How to prove the following statement in the Convex Optimization book (Boyd & Vandenberghe) with the help of composition rules

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It is written in the book that if $h(x)$ is convex then $f(x)=h(Ax+b)$ is also convex. Now according to the composition rules if we write $f(x)=h(g(x))$ where $g(x)=Ax+b$ then if

1- $h(x)$ is convex and nondecreasing and $g(x)$ is also convex then $f(x)$ will be convex. Although $g(x)$ is convex here, due to being affine, and $h(x)$ is convex by assumption but how can we be sure that $h(x)$ is nondecreasing?

2- $h(x)$ is convex and nonincreasing and $g(x)$ is concave then $f(x)$ will be convex. Although $g(x)$ is concave here, due to being affine, and $h(x)$ is convex by assumption but how can we be sure that $h(x)$ is nonincreasing?

Any help in this regard will be much appreciated. Thanks in advance.

optimization convex-analysis convex-optimization

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edited Jul 24 at 1:46

asked Jul 23 at 1:00

Frank Moses

1,092317

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  Neither of those composition rules apply. Affine composition is a different rule. This is quite important, because affine composition is true even if the outer function is non-monotonic.
  – Michael Grant
  Jul 23 at 1:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MichaelGrant thank you for your comment. I think the book does not provide any proof of the statement. However, it is used in several exercises. So what could be the easiest way to proof this statement?
  – Frank Moses
  Jul 23 at 1:18
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Try applying the definition of a convex function.
  – Rahul
  Jul 23 at 2:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Rahul thank you. I got the desired result.
  – Frank Moses
  Jul 23 at 2:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @LinAlg he is one of the two authors of the book
  – Frank Moses
  Jul 24 at 0:17
  

  
  
  
  

 | 
show 2 more comments

up vote
0
down vote

favorite

It is written in the book that if $h(x)$ is convex then $f(x)=h(Ax+b)$ is also convex. Now according to the composition rules if we write $f(x)=h(g(x))$ where $g(x)=Ax+b$ then if

1- $h(x)$ is convex and nondecreasing and $g(x)$ is also convex then $f(x)$ will be convex. Although $g(x)$ is convex here, due to being affine, and $h(x)$ is convex by assumption but how can we be sure that $h(x)$ is nondecreasing?

2- $h(x)$ is convex and nonincreasing and $g(x)$ is concave then $f(x)$ will be convex. Although $g(x)$ is concave here, due to being affine, and $h(x)$ is convex by assumption but how can we be sure that $h(x)$ is nonincreasing?

Any help in this regard will be much appreciated. Thanks in advance.

optimization convex-analysis convex-optimization

share|cite|improve this question

edited Jul 24 at 1:46

asked Jul 23 at 1:00

Frank Moses

1,092317

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Neither of those composition rules apply. Affine composition is a different rule. This is quite important, because affine composition is true even if the outer function is non-monotonic.
  – Michael Grant
  Jul 23 at 1:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MichaelGrant thank you for your comment. I think the book does not provide any proof of the statement. However, it is used in several exercises. So what could be the easiest way to proof this statement?
  – Frank Moses
  Jul 23 at 1:18
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Try applying the definition of a convex function.
  – Rahul
  Jul 23 at 2:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Rahul thank you. I got the desired result.
  – Frank Moses
  Jul 23 at 2:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @LinAlg he is one of the two authors of the book
  – Frank Moses
  Jul 24 at 0:17
  

  
  
  
  

 | 
show 2 more comments

up vote
0
down vote

favorite

up vote
0
down vote

favorite

It is written in the book that if $h(x)$ is convex then $f(x)=h(Ax+b)$ is also convex. Now according to the composition rules if we write $f(x)=h(g(x))$ where $g(x)=Ax+b$ then if

1- $h(x)$ is convex and nondecreasing and $g(x)$ is also convex then $f(x)$ will be convex. Although $g(x)$ is convex here, due to being affine, and $h(x)$ is convex by assumption but how can we be sure that $h(x)$ is nondecreasing?

2- $h(x)$ is convex and nonincreasing and $g(x)$ is concave then $f(x)$ will be convex. Although $g(x)$ is concave here, due to being affine, and $h(x)$ is convex by assumption but how can we be sure that $h(x)$ is nonincreasing?

