Finding Outer Normal of Supporting Hyperplane
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Let $mathcal{M}:={x in mathbb R^{2}: x_{2} geq |x_{1}| }$. Find all outer normals $y in mathbb R^{2}$ of supporting hyperplanes to $mathcal{M}$.
My ideas:
Let supporting hyperplane $mathcal{H}:={x in mathbb R^{2}:y^{T}x=alpha}$ with
$mathcal{M}capmathcal{H}neq varnothing$, and $mathcal{M} subseteq mathcal{H^{-}}$
Note: $sup_{x in mathcal{M}}y^{T}x leq alpha$
So $forall x in mathcal{M}: y^{T}x leq alpha$. This is especially true for $(x_{1},x_{1}) in mathcal{M}$, where $x_{1} geq 0$, it follows that:
$(y_{1}+y_{2})x_{1} leq alpha$
and for $(x_{1}, -x_{1}) in mathcal{M}$ where $x_{1} leq 0$, it follows that:
$(y_{1}-y_{2})x_{1} leq alpha$
I am unable to progress here to find the $(y_{1},y_{2})$ so that $y$ is an outer normal. Any ideas?
real-analysis optimization convex-optimization orthonormal
asked Nov 18 at 10:13
SABOY
521211
add a comment |
up vote
0
down vote
favorite
Let $mathcal{M}:={x in mathbb R^{2}: x_{2} geq |x_{1}| }$. Find all outer normals $y in mathbb R^{2}$ of supporting hyperplanes to $mathcal{M}$.
My ideas:
Let supporting hyperplane $mathcal{H}:={x in mathbb R^{2}:y^{T}x=alpha}$ with
$mathcal{M}capmathcal{H}neq varnothing$, and $mathcal{M} subseteq mathcal{H^{-}}$
Note: $sup_{x in mathcal{M}}y^{T}x leq alpha$
So $forall x in mathcal{M}: y^{T}x leq alpha$. This is especially true for $(x_{1},x_{1}) in mathcal{M}$, where $x_{1} geq 0$, it follows that:
$(y_{1}+y_{2})x_{1} leq alpha$
and for $(x_{1}, -x_{1}) in mathcal{M}$ where $x_{1} leq 0$, it follows that:
$(y_{1}-y_{2})x_{1} leq alpha$
I am unable to progress here to find the $(y_{1},y_{2})$ so that $y$ is an outer normal. Any ideas?
real-analysis optimization convex-optimization orthonormal
asked Nov 18 at 10:13
SABOY
521211
try taking the dual of that 'sup'
– LinAlg
Nov 18 at 14:39
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $mathcal{M}:={x in mathbb R^{2}: x_{2} geq |x_{1}| }$. Find all outer normals $y in mathbb R^{2}$ of supporting hyperplanes to $mathcal{M}$.
My ideas:
Let supporting hyperplane $mathcal{H}:={x in mathbb R^{2}:y^{T}x=alpha}$ with
$mathcal{M}capmathcal{H}neq varnothing$, and $mathcal{M} subseteq mathcal{H^{-}}$
Note: $sup_{x in mathcal{M}}y^{T}x leq alpha$
So $forall x in mathcal{M}: y^{T}x leq alpha$. This is especially true for $(x_{1},x_{1}) in mathcal{M}$, where $x_{1} geq 0$, it follows that:
$(y_{1}+y_{2})x_{1} leq alpha$
and for $(x_{1}, -x_{1}) in mathcal{M}$ where $x_{1} leq 0$, it follows that:
$(y_{1}-y_{2})x_{1} leq alpha$
I am unable to progress here to find the $(y_{1},y_{2})$ so that $y$ is an outer normal. Any ideas?
real-analysis optimization convex-optimization orthonormal
asked Nov 18 at 10:13
SABOY
521211
Let $mathcal{M}:={x in mathbb R^{2}: x_{2} geq |x_{1}| }$. Find all outer normals $y in mathbb R^{2}$ of supporting hyperplanes to $mathcal{M}$.
My ideas:
Let supporting hyperplane $mathcal{H}:={x in mathbb R^{2}:y^{T}x=alpha}$ with
$mathcal{M}capmathcal{H}neq varnothing$, and $mathcal{M} subseteq mathcal{H^{-}}$
Note: $sup_{x in mathcal{M}}y^{T}x leq alpha$
So $forall x in mathcal{M}: y^{T}x leq alpha$. This is especially true for $(x_{1},x_{1}) in mathcal{M}$, where $x_{1} geq 0$, it follows that:
$(y_{1}+y_{2})x_{1} leq alpha$
and for $(x_{1}, -x_{1}) in mathcal{M}$ where $x_{1} leq 0$, it follows that:
$(y_{1}-y_{2})x_{1} leq alpha$
I am unable to progress here to find the $(y_{1},y_{2})$ so that $y$ is an outer normal. Any ideas?
real-analysis optimization convex-optimization orthonormal
real-analysis optimization convex-optimization orthonormal
asked Nov 18 at 10:13
SABOY
521211
asked Nov 18 at 10:13
SABOY
521211
asked Nov 18 at 10:13
SABOY
521211
asked Nov 18 at 10:13
SABOY
521211
asked Nov 18 at 10:13
SABOY
521211
521211
try taking the dual of that 'sup'
– LinAlg
Nov 18 at 14:39
add a comment |
try taking the dual of that 'sup'
– LinAlg
Nov 18 at 14:39
try taking the dual of that 'sup'
– LinAlg
Nov 18 at 14:39
try taking the dual of that 'sup'
– LinAlg
Nov 18 at 14:39
add a comment |
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try taking the dual of that 'sup'
– LinAlg
Nov 18 at 14:39