Is this function uniformly continuous or not?
I've been recently introduced to the concept of uniform continuity and I'm having some trouble deciding whether a function is uniformly continuous or not.
The function is:
$$f(x)=begin{cases}dfrac{sin(x^5)}{x} & xneq0 \ 0 & x=0end{cases}$$
I mean, I know that if the derivative of the function is bounded, then we can conclude that it is uniformly continuous, but if it isn't, I really don't know how to proceed.
real-analysis
asked 2 hours ago
Sergamar
113
New contributor
Sergamar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
|
show 3 more comments
I've been recently introduced to the concept of uniform continuity and I'm having some trouble deciding whether a function is uniformly continuous or not.
The function is:
$$f(x)=begin{cases}dfrac{sin(x^5)}{x} & xneq0 \ 0 & x=0end{cases}$$
I mean, I know that if the derivative of the function is bounded, then we can conclude that it is uniformly continuous, but if it isn't, I really don't know how to proceed.
real-analysis
asked 2 hours ago
Sergamar
113
New contributor
Sergamar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
Uniformly continuous on what set?
– user587192
2 hours ago
I believe it must be on $mathbb R$
– fonini
2 hours ago
On all the real numbers
– Sergamar
2 hours ago
It can't be on $mathbb{R}$: this function not defined at $x=0$.
– user587192
2 hours ago
1
0 if x = 0, that expresion otherwise
– Sergamar
2 hours ago
|
show 3 more comments
I've been recently introduced to the concept of uniform continuity and I'm having some trouble deciding whether a function is uniformly continuous or not.
The function is:
$$f(x)=begin{cases}dfrac{sin(x^5)}{x} & xneq0 \ 0 & x=0end{cases}$$
I mean, I know that if the derivative of the function is bounded, then we can conclude that it is uniformly continuous, but if it isn't, I really don't know how to proceed.
real-analysis
asked 2 hours ago
Sergamar
113
New contributor
Sergamar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I've been recently introduced to the concept of uniform continuity and I'm having some trouble deciding whether a function is uniformly continuous or not.
The function is:
$$f(x)=begin{cases}dfrac{sin(x^5)}{x} & xneq0 \ 0 & x=0end{cases}$$
I mean, I know that if the derivative of the function is bounded, then we can conclude that it is uniformly continuous, but if it isn't, I really don't know how to proceed.
real-analysis
real-analysis
asked 2 hours ago
Sergamar
113
New contributor
Sergamar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Sergamar
113
New contributor
Sergamar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
asked 2 hours ago
Sergamar
113
New contributor
Sergamar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Sergamar
113
asked 2 hours ago
Sergamar
113
113
New contributor
Sergamar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Sergamar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Sergamar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
Uniformly continuous on what set?
– user587192
2 hours ago
I believe it must be on $mathbb R$
– fonini
2 hours ago
On all the real numbers
– Sergamar
2 hours ago
It can't be on $mathbb{R}$: this function not defined at $x=0$.
– user587192
2 hours ago
1
0 if x = 0, that expresion otherwise
– Sergamar
2 hours ago
|
show 3 more comments
2
Uniformly continuous on what set?
– user587192
2 hours ago
I believe it must be on $mathbb R$
– fonini
2 hours ago
On all the real numbers
– Sergamar
2 hours ago
It can't be on $mathbb{R}$: this function not defined at $x=0$.
– user587192
2 hours ago
1
0 if x = 0, that expresion otherwise
– Sergamar
2 hours ago
2
2
Uniformly continuous on what set?
– user587192
2 hours ago
Uniformly continuous on what set?
– user587192
2 hours ago
I believe it must be on $mathbb R$
– fonini
2 hours ago
I believe it must be on $mathbb R$
– fonini
2 hours ago
On all the real numbers
– Sergamar
2 hours ago
On all the real numbers
– Sergamar
2 hours ago
It can't be on $mathbb{R}$: this function not defined at $x=0$.
– user587192
2 hours ago
It can't be on $mathbb{R}$: this function not defined at $x=0$.
– user587192
2 hours ago
1
1
0 if x = 0, that expresion otherwise
– Sergamar
2 hours ago
0 if x = 0, that expresion otherwise
– Sergamar
2 hours ago
|
show 3 more comments
1 Answer
1
active
oldest
votes
The function $f$ is uniformly continuous on $mathbb{R}$.
Note that $frac{sin x^5}{x} = x^4 frac{sin x^5}{x^5} to 0 cdot 1 = 0$ as $x to 0$ (from the left or the right). Hence, $f$ is continuous and uniformly continuous on any compact interval $[-a,a]$.
Since $f(x) to 0 $ as $|x| to infty$ the function is also uniformly continuous on intervals $(-infty,-a]$ and $[a,infty)$. Any continuous function with a finite limit as $x to pm infty$ is uniformly continuous -- proved many times on this site.
It is a common misconception that a function with a derivative that is unbounded for large $|x|$ cannot be uniformly continuous. The condition that $|f'(x)| to +infty$ as $|x| to +infty$ guarantees non-uniform continuity.
answered 1 hour ago
RRL
48.7k42573
Could you show me some proof of the fact that if $|f'(x)|to infty$ as $|x|to infty$ then it collides with uniform convergence?
