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Piecewise convexity and global convexity

Does local convexity imply global convexity?Convexity of Exponential Composite FunctionFunction on convex set is convex if all rays are convexA lower semi-continuous convex function being not continuous on its domainInequality of convexity with the middle point onlyProving Convexity of Multivariate Function using Conditional Univariate Convexitycomposition of functions and concavityContinuity + strict convexity on interior implies strict convexity everywhere?

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5

$begingroup$

Let $f:mathbbRrightarrowmathbbR$ be continuous on $left[0,1right]$.
Consider $zinleft(0,1right)$ and suppose that $f$ is differentiable
and convex on $left[0,zright]$ and $left[z,1right]$. If $f'$
(i.e., $fracdfdx$) is continuous at $z$, then $f$ is convex
on $left[0,1right]$. - Is this proposition true? While I could not create a counter-example, I am finding it difficult to generate a clean proof as well.

real-analysis convex-analysis

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edited 9 hours ago

Stefan Egger

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asked 9 hours ago

TapasTapas

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• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  OK, I got your answers, and thank all of you for that. However, I would like to add something more to this question. To what extent, can I relax continuity of $f'$, and still have global convexity? Following your lines of proof, it seems to me that as long as the left-side derivative ($f'_-$) at $z$, which is well defined, is less than the right-side derivative ($f'_+$) at $z$, which is well defined as well, global convexity prevails. I do not need continuity of $f'$ precisely - Is this correct?
  $endgroup$
  – Tapas
  9 hours ago
  
  

  
  
  
  

add a comment | 

5

$begingroup$

Let $f:mathbbRrightarrowmathbbR$ be continuous on $left[0,1right]$.
Consider $zinleft(0,1right)$ and suppose that $f$ is differentiable
and convex on $left[0,zright]$ and $left[z,1right]$. If $f'$
(i.e., $fracdfdx$) is continuous at $z$, then $f$ is convex
on $left[0,1right]$. - Is this proposition true? While I could not create a counter-example, I am finding it difficult to generate a clean proof as well.

real-analysis convex-analysis

share|cite|improve this question

edited 9 hours ago

Stefan Egger

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asked 9 hours ago

TapasTapas

2844 bronze badges

New contributor

Tapas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  OK, I got your answers, and thank all of you for that. However, I would like to add something more to this question. To what extent, can I relax continuity of $f'$, and still have global convexity? Following your lines of proof, it seems to me that as long as the left-side derivative ($f'_-$) at $z$, which is well defined, is less than the right-side derivative ($f'_+$) at $z$, which is well defined as well, global convexity prevails. I do not need continuity of $f'$ precisely - Is this correct?
  $endgroup$
  – Tapas
  9 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

Let $f:mathbbRrightarrowmathbbR$ be continuous on $left[0,1right]$.
Consider $zinleft(0,1right)$ and suppose that $f$ is differentiable
and convex on $left[0,zright]$ and $left[z,1right]$. If $f'$
(i.e., $fracdfdx$) is continuous at $z$, then $f$ is convex
on $left[0,1right]$. - Is this proposition true? While I could not create a counter-example, I am finding it difficult to generate a clean proof as well.

real-analysis convex-analysis

share|cite|improve this question

edited 9 hours ago

Stefan Egger

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asked 9 hours ago

TapasTapas

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New contributor

Tapas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 9 hours ago

Stefan Egger

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asked 9 hours ago

TapasTapas

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New contributor

Tapas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|cite|improve this question

edited 9 hours ago

Stefan Egger

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asked 9 hours ago

TapasTapas

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edited 9 hours ago

Stefan Egger

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edited 9 hours ago

Stefan Egger

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asked 9 hours ago

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3 Answers
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2

$begingroup$

A differentiable function $f$ is convex on an interval $(a,b)$ if and only if its derivative $f'$ is increasing.

