Is $f(x) = frac{1}{x}$ uniformly continuous on $(1, infty)$? [closed]
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is the function $f(x) = 1/x$ uniformly continuous on $(1, infty)?$ I know that it is not uniformly continuous on $(0, infty)$, but now I'm restricting it even more to get rid of most of the bad parts. I think the answer is yes.
real-analysis uniform-continuity
edited Nov 16 at 19:37
José Carlos Santos
141k19111207
asked Nov 16 at 18:48
joseph
1068
closed as off-topic by Servaes, Nosrati, RRL, Trevor Gunn, Gibbs Nov 16 at 23:41
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Servaes, Nosrati, RRL, Trevor Gunn, Gibbs
If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
1
down vote
favorite
is the function $f(x) = 1/x$ uniformly continuous on $(1, infty)?$ I know that it is not uniformly continuous on $(0, infty)$, but now I'm restricting it even more to get rid of most of the bad parts. I think the answer is yes.
real-analysis uniform-continuity
edited Nov 16 at 19:37
José Carlos Santos
141k19111207
asked Nov 16 at 18:48
joseph
1068
closed as off-topic by Servaes, Nosrati, RRL, Trevor Gunn, Gibbs Nov 16 at 23:41
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Servaes, Nosrati, RRL, Trevor Gunn, Gibbs
If this question can be reworded to fit the rules in the help center, please edit the question.
2
It is. ${}{}{}{}$
– T. Bongers
Nov 16 at 18:48
1
Very roughly, "uniformly continuous" means "having bounded derivative".
– Lord Shark the Unknown
Nov 16 at 18:51
1
Indeed, the function $f(x)=1/x$ will be uniformly continuous on any interval of the form $(varepsilon,infty)$ for any $varepsilon>0$. The failure to be uniformly continuous only happens in a neighborhood of $0$, so if you prevent the variable from being able to get too close, you ensure uniform continuity (see Lord Shark the Unknown's comment, too).
– Clayton
Nov 16 at 18:57
1
Though as a caution on taking Lord Shark the Unknown's comment too literally - for example, $f(x) = sqrt{x}$ and $f(x) = x sin(1/x)$ are both uniformly continuous on $(0,1)$ even though the derivative is unbounded in both cases.
– Daniel Schepler
Nov 16 at 19:32
@LordSharktheUnknown NO! Roughly, “having bounded derivative” is equivalent to being Lipchitz. Uniform continuity is a much weaker condition, as Daniel points out and can be made precise using holder or sobolev spaces for instance.
– Shalop
Nov 16 at 23:06
|
show 4 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
is the function $f(x) = 1/x$ uniformly continuous on $(1, infty)?$ I know that it is not uniformly continuous on $(0, infty)$, but now I'm restricting it even more to get rid of most of the bad parts. I think the answer is yes.
real-analysis uniform-continuity
edited Nov 16 at 19:37
José Carlos Santos
141k19111207
asked Nov 16 at 18:48
joseph
1068
is the function $f(x) = 1/x$ uniformly continuous on $(1, infty)?$ I know that it is not uniformly continuous on $(0, infty)$, but now I'm restricting it even more to get rid of most of the bad parts. I think the answer is yes.
real-analysis uniform-continuity
real-analysis uniform-continuity
edited Nov 16 at 19:37
José Carlos Santos
141k19111207
asked Nov 16 at 18:48
joseph
1068
edited Nov 16 at 19:37
José Carlos Santos
141k19111207
asked Nov 16 at 18:48
joseph
1068
edited Nov 16 at 19:37
José Carlos Santos
141k19111207
edited Nov 16 at 19:37
José Carlos Santos
141k19111207
edited Nov 16 at 19:37
José Carlos Santos
141k19111207
141k19111207
asked Nov 16 at 18:48
joseph
1068
asked Nov 16 at 18:48
joseph
1068
asked Nov 16 at 18:48
joseph
1068
1068
closed as off-topic by Servaes, Nosrati, RRL, Trevor Gunn, Gibbs Nov 16 at 23:41
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Servaes, Nosrati, RRL, Trevor Gunn, Gibbs
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Servaes, Nosrati, RRL, Trevor Gunn, Gibbs Nov 16 at 23:41
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Servaes, Nosrati, RRL, Trevor Gunn, Gibbs
If this question can be reworded to fit the rules in the help center, please edit the question.
2
It is. ${}{}{}{}$
– T. Bongers
Nov 16 at 18:48
1
Very roughly, "uniformly continuous" means "having bounded derivative".
– Lord Shark the Unknown
Nov 16 at 18:51
1
Indeed, the function $f(x)=1/x$ will be uniformly continuous on any interval of the form $(varepsilon,infty)$ for any $varepsilon>0$. The failure to be uniformly continuous only happens in a neighborhood of $0$, so if you prevent the variable from being able to get too close, you ensure uniform continuity (see Lord Shark the Unknown's comment, too).
