If $f$ is convex function and $f$ is continuous on $(a,b)$.When is $f$ continuous at $a$ and $b$? [closed]
$begingroup$
Let $f:[a,b]to mathbb{R}$. We know that if $f$ is convex function then $f$ is continuous in $(a,b)$. Ιn which cases is $f$ continuous at $a$ and $b$?
real-analysis continuity convex-analysis
edited Dec 18 '18 at 22:41
Will Fisher
4,0381032
asked Dec 18 '18 at 22:23
Chris RafaelChris Rafael
106
$endgroup$
closed as off-topic by Did, Saad, KReiser, Lord_Farin, user91500 Dec 27 '18 at 9:09
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Saad, KReiser, Lord_Farin, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $f:[a,b]to mathbb{R}$. We know that if $f$ is convex function then $f$ is continuous in $(a,b)$. Ιn which cases is $f$ continuous at $a$ and $b$?
real-analysis continuity convex-analysis
edited Dec 18 '18 at 22:41
Will Fisher
4,0381032
asked Dec 18 '18 at 22:23
Chris RafaelChris Rafael
106
$endgroup$
closed as off-topic by Did, Saad, KReiser, Lord_Farin, user91500 Dec 27 '18 at 9:09
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Saad, KReiser, Lord_Farin, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $f:[a,b]to mathbb{R}$. We know that if $f$ is convex function then $f$ is continuous in $(a,b)$. Ιn which cases is $f$ continuous at $a$ and $b$?
real-analysis continuity convex-analysis
edited Dec 18 '18 at 22:41
Will Fisher
4,0381032
asked Dec 18 '18 at 22:23
Chris RafaelChris Rafael
106
$endgroup$
Let $f:[a,b]to mathbb{R}$. We know that if $f$ is convex function then $f$ is continuous in $(a,b)$. Ιn which cases is $f$ continuous at $a$ and $b$?
real-analysis continuity convex-analysis
real-analysis continuity convex-analysis
edited Dec 18 '18 at 22:41
Will Fisher
4,0381032
asked Dec 18 '18 at 22:23
Chris RafaelChris Rafael
106
edited Dec 18 '18 at 22:41
Will Fisher
4,0381032
asked Dec 18 '18 at 22:23
Chris RafaelChris Rafael
106
edited Dec 18 '18 at 22:41
Will Fisher
4,0381032
edited Dec 18 '18 at 22:41
Will Fisher
4,0381032
edited Dec 18 '18 at 22:41
Will Fisher
4,0381032
4,0381032
asked Dec 18 '18 at 22:23
Chris RafaelChris Rafael
106
asked Dec 18 '18 at 22:23
Chris RafaelChris Rafael
106
asked Dec 18 '18 at 22:23
Chris RafaelChris Rafael
106
106
closed as off-topic by Did, Saad, KReiser, Lord_Farin, user91500 Dec 27 '18 at 9:09
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Saad, KReiser, Lord_Farin, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Did, Saad, KReiser, Lord_Farin, user91500 Dec 27 '18 at 9:09
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Saad, KReiser, Lord_Farin, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I think the best you can do, without knowing more about $f$, is say that $f:[a,b]rightarrowmathbb{R}$ is continuous at $a$ if $f(a)=lim_{xto a^+}f(x)$ and likewise for $f(b)$. For example, consider $f:[0,1]rightarrowmathbb{R}$ s.t. $f(0)=1$, $f(x)=-sqrt{x}$ for $x>0$. Then f is convex on $[0,1]$ but $f$ is not continuous at its left endpoint.
answered Dec 18 '18 at 22:58
R.JacksonR.Jackson
1688
$endgroup$
$begingroup$
You can have convex functions that either are or aren’t continuous at their endpoints as both answers have shown so knowing only that a function is convex, the best you can do is say it’s continuous at its endpoints when it’s continuous at its endpoints.
$endgroup$
– R.Jackson
Dec 18 '18 at 23:19
$begingroup$
Yes , i undestrand what you mean.Anyway in your example you take f:[0,1)→R but we must show for f:[0,1]→R to take this as example
$endgroup$
– Chris Rafael
Dec 18 '18 at 23:30
$begingroup$
I up voted your answer because of that last statement you made in the comment!
$endgroup$
– RRL
Dec 18 '18 at 23:31
1
$begingroup$
@ChrisRafael right, my bad, it should be $[0,1]$
$endgroup$
– R.Jackson
Dec 18 '18 at 23:34
$begingroup$
Also, thanks! @RRL
$endgroup$
– R.Jackson
Dec 18 '18 at 23:35
|
show 1 more comment
$begingroup$
First, if $f$ is convex on $(a,b)$ then finite or infinite limits $lim_{x to a+} f(x)$ and $lim_{x to b-} f(x)$ always exist.
