Understanding minimization of jointly convex functions

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I am trying to understand partial minimization of jointly convex function in one variable.

The theorem which I am facing difficulties to understand is expressed as follows.

Let $f: R^n times R^m to (-inf,+infty] $ be a jointly convex function. Then the function $g(x) =$ inf$_yin R^mf(x,y)$ , $x in R^m$ is convex

Could you please help me to understand the theorem through,

1. What is the intuition of jointly convex function? Appreciate if you can explain it using $f(x,y) = x^2 + y^2$


2. What is the intuition of inf$_yin R^mf(x,y)$. My understanding is the maximum value of the lower bound of the function $f$ when we vary $y$ with fixed $x$ value. Is it correct?

Appreciate your insight.

convex-analysis convex-optimization

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asked Jul 27 at 19:29

Malintha

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favorite

I am trying to understand partial minimization of jointly convex function in one variable.

The theorem which I am facing difficulties to understand is expressed as follows.

Let $f: R^n times R^m to (-inf,+infty] $ be a jointly convex function. Then the function $g(x) =$ inf$_yin R^mf(x,y)$ , $x in R^m$ is convex

Could you please help me to understand the theorem through,

1. What is the intuition of jointly convex function? Appreciate if you can explain it using $f(x,y) = x^2 + y^2$


2. What is the intuition of inf$_yin R^mf(x,y)$. My understanding is the maximum value of the lower bound of the function $f$ when we vary $y$ with fixed $x$ value. Is it correct?

Appreciate your insight.

convex-analysis convex-optimization

share|cite|improve this question

asked Jul 27 at 19:29

Malintha

1276

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

I am trying to understand partial minimization of jointly convex function in one variable.

The theorem which I am facing difficulties to understand is expressed as follows.

Let $f: R^n times R^m to (-inf,+infty] $ be a jointly convex function. Then the function $g(x) =$ inf$_yin R^mf(x,y)$ , $x in R^m$ is convex

Could you please help me to understand the theorem through,

1. What is the intuition of jointly convex function? Appreciate if you can explain it using $f(x,y) = x^2 + y^2$


2. What is the intuition of inf$_yin R^mf(x,y)$. My understanding is the maximum value of the lower bound of the function $f$ when we vary $y$ with fixed $x$ value. Is it correct?

Appreciate your insight.

convex-analysis convex-optimization

share|cite|improve this question

asked Jul 27 at 19:29

Malintha

1276

I am trying to understand partial minimization of jointly convex function in one variable.

The theorem which I am facing difficulties to understand is expressed as follows.

Let $f: R^n times R^m to (-inf,+infty] $ be a jointly convex function. Then the function $g(x) =$ inf$_yin R^mf(x,y)$ , $x in R^m$ is convex

Could you please help me to understand the theorem through,

1. What is the intuition of jointly convex function? Appreciate if you can explain it using $f(x,y) = x^2 + y^2$


2. What is the intuition of inf$_yin R^mf(x,y)$. My understanding is the maximum value of the lower bound of the function $f$ when we vary $y$ with fixed $x$ value. Is it correct?

Appreciate your insight.

convex-analysis convex-optimization

share|cite|improve this question

asked Jul 27 at 19:29

Malintha

1276

share|cite|improve this question

share|cite|improve this question

asked Jul 27 at 19:29

Malintha

1276

asked Jul 27 at 19:29

Malintha

1276

asked Jul 27 at 19:29

Malintha

1276

1276

add a comment | 

add a comment | 

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1. That function is jointly convex because its Hessian is positive definite.
   
   In contrast, the function $g(x,y)=xcdot y$ is only separately convex, i.e., $xmapsto xcdot y$ and $ymapsto xcdot y$ are convex (even linear), no matter what the fixed parameter is. On the other hand, $g$ is not jointly convex: compute the Hessian matrix and discover it is not positive semidefinite.




2. Yes.

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answered Jul 28 at 5:15

max_zorn

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1. That function is jointly convex because its Hessian is positive definite.
   
   In contrast, the function $g(x,y)=xcdot y$ is only separately convex, i.e., $xmapsto xcdot y$ and $ymapsto xcdot y$ are convex (even linear), no matter what the fixed parameter is. On the other hand, $g$ is not jointly convex: compute the Hessian matrix and discover it is not positive semidefinite.




2. Yes.

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answered Jul 28 at 5:15

max_zorn

3,15051028

add a comment | 

up vote
0
down vote

1. That function is jointly convex because its Hessian is positive definite.
   
   In contrast, the function $g(x,y)=xcdot y$ is only separately convex, i.e., $xmapsto xcdot y$ and $ymapsto xcdot y$ are convex (even linear), no matter what the fixed parameter is. On the other hand, $g$ is not jointly convex: compute the Hessian matrix and discover it is not positive semidefinite.




2. Yes.

share|cite|improve this answer

answered Jul 28 at 5:15

max_zorn

3,15051028

add a comment | 

up vote
0
down vote

up vote
0
down vote

1. That function is jointly convex because its Hessian is positive definite.
   
   In contrast, the function $g(x,y)=xcdot y$ is only separately convex, i.e., $xmapsto xcdot y$ and $ymapsto xcdot y$ are convex (even linear), no matter what the fixed parameter is. On the other hand, $g$ is not jointly convex: compute the Hessian matrix and discover it is not positive semidefinite.




2. Yes.

share|cite|improve this answer

answered Jul 28 at 5:15

max_zorn

3,15051028

1. That function is jointly convex because its Hessian is positive definite.
   
   In contrast, the function $g(x,y)=xcdot y$ is only separately convex, i.e., $xmapsto xcdot y$ and $ymapsto xcdot y$ are convex (even linear), no matter what the fixed parameter is. On the other hand, $g$ is not jointly convex: compute the Hessian matrix and discover it is not positive semidefinite.




2. Yes.

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answered Jul 28 at 5:15

max_zorn

3,15051028

share|cite|improve this answer

share|cite|improve this answer

answered Jul 28 at 5:15

max_zorn

3,15051028

answered Jul 28 at 5:15

max_zorn

3,15051028

answered Jul 28 at 5:15

max_zorn

3,15051028

3,15051028

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