Why is strong duality holds even though the problem is not convex?












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$begingroup$


I write the question $5.29$ in Convex Optimization ,and it said



minimize $-3x_1^2+x_2^2+2x_3^2+2(x_1+x_2+x_3)$



subject to $x_1^2+x_2^2+x_3^2=1$



so strong duality holds even though the problem is not convex



I wonder that if we have to prove it is strong duality and not convex,how should we prove it?










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  • $begingroup$
    Perhaps you can read page 229 of the book (which is referred to from the question).
    $endgroup$
    – LinAlg
    Nov 26 '18 at 14:28










  • $begingroup$
    The problem actually becomes convex if we substitute $x_1^2=1-x_2^2-x_3^2$ and $x_1=-sqrt{1-x_2^2-x_3^2}$ into the objective function subject to $x_2^2+x_3^2le 1$.
    $endgroup$
    – A.Γ.
    Dec 7 '18 at 21:56
















0












$begingroup$


I write the question $5.29$ in Convex Optimization ,and it said



minimize $-3x_1^2+x_2^2+2x_3^2+2(x_1+x_2+x_3)$



subject to $x_1^2+x_2^2+x_3^2=1$



so strong duality holds even though the problem is not convex



I wonder that if we have to prove it is strong duality and not convex,how should we prove it?










share|cite|improve this question









$endgroup$












  • $begingroup$
    Perhaps you can read page 229 of the book (which is referred to from the question).
    $endgroup$
    – LinAlg
    Nov 26 '18 at 14:28










  • $begingroup$
    The problem actually becomes convex if we substitute $x_1^2=1-x_2^2-x_3^2$ and $x_1=-sqrt{1-x_2^2-x_3^2}$ into the objective function subject to $x_2^2+x_3^2le 1$.
    $endgroup$
    – A.Γ.
    Dec 7 '18 at 21:56














0












0








0





$begingroup$


I write the question $5.29$ in Convex Optimization ,and it said



minimize $-3x_1^2+x_2^2+2x_3^2+2(x_1+x_2+x_3)$



subject to $x_1^2+x_2^2+x_3^2=1$



so strong duality holds even though the problem is not convex



I wonder that if we have to prove it is strong duality and not convex,how should we prove it?










share|cite|improve this question









$endgroup$




I write the question $5.29$ in Convex Optimization ,and it said



minimize $-3x_1^2+x_2^2+2x_3^2+2(x_1+x_2+x_3)$



subject to $x_1^2+x_2^2+x_3^2=1$



so strong duality holds even though the problem is not convex



I wonder that if we have to prove it is strong duality and not convex,how should we prove it?







convex-analysis convex-optimization duality-theorems






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 26 '18 at 13:09









shineeleshineele

377




377












  • $begingroup$
    Perhaps you can read page 229 of the book (which is referred to from the question).
    $endgroup$
    – LinAlg
    Nov 26 '18 at 14:28










  • $begingroup$
    The problem actually becomes convex if we substitute $x_1^2=1-x_2^2-x_3^2$ and $x_1=-sqrt{1-x_2^2-x_3^2}$ into the objective function subject to $x_2^2+x_3^2le 1$.
    $endgroup$
    – A.Γ.
    Dec 7 '18 at 21:56


















  • $begingroup$
    Perhaps you can read page 229 of the book (which is referred to from the question).
    $endgroup$
    – LinAlg
    Nov 26 '18 at 14:28










  • $begingroup$
    The problem actually becomes convex if we substitute $x_1^2=1-x_2^2-x_3^2$ and $x_1=-sqrt{1-x_2^2-x_3^2}$ into the objective function subject to $x_2^2+x_3^2le 1$.
    $endgroup$
    – A.Γ.
    Dec 7 '18 at 21:56
















$begingroup$
Perhaps you can read page 229 of the book (which is referred to from the question).
$endgroup$
– LinAlg
Nov 26 '18 at 14:28




$begingroup$
Perhaps you can read page 229 of the book (which is referred to from the question).
$endgroup$
– LinAlg
Nov 26 '18 at 14:28












$begingroup$
The problem actually becomes convex if we substitute $x_1^2=1-x_2^2-x_3^2$ and $x_1=-sqrt{1-x_2^2-x_3^2}$ into the objective function subject to $x_2^2+x_3^2le 1$.
$endgroup$
– A.Γ.
Dec 7 '18 at 21:56




$begingroup$
The problem actually becomes convex if we substitute $x_1^2=1-x_2^2-x_3^2$ and $x_1=-sqrt{1-x_2^2-x_3^2}$ into the objective function subject to $x_2^2+x_3^2le 1$.
$endgroup$
– A.Γ.
Dec 7 '18 at 21:56










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