Boundedness of sublevelsets of strongly convex functions implies boundedness of second-order gradient
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In page 460 of Stephen Boyd's "Convex Optimization", he described a property of strongly convex functions:
"The inequality (9.8) (i.e. $f(y) geq f(x) + nabla f(x)^T (y - x) + fracm2 |y - x|_2^2$) implies that the sublevel sets contained in $S$ (i.e. $S = f(x) leq f(x^(0))$) are bounded, so in particular, $S$ is bounded. Therefore the maximum eigenvalue of $nabla^2 f(x)$, which is a continuous function of $x$ on $S$, is bounded on $S$"
I don't understand why the boundedness of $S$ implies the boundedness of $nabla^2 f(x)$.
Can anyone explain it for me ? Thank you for reading my question.
convex-analysis convex-optimization convex-geometry
edited Sep 8 at 14:41
Brian Borchers
5,24111119
asked Sep 8 at 12:12
HOANG GIANG
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In page 460 of Stephen Boyd's "Convex Optimization", he described a property of strongly convex functions:
"The inequality (9.8) (i.e. $f(y) geq f(x) + nabla f(x)^T (y - x) + fracm2 |y - x|_2^2$) implies that the sublevel sets contained in $S$ (i.e. $S = f(x) leq f(x^(0))$) are bounded, so in particular, $S$ is bounded. Therefore the maximum eigenvalue of $nabla^2 f(x)$, which is a continuous function of $x$ on $S$, is bounded on $S$"
I don't understand why the boundedness of $S$ implies the boundedness of $nabla^2 f(x)$.
Can anyone explain it for me ? Thank you for reading my question.
convex-analysis convex-optimization convex-geometry
edited Sep 8 at 14:41
Brian Borchers
5,24111119
asked Sep 8 at 12:12
HOANG GIANG
33
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In page 460 of Stephen Boyd's "Convex Optimization", he described a property of strongly convex functions:
"The inequality (9.8) (i.e. $f(y) geq f(x) + nabla f(x)^T (y - x) + fracm2 |y - x|_2^2$) implies that the sublevel sets contained in $S$ (i.e. $S = f(x) leq f(x^(0))$) are bounded, so in particular, $S$ is bounded. Therefore the maximum eigenvalue of $nabla^2 f(x)$, which is a continuous function of $x$ on $S$, is bounded on $S$"
I don't understand why the boundedness of $S$ implies the boundedness of $nabla^2 f(x)$.
Can anyone explain it for me ? Thank you for reading my question.
convex-analysis convex-optimization convex-geometry
edited Sep 8 at 14:41
Brian Borchers
5,24111119
asked Sep 8 at 12:12
HOANG GIANG
33
In page 460 of Stephen Boyd's "Convex Optimization", he described a property of strongly convex functions:
"The inequality (9.8) (i.e. $f(y) geq f(x) + nabla f(x)^T (y - x) + fracm2 |y - x|_2^2$) implies that the sublevel sets contained in $S$ (i.e. $S = f(x) leq f(x^(0))$) are bounded, so in particular, $S$ is bounded. Therefore the maximum eigenvalue of $nabla^2 f(x)$, which is a continuous function of $x$ on $S$, is bounded on $S$"
I don't understand why the boundedness of $S$ implies the boundedness of $nabla^2 f(x)$.
Can anyone explain it for me ? Thank you for reading my question.
convex-analysis convex-optimization convex-geometry
convex-analysis convex-optimization convex-geometry
edited Sep 8 at 14:41
Brian Borchers
5,24111119
asked Sep 8 at 12:12
HOANG GIANG
33
edited Sep 8 at 14:41
Brian Borchers
5,24111119
asked Sep 8 at 12:12
HOANG GIANG
33
edited Sep 8 at 14:41
Brian Borchers
5,24111119
edited Sep 8 at 14:41
Brian Borchers
5,24111119
edited Sep 8 at 14:41
Brian Borchers
5,24111119
5,24111119
asked Sep 8 at 12:12
HOANG GIANG
33
asked Sep 8 at 12:12
HOANG GIANG
33
asked Sep 8 at 12:12
HOANG GIANG
33
33
add a comment |Â
add a comment |Â
1 Answer
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Let
$g(x)=lambda_max(nabla^2f(x))=| nabla^2f(x) |_2$.
$g(x)$ is a continuous function.
