Mixing
Maximize $b^Tx$ subject to $x in C(a) := {x in mathbb R^p | |a+x|_2 le r,;|x|_infty le epsilon}$
$begingroup$
Let $r,epsilon > 0$ and $a, b in mathbb R^p$ with $|a|_2 le r$. Define the set $C(a) := {x in mathbb R^p | |x+a|_2 le r,;|x|_infty le epsilon}$. Note that $C(a)$ is non-empty (it contains the origin $x=0$, at least).
Question
(A) What is an analytic solution to the problem $sup_{x in C(a)} b^Tx$ ?
(B) Same question with $|a|_2 = r$ and $C(a) := {x in mathbb R^p | |x+a|_2 = r,;|x|_infty le epsilon}$
convex-analysis convex-optimization convex-geometry
asked Jan 20 at 16:46
dohmatobdohmatob
3,672629
$endgroup$
add a comment |
$begingroup$
Let $r,epsilon > 0$ and $a, b in mathbb R^p$ with $|a|_2 le r$. Define the set $C(a) := {x in mathbb R^p | |x+a|_2 le r,;|x|_infty le epsilon}$. Note that $C(a)$ is non-empty (it contains the origin $x=0$, at least).
Question
(A) What is an analytic solution to the problem $sup_{x in C(a)} b^Tx$ ?
(B) Same question with $|a|_2 = r$ and $C(a) := {x in mathbb R^p | |x+a|_2 = r,;|x|_infty le epsilon}$
convex-analysis convex-optimization convex-geometry
asked Jan 20 at 16:46
dohmatobdohmatob
3,672629
$endgroup$
1
$begingroup$
You realize that $C(a)$ can be empty?
$endgroup$
– SmileyCraft
Jan 20 at 17:03
$begingroup$
Indeed, good catch. Fixed (i.e you may assume $C(a)$ is non-empty).
$endgroup$
– dohmatob
Jan 20 at 17:07
1
$begingroup$
Can we assume that $|a|_2 leq r$?
$endgroup$
– littleO
Jan 20 at 17:17
$begingroup$
Hum, good point. Yes, you may.
$endgroup$
– dohmatob
Jan 20 at 17:22
$begingroup$
have you considered the dual problem yet?
$endgroup$
– LinAlg
Jan 22 at 18:54
add a comment |
$begingroup$
Let $r,epsilon > 0$ and $a, b in mathbb R^p$ with $|a|_2 le r$. Define the set $C(a) := {x in mathbb R^p | |x+a|_2 le r,;|x|_infty le epsilon}$. Note that $C(a)$ is non-empty (it contains the origin $x=0$, at least).
Question
(A) What is an analytic solution to the problem $sup_{x in C(a)} b^Tx$ ?
(B) Same question with $|a|_2 = r$ and $C(a) := {x in mathbb R^p | |x+a|_2 = r,;|x|_infty le epsilon}$
convex-analysis convex-optimization convex-geometry
asked Jan 20 at 16:46
dohmatobdohmatob
3,672629
$endgroup$
Let $r,epsilon > 0$ and $a, b in mathbb R^p$ with $|a|_2 le r$. Define the set $C(a) := {x in mathbb R^p | |x+a|_2 le r,;|x|_infty le epsilon}$. Note that $C(a)$ is non-empty (it contains the origin $x=0$, at least).
Question
(A) What is an analytic solution to the problem $sup_{x in C(a)} b^Tx$ ?
(B) Same question with $|a|_2 = r$ and $C(a) := {x in mathbb R^p | |x+a|_2 = r,;|x|_infty le epsilon}$
convex-analysis convex-optimization convex-geometry
convex-analysis convex-optimization convex-geometry
asked Jan 20 at 16:46
dohmatobdohmatob
3,672629
asked Jan 20 at 16:46
dohmatobdohmatob
3,672629
edited Jan 23 at 5:41
dohmatob
asked Jan 20 at 16:46
dohmatobdohmatob
3,672629
asked Jan 20 at 16:46
dohmatobdohmatob
3,672629
asked Jan 20 at 16:46
dohmatobdohmatob
3,672629
3,672629
1
$begingroup$
You realize that $C(a)$ can be empty?