Therefore, your assumptions imply that $f'$ is increasing on $(0,z)$ and on $(z,1)$. Now your assumption that $f'$ is continuous immediately gives you that $f'$ is increasing on $(0,1)$ and therefore convex.

share|cite|improve this answer

answered 9 hours ago

Stefan EggerStefan Egger

53111 silver badge1010 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for your answer. If I relax continuity of $f'$ at $z$, and just assume that the left-side derivative of $f$ at $z$ is less than the right-side derivative of $f$ at $z$, global convexity will still hold - is it right? In fact, I was wondering that is an "if and only if" condition to maintain global convexity.
  $endgroup$
  – Tapas
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, you are right it generalizes in the way you suggest it. But note that this does not follow from the proof I gave because the criterion I used needs that $f$ is differentiable in the whole interval. If you weaken this assumption you will have to proof it in a different way. I think it will boil down to using the definition of convexity and show it explicitly like one would show the theorem I used in my answer.
  $endgroup$
  – Stefan Egger
  8 hours ago
  
  

  
  
  
  

add a comment | 

3

$begingroup$

A differentiable function is convex in an interval $I$ if and only if $f'$ is increasing in $I$. Now we have that
$$f'(x)leq f'(z)leq f'(y)$$
for any $0leq x<z<yleq 1$. Can you take it from here and show that the proposition is true?

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

Robert ZRobert Z

109k1010 gold badges7777 silver badges152152 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

Yes it's true. $f$ is convex iff $f'$ is monotonically non-decreasing, but it works on $[0, z)$ and $(z, 1]$. But $f'$ is continuous at $z$ so it can't be less than 0 otherwise there would be a neighborhood where it is negative and thus not convex.

share|cite|improve this answer

answered 9 hours ago

aeruponsaerupons

4122 bronze badges

$endgroup$

add a comment | 

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2

$begingroup$

A differentiable function $f$ is convex on an interval $(a,b)$ if and only if its derivative $f'$ is increasing.

Therefore, your assumptions imply that $f'$ is increasing on $(0,z)$ and on $(z,1)$. Now your assumption that $f'$ is continuous immediately gives you that $f'$ is increasing on $(0,1)$ and therefore convex.

share|cite|improve this answer

answered 9 hours ago

Stefan EggerStefan Egger

53111 silver badge1010 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for your answer. If I relax continuity of $f'$ at $z$, and just assume that the left-side derivative of $f$ at $z$ is less than the right-side derivative of $f$ at $z$, global convexity will still hold - is it right? In fact, I was wondering that is an "if and only if" condition to maintain global convexity.
  $endgroup$
  – Tapas
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, you are right it generalizes in the way you suggest it. But note that this does not follow from the proof I gave because the criterion I used needs that $f$ is differentiable in the whole interval. If you weaken this assumption you will have to proof it in a different way. I think it will boil down to using the definition of convexity and show it explicitly like one would show the theorem I used in my answer.
  $endgroup$
  – Stefan Egger
  8 hours ago
  
  

  
  
  
  

add a comment | 

2

$begingroup$

A differentiable function $f$ is convex on an interval $(a,b)$ if and only if its derivative $f'$ is increasing.

Therefore, your assumptions imply that $f'$ is increasing on $(0,z)$ and on $(z,1)$. Now your assumption that $f'$ is continuous immediately gives you that $f'$ is increasing on $(0,1)$ and therefore convex.

share|cite|improve this answer

answered 9 hours ago

Stefan EggerStefan Egger

53111 silver badge1010 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for your answer. If I relax continuity of $f'$ at $z$, and just assume that the left-side derivative of $f$ at $z$ is less than the right-side derivative of $f$ at $z$, global convexity will still hold - is it right? In fact, I was wondering that is an "if and only if" condition to maintain global convexity.
  $endgroup$
  – Tapas
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, you are right it generalizes in the way you suggest it. But note that this does not follow from the proof I gave because the criterion I used needs that $f$ is differentiable in the whole interval. If you weaken this assumption you will have to proof it in a different way. I think it will boil down to using the definition of convexity and show it explicitly like one would show the theorem I used in my answer.
  $endgroup$
  – Stefan Egger
  8 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

A differentiable function $f$ is convex on an interval $(a,b)$ if and only if its derivative $f'$ is increasing.