$S$ is a closed and bounded subset of $R^n$ and thus compact.
By the extreme value theorem, $g(x)$ achieves its maximum value on $S$. Call it $M$.
Thus on $S$, $| nabla^2f(x) |_2$ is bounded by $M$.
answered Sep 8 at 15:04
Brian Borchers
5,24111119
1
To shows that S is bounded, see the earlier stack exchange question: math.stackexchange.com/questions/993357/…
– Brian Borchers
Sep 8 at 22:32
Thanks for your answer! Wish you a good day!
– HOANG GIANG
Sep 9 at 2:01
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Let
$g(x)=lambda_max(nabla^2f(x))=| nabla^2f(x) |_2$.
$g(x)$ is a continuous function.
$S$ is a closed and bounded subset of $R^n$ and thus compact.
By the extreme value theorem, $g(x)$ achieves its maximum value on $S$. Call it $M$.
Thus on $S$, $| nabla^2f(x) |_2$ is bounded by $M$.
answered Sep 8 at 15:04
Brian Borchers
5,24111119
1
To shows that S is bounded, see the earlier stack exchange question: math.stackexchange.com/questions/993357/…
– Brian Borchers
Sep 8 at 22:32
Thanks for your answer! Wish you a good day!
– HOANG GIANG
Sep 9 at 2:01
add a comment |Â
up vote
0
down vote
accepted
Let
$g(x)=lambda_max(nabla^2f(x))=| nabla^2f(x) |_2$.
$g(x)$ is a continuous function.
$S$ is a closed and bounded subset of $R^n$ and thus compact.
By the extreme value theorem, $g(x)$ achieves its maximum value on $S$. Call it $M$.
Thus on $S$, $| nabla^2f(x) |_2$ is bounded by $M$.
answered Sep 8 at 15:04
Brian Borchers
5,24111119
1
To shows that S is bounded, see the earlier stack exchange question: math.stackexchange.com/questions/993357/…
– Brian Borchers
Sep 8 at 22:32
Thanks for your answer! Wish you a good day!
– HOANG GIANG
Sep 9 at 2:01
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Let
$g(x)=lambda_max(nabla^2f(x))=| nabla^2f(x) |_2$.
$g(x)$ is a continuous function.
$S$ is a closed and bounded subset of $R^n$ and thus compact.
By the extreme value theorem, $g(x)$ achieves its maximum value on $S$. Call it $M$.
Thus on $S$, $| nabla^2f(x) |_2$ is bounded by $M$.
answered Sep 8 at 15:04
Brian Borchers
5,24111119
Let
$g(x)=lambda_max(nabla^2f(x))=| nabla^2f(x) |_2$.
$g(x)$ is a continuous function.
$S$ is a closed and bounded subset of $R^n$ and thus compact.
By the extreme value theorem, $g(x)$ achieves its maximum value on $S$. Call it $M$.
Thus on $S$, $| nabla^2f(x) |_2$ is bounded by $M$.
answered Sep 8 at 15:04
Brian Borchers
5,24111119
answered Sep 8 at 15:04
Brian Borchers
5,24111119
answered Sep 8 at 15:04
Brian Borchers
5,24111119
answered Sep 8 at 15:04
Brian Borchers
5,24111119
5,24111119
1
To shows that S is bounded, see the earlier stack exchange question: math.stackexchange.com/questions/993357/…
– Brian Borchers
Sep 8 at 22:32
Thanks for your answer! Wish you a good day!
– HOANG GIANG
Sep 9 at 2:01
add a comment |Â
1
To shows that S is bounded, see the earlier stack exchange question: math.stackexchange.com/questions/993357/…
– Brian Borchers
Sep 8 at 22:32
Thanks for your answer! Wish you a good day!
– HOANG GIANG
Sep 9 at 2:01
1
1
To shows that S is bounded, see the earlier stack exchange question: math.stackexchange.com/questions/993357/…
– Brian Borchers
Sep 8 at 22:32
To shows that S is bounded, see the earlier stack exchange question: math.stackexchange.com/questions/993357/…
– Brian Borchers
Sep 8 at 22:32
Thanks for your answer! Wish you a good day!
– HOANG GIANG
Sep 9 at 2:01
Thanks for your answer! Wish you a good day!
– HOANG GIANG
Sep 9 at 2:01
add a comment |Â
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