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Identity for multiple variables minimization

0

$begingroup$

I would like the prove the following identity:

$$minlimits_mathbfx,mathbfy[f(mathbfx)+g(mathbfy)]=minlimits_mathbfx[f(mathbfx)]+minlimits_mathbfy[g(mathbfy)]$$

At first, I supposed that this should be an inequality, because I tried proving this by doing:

beginalign
minlimits_mathbfx,mathbfy[f(mathbfx)+g(mathbfy)]geq minlimits_mathbfx,mathbfy[f(mathbfx)]+minlimits_mathbfx,mathbfy[g(mathbfy)]
endalign

But I am not sure if this is correct...

Could someone help me proving this by using some basic principles and properties? Also, are there any conditions/constraints over the functions so that the proposed equality holds?

inequality optimization maxima-minima

share|cite|improve this question

edited Apr 1 at 15:01

bertozzijr

asked Apr 1 at 14:39

bertozzijrbertozzijr

6331720

$endgroup$

add a comment | 

0

$begingroup$

I would like the prove the following identity:

$$minlimits_mathbfx,mathbfy[f(mathbfx)+g(mathbfy)]=minlimits_mathbfx[f(mathbfx)]+minlimits_mathbfy[g(mathbfy)]$$

At first, I supposed that this should be an inequality, because I tried proving this by doing:

beginalign
minlimits_mathbfx,mathbfy[f(mathbfx)+g(mathbfy)]geq minlimits_mathbfx,mathbfy[f(mathbfx)]+minlimits_mathbfx,mathbfy[g(mathbfy)]
endalign

But I am not sure if this is correct...

Could someone help me proving this by using some basic principles and properties? Also, are there any conditions/constraints over the functions so that the proposed equality holds?

inequality optimization maxima-minima

share|cite|improve this question

edited Apr 1 at 15:01

bertozzijr

asked Apr 1 at 14:39

bertozzijrbertozzijr

6331720

$endgroup$

add a comment | 

$begingroup$

I would like the prove the following identity:

$$minlimits_mathbfx,mathbfy[f(mathbfx)+g(mathbfy)]=minlimits_mathbfx[f(mathbfx)]+minlimits_mathbfy[g(mathbfy)]$$

At first, I supposed that this should be an inequality, because I tried proving this by doing:

beginalign
minlimits_mathbfx,mathbfy[f(mathbfx)+g(mathbfy)]geq minlimits_mathbfx,mathbfy[f(mathbfx)]+minlimits_mathbfx,mathbfy[g(mathbfy)]
endalign

But I am not sure if this is correct...

Could someone help me proving this by using some basic principles and properties? Also, are there any conditions/constraints over the functions so that the proposed equality holds?

inequality optimization maxima-minima

share|cite|improve this question

edited Apr 1 at 15:01

bertozzijr

asked Apr 1 at 14:39

bertozzijrbertozzijr

6331720

$endgroup$

share|cite|improve this question

edited Apr 1 at 15:01

bertozzijr

asked Apr 1 at 14:39

bertozzijrbertozzijr

6331720

share|cite|improve this question

edited Apr 1 at 15:01

bertozzijr

asked Apr 1 at 14:39

bertozzijrbertozzijr

6331720

asked Apr 1 at 14:39

bertozzijrbertozzijr

6331720

asked Apr 1 at 14:39

bertozzijrbertozzijr

6331720

1 Answer
1

active

oldest

votes

1

$begingroup$

You have
$$min_x,y (f(x)+g(y)) leq f(x)+g(y), quad textfor all,, x,y$$
which implies
$$min_x,y (f(x)+g(y)) leq min_x f(x)+ min_y g(y)$$
On the other hand
$$f(x)+g(y) geq min_x f(x) + min_y g(y), quad textfor all,, x,y$$
therefore
$$min_x,y (f(x)+g(y)) geq min_x f(x)+ min_y g(y)$$
So you have the equality.

share|cite|improve this answer

edited Apr 1 at 20:10

answered Apr 1 at 19:56

RaminRamin

112

$endgroup$

add a comment | 

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1

$begingroup$

You have
$$min_x,y (f(x)+g(y)) leq f(x)+g(y), quad textfor all,, x,y$$
which implies
$$min_x,y (f(x)+g(y)) leq min_x f(x)+ min_y g(y)$$
On the other hand
$$f(x)+g(y) geq min_x f(x) + min_y g(y), quad textfor all,, x,y$$
therefore
$$min_x,y (f(x)+g(y)) geq min_x f(x)+ min_y g(y)$$
So you have the equality.

share|cite|improve this answer

edited Apr 1 at 20:10

answered Apr 1 at 19:56

RaminRamin

112

$endgroup$

add a comment | 

1

$begingroup$

You have
$$min_x,y (f(x)+g(y)) leq f(x)+g(y), quad textfor all,, x,y$$
which implies
$$min_x,y (f(x)+g(y)) leq min_x f(x)+ min_y g(y)$$
On the other hand
$$f(x)+g(y) geq min_x f(x) + min_y g(y), quad textfor all,, x,y$$
therefore
$$min_x,y (f(x)+g(y)) geq min_x f(x)+ min_y g(y)$$
So you have the equality.

share|cite|improve this answer

edited Apr 1 at 20:10

answered Apr 1 at 19:56

RaminRamin

112

$endgroup$

add a comment | 

$begingroup$

You have
$$min_x,y (f(x)+g(y)) leq f(x)+g(y), quad textfor all,, x,y$$
which implies
$$min_x,y (f(x)+g(y)) leq min_x f(x)+ min_y g(y)$$
On the other hand
$$f(x)+g(y) geq min_x f(x) + min_y g(y), quad textfor all,, x,y$$
therefore
$$min_x,y (f(x)+g(y)) geq min_x f(x)+ min_y g(y)$$
So you have the equality.

share|cite|improve this answer

edited Apr 1 at 20:10

answered Apr 1 at 19:56

RaminRamin

112

$endgroup$

share|cite|improve this answer

edited Apr 1 at 20:10

answered Apr 1 at 19:56

RaminRamin

112

answered Apr 1 at 19:56

RaminRamin

112

answered Apr 1 at 19:56

RaminRamin

112