How to prove that continuous convex functional on normed vector space must be lower bounded by some...
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Let $X$ be a normed vector space. Let $f:Xmapsto mathbb{R}$ be a continuous convex function. How to show that there exists a continuous linear functional $l$ and a constant $cinmathbb{R}$ such that $f(x)>l(x)+c$ holds for all $xin X$? I think it could be proved by hyperplane separation theorem. But I don't know how to start it.
functional-analysis convex-analysis normed-spaces convex-geometry hahn-banach-theorem
asked Dec 31 '18 at 16:21
Lin XueleiLin Xuelei
16011
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add a comment |
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Let $X$ be a normed vector space. Let $f:Xmapsto mathbb{R}$ be a continuous convex function. How to show that there exists a continuous linear functional $l$ and a constant $cinmathbb{R}$ such that $f(x)>l(x)+c$ holds for all $xin X$? I think it could be proved by hyperplane separation theorem. But I don't know how to start it.
functional-analysis convex-analysis normed-spaces convex-geometry hahn-banach-theorem
asked Dec 31 '18 at 16:21
Lin XueleiLin Xuelei
16011
$endgroup$
2
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The comments at this similar question math.stackexchange.com/q/2734823/42969 may be helpful.
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– Martin R
Dec 31 '18 at 16:30
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Thanks. But it's not.
$endgroup$
– Lin Xuelei
Jan 1 at 1:44
add a comment |
$begingroup$
Let $X$ be a normed vector space. Let $f:Xmapsto mathbb{R}$ be a continuous convex function. How to show that there exists a continuous linear functional $l$ and a constant $cinmathbb{R}$ such that $f(x)>l(x)+c$ holds for all $xin X$? I think it could be proved by hyperplane separation theorem. But I don't know how to start it.
functional-analysis convex-analysis normed-spaces convex-geometry hahn-banach-theorem
asked Dec 31 '18 at 16:21
Lin XueleiLin Xuelei
16011
$endgroup$
Let $X$ be a normed vector space. Let $f:Xmapsto mathbb{R}$ be a continuous convex function. How to show that there exists a continuous linear functional $l$ and a constant $cinmathbb{R}$ such that $f(x)>l(x)+c$ holds for all $xin X$? I think it could be proved by hyperplane separation theorem. But I don't know how to start it.
functional-analysis convex-analysis normed-spaces convex-geometry hahn-banach-theorem
functional-analysis convex-analysis normed-spaces convex-geometry hahn-banach-theorem
asked Dec 31 '18 at 16:21
Lin XueleiLin Xuelei
16011
asked Dec 31 '18 at 16:21
Lin XueleiLin Xuelei
16011
edited Jan 1 at 13:45
Lin Xuelei
asked Dec 31 '18 at 16:21
Lin XueleiLin Xuelei
16011
asked Dec 31 '18 at 16:21
Lin XueleiLin Xuelei
16011
asked Dec 31 '18 at 16:21
Lin XueleiLin Xuelei
16011
16011
2
$begingroup$
The comments at this similar question math.stackexchange.com/q/2734823/42969 may be helpful.
$endgroup$
– Martin R
Dec 31 '18 at 16:30
$begingroup$
Thanks. But it's not.
$endgroup$
– Lin Xuelei
Jan 1 at 1:44
add a comment |
2
$begingroup$
The comments at this similar question math.stackexchange.com/q/2734823/42969 may be helpful.
$endgroup$
– Martin R
Dec 31 '18 at 16:30
$begingroup$
Thanks. But it's not.
$endgroup$
– Lin Xuelei
Jan 1 at 1:44
2
2
$begingroup$
The comments at this similar question math.stackexchange.com/q/2734823/42969 may be helpful.
$endgroup$
– Martin R
Dec 31 '18 at 16:30
$begingroup$
The comments at this similar question math.stackexchange.com/q/2734823/42969 may be helpful.
$endgroup$
– Martin R
Dec 31 '18 at 16:30
$begingroup$
Thanks. But it's not.
$endgroup$
– Lin Xuelei
Jan 1 at 1:44
$begingroup$
Thanks. But it's not.
$endgroup$
– Lin Xuelei
Jan 1 at 1:44
add a comment |
1 Answer
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The epigraph $operatorname{epi}f$ is closed and convex set in $XtimesBbb R$ and $(x,f(x)-epsilon)notinoperatorname{epi}f$ is compact for any $epsilon>0$ and $xin X$. By the separation theorem there exists a non-zero $(phi,psi)in(XtimesBbb R)'=X'timesBbb R$ such that the hyperplane $(phi,psi)cdot(x,y)=phi(x)+psi y=alpha$ separates $operatorname{epi}f$ and the point. We can choose the sign such that $operatorname{epi}f$ belongs to the half-space $phi(x)+psi y>alpha$. Clearly $psine 0$ since $phi(x)>alpha$ is not possible for all $xin X$ and $phine 0$. Hence, $psi>0$ otherwise $yto+infty$ (still in $operatorname{epi}f$) will give us a contradiction with $phi(x)+psi y>alpha$. Finally $(x,f(x))inoperatorname{epi}f$ then $phi(x)+psi f(x)>alpha$. Divide by $psi$ and move the things around.
answered Jan 1 at 20:32
A.Γ.A.Γ.
23k32657
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thanks. I never thought dual of product space has such a precise form.
