21 at 23:14
1
1
$begingroup$
@hopefully Yes, you are right.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 0:12
$begingroup$
@hopefully Yes, you are right.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 0:12
|
show 8 more comments
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%2f3081335%2fdetermining-whether-the-sequence-f-n-converges-uniformly-on-the-set-a%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
