Is $g$ continuous? [closed] Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Prove that function $f$ is continuous at $x = x_0$Example of measure that is discontinuous at $emptyset$For what value(s) of $k$ is $f$ continuous on $mathbbR$Integral on continuous functions metric space, is it right?Show that a modified Dirichlet function is continuous at zeroCounterexample for functional sequence $f_n$ that converges uniformly on $E=[0;A]$ but not on $E=[0;+infty)$Verification of proof of two continuous functionIs this function $f(x) smooth despite it not being continuous?Prove that if $F$ and $G$ are both continuous and $F(a)=G(a)$ then the function $H$ is continuousIdentity criterion and the antiderivative
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Is $g$ continuous? [closed]
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Prove that function $f$ is continuous at $x = x_0$Example of measure that is discontinuous at $emptyset$For what value(s) of $k$ is $f$ continuous on $mathbbR$Integral on continuous functions metric space, is it right?Show that a modified Dirichlet function is continuous at zeroCounterexample for functional sequence $f_n$ that converges uniformly on $E=[0;A]$ but not on $E=[0;+infty)$Verification of proof of two continuous functionIs this function $f(x) smooth despite it not being continuous?Prove that if $F$ and $G$ are both continuous and $F(a)=G(a)$ then the function $H$ is continuousIdentity criterion and the antiderivative
$begingroup$
Suppose that $f$ is continuous on $mathbbR, lim_x rightarrow infty f(x) = - infty $ and $ lim_x rightarrow inftyf(x) = + infty $. define $g$ be setting
$$g(x)= supt : f(t) <x textfor x in mathbbR$$
Now my question is that
Is $g$ continuous ?
my attempt : No , $g $ need not be continious
take $f(x) = begincases x, & x < 1,\
-x+2,& 1 le x le 2, \
x-2, & x > 2endcases$.
Then $g(x) = begincases x, & x le 0,\
x+2, & x > 0endcases$.
graph :
real-analysis
$endgroup$
closed as unclear what you're asking by T. Bongers, RRL, Saad, Jyrki Lahtonen, Parcly Taxel Mar 28 at 0:59
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 4 more comments
$begingroup$
Suppose that $f$ is continuous on $mathbbR, lim_x rightarrow infty f(x) = - infty $ and $ lim_x rightarrow inftyf(x) = + infty $. define $g$ be setting
$$g(x)= supt : f(t) <x textfor x in mathbbR$$
Now my question is that
Is $g$ continuous ?
my attempt : No , $g $ need not be continious
take $f(x) = begincases x, & x < 1,\
-x+2,& 1 le x le 2, \
x-2, & x > 2endcases$.
Then $g(x) = begincases x, & x le 0,\
x+2, & x > 0endcases$.
graph :
real-analysis
$endgroup$
closed as unclear what you're asking by T. Bongers, RRL, Saad, Jyrki Lahtonen, Parcly Taxel Mar 28 at 0:59
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
It has to hold for all such functions $f$, not simply a particular one like you have.
$endgroup$
– Eevee Trainer
Mar 27 at 4:11
$begingroup$
okss @EeveeTrainer..im thinking
$endgroup$
– jasmine
Mar 27 at 4:12
2
$begingroup$
Your $f$ does not satisfy the conditions...
$endgroup$
– copper.hat
Mar 27 at 4:13
2
$begingroup$
To be fair, @copper.hat, no function (as far as I know) satisfies the conditions as written. :p Probably a typo somewhere.
$endgroup$
– Eevee Trainer
Mar 27 at 4:14
1
$begingroup$
@jasmine As $x to -infty$, $f(x) = |x| to +infty$. Consider increasingly negative values of $x$ - $-10, -100, -1000, -10000$, $f$ maps each to their positive version. Not that this helps much since you need to generalize beyond a specific $f$ but I figured I'd point that out.
$endgroup$
– Eevee Trainer
Mar 27 at 4:15
|
show 4 more comments
$begingroup$
Suppose that $f$ is continuous on $mathbbR, lim_x rightarrow infty f(x) = - infty $ and $ lim_x rightarrow inftyf(x) = + infty $. define $g$ be setting
$$g(x)= supt : f(t) <x textfor x in mathbbR$$
Now my question is that
Is $g$ continuous ?
my attempt : No , $g $ need not be continious
take $f(x) = begincases x, & x < 1,\
-x+2,& 1 le x le 2, \
x-2, & x > 2endcases$.