– Clayton
Nov 16 at 18:57
1
Though as a caution on taking Lord Shark the Unknown's comment too literally - for example, $f(x) = sqrt{x}$ and $f(x) = x sin(1/x)$ are both uniformly continuous on $(0,1)$ even though the derivative is unbounded in both cases.
– Daniel Schepler
Nov 16 at 19:32
@LordSharktheUnknown NO! Roughly, “having bounded derivative” is equivalent to being Lipchitz. Uniform continuity is a much weaker condition, as Daniel points out and can be made precise using holder or sobolev spaces for instance.
– Shalop
Nov 16 at 23:06
|
show 4 more comments
2
It is. ${}{}{}{}$
– T. Bongers
Nov 16 at 18:48
1
Very roughly, "uniformly continuous" means "having bounded derivative".
– Lord Shark the Unknown
Nov 16 at 18:51
1
Indeed, the function $f(x)=1/x$ will be uniformly continuous on any interval of the form $(varepsilon,infty)$ for any $varepsilon>0$. The failure to be uniformly continuous only happens in a neighborhood of $0$, so if you prevent the variable from being able to get too close, you ensure uniform continuity (see Lord Shark the Unknown's comment, too).
– Clayton
Nov 16 at 18:57
1
Though as a caution on taking Lord Shark the Unknown's comment too literally - for example, $f(x) = sqrt{x}$ and $f(x) = x sin(1/x)$ are both uniformly continuous on $(0,1)$ even though the derivative is unbounded in both cases.
– Daniel Schepler
Nov 16 at 19:32
@LordSharktheUnknown NO! Roughly, “having bounded derivative” is equivalent to being Lipchitz. Uniform continuity is a much weaker condition, as Daniel points out and can be made precise using holder or sobolev spaces for instance.
– Shalop
Nov 16 at 23:06
2
2
It is. ${}{}{}{}$
– T. Bongers
Nov 16 at 18:48
It is. ${}{}{}{}$
– T. Bongers
Nov 16 at 18:48
1
1
Very roughly, "uniformly continuous" means "having bounded derivative".
– Lord Shark the Unknown
Nov 16 at 18:51
Very roughly, "uniformly continuous" means "having bounded derivative".
– Lord Shark the Unknown
Nov 16 at 18:51
1
1
Indeed, the function $f(x)=1/x$ will be uniformly continuous on any interval of the form $(varepsilon,infty)$ for any $varepsilon>0$. The failure to be uniformly continuous only happens in a neighborhood of $0$, so if you prevent the variable from being able to get too close, you ensure uniform continuity (see Lord Shark the Unknown's comment, too).
– Clayton
Nov 16 at 18:57
Indeed, the function $f(x)=1/x$ will be uniformly continuous on any interval of the form $(varepsilon,infty)$ for any $varepsilon>0$. The failure to be uniformly continuous only happens in a neighborhood of $0$, so if you prevent the variable from being able to get too close, you ensure uniform continuity (see Lord Shark the Unknown's comment, too).
– Clayton
Nov 16 at 18:57
1
1
Though as a caution on taking Lord Shark the Unknown's comment too literally - for example, $f(x) = sqrt{x}$ and $f(x) = x sin(1/x)$ are both uniformly continuous on $(0,1)$ even though the derivative is unbounded in both cases.
– Daniel Schepler
Nov 16 at 19:32
Though as a caution on taking Lord Shark the Unknown's comment too literally - for example, $f(x) = sqrt{x}$ and $f(x) = x sin(1/x)$ are both uniformly continuous on $(0,1)$ even though the derivative is unbounded in both cases.
– Daniel Schepler
Nov 16 at 19:32
@LordSharktheUnknown NO! Roughly, “having bounded derivative” is equivalent to being Lipchitz. Uniform continuity is a much weaker condition, as Daniel points out and can be made precise using holder or sobolev spaces for instance.
– Shalop
Nov 16 at 23:06
@LordSharktheUnknown NO! Roughly, “having bounded derivative” is equivalent to being Lipchitz. Uniform continuity is a much weaker condition, as Daniel points out and can be made precise using holder or sobolev spaces for instance.
