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 7 hours ago
Sergamar
163
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 7 hours ago
Sergamar
163
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
7 hours ago
I believe it must be on $mathbb R$
– fonini
7 hours ago
On all the real numbers
– Sergamar
7 hours ago
It can't be on $mathbb{R}$: this function not defined at $x=0$.
– user587192
7 hours ago
1
0 if x = 0, that expresion otherwise
– Sergamar
7 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 7 hours ago
Sergamar
163
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 7 hours ago
Sergamar
163
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 7 hours ago
Sergamar
163
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 7 hours ago
asked 7 hours ago
Sergamar
163
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 7 hours ago
Sergamar
163
asked 7 hours ago
Sergamar
163
163
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
7 hours ago
I believe it must be on $mathbb R$
– fonini
7 hours ago
On all the real numbers
– Sergamar
7 hours ago
It can't be on $mathbb{R}$: this function not defined at $x=0$.
– user587192
7 hours ago
1
0 if x = 0, that expresion otherwise
– Sergamar
7 hours ago
|
show 3 more comments
2
Uniformly continuous on what set?
– user587192
7 hours ago
I believe it must be on $mathbb R$
– fonini
7 hours ago
On all the real numbers
– Sergamar
7 hours ago
It can't be on $mathbb{R}$: this function not defined at $x=0$.
– user587192
7 hours ago
1
0 if x = 0, that expresion otherwise
– Sergamar
7 hours ago
2
2
Uniformly continuous on what set?
– user587192
7 hours ago
Uniformly continuous on what set?
– user587192
7 hours ago
I believe it must be on $mathbb R$
– fonini
7 hours ago
I believe it must be on $mathbb R$
– fonini
7 hours ago
On all the real numbers
– Sergamar
7 hours ago
On all the real numbers
– Sergamar
7 hours ago
It can't be on $mathbb{R}$: this function not defined at $x=0$.
– user587192
7 hours ago
It can't be on $mathbb{R}$: this function not defined at $x=0$.
– user587192
7 hours ago
1
1
0 if x = 0, that expresion otherwise
– Sergamar
7 hours ago
0 if x = 0, that expresion otherwise
– Sergamar
7 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 6 hours 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
6 hours 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
6 hours ago
@Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity?
– ImNotTheGuy
6 hours ago
@RRL oh, thank you, it was a typo and it got me confused
– Jakobian
6 hours ago
@Jakobian It got me confused too, lol.
– ImNotTheGuy
6 hours ago
|
show 1 more comment
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1 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 6 hours 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
6 hours 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
6 hours ago
@Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity?
– ImNotTheGuy
6 hours ago
@RRL oh, thank you, it was a typo and it got me confused
– Jakobian
6 hours ago
@Jakobian It got me confused too, lol.
– ImNotTheGuy
6 hours 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 6 hours 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
6 hours 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
6 hours ago
@Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity?
– ImNotTheGuy
6 hours ago
@RRL oh, thank you, it was a typo and it got me confused
– Jakobian
6 hours ago
@Jakobian It got me confused too, lol.
– ImNotTheGuy
6 hours 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 6 hours 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 6 hours ago
RRL
48.7k42573
edited 5 hours ago
answered 6 hours ago
RRL
48.7k42573
answered 6 hours ago
RRL
48.7k42573
answered 6 hours 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
6 hours 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
6 hours ago
@Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity?
– ImNotTheGuy
6 hours ago
@RRL oh, thank you, it was a typo and it got me confused
– Jakobian
6 hours ago
@Jakobian It got me confused too, lol.
– ImNotTheGuy
6 hours 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
6 hours 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
6 hours ago
@Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity?
– ImNotTheGuy
6 hours ago
@RRL oh, thank you, it was a typo and it got me confused
– Jakobian
6 hours ago
@Jakobian It got me confused too, lol.
– ImNotTheGuy
6 hours 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
6 hours 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
6 hours 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
6 hours 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
6 hours ago
@Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity?
– ImNotTheGuy
6 hours ago
@Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity?
– ImNotTheGuy
6 hours ago
@RRL oh, thank you, it was a typo and it got me confused
– Jakobian
6 hours ago
@RRL oh, thank you, it was a typo and it got me confused
– Jakobian
6 hours ago
@Jakobian It got me confused too, lol.
– ImNotTheGuy
6 hours ago
@Jakobian It got me confused too, lol.
– ImNotTheGuy
6 hours 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.
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2
Uniformly continuous on what set?
– user587192
7 hours ago
I believe it must be on $mathbb R$
– fonini
7 hours ago
On all the real numbers
– Sergamar
7 hours ago
It can't be on $mathbb{R}$: this function not defined at $x=0$.
– user587192
7 hours ago
1
0 if x = 0, that expresion otherwise
– Sergamar
7 hours ago