Then $g(x) = begincases x, & x le 0,\
x+2, & x > 0endcases$.
graph :
real-analysis
$endgroup$
Suppose that $f$ is continuous on $mathbbR, lim_x rightarrow infty f(x) = - infty $ and $ lim_x rightarrow inftyf(x) = + infty $. define $g$ be setting
$$g(x)= supt : f(t) <x textfor x in mathbbR$$
Now my question is that
Is $g$ continuous ?
my attempt : No , $g $ need not be continious
take $f(x) = begincases x, & x < 1,\
-x+2,& 1 le x le 2, \
x-2, & x > 2endcases$.
Then $g(x) = begincases x, & x le 0,\
x+2, & x > 0endcases$.
graph :
real-analysis
real-analysis
edited Mar 27 at 5:59
jasmine
asked Mar 27 at 4:09
jasminejasmine
1,980420
1,980420
closed as unclear what you're asking by T. Bongers, RRL, Saad, Jyrki Lahtonen, Parcly Taxel Mar 28 at 0:59
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by T. Bongers, RRL, Saad, Jyrki Lahtonen, Parcly Taxel Mar 28 at 0:59
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
It has to hold for all such functions $f$, not simply a particular one like you have.
$endgroup$
– Eevee Trainer
Mar 27 at 4:11
$begingroup$
okss @EeveeTrainer..im thinking
$endgroup$
– jasmine
Mar 27 at 4:12
2
$begingroup$
Your $f$ does not satisfy the conditions...
$endgroup$
– copper.hat
Mar 27 at 4:13
2
$begingroup$
To be fair, @copper.hat, no function (as far as I know) satisfies the conditions as written. :p Probably a typo somewhere.
$endgroup$
– Eevee Trainer
Mar 27 at 4:14
1
$begingroup$
@jasmine As $x to -infty$, $f(x) = |x| to +infty$. Consider increasingly negative values of $x$ - $-10, -100, -1000, -10000$, $f$ maps each to their positive version. Not that this helps much since you need to generalize beyond a specific $f$ but I figured I'd point that out.
$endgroup$
– Eevee Trainer
Mar 27 at 4:15
|
show 4 more comments
1
$begingroup$
It has to hold for all such functions $f$, not simply a particular one like you have.
$endgroup$
– Eevee Trainer
Mar 27 at 4:11
$begingroup$
okss @EeveeTrainer..im thinking
$endgroup$
– jasmine
Mar 27 at 4:12
2
$begingroup$
Your $f$ does not satisfy the conditions...
$endgroup$
– copper.hat
Mar 27 at 4:13
2
$begingroup$
To be fair, @copper.hat, no function (as far as I know) satisfies the conditions as written. :p Probably a typo somewhere.
$endgroup$
– Eevee Trainer
Mar 27 at 4:14
1
$begingroup$
@jasmine As $x to -infty$, $f(x) = |x| to +infty$. Consider increasingly negative values of $x$ - $-10, -100, -1000, -10000$, $f$ maps each to their positive version. Not that this helps much since you need to generalize beyond a specific $f$ but I figured I'd point that out.
$endgroup$
– Eevee Trainer
Mar 27 at 4:15
1
1
$begingroup$
It has to hold for all such functions $f$, not simply a particular one like you have.
$endgroup$
– Eevee Trainer
Mar 27 at 4:11
$begingroup$
It has to hold for all such functions $f$, not simply a particular one like you have.
$endgroup$
– Eevee Trainer
Mar 27 at 4:11
$begingroup$
okss @EeveeTrainer..im thinking
$endgroup$
– jasmine
Mar 27 at 4:12
$begingroup$
okss @EeveeTrainer..im thinking
$endgroup$
– jasmine
Mar 27 at 4:12
2
2
$begingroup$
Your $f$ does not satisfy the conditions...
$endgroup$
– copper.hat
Mar 27 at 4:13
$begingroup$
Your $f$ does not satisfy the conditions...
$endgroup$
– copper.hat
Mar 27 at 4:13
2
2
$begingroup$
To be fair, @copper.hat, no function (as far as I know) satisfies the conditions as written. :p Probably a typo somewhere.
$endgroup$
– Eevee Trainer
Mar 27 at 4:14
$begingroup$
To be fair, @copper.hat, no function (as far as I know) satisfies the conditions as written. :p Probably a typo somewhere.