$endgroup$
– SmileyCraft
Jan 20 at 17:03
$begingroup$
Indeed, good catch. Fixed (i.e you may assume $C(a)$ is non-empty).
$endgroup$
– dohmatob
Jan 20 at 17:07
1
$begingroup$
Can we assume that $|a|_2 leq r$?
$endgroup$
– littleO
Jan 20 at 17:17
$begingroup$
Hum, good point. Yes, you may.
$endgroup$
– dohmatob
Jan 20 at 17:22
$begingroup$
have you considered the dual problem yet?
$endgroup$
– LinAlg
Jan 22 at 18:54
add a comment |
1
$begingroup$
You realize that $C(a)$ can be empty?
$endgroup$
– SmileyCraft
Jan 20 at 17:03
$begingroup$
Indeed, good catch. Fixed (i.e you may assume $C(a)$ is non-empty).
$endgroup$
– dohmatob
Jan 20 at 17:07
1
$begingroup$
Can we assume that $|a|_2 leq r$?
$endgroup$
– littleO
Jan 20 at 17:17
$begingroup$
Hum, good point. Yes, you may.
$endgroup$
– dohmatob
Jan 20 at 17:22
$begingroup$
have you considered the dual problem yet?
$endgroup$
– LinAlg
Jan 22 at 18:54
1
1
$begingroup$
You realize that $C(a)$ can be empty?
$endgroup$
– SmileyCraft
Jan 20 at 17:03
$begingroup$
You realize that $C(a)$ can be empty?
$endgroup$
– SmileyCraft
Jan 20 at 17:03
$begingroup$
Indeed, good catch. Fixed (i.e you may assume $C(a)$ is non-empty).
$endgroup$
– dohmatob
Jan 20 at 17:07
$begingroup$
Indeed, good catch. Fixed (i.e you may assume $C(a)$ is non-empty).
$endgroup$
– dohmatob
Jan 20 at 17:07
1
1
$begingroup$
Can we assume that $|a|_2 leq r$?
$endgroup$
– littleO
Jan 20 at 17:17
$begingroup$
Can we assume that $|a|_2 leq r$?
$endgroup$
– littleO
Jan 20 at 17:17
$begingroup$
Hum, good point. Yes, you may.
$endgroup$
– dohmatob
Jan 20 at 17:22
$begingroup$
Hum, good point. Yes, you may.
$endgroup$
– dohmatob
Jan 20 at 17:22
$begingroup$
have you considered the dual problem yet?
$endgroup$
– LinAlg
Jan 22 at 18:54
$begingroup$
have you considered the dual problem yet?
$endgroup$
– LinAlg
Jan 22 at 18:54
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The general answer will be very complicated. But here is a special case. You can see $C(a)$ as the intersection of two balls: $X=B_{|cdot|_2}(-a, r)$ and $Y=B_{|cdot|_infty}(0, epsilon)$. If $Ysubseteq X$ then we can forget about $X$. For every $1leq ileq d$ let $s_iin{-1,0,1}$ denote the sign of $a_i$. Then $x=(s_1epsilon,s_2epsilon,...,s_depsilon)$ is optimal.
answered Jan 20 at 17:23
SmileyCraftSmileyCraft
3,526518
$endgroup$
$begingroup$
Yes, things are trivial if $X subseteq Y$ or $Y subseteq X$. The problem only becomes remotely interesting when neither one of $X$ or $Y$ is contained in the other.
$endgroup$
– dohmatob
Jan 20 at 17:38
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
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votes
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votes
$begingroup$
The general answer will be very complicated. But here is a special case. You can see $C(a)$ as the intersection of two balls: $X=B_{|cdot|_2}(-a, r)$ and $Y=B_{|cdot|_infty}(0, epsilon)$. If $Ysubseteq X$ then we can forget about $X$. For every $1leq ileq d$ let $s_iin{-1,0,1}$ denote the sign of $a_i$. Then $x=(s_1epsilon,s_2epsilon,...,s_depsilon)$ is optimal.
answered Jan 20 at 17:23
SmileyCraftSmileyCraft
3,526518
$endgroup$
$begingroup$
Yes, things are trivial if $X subseteq Y$ or $Y subseteq X$. The problem only becomes remotely interesting when neither one of $X$ or $Y$ is contained in the other.
