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 '18 at 1:15
SergamarSergamar
185
|
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 '18 at 1:15
SergamarSergamar
185
2
Uniformly continuous on what set?
– user587192
Dec 26 '18 at 1:16
I believe it must be on $mathbb R$
– fonini
Dec 26 '18 at 1:17
1
0 if x = 0, that expresion otherwise
– Sergamar
Dec 26 '18 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 '18 at 1:44
1
Here is a topic about that theorem
– Jakobian
Dec 26 '18 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 '18 at 1:15
SergamarSergamar
185
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 '18 at 1:15
SergamarSergamar
185
asked Dec 26 '18 at 1:15
SergamarSergamar
185
edited Dec 26 '18 at 1:22
Sergamar
asked Dec 26 '18 at 1:15
SergamarSergamar
185
asked Dec 26 '18 at 1:15
SergamarSergamar
185
asked Dec 26 '18 at 1:15
SergamarSergamar
185
185
2
Uniformly continuous on what set?
– user587192
Dec 26 '18 at 1:16
I believe it must be on $mathbb R$
– fonini
Dec 26 '18 at 1:17
1
0 if x = 0, that expresion otherwise
– Sergamar
Dec 26 '18 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 '18 at 1:44
1
Here is a topic about that theorem
– Jakobian
Dec 26 '18 at 1:50
|
show 3 more comments
2
Uniformly continuous on what set?
– user587192
Dec 26 '18 at 1:16
I believe it must be on $mathbb R$
– fonini
Dec 26 '18 at 1:17
1
0 if x = 0, that expresion otherwise
– Sergamar
Dec 26 '18 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 '18 at 1:44
1
Here is a topic about that theorem
– Jakobian
Dec 26 '18 at 1:50
2
2
Uniformly continuous on what set?
– user587192
Dec 26 '18 at 1:16
Uniformly continuous on what set?
– user587192
Dec 26 '18 at 1:16
I believe it must be on $mathbb R$
– fonini
Dec 26 '18 at 1:17
I believe it must be on $mathbb R$
– fonini
Dec 26 '18 at 1:17
1
1
0 if x = 0, that expresion otherwise
– Sergamar
Dec 26 '18 at 1:18
0 if x = 0, that expresion otherwise
– Sergamar
Dec 26 '18 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 '18 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 '18 at 1:44
1
1
Here is a topic about that theorem
– Jakobian
Dec 26 '18 at 1:50
Here is a topic about that theorem
– Jakobian
Dec 26 '18 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 '18 at 1:50
RRLRRL
49.3k42573
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 '18 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 '18 at 2:07
@Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity?
– ImNotTheGuy
Dec 26 '18 at 2:09
@RRL oh, thank you, it was a typo and it got me confused
– Jakobian
Dec 26 '18 at 2:10
@Jakobian It got me confused too, lol.
– ImNotTheGuy
Dec 26 '18 at 2:10
|
show 1 more comment
Your 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 '18 at 1:50
RRLRRL
49.3k42573
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 '18 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 '18 at 2:07
@Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity?
– ImNotTheGuy
Dec 26 '18 at 2:09
@RRL oh, thank you, it was a typo and it got me confused
– Jakobian
Dec 26 '18 at 2:10
@Jakobian It got me confused too, lol.
– ImNotTheGuy
Dec 26 '18 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 '18 at 1:50
RRLRRL
49.3k42573
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 '18 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 '18 at 2:07
@Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity?
– ImNotTheGuy
Dec 26 '18 at 2:09
@RRL oh, thank you, it was a typo and it got me confused
– Jakobian
Dec 26 '18 at 2:10
@Jakobian It got me confused too, lol.
– ImNotTheGuy
Dec 26 '18 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 '18 at 1:50
RRLRRL
49.3k42573
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 '18 at 1:50
RRLRRL
49.3k42573
edited Dec 26 '18 at 2:32
answered Dec 26 '18 at 1:50
RRLRRL
49.3k42573
answered Dec 26 '18 at 1:50
RRLRRL
49.3k42573
answered Dec 26 '18 at 1:50
RRLRRL
49.3k42573
49.3k42573
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 '18 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 '18 at 2:07
@Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity?
– ImNotTheGuy
Dec 26 '18 at 2:09
@RRL oh, thank you, it was a typo and it got me confused
– Jakobian
Dec 26 '18 at 2:10
@Jakobian It got me confused too, lol.
– ImNotTheGuy
Dec 26 '18 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 '18 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 '18 at 2:07
@Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity?
– ImNotTheGuy
Dec 26 '18 at 2:09
@RRL oh, thank you, it was a typo and it got me confused
– Jakobian
Dec 26 '18 at 2:10
@Jakobian It got me confused too, lol.
– ImNotTheGuy
Dec 26 '18 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 '18 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 '18 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 '18 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 '18 at 2:07
@Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity?
– ImNotTheGuy
Dec 26 '18 at 2:09
@Jakobian Where did uniform convergence come into this question? I thought we were talking about continuity?
– ImNotTheGuy
Dec 26 '18 at 2:09
@RRL oh, thank you, it was a typo and it got me confused
– Jakobian
Dec 26 '18 at 2:10
@RRL oh, thank you, it was a typo and it got me confused
– Jakobian
Dec 26 '18 at 2:10
@Jakobian It got me confused too, lol.
– ImNotTheGuy
Dec 26 '18 at 2:10
@Jakobian It got me confused too, lol.
– ImNotTheGuy
Dec 26 '18 at 2:10
|
show 1 more comment
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2
Uniformly continuous on what set?
– user587192
Dec 26 '18 at 1:16
I believe it must be on $mathbb R$
– fonini
Dec 26 '18 at 1:17
1
0 if x = 0, that expresion otherwise
– Sergamar
Dec 26 '18 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 '18 at 1:44
1
Here is a topic about that theorem
– Jakobian
Dec 26 '18 at 1:50