Question about Lagrange Duality and saddle points
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Consider two Hilbert spaces $mathcal{H}$ and $mathcal{G}$ and an extended real valued function $f:mathcal{H}tomathbb{R}cup{+infty}$ with $finGamma_0(mathcal{H})$ the set of proper, lower semicontinuous, convex functions. We look at the following problem (we will refer to this as the primal),
$$min f(x) : Ax=0$$
where $A:mathcal{H}tomathcal{G}$ is a linear operator.
We can form the classical Lagrangian for our problem,
$$mathcal{L}(x,mu) = f(x) + langle mu, Axrangle$$
and state the dual problem,
$$max_mumin_xmathcal{L}(x,mu)$$
Let us also assume that strong duality holds. My question, then, is the following:
Imagine you have a particular solution $bar{mu}$ to the dual problem and you also have a solution $x^*inargminlimits_{x}mathcal{L}(x,bar{mu})$. Is it true that $Ax^*=0$, i.e. $x^*$ is feasible and thus $(x^*,bar{mu})$ is a saddle point/primal dual pair?
functional-analysis convex-analysis lagrange-multiplier
asked Dec 14 '18 at 11:06
Tony S.F.Tony S.F.
3,30821028
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add a comment |
$begingroup$
Consider two Hilbert spaces $mathcal{H}$ and $mathcal{G}$ and an extended real valued function $f:mathcal{H}tomathbb{R}cup{+infty}$ with $finGamma_0(mathcal{H})$ the set of proper, lower semicontinuous, convex functions. We look at the following problem (we will refer to this as the primal),
$$min f(x) : Ax=0$$
where $A:mathcal{H}tomathcal{G}$ is a linear operator.
We can form the classical Lagrangian for our problem,
$$mathcal{L}(x,mu) = f(x) + langle mu, Axrangle$$
and state the dual problem,
$$max_mumin_xmathcal{L}(x,mu)$$
Let us also assume that strong duality holds. My question, then, is the following:
Imagine you have a particular solution $bar{mu}$ to the dual problem and you also have a solution $x^*inargminlimits_{x}mathcal{L}(x,bar{mu})$. Is it true that $Ax^*=0$, i.e. $x^*$ is feasible and thus $(x^*,bar{mu})$ is a saddle point/primal dual pair?
functional-analysis convex-analysis lagrange-multiplier
asked Dec 14 '18 at 11:06
Tony S.F.Tony S.F.
3,30821028
$endgroup$
1
$begingroup$
If $x^*$ is a unique minimizer of $L(x,barmu)$ then yes, otherwise no.
$endgroup$
– A.Γ.
Dec 14 '18 at 19:54
add a comment |
$begingroup$
Consider two Hilbert spaces $mathcal{H}$ and $mathcal{G}$ and an extended real valued function $f:mathcal{H}tomathbb{R}cup{+infty}$ with $finGamma_0(mathcal{H})$ the set of proper, lower semicontinuous, convex functions. We look at the following problem (we will refer to this as the primal),
$$min f(x) : Ax=0$$
where $A:mathcal{H}tomathcal{G}$ is a linear operator.
We can form the classical Lagrangian for our problem,
$$mathcal{L}(x,mu) = f(x) + langle mu, Axrangle$$
and state the dual problem,
$$max_mumin_xmathcal{L}(x,mu)$$
Let us also assume that strong duality holds. My question, then, is the following:
Imagine you have a particular solution $bar{mu}$ to the dual problem and you also have a solution $x^*inargminlimits_{x}mathcal{L}(x,bar{mu})$. Is it true that $Ax^*=0$, i.e. $x^*$ is feasible and thus $(x^*,bar{mu})$ is a saddle point/primal dual pair?
functional-analysis convex-analysis lagrange-multiplier
asked Dec 14 '18 at 11:06
Tony S.F.Tony S.F.
3,30821028
$endgroup$
Consider two Hilbert spaces $mathcal{H}$ and $mathcal{G}$ and an extended real valued function $f:mathcal{H}tomathbb{R}cup{+infty}$ with $finGamma_0(mathcal{H})$ the set of proper, lower semicontinuous, convex functions. We look at the following problem (we will refer to this as the primal),
$$min f(x) : Ax=0$$
where $A:mathcal{H}tomathcal{G}$ is a linear operator.
We can form the classical Lagrangian for our problem,
$$mathcal{L}(x,mu) = f(x) + langle mu, Axrangle$$
and state the dual problem,
$$max_mumin_xmathcal{L}(x,mu)$$
Let us also assume that strong duality holds. My question, then, is the following:
Imagine you have a particular solution $bar{mu}$ to the dual problem and you also have a solution $x^*inargminlimits_{x}mathcal{L}(x,bar{mu})$. Is it true that $Ax^*=0$, i.e. $x^*$ is feasible and thus $(x^*,bar{mu})$ is a saddle point/primal dual pair?
functional-analysis convex-analysis lagrange-multiplier
functional-analysis convex-analysis lagrange-multiplier
asked Dec 14 '18 at 11:06
Tony S.F.Tony S.F.
3,30821028
asked Dec 14 '18 at 11:06
Tony S.F.Tony S.F.
3,30821028
asked Dec 14 '18 at 11:06
Tony S.F.Tony S.F.
3,30821028
asked Dec 14 '18 at 11:06
Tony S.F.Tony S.F.
