How can a function increase/decrease in a point?
$begingroup$
I am working on exercise 4.6 in the book "Convex Optimization", which is as follows: Quiz description
I don't understand what the author meant by saying a function h (similarly $f_0$, $f_i$) is monotonically decreasing in $x_r$ which is an element of the vector $x$. Should a function increase/decrease over an interval? Could anyone explain this to me?
I assume that the author meant to be on the whole domain $R^n$, then my approach is: Let $x^*$ be the optimal point and assume that $h(x) < 0$. Since $f_0$ increases, then $x^*$ must be the minimum point of the feasible domain.
For $y > x^*$, then we have: $f_i(y) < f_i(x*) leq 0$, $h(y) < h(x^*) < 0$.
I have got stuck here and don't know how to keep going with it. Could someone help me out at this point?
convex-optimization
asked Dec 24 '18 at 3:50
hoangtuansuhoangtuansu
84
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add a comment |
$begingroup$
I am working on exercise 4.6 in the book "Convex Optimization", which is as follows: Quiz description
I don't understand what the author meant by saying a function h (similarly $f_0$, $f_i$) is monotonically decreasing in $x_r$ which is an element of the vector $x$. Should a function increase/decrease over an interval? Could anyone explain this to me?
I assume that the author meant to be on the whole domain $R^n$, then my approach is: Let $x^*$ be the optimal point and assume that $h(x) < 0$. Since $f_0$ increases, then $x^*$ must be the minimum point of the feasible domain.
For $y > x^*$, then we have: $f_i(y) < f_i(x*) leq 0$, $h(y) < h(x^*) < 0$.
I have got stuck here and don't know how to keep going with it. Could someone help me out at this point?
convex-optimization
asked Dec 24 '18 at 3:50
hoangtuansuhoangtuansu
84
$endgroup$
add a comment |
$begingroup$
I am working on exercise 4.6 in the book "Convex Optimization", which is as follows: Quiz description
I don't understand what the author meant by saying a function h (similarly $f_0$, $f_i$) is monotonically decreasing in $x_r$ which is an element of the vector $x$. Should a function increase/decrease over an interval? Could anyone explain this to me?
I assume that the author meant to be on the whole domain $R^n$, then my approach is: Let $x^*$ be the optimal point and assume that $h(x) < 0$. Since $f_0$ increases, then $x^*$ must be the minimum point of the feasible domain.
For $y > x^*$, then we have: $f_i(y) < f_i(x*) leq 0$, $h(y) < h(x^*) < 0$.
I have got stuck here and don't know how to keep going with it. Could someone help me out at this point?
convex-optimization
asked Dec 24 '18 at 3:50
hoangtuansuhoangtuansu
84
$endgroup$
I am working on exercise 4.6 in the book "Convex Optimization", which is as follows: Quiz description
I don't understand what the author meant by saying a function h (similarly $f_0$, $f_i$) is monotonically decreasing in $x_r$ which is an element of the vector $x$. Should a function increase/decrease over an interval? Could anyone explain this to me?
I assume that the author meant to be on the whole domain $R^n$, then my approach is: Let $x^*$ be the optimal point and assume that $h(x) < 0$. Since $f_0$ increases, then $x^*$ must be the minimum point of the feasible domain.
For $y > x^*$, then we have: $f_i(y) < f_i(x*) leq 0$, $h(y) < h(x^*) < 0$.
I have got stuck here and don't know how to keep going with it. Could someone help me out at this point?
convex-optimization
convex-optimization
asked Dec 24 '18 at 3:50
hoangtuansuhoangtuansu
84
asked Dec 24 '18 at 3:50
hoangtuansuhoangtuansu
84
edited Dec 24 '18 at 4:02
hoangtuansu
asked Dec 24 '18 at 3:50
hoangtuansuhoangtuansu
84
asked Dec 24 '18 at 3:50
hoangtuansuhoangtuansu
84
asked Dec 24 '18 at 3:50
hoangtuansuhoangtuansu
84
84
add a comment |
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1 Answer
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It is true that $x_r$ is a component (I would not say "element") of the vector $x.$
But $x$ is not a fixed vector, and its components, in particular a selected component $x_r,$ can range over $mathbb R.$ Surely that is enough of an interval.
answered Dec 24 '18 at 4:28
David KDavid K
54.9k344120
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1 Answer
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1 Answer
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$begingroup$
It is true that $x_r$ is a component (I would not say "element") of the vector $x.$
But $x$ is not a fixed vector, and its components, in particular a selected component $x_r,$ can range over $mathbb R.$ Surely that is enough of an interval.
answered Dec 24 '18 at 4:28
David KDavid K
54.9k344120
$endgroup$
add a comment |
$begingroup$
It is true that $x_r$ is a component (I would not say "element") of the vector $x.$
But $x$ is not a fixed vector, and its components, in particular a selected component $x_r,$ can range over $mathbb R.$ Surely that is enough of an interval.
answered Dec 24 '18 at 4:28
David KDavid K
54.9k344120
$endgroup$
add a comment |
$begingroup$
It is true that $x_r$ is a component (I would not say "element") of the vector $x.$
But $x$ is not a fixed vector, and its components, in particular a selected component $x_r,$ can range over $mathbb R.$ Surely that is enough of an interval.
answered Dec 24 '18 at 4:28
David KDavid K
54.9k344120
$endgroup$
It is true that $x_r$ is a component (I would not say "element") of the vector $x.$
But $x$ is not a fixed vector, and its components, in particular a selected component $x_r,$ can range over $mathbb R.$ Surely that is enough of an interval.
answered Dec 24 '18 at 4:28
David KDavid K
54.9k344120
answered Dec 24 '18 at 4:28
David KDavid K
54.9k344120
answered Dec 24 '18 at 4:28
David KDavid K
54.9k344120
answered Dec 24 '18 at 4:28
David KDavid K
54.9k344120
54.9k344120
add a comment |
add a comment |
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