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Composite of 2 bounded functions is bounded [closed]

-3

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Prove that if function f with domain A and codomain B is bounded, and function g with domain B and codomain C is bounded, then g composite f with domain A and codomain C is bounded.

functions discrete-mathematics

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asked Mar 24 at 17:30

RumiRumi

615

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closed as off-topic by José Carlos Santos, T. Bongers, John Douma, verret, GNUSupporter 8964民主女神 地下教會 Mar 24 at 23:39

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." – José Carlos Santos, T. Bongers, John Douma, verret, GNUSupporter 8964民主女神 地下教會

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Done! What else do you want me to do?
  $endgroup$
  – José Carlos Santos
  Mar 24 at 17:32
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  It is sufficient that $g$ is bounded. The assumption of boundedness on $f$ is redundant.
  $endgroup$
  – amsmath
  Mar 24 at 17:44
  

  
  
  
  

add a comment | 

-3

$begingroup$

Prove that if function f with domain A and codomain B is bounded, and function g with domain B and codomain C is bounded, then g composite f with domain A and codomain C is bounded.

functions discrete-mathematics

share|cite|improve this question

asked Mar 24 at 17:30

RumiRumi

615

$endgroup$

closed as off-topic by José Carlos Santos, T. Bongers, John Douma, verret, GNUSupporter 8964民主女神 地下教會 Mar 24 at 23:39

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." – José Carlos Santos, T. Bongers, John Douma, verret, GNUSupporter 8964民主女神 地下教會

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Done! What else do you want me to do?
  $endgroup$
  – José Carlos Santos
  Mar 24 at 17:32
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  It is sufficient that $g$ is bounded. The assumption of boundedness on $f$ is redundant.
  $endgroup$
  – amsmath
  Mar 24 at 17:44
  

  
  
  
  

add a comment | 

$begingroup$

Prove that if function f with domain A and codomain B is bounded, and function g with domain B and codomain C is bounded, then g composite f with domain A and codomain C is bounded.

functions discrete-mathematics

share|cite|improve this question

asked Mar 24 at 17:30

RumiRumi

615

$endgroup$

share|cite|improve this question

asked Mar 24 at 17:30

RumiRumi

615

share|cite|improve this question

asked Mar 24 at 17:30

RumiRumi

615

asked Mar 24 at 17:30

RumiRumi

615

asked Mar 24 at 17:30

RumiRumi

615

1 Answer
1

active

oldest

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1

$begingroup$

It's always good to go back to the definitions.

If $f$ is bounded, it means that $|f(x)| < M$ for some $M>0$ for all $x$ in its domain.

If $g$ is bounded, it means that $|g(x)| < N$ for some $N>0$ for all $x$ in its domain.

To prove that $fcirc g$ is bounded, one must show that $|(fcirc g)(x)|<K$ for some $K>0$ for all $x$ in its domain.

Do you see how to proceed from here? If you give it a shot, I'll guide you along.

share|cite|improve this answer

answered Mar 24 at 17:36

NicNic8NicNic8

4,64831123

$endgroup$

add a comment | 

1

$begingroup$

It's always good to go back to the definitions.

If $f$ is bounded, it means that $|f(x)| < M$ for some $M>0$ for all $x$ in its domain.

If $g$ is bounded, it means that $|g(x)| < N$ for some $N>0$ for all $x$ in its domain.

To prove that $fcirc g$ is bounded, one must show that $|(fcirc g)(x)|<K$ for some $K>0$ for all $x$ in its domain.

Do you see how to proceed from here? If you give it a shot, I'll guide you along.

share|cite|improve this answer

answered Mar 24 at 17:36

NicNic8NicNic8

4,64831123

$endgroup$

add a comment | 

1

$begingroup$

It's always good to go back to the definitions.

If $f$ is bounded, it means that $|f(x)| < M$ for some $M>0$ for all $x$ in its domain.

If $g$ is bounded, it means that $|g(x)| < N$ for some $N>0$ for all $x$ in its domain.

To prove that $fcirc g$ is bounded, one must show that $|(fcirc g)(x)|<K$ for some $K>0$ for all $x$ in its domain.

Do you see how to proceed from here? If you give it a shot, I'll guide you along.

share|cite|improve this answer

answered Mar 24 at 17:36

NicNic8NicNic8

4,64831123

$endgroup$

add a comment | 

$begingroup$

It's always good to go back to the definitions.

If $f$ is bounded, it means that $|f(x)| < M$ for some $M>0$ for all $x$ in its domain.

If $g$ is bounded, it means that $|g(x)| < N$ for some $N>0$ for all $x$ in its domain.

To prove that $fcirc g$ is bounded, one must show that $|(fcirc g)(x)|<K$ for some $K>0$ for all $x$ in its domain.

Do you see how to proceed from here? If you give it a shot, I'll guide you along.

share|cite|improve this answer

answered Mar 24 at 17:36

NicNic8NicNic8

4,64831123

$endgroup$

share|cite|improve this answer

answered Mar 24 at 17:36

NicNic8NicNic8

4,64831123

answered Mar 24 at 17:36

NicNic8NicNic8

4,64831123

answered Mar 24 at 17:36

NicNic8NicNic8

4,64831123