Consider the functions $f:x mapsto x^2$ on $[0,1]$ and
$$g(x) = begin{cases}f(x), & 0 leqslant x < 1,\ 2, & x= 1 end{cases}$$
Both are convex, but $f$ is continuous at $x=1$ and $g$ is not.
What does this tell you?
The one-sided limit at an endpoint is always a finite number or $pm infty$. The limit cannot fail to exist (as with an oscillating function). As long as limits are not infinite, the function may or may not be continuous at the endpoints, but it can be made continuous by changing values to match the one-sided limits.
answered Dec 18 '18 at 22:55
RRLRRL
51.3k42573
$endgroup$
$begingroup$
Can you tell me one exaple with only f ?
$endgroup$
– Chris Rafael
Dec 18 '18 at 23:27
$begingroup$
What do you mean by "example with only $f$"?
$endgroup$
– RRL
Dec 18 '18 at 23:32
$begingroup$
An example with only one function
$endgroup$
– Chris Rafael
Dec 18 '18 at 23:33
$begingroup$
Your question was in which cases is $f$ continuous at $a$ and $b$. Between the two posts here you have the answer to that. If $f(x) to pm infty$ as $x to b-$ then it is not only discontinuous at $x = b$, but the discontinuity can never be removed. If $f(x) to c neq pm infty$, then $f$ may or may not be continuous at $b$. It will be continuous if $f(b) = c$.
$endgroup$
– RRL
Dec 18 '18 at 23:41
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think the best you can do, without knowing more about $f$, is say that $f:[a,b]rightarrowmathbb{R}$ is continuous at $a$ if $f(a)=lim_{xto a^+}f(x)$ and likewise for $f(b)$. For example, consider $f:[0,1]rightarrowmathbb{R}$ s.t. $f(0)=1$, $f(x)=-sqrt{x}$ for $x>0$. Then f is convex on $[0,1]$ but $f$ is not continuous at its left endpoint.
answered Dec 18 '18 at 22:58
R.JacksonR.Jackson
1688
$endgroup$
$begingroup$
You can have convex functions that either are or aren’t continuous at their endpoints as both answers have shown so knowing only that a function is convex, the best you can do is say it’s continuous at its endpoints when it’s continuous at its endpoints.
$endgroup$
– R.Jackson
Dec 18 '18 at 23:19
$begingroup$
Yes , i undestrand what you mean.Anyway in your example you take f:[0,1)→R but we must show for f:[0,1]→R to take this as example
$endgroup$
– Chris Rafael
Dec 18 '18 at 23:30
$begingroup$
I up voted your answer because of that last statement you made in the comment!
$endgroup$
– RRL
Dec 18 '18 at 23:31
1
$begingroup$
@ChrisRafael right, my bad, it should be $[0,1]$
$endgroup$
– R.Jackson
Dec 18 '18 at 23:34
$begingroup$
Also, thanks! @RRL
$endgroup$
– R.Jackson
Dec 18 '18 at 23:35
|
show 1 more comment
$begingroup$
I think the best you can do, without knowing more about $f$, is say that $f:[a,b]rightarrowmathbb{R}$ is continuous at $a$ if $f(a)=lim_{xto a^+}f(x)$ and likewise for $f(b)$. For example, consider $f:[0,1]rightarrowmathbb{R}$ s.t. $f(0)=1$, $f(x)=-sqrt{x}$ for $x>0$. Then f is convex on $[0,1]$ but $f$ is not continuous at its left endpoint.
answered Dec 18 '18 at 22:58
R.JacksonR.Jackson
1688
$endgroup$
$begingroup$
You can have convex functions that either are or aren’t continuous at their endpoints as both answers have shown so knowing only that a function is convex, the best you can do is say it’s continuous at its endpoints when it’s continuous at its endpoints.
$endgroup$
– R.Jackson
Dec 18 '18 at 23:19
$begingroup$
Yes , i undestrand what you mean.Anyway in your example you take f:[0,1)→R but we must show for f:[0,1]→R to take this as example
$endgroup$
– Chris Rafael
Dec 18 '18 at 23:30
$begingroup$
I up voted your answer because of that last statement you made in the comment!
