Proof of basic variational principle.
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The normalized duality mapping $J$ from a real Banach space $X$ to its dual space $X^*$ is defined as:
$$J(x)={x^* in X^*: (x^*,x)=|x|^2=|x^*|^2}.$$
Where $(x^*,x)$ denotes the duality pairing. And $J$ is surjective, single valued and injective if $X$ is uniformly smooth and uniformly convex.
The Lyapunov functional $W(x,y)$ is defined as:
$$W(x,y)=|x|^2-2(Jx,y)+|y|^2$$ this $W$ is always positive.
And the generalized projection $pi_K$ from $X$ to a closed convex subset $K subset X$ is defined as:
$$pi_{K}(x)=x'; x': W(x,x')=inf_{y in K}W(x,y)$$
I want to prove the following:
$pi_K(x)=x'$ is a generalized projection of $x$ if and only if $(Jx-Jx',x'-y)geq 0, forall y in K$ (This is called Basic Variational Principle).
I tried to prove $pi_K(x)=x'implies (Jx-Jx',x'-y)geq 0, forall y in K.$ However, during my proof I get $W(x,y)leq 0$ (but $W$ is always positive).
functional-analysis dual-maps
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up vote
-1
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The normalized duality mapping $J$ from a real Banach space $X$ to its dual space $X^*$ is defined as:
$$J(x)={x^* in X^*: (x^*,x)=|x|^2=|x^*|^2}.$$
Where $(x^*,x)$ denotes the duality pairing. And $J$ is surjective, single valued and injective if $X$ is uniformly smooth and uniformly convex.
The Lyapunov functional $W(x,y)$ is defined as:
$$W(x,y)=|x|^2-2(Jx,y)+|y|^2$$ this $W$ is always positive.
And the generalized projection $pi_K$ from $X$ to a closed convex subset $K subset X$ is defined as:
$$pi_{K}(x)=x'; x': W(x,x')=inf_{y in K}W(x,y)$$
I want to prove the following:
$pi_K(x)=x'$ is a generalized projection of $x$ if and only if $(Jx-Jx',x'-y)geq 0, forall y in K$ (This is called Basic Variational Principle).
I tried to prove $pi_K(x)=x'implies (Jx-Jx',x'-y)geq 0, forall y in K.$ However, during my proof I get $W(x,y)leq 0$ (but $W$ is always positive).
functional-analysis dual-maps
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
The normalized duality mapping $J$ from a real Banach space $X$ to its dual space $X^*$ is defined as:
$$J(x)={x^* in X^*: (x^*,x)=|x|^2=|x^*|^2}.$$
Where $(x^*,x)$ denotes the duality pairing. And $J$ is surjective, single valued and injective if $X$ is uniformly smooth and uniformly convex.
The Lyapunov functional $W(x,y)$ is defined as:
$$W(x,y)=|x|^2-2(Jx,y)+|y|^2$$ this $W$ is always positive.
And the generalized projection $pi_K$ from $X$ to a closed convex subset $K subset X$ is defined as:
$$pi_{K}(x)=x'; x': W(x,x')=inf_{y in K}W(x,y)$$
I want to prove the following:
$pi_K(x)=x'$ is a generalized projection of $x$ if and only if $(Jx-Jx',x'-y)geq 0, forall y in K$ (This is called Basic Variational Principle).
I tried to prove $pi_K(x)=x'implies (Jx-Jx',x'-y)geq 0, forall y in K.$ However, during my proof I get $W(x,y)leq 0$ (but $W$ is always positive).
functional-analysis dual-maps
The normalized duality mapping $J$ from a real Banach space $X$ to its dual space $X^*$ is defined as:
$$J(x)={x^* in X^*: (x^*,x)=|x|^2=|x^*|^2}.$$
Where $(x^*,x)$ denotes the duality pairing. And $J$ is surjective, single valued and injective if $X$ is uniformly smooth and uniformly convex.
The Lyapunov functional $W(x,y)$ is defined as:
$$W(x,y)=|x|^2-2(Jx,y)+|y|^2$$ this $W$ is always positive.
And the generalized projection $pi_K$ from $X$ to a closed convex subset $K subset X$ is defined as:
$$pi_{K}(x)=x'; x': W(x,x')=inf_{y in K}W(x,y)$$
I want to prove the following:
$pi_K(x)=x'$ is a generalized projection of $x$ if and only if $(Jx-Jx',x'-y)geq 0, forall y in K$ (This is called Basic Variational Principle).
I tried to prove $pi_K(x)=x'implies (Jx-Jx',x'-y)geq 0, forall y in K.$ However, during my proof I get $W(x,y)leq 0$ (but $W$ is always positive).
functional-analysis dual-maps
functional-analysis dual-maps
edited Nov 18 at 10:38
asked Nov 18 at 6:41
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