– Jakobian
1 hour ago
2
@Jakobian: Sure. $|f(x) - f(y)| = |f'(xi)||x - y|$ by the MVT. For any $delta > 0$ take $y = x + delta/2$ so that $|x-y| < delta$ and $|f(x) - f(y)| = |f'(xi)|delta/2$ Now this stays bigger than $1$ for some $x$ and $xi > x$ sufficiently large if $|f'(x)| to +infty$.
– RRL
1 hour ago
@Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity?
– ImNotTheGuy
1 hour ago
@RRL oh, thank you, it was a typo and it got me confused
– Jakobian
1 hour ago
@Jakobian It got me confused too, lol.
– ImNotTheGuy
1 hour ago
|
show 1 more 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
});
}
});
Sergamar is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3052560%2fis-this-function-uniformly-continuous-or-not%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
The function $f$ is uniformly continuous on $mathbb{R}$.
Note that $frac{sin x^5}{x} = x^4 frac{sin x^5}{x^5} to 0 cdot 1 = 0$ as $x to 0$ (from the left or the right). Hence, $f$ is continuous and uniformly continuous on any compact interval $[-a,a]$.
Since $f(x) to 0 $ as $|x| to infty$ the function is also uniformly continuous on intervals $(-infty,-a]$ and $[a,infty)$. Any continuous function with a finite limit as $x to pm infty$ is uniformly continuous -- proved many times on this site.
It is a common misconception that a function with a derivative that is unbounded for large $|x|$ cannot be uniformly continuous. The condition that $|f'(x)| to +infty$ as $|x| to +infty$ guarantees non-uniform continuity.
answered 1 hour ago
RRL
48.7k42573
Could you show me some proof of the fact that if $|f'(x)|to infty$ as $|x|to infty$ then it collides with uniform convergence?
– Jakobian
1 hour ago
2
@Jakobian: Sure. $|f(x) - f(y)| = |f'(xi)||x - y|$ by the MVT. For any $delta > 0$ take $y = x + delta/2$ so that $|x-y| < delta$ and $|f(x) - f(y)| = |f'(xi)|delta/2$ Now this stays bigger than $1$ for some $x$ and $xi > x$ sufficiently large if $|f'(x)| to +infty$.
– RRL
1 hour ago
@Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity?
– ImNotTheGuy
1 hour ago
@RRL oh, thank you, it was a typo and it got me confused
– Jakobian
1 hour ago
@Jakobian It got me confused too, lol.
– ImNotTheGuy
1 hour ago
|
show 1 more comment
The function $f$ is uniformly continuous on $mathbb{R}$.
Note that $frac{sin x^5}{x} = x^4 frac{sin x^5}{x^5} to 0 cdot 1 = 0$ as $x to 0$ (from the left or the right). Hence, $f$ is continuous and uniformly continuous on any compact interval $[-a,a]$.
Since $f(x) to 0 $ as $|x| to infty$ the function is also uniformly continuous on intervals $(-infty,-a]$ and $[a,infty)$. Any continuous function with a finite limit as $x to pm infty$ is uniformly continuous -- proved many times on this site.
It is a common misconception that a function with a derivative that is unbounded for large $|x|$ cannot be uniformly continuous. The condition that $|f'(x)| to +infty$ as $|x| to +infty$ guarantees non-uniform continuity.
answered 1 hour ago
RRL
48.7k42573
Could you show me some proof of the fact that if $|f'(x)|to infty$ as $|x|to infty$ then it collides with uniform convergence?
– Jakobian
1 hour ago
2
@Jakobian: Sure. $|f(x) - f(y)| = |f'(xi)||x - y|$ by the MVT. For any $delta > 0$ take $y = x + delta/2$ so that $|x-y| < delta$ and $|f(x) - f(y)| = |f'(xi)|delta/2$ Now this stays bigger than $1$ for some $x$ and $xi > x$ sufficiently large if $|f'(x)| to +infty$.
– RRL
1 hour ago
@Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity?
– ImNotTheGuy
1 hour ago
@RRL oh, thank you, it was a typo and it got me confused
– Jakobian
1 hour ago
@Jakobian It got me confused too, lol.
– ImNotTheGuy
1 hour ago
|
show 1 more comment
The function $f$ is uniformly continuous on $mathbb{R}$.
Note that $frac{sin x^5}{x} = x^4 frac{sin x^5}{x^5} to 0 cdot 1 = 0$ as $x to 0$ (from the left or the right). Hence, $f$ is continuous and uniformly continuous on any compact interval $[-a,a]$.
Since $f(x) to 0 $ as $|x| to infty$ the function is also uniformly continuous on intervals $(-infty,-a]$ and $[a,infty)$. Any continuous function with a finite limit as $x to pm infty$ is uniformly continuous -- proved many times on this site.
It is a common misconception that a function with a derivative that is unbounded for large $|x|$ cannot be uniformly continuous. The condition that $|f'(x)| to +infty$ as $|x| to +infty$ guarantees non-uniform continuity.
answered 1 hour ago
RRL
48.7k42573
The function $f$ is uniformly continuous on $mathbb{R}$.
