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Is $g$ continuous? [closed]

-7

$begingroup$

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

share|cite|improve this question

edited Mar 27 at 5:59

jasmine

asked Mar 27 at 4:09

jasminejasmine

1,980420

$endgroup$

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

-7

$begingroup$

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

share|cite|improve this question

edited Mar 27 at 5:59

jasmine

asked Mar 27 at 4:09

jasminejasmine

1,980420

$endgroup$

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

share|cite|improve this question

edited Mar 27 at 5:59

jasmine

asked Mar 27 at 4:09

jasminejasmine

1,980420

$endgroup$

share|cite|improve this question

edited Mar 27 at 5:59

jasmine

asked Mar 27 at 4:09

jasminejasmine

1,980420

share|cite|improve this question

edited Mar 27 at 5:59

jasmine

asked Mar 27 at 4:09

jasminejasmine

1,980420

asked Mar 27 at 4:09

jasminejasmine

1,980420

asked Mar 27 at 4:09

jasminejasmine

1,980420

2 Answers
2

active

oldest

votes

1

$begingroup$

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 $x >0$.

share|cite|improve this answer

edited Mar 27 at 12:52

answered Mar 27 at 4:19

copper.hatcopper.hat

128k561161

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 | 

0

$begingroup$

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 :

share|cite|improve this answer

answered Mar 27 at 5:55

jasminejasmine

1,980420

$endgroup$

add a comment | 

1

$begingroup$

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 $x >0$.

share|cite|improve this answer

edited Mar 27 at 12:52

answered Mar 27 at 4:19

copper.hatcopper.hat

128k561161

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 | 

1

$begingroup$

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 $x >0$.

share|cite|improve this answer

edited Mar 27 at 12:52

answered Mar 27 at 4:19

copper.hatcopper.hat

128k561161

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 $x >0$.

share|cite|improve this answer

edited Mar 27 at 12:52

answered Mar 27 at 4:19

copper.hatcopper.hat

128k561161

$endgroup$

share|cite|improve this answer

edited Mar 27 at 12:52

answered Mar 27 at 4:19

copper.hatcopper.hat

128k561161

answered Mar 27 at 4:19

copper.hatcopper.hat

128k561161

answered Mar 27 at 4:19

copper.hatcopper.hat

128k561161

0

$begingroup$

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 :

share|cite|improve this answer

answered Mar 27 at 5:55

jasminejasmine

1,980420

$endgroup$

add a comment | 

0

$begingroup$

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 :

share|cite|improve this answer

answered Mar 27 at 5:55

jasminejasmine

1,980420

$endgroup$

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 :

share|cite|improve this answer

answered Mar 27 at 5:55

jasminejasmine

1,980420

$endgroup$

share|cite|improve this answer

answered Mar 27 at 5:55

jasminejasmine

1,980420

answered Mar 27 at 5:55

jasminejasmine

1,980420

answered Mar 27 at 5:55

jasminejasmine

1,980420