$endgroup$
– Eevee Trainer
Mar 27 at 4:14
1
1
$begingroup$
@jasmine As $x to -infty$, $f(x) = |x| to +infty$. Consider increasingly negative values of $x$ - $-10, -100, -1000, -10000$, $f$ maps each to their positive version. Not that this helps much since you need to generalize beyond a specific $f$ but I figured I'd point that out.
$endgroup$
– Eevee Trainer
Mar 27 at 4:15
$begingroup$
@jasmine As $x to -infty$, $f(x) = |x| to +infty$. Consider increasingly negative values of $x$ - $-10, -100, -1000, -10000$, $f$ maps each to their positive version. Not that this helps much since you need to generalize beyond a specific $f$ but I figured I'd point that out.
$endgroup$
– Eevee Trainer
Mar 27 at 4:15
|
show 4 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Let $f(x) = begincases x+1, & x < -1,\
0,& -1 le x le 1, \
x-1, & x > 1endcases$.
Note that $ f(t) <0 = (-infty,-1)$.
Then $g(0)= -1$, but $g(x) = x+1$ for $x >0$.
$endgroup$
$begingroup$
how $g(0) =-1$ ?.im not getting @copper.hat
$endgroup$
– jasmine
Mar 27 at 4:35
1
$begingroup$
$g(0) =0$ ?? its is correct ?
$endgroup$
– jasmine
Mar 27 at 4:36
2
$begingroup$
No, $g(0)=-1$. $$
$endgroup$
– copper.hat
Mar 27 at 11:49
$begingroup$
Why the downvote?
$endgroup$
– copper.hat
Mar 27 at 12:51
add a comment |
$begingroup$
No , $g $ need not be continious
take $f(x) = begincases x, & x < 1,\
-x+2,& 1 le x le 2, \
x-2, & x > 2endcases$.
Then $g(x) = begincases x, & x le 0,\
x+2, & x > 0endcases$.
graph :
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $f(x) = begincases x+1, & x < -1,\
0,& -1 le x le 1, \
x-1, & x > 1endcases$.
Note that $ f(t) <0 = (-infty,-1)$.
Then $g(0)= -1$, but $g(x) = x+1$ for $x >0$.
$endgroup$
$begingroup$
how $g(0) =-1$ ?.im not getting @copper.hat
$endgroup$
– jasmine
Mar 27 at 4:35
1
$begingroup$
$g(0) =0$ ?? its is correct ?
$endgroup$
– jasmine
Mar 27 at 4:36
2
$begingroup$
No, $g(0)=-1$. $$
$endgroup$
– copper.hat
Mar 27 at 11:49
$begingroup$
Why the downvote?
$endgroup$
– copper.hat
Mar 27 at 12:51
add a comment |
$begingroup$
Let $f(x) = begincases x+1, & x < -1,\
0,& -1 le x le 1, \
x-1, & x > 1endcases$.
Note that $ f(t) <0 = (-infty,-1)$.
Then $g(0)= -1$, but $g(x) = x+1$ for $x >0$.
$endgroup$
$begingroup$
how $g(0) =-1$ ?.im not getting @copper.hat
$endgroup$
– jasmine
Mar 27 at 4:35
1
$begingroup$
$g(0) =0$ ?? its is correct ?
$endgroup$
– jasmine
Mar 27 at 4:36
2
$begingroup$
No, $g(0)=-1$. $$
$endgroup$
– copper.hat
Mar 27 at 11:49
$begingroup$
Why the downvote?
$endgroup$
– copper.hat
Mar 27 at 12:51
add a comment |
$begingroup$
Let $f(x) = begincases x+1, & x < -1,\
0,& -1 le x le 1, \
x-1, & x > 1endcases$.
Note that $ f(t) <0 = (-infty,-1)$.
Then $g(0)= -1$, but $g(x) = x+1$ for $x >0$.
$endgroup$
Let $f(x) = begincases x+1, & x < -1,\
0,& -1 le x le 1, \
x-1, & x > 1endcases$.
Note that $ f(t) <0 = (-infty,-1)$.