$endgroup$
– RRL
Dec 18 '18 at 23:31
1
$begingroup$
@ChrisRafael right, my bad, it should be $[0,1]$
$endgroup$
– R.Jackson
Dec 18 '18 at 23:34
$begingroup$
Also, thanks! @RRL
$endgroup$
– R.Jackson
Dec 18 '18 at 23:35
|
show 1 more comment
$begingroup$
I think the best you can do, without knowing more about $f$, is say that $f:[a,b]rightarrowmathbb{R}$ is continuous at $a$ if $f(a)=lim_{xto a^+}f(x)$ and likewise for $f(b)$. For example, consider $f:[0,1]rightarrowmathbb{R}$ s.t. $f(0)=1$, $f(x)=-sqrt{x}$ for $x>0$. Then f is convex on $[0,1]$ but $f$ is not continuous at its left endpoint.
answered Dec 18 '18 at 22:58
R.JacksonR.Jackson
1688
$endgroup$
I think the best you can do, without knowing more about $f$, is say that $f:[a,b]rightarrowmathbb{R}$ is continuous at $a$ if $f(a)=lim_{xto a^+}f(x)$ and likewise for $f(b)$. For example, consider $f:[0,1]rightarrowmathbb{R}$ s.t. $f(0)=1$, $f(x)=-sqrt{x}$ for $x>0$. Then f is convex on $[0,1]$ but $f$ is not continuous at its left endpoint.
answered Dec 18 '18 at 22:58
R.JacksonR.Jackson
1688
edited Dec 18 '18 at 23:34
answered Dec 18 '18 at 22:58
R.JacksonR.Jackson
1688
answered Dec 18 '18 at 22:58
R.JacksonR.Jackson
1688
answered Dec 18 '18 at 22:58
R.JacksonR.Jackson
1688
1688
$begingroup$
You can have convex functions that either are or aren’t continuous at their endpoints as both answers have shown so knowing only that a function is convex, the best you can do is say it’s continuous at its endpoints when it’s continuous at its endpoints.
$endgroup$
– R.Jackson
Dec 18 '18 at 23:19
$begingroup$
Yes , i undestrand what you mean.Anyway in your example you take f:[0,1)→R but we must show for f:[0,1]→R to take this as example
$endgroup$
– Chris Rafael
Dec 18 '18 at 23:30
$begingroup$
I up voted your answer because of that last statement you made in the comment!
$endgroup$
– RRL
Dec 18 '18 at 23:31
1
$begingroup$
@ChrisRafael right, my bad, it should be $[0,1]$
$endgroup$
– R.Jackson
Dec 18 '18 at 23:34
$begingroup$
Also, thanks! @RRL
$endgroup$
– R.Jackson
Dec 18 '18 at 23:35
|
show 1 more comment
$begingroup$
You can have convex functions that either are or aren’t continuous at their endpoints as both answers have shown so knowing only that a function is convex, the best you can do is say it’s continuous at its endpoints when it’s continuous at its endpoints.
$endgroup$
– R.Jackson
Dec 18 '18 at 23:19
$begingroup$
Yes , i undestrand what you mean.Anyway in your example you take f:[0,1)→R but we must show for f:[0,1]→R to take this as example
$endgroup$
– Chris Rafael
Dec 18 '18 at 23:30
$begingroup$
I up voted your answer because of that last statement you made in the comment!
$endgroup$
– RRL
Dec 18 '18 at 23:31
1
$begingroup$
@ChrisRafael right, my bad, it should be $[0,1]$
$endgroup$
– R.Jackson
Dec 18 '18 at 23:34
$begingroup$
Also, thanks! @RRL
$endgroup$
– R.Jackson
Dec 18 '18 at 23:35
$begingroup$
You can have convex functions that either are or aren’t continuous at their endpoints as both answers have shown so knowing only that a function is convex, the best you can do is say it’s continuous at its endpoints when it’s continuous at its endpoints.
$endgroup$
– R.Jackson
Dec 18 '18 at 23:19
$begingroup$
You can have convex functions that either are or aren’t continuous at their endpoints as both answers have shown so knowing only that a function is convex, the best you can do is say it’s continuous at its endpoints when it’s continuous at its endpoints.
$endgroup$
– R.Jackson
Dec 18 '18 at 23:19
$begingroup$
Yes , i undestrand what you mean.Anyway in your example you take f:[0,1)→R but we must show for f:[0,1]→R to take this as example
$endgroup$
– Chris Rafael
Dec 18 '18 at 23:30
$begingroup$
Yes , i undestrand what you mean.Anyway in your example you take f:[0,1)→R but we must show for f:[0,1]→R to take this as example
$endgroup$
– Chris Rafael
Dec 18 '18 at 23:30
$begingroup$
I up voted your answer because of that last statement you made in the comment!
$endgroup$
– RRL
Dec 18 '18 at 23:31
$begingroup$
I up voted your answer because of that last statement you made in the comment!
