How to prove normal cones are closed?
$begingroup$
Let $C$ be a convex subset of $mathbb{R}^{n}$ and let $bar{x} in C$. Then the normal cone $N_{C}(bar{x})$ is closed and convex. Here, we're defining the normal cone as follows:
$$N_{C}(bar{x}) = {v in mathbb{R}^{n} vert langle v, x - bar{x} rangle le 0, forall x in C }.$$
Proving convexity is straightforward, as is proving $N_{C}(bar{x})$ is closed when $C$ is open ($i.e.$ every $x in C$ is an interior point). However, I'm not sure how to prove that $N_{C}(bar{x})$ is closed more generally?
convex-analysis convex-geometry
asked Jan 9 at 23:19
StevenAStevenA
755
$endgroup$
add a comment |
$begingroup$
Let $C$ be a convex subset of $mathbb{R}^{n}$ and let $bar{x} in C$. Then the normal cone $N_{C}(bar{x})$ is closed and convex. Here, we're defining the normal cone as follows:
$$N_{C}(bar{x}) = {v in mathbb{R}^{n} vert langle v, x - bar{x} rangle le 0, forall x in C }.$$
Proving convexity is straightforward, as is proving $N_{C}(bar{x})$ is closed when $C$ is open ($i.e.$ every $x in C$ is an interior point). However, I'm not sure how to prove that $N_{C}(bar{x})$ is closed more generally?
convex-analysis convex-geometry
asked Jan 9 at 23:19
StevenAStevenA
755
$endgroup$
add a comment |
$begingroup$
Let $C$ be a convex subset of $mathbb{R}^{n}$ and let $bar{x} in C$. Then the normal cone $N_{C}(bar{x})$ is closed and convex. Here, we're defining the normal cone as follows:
$$N_{C}(bar{x}) = {v in mathbb{R}^{n} vert langle v, x - bar{x} rangle le 0, forall x in C }.$$
Proving convexity is straightforward, as is proving $N_{C}(bar{x})$ is closed when $C$ is open ($i.e.$ every $x in C$ is an interior point). However, I'm not sure how to prove that $N_{C}(bar{x})$ is closed more generally?
convex-analysis convex-geometry
asked Jan 9 at 23:19
StevenAStevenA
755
$endgroup$
Let $C$ be a convex subset of $mathbb{R}^{n}$ and let $bar{x} in C$. Then the normal cone $N_{C}(bar{x})$ is closed and convex. Here, we're defining the normal cone as follows:
$$N_{C}(bar{x}) = {v in mathbb{R}^{n} vert langle v, x - bar{x} rangle le 0, forall x in C }.$$
Proving convexity is straightforward, as is proving $N_{C}(bar{x})$ is closed when $C$ is open ($i.e.$ every $x in C$ is an interior point). However, I'm not sure how to prove that $N_{C}(bar{x})$ is closed more generally?
convex-analysis convex-geometry
convex-analysis convex-geometry
asked Jan 9 at 23:19
StevenAStevenA
755
asked Jan 9 at 23:19
StevenAStevenA
755
asked Jan 9 at 23:19
StevenAStevenA
755
asked Jan 9 at 23:19
StevenAStevenA
755
asked Jan 9 at 23:19
StevenAStevenA
755
755
add a comment |
add a comment |
1 Answer
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$begingroup$
Write $N_C(bar{x}) = cap_{x in C} {v | langle v, x-bar{x} rangle le 0 }$.
Hence $N_C(bar{x})$ is the intersection of closed hyperplanes which is closed (the
function $v mapsto langle v, x-bar{x} rangle$ is continuous).
This approach also shows that $N_C(bar{x})$ is convex.
answered Jan 9 at 23:22
copper.hatcopper.hat
127k559160
$endgroup$
1
$begingroup$
....and now I see why this proof has been omitted from so many books that I've looked through. Thanks @copper.hat.
$endgroup$
– StevenA
Jan 9 at 23:25
$begingroup$
@StevenA: There are many cute tricks, after a while you will add them to your quiver :-).
