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 Dec 26 at 1:15
Sergamar
185
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 Dec 26 at 1:15
Sergamar
185
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
Dec 26 at 1:16
I believe it must be on $mathbb R$
– fonini
Dec 26 at 1:17
1
0 if x = 0, that expresion otherwise
– Sergamar
Dec 26 at 1:18
1
I've found a theorem that states: If a function $f$ is continuous on $mathbb{R_+}$, and $lim_{xtoinfty} f(x)$ exists, then that function is uniformly continuous on $mathbb{R_+}$. So that function is indeed uniformly continuous.
– Jakobian
Dec 26 at 1:44
1
Here is a topic about that theorem
– Jakobian
Dec 26 at 1:50
|
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 Dec 26 at 1:15
Sergamar
185
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 Dec 26 at 1:15
Sergamar
185
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 Dec 26 at 1:15
Sergamar
185
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 Dec 26 at 1:22
asked Dec 26 at 1:15
Sergamar
185
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 Dec 26 at 1:15
Sergamar
185
asked Dec 26 at 1:15
Sergamar
185
185
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
Dec 26 at 1:16
I believe it must be on $mathbb R$
– fonini
Dec 26 at 1:17
1
0 if x = 0, that expresion otherwise
– Sergamar
Dec 26 at 1:18
1
I've found a theorem that states: If a function $f$ is continuous on $mathbb{R_+}$, and $lim_{xtoinfty} f(x)$ exists, then that function is uniformly continuous on $mathbb{R_+}$. So that function is indeed uniformly continuous.
– Jakobian
Dec 26 at 1:44
1
Here is a topic about that theorem
– Jakobian
Dec 26 at 1:50
|
show 3 more comments
2
Uniformly continuous on what set?
– user587192
Dec 26 at 1:16
I believe it must be on $mathbb R$
– fonini
Dec 26 at 1:17
1
0 if x = 0, that expresion otherwise
– Sergamar
Dec 26 at 1:18
1
I've found a theorem that states: If a function $f$ is continuous on $mathbb{R_+}$, and $lim_{xtoinfty} f(x)$ exists, then that function is uniformly continuous on $mathbb{R_+}$. So that function is indeed uniformly continuous.
– Jakobian
Dec 26 at 1:44
1
Here is a topic about that theorem
– Jakobian
Dec 26 at 1:50
2
2
Uniformly continuous on what set?
– user587192
Dec 26 at 1:16
Uniformly continuous on what set?
– user587192
Dec 26 at 1:16
I believe it must be on $mathbb R$
– fonini
Dec 26 at 1:17
I believe it must be on $mathbb R$
– fonini
Dec 26 at 1:17
1
1
0 if x = 0, that expresion otherwise
– Sergamar
Dec 26 at 1:18
0 if x = 0, that expresion otherwise
– Sergamar
Dec 26 at 1:18
1
1
I've found a theorem that states: If a function $f$ is continuous on $mathbb{R_+}$, and $lim_{xtoinfty} f(x)$ exists, then that function is uniformly continuous on $mathbb{R_+}$. So that function is indeed uniformly continuous.
– Jakobian
Dec 26 at 1:44
I've found a theorem that states: If a function $f$ is continuous on $mathbb{R_+}$, and $lim_{xtoinfty} f(x)$ exists, then that function is uniformly continuous on $mathbb{R_+}$. So that function is indeed uniformly continuous.
– Jakobian
Dec 26 at 1:44
1
1
Here is a topic about that theorem
– Jakobian
Dec 26 at 1:50
Here is a topic about that theorem
– Jakobian
Dec 26 at 1:50
|
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 Dec 26 at 1:50
RRL
48.9k42573
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
Dec 26 at 1:59
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
Dec 26 at 2:07
@Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity?
– ImNotTheGuy
Dec 26 at 2:09
@RRL oh, thank you, it was a typo and it got me confused
– Jakobian
Dec 26 at 2:10
@Jakobian It got me confused too, lol.
– ImNotTheGuy
Dec 26 at 2:10
|
show 1 more comment
Your Answer
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1 Answer
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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 Dec 26 at 1:50
RRL
48.9k42573
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
Dec 26 at 1:59
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
Dec 26 at 2:07
@Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity?
– ImNotTheGuy
Dec 26 at 2:09
@RRL oh, thank you, it was a typo and it got me confused
– Jakobian
Dec 26 at 2:10
@Jakobian It got me confused too, lol.
– ImNotTheGuy
Dec 26 at 2:10
|
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 Dec 26 at 1:50
RRL
48.9k42573
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
Dec 26 at 1:59
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
Dec 26 at 2:07
@Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity?
– ImNotTheGuy
Dec 26 at 2:09
@RRL oh, thank you, it was a typo and it got me confused
– Jakobian
Dec 26 at 2:10
@Jakobian It got me confused too, lol.
– ImNotTheGuy
Dec 26 at 2:10
|
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 Dec 26 at 1:50
RRL
48.9k42573
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 Dec 26 at 1:50
RRL
48.9k42573
edited Dec 26 at 2:32
answered Dec 26 at 1:50
RRL
48.9k42573
answered Dec 26 at 1:50
RRL
48.9k42573
answered Dec 26 at 1:50
RRL
48.9k42573
48.9k42573
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
Dec 26 at 1:59
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
Dec 26 at 2:07
@Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity?
– ImNotTheGuy
Dec 26 at 2:09
@RRL oh, thank you, it was a typo and it got me confused
– Jakobian
Dec 26 at 2:10
@Jakobian It got me confused too, lol.
– ImNotTheGuy
Dec 26 at 2:10
|
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
Dec 26 at 1:59
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
Dec 26 at 2:07
@Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity?
– ImNotTheGuy
Dec 26 at 2:09
@RRL oh, thank you, it was a typo and it got me confused
– Jakobian
Dec 26 at 2:10
@Jakobian It got me confused too, lol.
– ImNotTheGuy
Dec 26 at 2:10
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
Dec 26 at 1:59
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
Dec 26 at 1:59
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
Dec 26 at 2:07
@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
Dec 26 at 2:07
@Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity?
– ImNotTheGuy
Dec 26 at 2:09
@Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity?
– ImNotTheGuy
Dec 26 at 2:09
@RRL oh, thank you, it was a typo and it got me confused
– Jakobian
Dec 26 at 2:10
@RRL oh, thank you, it was a typo and it got me confused
– Jakobian
Dec 26 at 2:10
@Jakobian It got me confused too, lol.
– ImNotTheGuy
Dec 26 at 2:10
@Jakobian It got me confused too, lol.
– ImNotTheGuy
Dec 26 at 2:10
|
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.
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2
Uniformly continuous on what set?
– user587192
Dec 26 at 1:16
I believe it must be on $mathbb R$
– fonini
Dec 26 at 1:17
1
0 if x = 0, that expresion otherwise
– Sergamar
Dec 26 at 1:18
1
I've found a theorem that states: If a function $f$ is continuous on $mathbb{R_+}$, and $lim_{xtoinfty} f(x)$ exists, then that function is uniformly continuous on $mathbb{R_+}$. So that function is indeed uniformly continuous.
– Jakobian
Dec 26 at 1:44
1
Here is a topic about that theorem
– Jakobian
Dec 26 at 1:50