3,30821028
asked Dec 14 '18 at 11:06
Tony S.F.Tony S.F.
3,30821028
3,30821028
1
$begingroup$
If $x^*$ is a unique minimizer of $L(x,barmu)$ then yes, otherwise no.
$endgroup$
– A.Γ.
Dec 14 '18 at 19:54
add a comment |
1
$begingroup$
If $x^*$ is a unique minimizer of $L(x,barmu)$ then yes, otherwise no.
$endgroup$
– A.Γ.
Dec 14 '18 at 19:54
1
1
$begingroup$
If $x^*$ is a unique minimizer of $L(x,barmu)$ then yes, otherwise no.
$endgroup$
– A.Γ.
Dec 14 '18 at 19:54
$begingroup$
If $x^*$ is a unique minimizer of $L(x,barmu)$ then yes, otherwise no.
$endgroup$
– A.Γ.
Dec 14 '18 at 19:54
add a comment |
1 Answer
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No, not necessarily. Take $min_{x} { x : x = 0 }$. The Lagrangian is $L(x,mu)=x+mu x$, the unique saddle point is $(x,mu)=(0,-1)$, while any $x$ minimizes the Lagrangian.
answered Dec 14 '18 at 19:04
LinAlgLinAlg
9,9341521
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I was just about to post the example $min xcolon x=0$ :)
$endgroup$
– A.Γ.
Dec 14 '18 at 19:55
$begingroup$
@A.Γ. that is actually a better example because the rhs in the question is 0.
$endgroup$
– LinAlg
Dec 14 '18 at 20:14
add a comment |
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No, not necessarily. Take $min_{x} { x : x = 0 }$. The Lagrangian is $L(x,mu)=x+mu x$, the unique saddle point is $(x,mu)=(0,-1)$, while any $x$ minimizes the Lagrangian.
answered Dec 14 '18 at 19:04
LinAlgLinAlg
9,9341521
$endgroup$
$begingroup$
I was just about to post the example $min xcolon x=0$ :)
$endgroup$
– A.Γ.
Dec 14 '18 at 19:55
$begingroup$
@A.Γ. that is actually a better example because the rhs in the question is 0.
$endgroup$
– LinAlg
Dec 14 '18 at 20:14
add a comment |
$begingroup$
No, not necessarily. Take $min_{x} { x : x = 0 }$. The Lagrangian is $L(x,mu)=x+mu x$, the unique saddle point is $(x,mu)=(0,-1)$, while any $x$ minimizes the Lagrangian.
answered Dec 14 '18 at 19:04
LinAlgLinAlg
9,9341521
$endgroup$
$begingroup$
I was just about to post the example $min xcolon x=0$ :)
$endgroup$
– A.Γ.
Dec 14 '18 at 19:55
$begingroup$
@A.Γ. that is actually a better example because the rhs in the question is 0.
$endgroup$
– LinAlg
Dec 14 '18 at 20:14
add a comment |
$begingroup$
No, not necessarily. Take $min_{x} { x : x = 0 }$. The Lagrangian is $L(x,mu)=x+mu x$, the unique saddle point is $(x,mu)=(0,-1)$, while any $x$ minimizes the Lagrangian.
answered Dec 14 '18 at 19:04
LinAlgLinAlg
9,9341521
$endgroup$
No, not necessarily. Take $min_{x} { x : x = 0 }$. The Lagrangian is $L(x,mu)=x+mu x$, the unique saddle point is $(x,mu)=(0,-1)$, while any $x$ minimizes the Lagrangian.
answered Dec 14 '18 at 19:04
LinAlgLinAlg
9,9341521
edited Dec 14 '18 at 20:14
answered Dec 14 '18 at 19:04
LinAlgLinAlg
9,9341521
answered Dec 14 '18 at 19:04
LinAlgLinAlg
9,9341521
answered Dec 14 '18 at 19:04
LinAlgLinAlg
9,9341521
9,9341521
$begingroup$
I was just about to post the example $min xcolon x=0$ :)
$endgroup$
– A.Γ.
Dec 14 '18 at 19:55
$begingroup$
@A.Γ. that is actually a better example because the rhs in the question is 0.
$endgroup$
– LinAlg
Dec 14 '18 at 20:14
add a comment |
$begingroup$
I was just about to post the example $min xcolon x=0$ :)
$endgroup$
– A.Γ.
Dec 14 '18 at 19:55
$begingroup$
@A.Γ. that is actually a better example because the rhs in the question is 0.
$endgroup$
– LinAlg
Dec 14 '18 at 20:14
$begingroup$
I was just about to post the example $min xcolon x=0$ :)
$endgroup$
– A.Γ.
Dec 14 '18 at 19:55
$begingroup$
I was just about to post the example $min xcolon x=0$ :)
$endgroup$
– A.Γ.
Dec 14 '18 at 19:55
$begingroup$
@A.Γ. that is actually a better example because the rhs in the question is 0.
$endgroup$
– LinAlg
Dec 14 '18 at 20:14
$begingroup$
@A.Γ. that is actually a better example because the rhs in the question is 0.
$endgroup$
– LinAlg
Dec 14 '18 at 20:14
add a comment |
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If $x^*$ is a unique minimizer of $L(x,barmu)$ then yes, otherwise no.
$endgroup$
– A.Γ.
Dec 14 '18 at 19:54