– Shalop
Nov 16 at 23:06
|
show 4 more comments
2 Answers
2
active
oldest
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up vote
6
down vote
accepted
If $x,yin(1,infty)$, then$$leftlvertfrac1x-frac1yrightrvert=frac{lvert x-yrvert}{xy}<lvert x-yrvert.$$So, for each $varepsilon>0$, just take $delta=varepsilon$ and$$lvert x-yrvert<deltaimpliesleftlvertfrac1x-frac1yrightrvert<varepsilon.$$
answered Nov 16 at 18:57
José Carlos Santos
141k19111207
add a comment |
up vote
0
down vote
$f$ is differentiable at $(1,+infty)$ and
$$|f'(x)|=|frac{-1}{x^2}|le 1$$
$$f' text{ is bounded } implies$$
$$f text{ is Lipschitz} implies$$
$$f text{ is uniformly continuous}.$$
answered Nov 16 at 19:35
hamam_Abdallah
36.8k21533
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
If $x,yin(1,infty)$, then$$leftlvertfrac1x-frac1yrightrvert=frac{lvert x-yrvert}{xy}<lvert x-yrvert.$$So, for each $varepsilon>0$, just take $delta=varepsilon$ and$$lvert x-yrvert<deltaimpliesleftlvertfrac1x-frac1yrightrvert<varepsilon.$$
answered Nov 16 at 18:57
José Carlos Santos
141k19111207
add a comment |
up vote
6
down vote
accepted
If $x,yin(1,infty)$, then$$leftlvertfrac1x-frac1yrightrvert=frac{lvert x-yrvert}{xy}<lvert x-yrvert.$$So, for each $varepsilon>0$, just take $delta=varepsilon$ and$$lvert x-yrvert<deltaimpliesleftlvertfrac1x-frac1yrightrvert<varepsilon.$$
answered Nov 16 at 18:57
José Carlos Santos
141k19111207
add a comment |
up vote
6
down vote
accepted
up vote
6
down vote
accepted
If $x,yin(1,infty)$, then$$leftlvertfrac1x-frac1yrightrvert=frac{lvert x-yrvert}{xy}<lvert x-yrvert.$$So, for each $varepsilon>0$, just take $delta=varepsilon$ and$$lvert x-yrvert<deltaimpliesleftlvertfrac1x-frac1yrightrvert<varepsilon.$$
answered Nov 16 at 18:57
José Carlos Santos
141k19111207
If $x,yin(1,infty)$, then$$leftlvertfrac1x-frac1yrightrvert=frac{lvert x-yrvert}{xy}<lvert x-yrvert.$$So, for each $varepsilon>0$, just take $delta=varepsilon$ and$$lvert x-yrvert<deltaimpliesleftlvertfrac1x-frac1yrightrvert<varepsilon.$$
answered Nov 16 at 18:57
José Carlos Santos
141k19111207
answered Nov 16 at 18:57
José Carlos Santos
141k19111207
answered Nov 16 at 18:57
José Carlos Santos
141k19111207
answered Nov 16 at 18:57
José Carlos Santos
141k19111207
141k19111207
add a comment |
add a comment |
up vote
0
down vote
$f$ is differentiable at $(1,+infty)$ and
$$|f'(x)|=|frac{-1}{x^2}|le 1$$
$$f' text{ is bounded } implies$$
$$f text{ is Lipschitz} implies$$
$$f text{ is uniformly continuous}.$$
answered Nov 16 at 19:35
hamam_Abdallah
36.8k21533
add a comment |
up vote
0
down vote
$f$ is differentiable at $(1,+infty)$ and
$$|f'(x)|=|frac{-1}{x^2}|le 1$$
$$f' text{ is bounded } implies$$
$$f text{ is Lipschitz} implies$$
$$f text{ is uniformly continuous}.$$
answered Nov 16 at 19:35
hamam_Abdallah
36.8k21533
add a comment |
up vote
0
down vote
up vote
0
down vote
$f$ is differentiable at $(1,+infty)$ and
$$|f'(x)|=|frac{-1}{x^2}|le 1$$
$$f' text{ is bounded } implies$$
$$f text{ is Lipschitz} implies$$
$$f text{ is uniformly continuous}.$$
answered Nov 16 at 19:35
hamam_Abdallah
36.8k21533
$f$ is differentiable at $(1,+infty)$ and
$$|f'(x)|=|frac{-1}{x^2}|le 1$$
$$f' text{ is bounded } implies$$
$$f text{ is Lipschitz} implies$$
$$f text{ is uniformly continuous}.$$
answered Nov 16 at 19:35
hamam_Abdallah
36.8k21533
answered Nov 16 at 19:35
hamam_Abdallah
36.8k21533
answered Nov 16 at 19:35
hamam_Abdallah
36.8k21533
answered Nov 16 at 19:35
hamam_Abdallah
36.8k21533
36.8k21533
add a comment |
add a comment |
2
It is. ${}{}{}{}$
– T. Bongers
Nov 16 at 18:48
1
Very roughly, "uniformly continuous" means "having bounded derivative".
– Lord Shark the Unknown
Nov 16 at 18:51
1
Indeed, the function $f(x)=1/x$ will be uniformly continuous on any interval of the form $(varepsilon,infty)$ for any $varepsilon>0$. The failure to be uniformly continuous only happens in a neighborhood of $0$, so if you prevent the variable from being able to get too close, you ensure uniform continuity (see Lord Shark the Unknown's comment, too).
– Clayton
Nov 16 at 18:57
1
Though as a caution on taking Lord Shark the Unknown's comment too literally - for example, $f(x) = sqrt{x}$ and $f(x) = x sin(1/x)$ are both uniformly continuous on $(0,1)$ even though the derivative is unbounded in both cases.
– Daniel Schepler
Nov 16 at 19:32
@LordSharktheUnknown NO! Roughly, “having bounded derivative” is equivalent to being Lipchitz. Uniform continuity is a much weaker condition, as Daniel points out and can be made precise using holder or sobolev spaces for instance.
– Shalop
Nov 16 at 23:06