$endgroup$
– dohmatob
Jan 20 at 17:38
add a comment |
$begingroup$
The general answer will be very complicated. But here is a special case. You can see $C(a)$ as the intersection of two balls: $X=B_{|cdot|_2}(-a, r)$ and $Y=B_{|cdot|_infty}(0, epsilon)$. If $Ysubseteq X$ then we can forget about $X$. For every $1leq ileq d$ let $s_iin{-1,0,1}$ denote the sign of $a_i$. Then $x=(s_1epsilon,s_2epsilon,...,s_depsilon)$ is optimal.
answered Jan 20 at 17:23
SmileyCraftSmileyCraft
3,526518
$endgroup$
$begingroup$
Yes, things are trivial if $X subseteq Y$ or $Y subseteq X$. The problem only becomes remotely interesting when neither one of $X$ or $Y$ is contained in the other.
$endgroup$
– dohmatob
Jan 20 at 17:38
add a comment |
$begingroup$
The general answer will be very complicated. But here is a special case. You can see $C(a)$ as the intersection of two balls: $X=B_{|cdot|_2}(-a, r)$ and $Y=B_{|cdot|_infty}(0, epsilon)$. If $Ysubseteq X$ then we can forget about $X$. For every $1leq ileq d$ let $s_iin{-1,0,1}$ denote the sign of $a_i$. Then $x=(s_1epsilon,s_2epsilon,...,s_depsilon)$ is optimal.
answered Jan 20 at 17:23
SmileyCraftSmileyCraft
3,526518
$endgroup$
The general answer will be very complicated. But here is a special case. You can see $C(a)$ as the intersection of two balls: $X=B_{|cdot|_2}(-a, r)$ and $Y=B_{|cdot|_infty}(0, epsilon)$. If $Ysubseteq X$ then we can forget about $X$. For every $1leq ileq d$ let $s_iin{-1,0,1}$ denote the sign of $a_i$. Then $x=(s_1epsilon,s_2epsilon,...,s_depsilon)$ is optimal.
answered Jan 20 at 17:23
SmileyCraftSmileyCraft
3,526518
answered Jan 20 at 17:23
SmileyCraftSmileyCraft
3,526518
answered Jan 20 at 17:23
SmileyCraftSmileyCraft
3,526518
answered Jan 20 at 17:23
SmileyCraftSmileyCraft
3,526518
3,526518
$begingroup$
Yes, things are trivial if $X subseteq Y$ or $Y subseteq X$. The problem only becomes remotely interesting when neither one of $X$ or $Y$ is contained in the other.
$endgroup$
– dohmatob
Jan 20 at 17:38
add a comment |
$begingroup$
Yes, things are trivial if $X subseteq Y$ or $Y subseteq X$. The problem only becomes remotely interesting when neither one of $X$ or $Y$ is contained in the other.
$endgroup$
– dohmatob
Jan 20 at 17:38
$begingroup$
Yes, things are trivial if $X subseteq Y$ or $Y subseteq X$. The problem only becomes remotely interesting when neither one of $X$ or $Y$ is contained in the other.
$endgroup$
– dohmatob
Jan 20 at 17:38
$begingroup$
Yes, things are trivial if $X subseteq Y$ or $Y subseteq X$. The problem only becomes remotely interesting when neither one of $X$ or $Y$ is contained in the other.
$endgroup$
– dohmatob
Jan 20 at 17:38
add a comment |
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1
$begingroup$
You realize that $C(a)$ can be empty?
$endgroup$
– SmileyCraft
Jan 20 at 17:03
$begingroup$
Indeed, good catch. Fixed (i.e you may assume $C(a)$ is non-empty).
$endgroup$
– dohmatob
Jan 20 at 17:07
1
$begingroup$
Can we assume that $|a|_2 leq r$?
$endgroup$
– littleO
Jan 20 at 17:17
$begingroup$
Hum, good point. Yes, you may.
$endgroup$
– dohmatob
Jan 20 at 17:22
$begingroup$
have you considered the dual problem yet?
$endgroup$
– LinAlg
Jan 22 at 18:54