$endgroup$
– copper.hat
Jan 9 at 23:28
add a comment |
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1 Answer
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$begingroup$
Write $N_C(bar{x}) = cap_{x in C} {v | langle v, x-bar{x} rangle le 0 }$.
Hence $N_C(bar{x})$ is the intersection of closed hyperplanes which is closed (the
function $v mapsto langle v, x-bar{x} rangle$ is continuous).
This approach also shows that $N_C(bar{x})$ is convex.
answered Jan 9 at 23:22
copper.hatcopper.hat
127k559160
$endgroup$
1
$begingroup$
....and now I see why this proof has been omitted from so many books that I've looked through. Thanks @copper.hat.
$endgroup$
– StevenA
Jan 9 at 23:25
$begingroup$
@StevenA: There are many cute tricks, after a while you will add them to your quiver :-).
$endgroup$
– copper.hat
Jan 9 at 23:28
add a comment |
$begingroup$
Write $N_C(bar{x}) = cap_{x in C} {v | langle v, x-bar{x} rangle le 0 }$.
Hence $N_C(bar{x})$ is the intersection of closed hyperplanes which is closed (the
function $v mapsto langle v, x-bar{x} rangle$ is continuous).
This approach also shows that $N_C(bar{x})$ is convex.
answered Jan 9 at 23:22
copper.hatcopper.hat
127k559160
$endgroup$
1
$begingroup$
....and now I see why this proof has been omitted from so many books that I've looked through. Thanks @copper.hat.
$endgroup$
– StevenA
Jan 9 at 23:25
$begingroup$
@StevenA: There are many cute tricks, after a while you will add them to your quiver :-).
$endgroup$
– copper.hat
Jan 9 at 23:28
add a comment |
$begingroup$
Write $N_C(bar{x}) = cap_{x in C} {v | langle v, x-bar{x} rangle le 0 }$.
Hence $N_C(bar{x})$ is the intersection of closed hyperplanes which is closed (the
function $v mapsto langle v, x-bar{x} rangle$ is continuous).
This approach also shows that $N_C(bar{x})$ is convex.
answered Jan 9 at 23:22
copper.hatcopper.hat
127k559160
$endgroup$
Write $N_C(bar{x}) = cap_{x in C} {v | langle v, x-bar{x} rangle le 0 }$.
Hence $N_C(bar{x})$ is the intersection of closed hyperplanes which is closed (the
function $v mapsto langle v, x-bar{x} rangle$ is continuous).
This approach also shows that $N_C(bar{x})$ is convex.
answered Jan 9 at 23:22
copper.hatcopper.hat
127k559160
answered Jan 9 at 23:22
copper.hatcopper.hat
127k559160
answered Jan 9 at 23:22
copper.hatcopper.hat
127k559160
answered Jan 9 at 23:22
copper.hatcopper.hat
127k559160
127k559160
1
$begingroup$
....and now I see why this proof has been omitted from so many books that I've looked through. Thanks @copper.hat.
$endgroup$
– StevenA
Jan 9 at 23:25
$begingroup$
@StevenA: There are many cute tricks, after a while you will add them to your quiver :-).
$endgroup$
– copper.hat
Jan 9 at 23:28
add a comment |
1
$begingroup$
....and now I see why this proof has been omitted from so many books that I've looked through. Thanks @copper.hat.
$endgroup$
– StevenA
Jan 9 at 23:25
$begingroup$
@StevenA: There are many cute tricks, after a while you will add them to your quiver :-).
$endgroup$
– copper.hat
Jan 9 at 23:28
1
1
$begingroup$
....and now I see why this proof has been omitted from so many books that I've looked through. Thanks @copper.hat.
$endgroup$
– StevenA
Jan 9 at 23:25
$begingroup$
....and now I see why this proof has been omitted from so many books that I've looked through. Thanks @copper.hat.
$endgroup$
– StevenA
Jan 9 at 23:25
$begingroup$
@StevenA: There are many cute tricks, after a while you will add them to your quiver :-).
$endgroup$
– copper.hat
Jan 9 at 23:28
$begingroup$
@StevenA: There are many cute tricks, after a while you will add them to your quiver :-).
$endgroup$
– copper.hat
Jan 9 at 23:28
add a comment |
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