$endgroup$
– RRL
Dec 18 '18 at 23:31
1
1
$begingroup$
@ChrisRafael right, my bad, it should be $[0,1]$
$endgroup$
– R.Jackson
Dec 18 '18 at 23:34
$begingroup$
@ChrisRafael right, my bad, it should be $[0,1]$
$endgroup$
– R.Jackson
Dec 18 '18 at 23:34
$begingroup$
Also, thanks! @RRL
$endgroup$
– R.Jackson
Dec 18 '18 at 23:35
$begingroup$
Also, thanks! @RRL
$endgroup$
– R.Jackson
Dec 18 '18 at 23:35
|
show 1 more comment
$begingroup$
First, if $f$ is convex on $(a,b)$ then finite or infinite limits $lim_{x to a+} f(x)$ and $lim_{x to b-} f(x)$ always exist.
Consider the functions $f:x mapsto x^2$ on $[0,1]$ and
$$g(x) = begin{cases}f(x), & 0 leqslant x < 1,\ 2, & x= 1 end{cases}$$
Both are convex, but $f$ is continuous at $x=1$ and $g$ is not.
What does this tell you?
The one-sided limit at an endpoint is always a finite number or $pm infty$. The limit cannot fail to exist (as with an oscillating function). As long as limits are not infinite, the function may or may not be continuous at the endpoints, but it can be made continuous by changing values to match the one-sided limits.
answered Dec 18 '18 at 22:55
RRLRRL
51.3k42573
$endgroup$
$begingroup$
Can you tell me one exaple with only f ?
$endgroup$
– Chris Rafael
Dec 18 '18 at 23:27
$begingroup$
What do you mean by "example with only $f$"?
$endgroup$
– RRL
Dec 18 '18 at 23:32
$begingroup$
An example with only one function
$endgroup$
– Chris Rafael
Dec 18 '18 at 23:33
$begingroup$
Your question was in which cases is $f$ continuous at $a$ and $b$. Between the two posts here you have the answer to that. If $f(x) to pm infty$ as $x to b-$ then it is not only discontinuous at $x = b$, but the discontinuity can never be removed. If $f(x) to c neq pm infty$, then $f$ may or may not be continuous at $b$. It will be continuous if $f(b) = c$.
$endgroup$
– RRL
Dec 18 '18 at 23:41
add a comment |
$begingroup$
First, if $f$ is convex on $(a,b)$ then finite or infinite limits $lim_{x to a+} f(x)$ and $lim_{x to b-} f(x)$ always exist.
Consider the functions $f:x mapsto x^2$ on $[0,1]$ and
$$g(x) = begin{cases}f(x), & 0 leqslant x < 1,\ 2, & x= 1 end{cases}$$
Both are convex, but $f$ is continuous at $x=1$ and $g$ is not.
What does this tell you?
The one-sided limit at an endpoint is always a finite number or $pm infty$. The limit cannot fail to exist (as with an oscillating function). As long as limits are not infinite, the function may or may not be continuous at the endpoints, but it can be made continuous by changing values to match the one-sided limits.
answered Dec 18 '18 at 22:55
RRLRRL
51.3k42573
$endgroup$
$begingroup$
Can you tell me one exaple with only f ?
$endgroup$
– Chris Rafael
Dec 18 '18 at 23:27
$begingroup$
What do you mean by "example with only $f$"?
$endgroup$
– RRL
Dec 18 '18 at 23:32
$begingroup$
An example with only one function
$endgroup$
– Chris Rafael
Dec 18 '18 at 23:33
$begingroup$
Your question was in which cases is $f$ continuous at $a$ and $b$. Between the two posts here you have the answer to that. If $f(x) to pm infty$ as $x to b-$ then it is not only discontinuous at $x = b$, but the discontinuity can never be removed. If $f(x) to c neq pm infty$, then $f$ may or may not be continuous at $b$. It will be continuous if $f(b) = c$.
$endgroup$
– RRL
Dec 18 '18 at 23:41
add a comment |
$begingroup$
First, if $f$ is convex on $(a,b)$ then finite or infinite limits $lim_{x to a+} f(x)$ and $lim_{x to b-} f(x)$ always exist.
Consider the functions $f:x mapsto x^2$ on $[0,1]$ and
$$g(x) = begin{cases}f(x), & 0 leqslant x < 1,\ 2, & x= 1 end{cases}$$
Both are convex, but $f$ is continuous at $x=1$ and $g$ is not.
What does this tell you?
