A convex function problem
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$f(x)$ is a convex function, and x $in$ $R^n$, $f(x)$ $in$ $R$.
$x^*$= arg min $f(x)$.
from above two points, can we say that <$f'(x)$, $x-x^*$> $ge$ $f(x)-f(x^*)$ ?
additional remark，I think the first order Taylor approximation can prove this, but I failed.
anyone can help me? thanks in advance.
convex-optimization
asked Nov 13 at 5:36
user8234641
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up vote
0
down vote
favorite
$f(x)$ is a convex function, and x $in$ $R^n$, $f(x)$ $in$ $R$.
$x^*$= arg min $f(x)$.
from above two points, can we say that <$f'(x)$, $x-x^*$> $ge$ $f(x)-f(x^*)$ ?
additional remark，I think the first order Taylor approximation can prove this, but I failed.
anyone can help me? thanks in advance.
convex-optimization
asked Nov 13 at 5:36
user8234641
1
Could you clarify what you mean by <$f'(x)$, $x-x^*$> $ge$ $f(x)-f(x^*)$ what is <$f'(x)$? Do you want to show $x-x^*ge f(x)-f(x^*)$? If so how can you compare $x-x^*$ to $f(x)-f(x^*)$ the former is in $R^n$ and the later in $R$
– Abe
Nov 13 at 6:38
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$f(x)$ is a convex function, and x $in$ $R^n$, $f(x)$ $in$ $R$.
$x^*$= arg min $f(x)$.
from above two points, can we say that <$f'(x)$, $x-x^*$> $ge$ $f(x)-f(x^*)$ ?
additional remark，I think the first order Taylor approximation can prove this, but I failed.
anyone can help me? thanks in advance.
convex-optimization
asked Nov 13 at 5:36
user8234641
1
$f(x)$ is a convex function, and x $in$ $R^n$, $f(x)$ $in$ $R$.
$x^*$= arg min $f(x)$.
from above two points, can we say that <$f'(x)$, $x-x^*$> $ge$ $f(x)-f(x^*)$ ?
additional remark，I think the first order Taylor approximation can prove this, but I failed.
anyone can help me? thanks in advance.
convex-optimization
convex-optimization
asked Nov 13 at 5:36
user8234641
1
asked Nov 13 at 5:36
user8234641
1
asked Nov 13 at 5:36
user8234641
1
asked Nov 13 at 5:36
user8234641
1
asked Nov 13 at 5:36
user8234641
1
1
Could you clarify what you mean by <$f'(x)$, $x-x^*$> $ge$ $f(x)-f(x^*)$ what is <$f'(x)$? Do you want to show $x-x^*ge f(x)-f(x^*)$? If so how can you compare $x-x^*$ to $f(x)-f(x^*)$ the former is in $R^n$ and the later in $R$
– Abe
Nov 13 at 6:38
add a comment |
Could you clarify what you mean by <$f'(x)$, $x-x^*$> $ge$ $f(x)-f(x^*)$ what is <$f'(x)$? Do you want to show $x-x^*ge f(x)-f(x^*)$? If so how can you compare $x-x^*$ to $f(x)-f(x^*)$ the former is in $R^n$ and the later in $R$
– Abe
Nov 13 at 6:38
Could you clarify what you mean by <$f'(x)$, $x-x^*$> $ge$ $f(x)-f(x^*)$ what is <$f'(x)$? Do you want to show $x-x^*ge f(x)-f(x^*)$? If so how can you compare $x-x^*$ to $f(x)-f(x^*)$ the former is in $R^n$ and the later in $R$
– Abe
Nov 13 at 6:38
Could you clarify what you mean by <$f'(x)$, $x-x^*$> $ge$ $f(x)-f(x^*)$ what is <$f'(x)$? Do you want to show $x-x^*ge f(x)-f(x^*)$? If so how can you compare $x-x^*$ to $f(x)-f(x^*)$ the former is in $R^n$ and the later in $R$
– Abe
Nov 13 at 6:38
add a comment |
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Could you clarify what you mean by <$f'(x)$, $x-x^*$> $ge$ $f(x)-f(x^*)$ what is <$f'(x)$? Do you want to show $x-x^*ge f(x)-f(x^*)$? If so how can you compare $x-x^*$ to $f(x)-f(x^*)$ the former is in $R^n$ and the later in $R$
– Abe
Nov 13 at 6:38