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– Lin Xuelei
Jan 2 at 10:17
add a comment |
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The epigraph $operatorname{epi}f$ is closed and convex set in $XtimesBbb R$ and $(x,f(x)-epsilon)notinoperatorname{epi}f$ is compact for any $epsilon>0$ and $xin X$. By the separation theorem there exists a non-zero $(phi,psi)in(XtimesBbb R)'=X'timesBbb R$ such that the hyperplane $(phi,psi)cdot(x,y)=phi(x)+psi y=alpha$ separates $operatorname{epi}f$ and the point. We can choose the sign such that $operatorname{epi}f$ belongs to the half-space $phi(x)+psi y>alpha$. Clearly $psine 0$ since $phi(x)>alpha$ is not possible for all $xin X$ and $phine 0$. Hence, $psi>0$ otherwise $yto+infty$ (still in $operatorname{epi}f$) will give us a contradiction with $phi(x)+psi y>alpha$. Finally $(x,f(x))inoperatorname{epi}f$ then $phi(x)+psi f(x)>alpha$. Divide by $psi$ and move the things around.
answered Jan 1 at 20:32
A.Γ.A.Γ.
23k32657
$endgroup$
$begingroup$
thanks. I never thought dual of product space has such a precise form.
$endgroup$
– Lin Xuelei
Jan 2 at 10:17
add a comment |
$begingroup$
The epigraph $operatorname{epi}f$ is closed and convex set in $XtimesBbb R$ and $(x,f(x)-epsilon)notinoperatorname{epi}f$ is compact for any $epsilon>0$ and $xin X$. By the separation theorem there exists a non-zero $(phi,psi)in(XtimesBbb R)'=X'timesBbb R$ such that the hyperplane $(phi,psi)cdot(x,y)=phi(x)+psi y=alpha$ separates $operatorname{epi}f$ and the point. We can choose the sign such that $operatorname{epi}f$ belongs to the half-space $phi(x)+psi y>alpha$. Clearly $psine 0$ since $phi(x)>alpha$ is not possible for all $xin X$ and $phine 0$. Hence, $psi>0$ otherwise $yto+infty$ (still in $operatorname{epi}f$) will give us a contradiction with $phi(x)+psi y>alpha$. Finally $(x,f(x))inoperatorname{epi}f$ then $phi(x)+psi f(x)>alpha$. Divide by $psi$ and move the things around.
answered Jan 1 at 20:32
A.Γ.A.Γ.
23k32657
$endgroup$
$begingroup$
thanks. I never thought dual of product space has such a precise form.
$endgroup$
– Lin Xuelei
Jan 2 at 10:17
add a comment |
$begingroup$
The epigraph $operatorname{epi}f$ is closed and convex set in $XtimesBbb R$ and $(x,f(x)-epsilon)notinoperatorname{epi}f$ is compact for any $epsilon>0$ and $xin X$. By the separation theorem there exists a non-zero $(phi,psi)in(XtimesBbb R)'=X'timesBbb R$ such that the hyperplane $(phi,psi)cdot(x,y)=phi(x)+psi y=alpha$ separates $operatorname{epi}f$ and the point. We can choose the sign such that $operatorname{epi}f$ belongs to the half-space $phi(x)+psi y>alpha$. Clearly $psine 0$ since $phi(x)>alpha$ is not possible for all $xin X$ and $phine 0$. Hence, $psi>0$ otherwise $yto+infty$ (still in $operatorname{epi}f$) will give us a contradiction with $phi(x)+psi y>alpha$. Finally $(x,f(x))inoperatorname{epi}f$ then $phi(x)+psi f(x)>alpha$. Divide by $psi$ and move the things around.
answered Jan 1 at 20:32
A.Γ.A.Γ.
23k32657
$endgroup$
The epigraph $operatorname{epi}f$ is closed and convex set in $XtimesBbb R$ and $(x,f(x)-epsilon)notinoperatorname{epi}f$ is compact for any $epsilon>0$ and $xin X$. By the separation theorem there exists a non-zero $(phi,psi)in(XtimesBbb R)'=X'timesBbb R$ such that the hyperplane $(phi,psi)cdot(x,y)=phi(x)+psi y=alpha$ separates $operatorname{epi}f$ and the point. We can choose the sign such that $operatorname{epi}f$ belongs to the half-space $phi(x)+psi y>alpha$. Clearly $psine 0$ since $phi(x)>alpha$ is not possible for all $xin X$ and $phine 0$. Hence, $psi>0$ otherwise $yto+infty$ (still in $operatorname{epi}f$) will give us a contradiction with $phi(x)+psi y>alpha$. Finally $(x,f(x))inoperatorname{epi}f$ then $phi(x)+psi f(x)>alpha$. Divide by $psi$ and move the things around.
answered Jan 1 at 20:32
A.Γ.A.Γ.
23k32657
answered Jan 1 at 20:32
A.Γ.A.Γ.
23k32657
answered Jan 1 at 20:32
A.Γ.A.Γ.
23k32657
answered Jan 1 at 20:32
A.Γ.A.Γ.
23k32657
23k32657
$begingroup$
thanks. I never thought dual of product space has such a precise form.
$endgroup$
– Lin Xuelei
Jan 2 at 10:17
add a comment |
$begingroup$
thanks. I never thought dual of product space has such a precise form.
$endgroup$
– Lin Xuelei
Jan 2 at 10:17
$begingroup$
thanks. I never thought dual of product space has such a precise form.
$endgroup$
– Lin Xuelei
Jan 2 at 10:17
$begingroup$
thanks. I never thought dual of product space has such a precise form.
$endgroup$
– Lin Xuelei
Jan 2 at 10:17
add a comment |
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2
$begingroup$
The comments at this similar question math.stackexchange.com/q/2734823/42969 may be helpful.
$endgroup$
– Martin R
Dec 31 '18 at 16:30
$begingroup$
Thanks. But it's not.
$endgroup$
– Lin Xuelei
Jan 1 at 1:44