Note that $frac{sin x^5}{x} = x^4 frac{sin x^5}{x^5} to 0 cdot 1 = 0$ as $x to 0$ (from the left or the right). Hence, $f$ is continuous and uniformly continuous on any compact interval $[-a,a]$.
Since $f(x) to 0 $ as $|x| to infty$ the function is also uniformly continuous on intervals $(-infty,-a]$ and $[a,infty)$. Any continuous function with a finite limit as $x to pm infty$ is uniformly continuous -- proved many times on this site.
It is a common misconception that a function with a derivative that is unbounded for large $|x|$ cannot be uniformly continuous. The condition that $|f'(x)| to +infty$ as $|x| to +infty$ guarantees non-uniform continuity.
answered 1 hour ago
RRL
48.7k42573
edited 51 mins ago
answered 1 hour ago
RRL
48.7k42573
answered 1 hour ago
RRL
48.7k42573
answered 1 hour ago
RRL
48.7k42573
48.7k42573
Could you show me some proof of the fact that if $|f'(x)|to infty$ as $|x|to infty$ then it collides with uniform convergence?
– Jakobian
1 hour ago
2
@Jakobian: Sure. $|f(x) - f(y)| = |f'(xi)||x - y|$ by the MVT. For any $delta > 0$ take $y = x + delta/2$ so that $|x-y| < delta$ and $|f(x) - f(y)| = |f'(xi)|delta/2$ Now this stays bigger than $1$ for some $x$ and $xi > x$ sufficiently large if $|f'(x)| to +infty$.
– RRL
1 hour ago
@Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity?
– ImNotTheGuy
1 hour ago
@RRL oh, thank you, it was a typo and it got me confused
– Jakobian
1 hour ago
@Jakobian It got me confused too, lol.
– ImNotTheGuy
1 hour ago
|
show 1 more comment
Could you show me some proof of the fact that if $|f'(x)|to infty$ as $|x|to infty$ then it collides with uniform convergence?
– Jakobian
1 hour ago
2
@Jakobian: Sure. $|f(x) - f(y)| = |f'(xi)||x - y|$ by the MVT. For any $delta > 0$ take $y = x + delta/2$ so that $|x-y| < delta$ and $|f(x) - f(y)| = |f'(xi)|delta/2$ Now this stays bigger than $1$ for some $x$ and $xi > x$ sufficiently large if $|f'(x)| to +infty$.
– RRL
1 hour ago
@Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity?
– ImNotTheGuy
1 hour ago
@RRL oh, thank you, it was a typo and it got me confused
– Jakobian
1 hour ago
@Jakobian It got me confused too, lol.
– ImNotTheGuy
1 hour ago
Could you show me some proof of the fact that if $|f'(x)|to infty$ as $|x|to infty$ then it collides with uniform convergence?
– Jakobian
1 hour ago
Could you show me some proof of the fact that if $|f'(x)|to infty$ as $|x|to infty$ then it collides with uniform convergence?
– Jakobian
1 hour ago
2
2
@Jakobian: Sure. $|f(x) - f(y)| = |f'(xi)||x - y|$ by the MVT. For any $delta > 0$ take $y = x + delta/2$ so that $|x-y| < delta$ and $|f(x) - f(y)| = |f'(xi)|delta/2$ Now this stays bigger than $1$ for some $x$ and $xi > x$ sufficiently large if $|f'(x)| to +infty$.
– RRL
1 hour ago
@Jakobian: Sure. $|f(x) - f(y)| = |f'(xi)||x - y|$ by the MVT. For any $delta > 0$ take $y = x + delta/2$ so that $|x-y| < delta$ and $|f(x) - f(y)| = |f'(xi)|delta/2$ Now this stays bigger than $1$ for some $x$ and $xi > x$ sufficiently large if $|f'(x)| to +infty$.
– RRL
1 hour ago
@Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity?
– ImNotTheGuy
1 hour ago
@Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity?
– ImNotTheGuy
1 hour ago
@RRL oh, thank you, it was a typo and it got me confused
– Jakobian
1 hour ago
@RRL oh, thank you, it was a typo and it got me confused
– Jakobian
1 hour ago
@Jakobian It got me confused too, lol.
– ImNotTheGuy
1 hour ago
@Jakobian It got me confused too, lol.
– ImNotTheGuy
1 hour ago
|
show 1 more comment
Sergamar is a new contributor. Be nice, and check out our Code of Conduct.
Sergamar is a new contributor. Be nice, and check out our Code of Conduct.
Sergamar is a new contributor. Be nice, and check out our Code of Conduct.
Sergamar is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3052560%2fis-this-function-uniformly-continuous-or-not%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
2
Uniformly continuous on what set?
– user587192
2 hours ago
I believe it must be on $mathbb R$
– fonini
2 hours ago
On all the real numbers
– Sergamar
2 hours ago
It can't be on $mathbb{R}$: this function not defined at $x=0$.
– user587192
2 hours ago
1
0 if x = 0, that expresion otherwise
– Sergamar
2 hours ago