Then $g(0)= -1$, but $g(x) = x+1$ for $x >0$.
edited Mar 27 at 12:52
answered Mar 27 at 4:19
copper.hatcopper.hat
128k561161
128k561161
$begingroup$
how $g(0) =-1$ ?.im not getting @copper.hat
$endgroup$
– jasmine
Mar 27 at 4:35
1
$begingroup$
$g(0) =0$ ?? its is correct ?
$endgroup$
– jasmine
Mar 27 at 4:36
2
$begingroup$
No, $g(0)=-1$. $$
$endgroup$
– copper.hat
Mar 27 at 11:49
$begingroup$
Why the downvote?
$endgroup$
– copper.hat
Mar 27 at 12:51
add a comment |
$begingroup$
how $g(0) =-1$ ?.im not getting @copper.hat
$endgroup$
– jasmine
Mar 27 at 4:35
1
$begingroup$
$g(0) =0$ ?? its is correct ?
$endgroup$
– jasmine
Mar 27 at 4:36
2
$begingroup$
No, $g(0)=-1$. $$
$endgroup$
– copper.hat
Mar 27 at 11:49
$begingroup$
Why the downvote?
$endgroup$
– copper.hat
Mar 27 at 12:51
$begingroup$
how $g(0) =-1$ ?.im not getting @copper.hat
$endgroup$
– jasmine
Mar 27 at 4:35
$begingroup$
how $g(0) =-1$ ?.im not getting @copper.hat
$endgroup$
– jasmine
Mar 27 at 4:35
1
1
$begingroup$
$g(0) =0$ ?? its is correct ?
$endgroup$
– jasmine
Mar 27 at 4:36
$begingroup$
$g(0) =0$ ?? its is correct ?
$endgroup$
– jasmine
Mar 27 at 4:36
2
2
$begingroup$
No, $g(0)=-1$. $$
$endgroup$
– copper.hat
Mar 27 at 11:49
$begingroup$
No, $g(0)=-1$. $$
$endgroup$
– copper.hat
Mar 27 at 11:49
$begingroup$
Why the downvote?
$endgroup$
– copper.hat
Mar 27 at 12:51
$begingroup$
Why the downvote?
$endgroup$
– copper.hat
Mar 27 at 12:51
add a comment |
$begingroup$
No , $g $ need not be continious
take $f(x) = begincases x, & x < 1,\
-x+2,& 1 le x le 2, \
x-2, & x > 2endcases$.
Then $g(x) = begincases x, & x le 0,\
x+2, & x > 0endcases$.
graph :
$endgroup$
add a comment |
$begingroup$
No , $g $ need not be continious
take $f(x) = begincases x, & x < 1,\
-x+2,& 1 le x le 2, \
x-2, & x > 2endcases$.
Then $g(x) = begincases x, & x le 0,\
x+2, & x > 0endcases$.
graph :
$endgroup$
add a comment |
$begingroup$
No , $g $ need not be continious
take $f(x) = begincases x, & x < 1,\
-x+2,& 1 le x le 2, \
x-2, & x > 2endcases$.
Then $g(x) = begincases x, & x le 0,\
x+2, & x > 0endcases$.
graph :
$endgroup$
No , $g $ need not be continious
take $f(x) = begincases x, & x < 1,\
-x+2,& 1 le x le 2, \
x-2, & x > 2endcases$.
Then $g(x) = begincases x, & x le 0,\
x+2, & x > 0endcases$.
graph :
answered Mar 27 at 5:55
jasminejasmine
1,980420
1,980420
add a comment |
add a comment |
1
$begingroup$
It has to hold for all such functions $f$, not simply a particular one like you have.
$endgroup$
– Eevee Trainer
Mar 27 at 4:11
$begingroup$
okss @EeveeTrainer..im thinking
$endgroup$
– jasmine
Mar 27 at 4:12
2
$begingroup$
Your $f$ does not satisfy the conditions...
$endgroup$
– copper.hat
Mar 27 at 4:13
2
$begingroup$
To be fair, @copper.hat, no function (as far as I know) satisfies the conditions as written. :p Probably a typo somewhere.
$endgroup$
– Eevee Trainer
Mar 27 at 4:14
1
$begingroup$
@jasmine As $x to -infty$, $f(x) = |x| to +infty$. Consider increasingly negative values of $x$ - $-10, -100, -1000, -10000$, $f$ maps each to their positive version. Not that this helps much since you need to generalize beyond a specific $f$ but I figured I'd point that out.
$endgroup$
– Eevee Trainer
Mar 27 at 4:15