Supporting Hyperplanes to the intersection of half-spaces
Let $H(a,r)$ be a hyperplane through point $a$ with normal vector $r$, and let $H(a,r)^{+}$ be the corresponding nonnegative halfspace ,and $H(a,r)^{++}$ the positive halfspace. I have a set $Omega$ of these hyperplanes, not necessarily finite.
Consider $A equiv bigcap_Omega H(a,r)^{+}$. I know it is certainly possible that there are $H(a,r) in Omega$ that are not supporting to $A$.
But is it possible to have a point $x$ on the boundary of $A$ that is not supported by any $H(a,r) in Omega$?
I suspect not. My thinking is as follows. A point $x$ on the boundary of $A$ must lie in $H(a,r)^{+}$ for every $H(a,r)in Omega$ by construction. If $x in H(a,r)^{++}$ for all $H(a,r) in Omega$, then it cannot be on the boundary. Therefore, it must lie on at least one $H(a,r)$, and this one would be supporting.
convex-analysis
asked Nov 29 '18 at 16:06
Mathew KnudsonMathew Knudson
84
add a comment |
Let $H(a,r)$ be a hyperplane through point $a$ with normal vector $r$, and let $H(a,r)^{+}$ be the corresponding nonnegative halfspace ,and $H(a,r)^{++}$ the positive halfspace. I have a set $Omega$ of these hyperplanes, not necessarily finite.
Consider $A equiv bigcap_Omega H(a,r)^{+}$. I know it is certainly possible that there are $H(a,r) in Omega$ that are not supporting to $A$.
But is it possible to have a point $x$ on the boundary of $A$ that is not supported by any $H(a,r) in Omega$?
I suspect not. My thinking is as follows. A point $x$ on the boundary of $A$ must lie in $H(a,r)^{+}$ for every $H(a,r)in Omega$ by construction. If $x in H(a,r)^{++}$ for all $H(a,r) in Omega$, then it cannot be on the boundary. Therefore, it must lie on at least one $H(a,r)$, and this one would be supporting.
convex-analysis
asked Nov 29 '18 at 16:06
Mathew KnudsonMathew Knudson
84
add a comment |
Let $H(a,r)$ be a hyperplane through point $a$ with normal vector $r$, and let $H(a,r)^{+}$ be the corresponding nonnegative halfspace ,and $H(a,r)^{++}$ the positive halfspace. I have a set $Omega$ of these hyperplanes, not necessarily finite.
Consider $A equiv bigcap_Omega H(a,r)^{+}$. I know it is certainly possible that there are $H(a,r) in Omega$ that are not supporting to $A$.
But is it possible to have a point $x$ on the boundary of $A$ that is not supported by any $H(a,r) in Omega$?
I suspect not. My thinking is as follows. A point $x$ on the boundary of $A$ must lie in $H(a,r)^{+}$ for every $H(a,r)in Omega$ by construction. If $x in H(a,r)^{++}$ for all $H(a,r) in Omega$, then it cannot be on the boundary. Therefore, it must lie on at least one $H(a,r)$, and this one would be supporting.
convex-analysis
asked Nov 29 '18 at 16:06
Mathew KnudsonMathew Knudson
84
Let $H(a,r)$ be a hyperplane through point $a$ with normal vector $r$, and let $H(a,r)^{+}$ be the corresponding nonnegative halfspace ,and $H(a,r)^{++}$ the positive halfspace. I have a set $Omega$ of these hyperplanes, not necessarily finite.
Consider $A equiv bigcap_Omega H(a,r)^{+}$. I know it is certainly possible that there are $H(a,r) in Omega$ that are not supporting to $A$.
But is it possible to have a point $x$ on the boundary of $A$ that is not supported by any $H(a,r) in Omega$?
I suspect not. My thinking is as follows. A point $x$ on the boundary of $A$ must lie in $H(a,r)^{+}$ for every $H(a,r)in Omega$ by construction. If $x in H(a,r)^{++}$ for all $H(a,r) in Omega$, then it cannot be on the boundary. Therefore, it must lie on at least one $H(a,r)$, and this one would be supporting.
convex-analysis
convex-analysis
asked Nov 29 '18 at 16:06
Mathew KnudsonMathew Knudson
84
asked Nov 29 '18 at 16:06
Mathew KnudsonMathew Knudson
84
asked Nov 29 '18 at 16:06
Mathew KnudsonMathew Knudson
84
asked Nov 29 '18 at 16:06
Mathew KnudsonMathew Knudson
84
asked Nov 29 '18 at 16:06
Mathew KnudsonMathew Knudson
84
84
add a comment |
add a comment |
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