Therefore, your assumptions imply that $f'$ is increasing on $(0,z)$ and on $(z,1)$. Now your assumption that $f'$ is continuous immediately gives you that $f'$ is increasing on $(0,1)$ and therefore convex.

share|cite|improve this answer

answered 9 hours ago

Stefan EggerStefan Egger

53111 silver badge1010 bronze badges

$endgroup$

share|cite|improve this answer

answered 9 hours ago

Stefan EggerStefan Egger

53111 silver badge1010 bronze badges

answered 9 hours ago

Stefan EggerStefan Egger

53111 silver badge1010 bronze badges

answered 9 hours ago

Stefan EggerStefan Egger

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3

$begingroup$

A differentiable function is convex in an interval $I$ if and only if $f'$ is increasing in $I$. Now we have that
$$f'(x)leq f'(z)leq f'(y)$$
for any $0leq x<z<yleq 1$. Can you take it from here and show that the proposition is true?

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

Robert ZRobert Z

109k1010 gold badges7777 silver badges152152 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

A differentiable function is convex in an interval $I$ if and only if $f'$ is increasing in $I$. Now we have that
$$f'(x)leq f'(z)leq f'(y)$$
for any $0leq x<z<yleq 1$. Can you take it from here and show that the proposition is true?

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

Robert ZRobert Z

109k1010 gold badges7777 silver badges152152 bronze badges

$endgroup$

add a comment | 

$begingroup$

A differentiable function is convex in an interval $I$ if and only if $f'$ is increasing in $I$. Now we have that
$$f'(x)leq f'(z)leq f'(y)$$
for any $0leq x<z<yleq 1$. Can you take it from here and show that the proposition is true?

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

Robert ZRobert Z

109k1010 gold badges7777 silver badges152152 bronze badges

$endgroup$

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

Robert ZRobert Z

109k1010 gold badges7777 silver badges152152 bronze badges

answered 9 hours ago

Robert ZRobert Z

109k1010 gold badges7777 silver badges152152 bronze badges

answered 9 hours ago

Robert ZRobert Z

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1

$begingroup$

Yes it's true. $f$ is convex iff $f'$ is monotonically non-decreasing, but it works on $[0, z)$ and $(z, 1]$. But $f'$ is continuous at $z$ so it can't be less than 0 otherwise there would be a neighborhood where it is negative and thus not convex.

share|cite|improve this answer

answered 9 hours ago

aeruponsaerupons

4122 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

Yes it's true. $f$ is convex iff $f'$ is monotonically non-decreasing, but it works on $[0, z)$ and $(z, 1]$. But $f'$ is continuous at $z$ so it can't be less than 0 otherwise there would be a neighborhood where it is negative and thus not convex.

share|cite|improve this answer

answered 9 hours ago

aeruponsaerupons

4122 bronze badges

$endgroup$

add a comment | 

$begingroup$

Yes it's true. $f$ is convex iff $f'$ is monotonically non-decreasing, but it works on $[0, z)$ and $(z, 1]$. But $f'$ is continuous at $z$ so it can't be less than 0 otherwise there would be a neighborhood where it is negative and thus not convex.

share|cite|improve this answer

answered 9 hours ago

aeruponsaerupons

4122 bronze badges

$endgroup$

share|cite|improve this answer

answered 9 hours ago

aeruponsaerupons

4122 bronze badges

answered 9 hours ago

aeruponsaerupons

4122 bronze badges

answered 9 hours ago

aeruponsaerupons

4122 bronze badges