The one-sided limit at an endpoint is always a finite number or $pm infty$. The limit cannot fail to exist (as with an oscillating function). As long as limits are not infinite, the function may or may not be continuous at the endpoints, but it can be made continuous by changing values to match the one-sided limits.
answered Dec 18 '18 at 22:55
RRLRRL
51.3k42573
$endgroup$
First, if $f$ is convex on $(a,b)$ then finite or infinite limits $lim_{x to a+} f(x)$ and $lim_{x to b-} f(x)$ always exist.
Consider the functions $f:x mapsto x^2$ on $[0,1]$ and
$$g(x) = begin{cases}f(x), & 0 leqslant x < 1,\ 2, & x= 1 end{cases}$$
Both are convex, but $f$ is continuous at $x=1$ and $g$ is not.
What does this tell you?
The one-sided limit at an endpoint is always a finite number or $pm infty$. The limit cannot fail to exist (as with an oscillating function). As long as limits are not infinite, the function may or may not be continuous at the endpoints, but it can be made continuous by changing values to match the one-sided limits.
answered Dec 18 '18 at 22:55
RRLRRL
51.3k42573
edited Dec 18 '18 at 23:16
answered Dec 18 '18 at 22:55
RRLRRL
51.3k42573
answered Dec 18 '18 at 22:55
RRLRRL
51.3k42573
answered Dec 18 '18 at 22:55
RRLRRL
51.3k42573
51.3k42573
$begingroup$
Can you tell me one exaple with only f ?
$endgroup$
– Chris Rafael
Dec 18 '18 at 23:27
$begingroup$
What do you mean by "example with only $f$"?
$endgroup$
– RRL
Dec 18 '18 at 23:32
$begingroup$
An example with only one function
$endgroup$
– Chris Rafael
Dec 18 '18 at 23:33
$begingroup$
Your question was in which cases is $f$ continuous at $a$ and $b$. Between the two posts here you have the answer to that. If $f(x) to pm infty$ as $x to b-$ then it is not only discontinuous at $x = b$, but the discontinuity can never be removed. If $f(x) to c neq pm infty$, then $f$ may or may not be continuous at $b$. It will be continuous if $f(b) = c$.
$endgroup$
– RRL
Dec 18 '18 at 23:41
add a comment |
$begingroup$
Can you tell me one exaple with only f ?
$endgroup$
– Chris Rafael
Dec 18 '18 at 23:27
$begingroup$
What do you mean by "example with only $f$"?
$endgroup$
– RRL
Dec 18 '18 at 23:32
$begingroup$
An example with only one function
$endgroup$
– Chris Rafael
Dec 18 '18 at 23:33
$begingroup$
Your question was in which cases is $f$ continuous at $a$ and $b$. Between the two posts here you have the answer to that. If $f(x) to pm infty$ as $x to b-$ then it is not only discontinuous at $x = b$, but the discontinuity can never be removed. If $f(x) to c neq pm infty$, then $f$ may or may not be continuous at $b$. It will be continuous if $f(b) = c$.
$endgroup$
– RRL
Dec 18 '18 at 23:41
$begingroup$
Can you tell me one exaple with only f ?
$endgroup$
– Chris Rafael
Dec 18 '18 at 23:27
$begingroup$
Can you tell me one exaple with only f ?
$endgroup$
– Chris Rafael
Dec 18 '18 at 23:27
$begingroup$
What do you mean by "example with only $f$"?
$endgroup$
– RRL
Dec 18 '18 at 23:32
$begingroup$
What do you mean by "example with only $f$"?
$endgroup$
– RRL
Dec 18 '18 at 23:32
$begingroup$
An example with only one function
$endgroup$
– Chris Rafael
Dec 18 '18 at 23:33
$begingroup$
An example with only one function
$endgroup$
– Chris Rafael
Dec 18 '18 at 23:33
$begingroup$
Your question was in which cases is $f$ continuous at $a$ and $b$. Between the two posts here you have the answer to that. If $f(x) to pm infty$ as $x to b-$ then it is not only discontinuous at $x = b$, but the discontinuity can never be removed. If $f(x) to c neq pm infty$, then $f$ may or may not be continuous at $b$. It will be continuous if $f(b) = c$.
$endgroup$
– RRL
Dec 18 '18 at 23:41
$begingroup$
Your question was in which cases is $f$ continuous at $a$ and $b$. Between the two posts here you have the answer to that. If $f(x) to pm infty$ as $x to b-$ then it is not only discontinuous at $x = b$, but the discontinuity can never be removed. If $f(x) to c neq pm infty$, then $f$ may or may not be continuous at $b$. It will be continuous if $f(b) = c$.
$endgroup$
– RRL
Dec 18 '18 at 23:41
add a comment |