Any help in this regard will be much appreciated. Thanks in advance.

optimization convex-analysis convex-optimization

share|cite|improve this question

edited Jul 24 at 1:46

asked Jul 23 at 1:00

Frank Moses

1,092317

It is written in the book that if $h(x)$ is convex then $f(x)=h(Ax+b)$ is also convex. Now according to the composition rules if we write $f(x)=h(g(x))$ where $g(x)=Ax+b$ then if

1- $h(x)$ is convex and nondecreasing and $g(x)$ is also convex then $f(x)$ will be convex. Although $g(x)$ is convex here, due to being affine, and $h(x)$ is convex by assumption but how can we be sure that $h(x)$ is nondecreasing?

2- $h(x)$ is convex and nonincreasing and $g(x)$ is concave then $f(x)$ will be convex. Although $g(x)$ is concave here, due to being affine, and $h(x)$ is convex by assumption but how can we be sure that $h(x)$ is nonincreasing?

Any help in this regard will be much appreciated. Thanks in advance.

optimization convex-analysis convex-optimization

share|cite|improve this question

edited Jul 24 at 1:46

asked Jul 23 at 1:00

Frank Moses

1,092317

share|cite|improve this question

share|cite|improve this question

edited Jul 24 at 1:46

edited Jul 24 at 1:46

edited Jul 24 at 1:46

asked Jul 23 at 1:00

Frank Moses

1,092317

asked Jul 23 at 1:00

Frank Moses

1,092317

asked Jul 23 at 1:00

Frank Moses

1,092317

1,092317

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Neither of those composition rules apply. Affine composition is a different rule. This is quite important, because affine composition is true even if the outer function is non-monotonic.
  – Michael Grant
  Jul 23 at 1:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MichaelGrant thank you for your comment. I think the book does not provide any proof of the statement. However, it is used in several exercises. So what could be the easiest way to proof this statement?
  – Frank Moses
  Jul 23 at 1:18
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Try applying the definition of a convex function.
  – Rahul
  Jul 23 at 2:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Rahul thank you. I got the desired result.
  – Frank Moses
  Jul 23 at 2:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @LinAlg he is one of the two authors of the book
  – Frank Moses
  Jul 24 at 0:17
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Neither of those composition rules apply. Affine composition is a different rule. This is quite important, because affine composition is true even if the outer function is non-monotonic.
  – Michael Grant
  Jul 23 at 1:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MichaelGrant thank you for your comment. I think the book does not provide any proof of the statement. However, it is used in several exercises. So what could be the easiest way to proof this statement?
  – Frank Moses
  Jul 23 at 1:18
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Try applying the definition of a convex function.
  – Rahul
  Jul 23 at 2:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Rahul thank you. I got the desired result.
  – Frank Moses
  Jul 23 at 2:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @LinAlg he is one of the two authors of the book
  – Frank Moses
  Jul 24 at 0:17
  

  
  
  
  

2

2

Neither of those composition rules apply. Affine composition is a different rule. This is quite important, because affine composition is true even if the outer function is non-monotonic.
– Michael Grant
Jul 23 at 1:06

Neither of those composition rules apply. Affine composition is a different rule. This is quite important, because affine composition is true even if the outer function is non-monotonic.
– Michael Grant
Jul 23 at 1:06

@MichaelGrant thank you for your comment. I think the book does not provide any proof of the statement. However, it is used in several exercises. So what could be the easiest way to proof this statement?
– Frank Moses
Jul 23 at 1:18

@MichaelGrant thank you for your comment. I think the book does not provide any proof of the statement. However, it is used in several exercises. So what could be the easiest way to proof this statement?
– Frank Moses
Jul 23 at 1:18

2

2

Try applying the definition of a convex function.
– Rahul
Jul 23 at 2:01

Try applying the definition of a convex function.
– Rahul
Jul 23 at 2:01

@Rahul thank you. I got the desired result.
– Frank Moses
Jul 23 at 2:37

@Rahul thank you. I got the desired result.
– Frank Moses
Jul 23 at 2:37

@LinAlg he is one of the two authors of the book
– Frank Moses
Jul 24 at 0:17

@LinAlg he is one of the two authors of the book
– Frank Moses
Jul 24 at 0:17

 | 
show 2 more comments

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