Boyd & Vandenberghe, problem 2.27 — converse supporting hyperplane theorem

Question

In the solution manual of Boyd & Vandenberghe's Convex Optimization, we have the solution for problem 2.27. I have the following queries.

1. Should the first sentence in the solution be "Let $H$ be the intersection of all the halfspaces that contain $C$" instead of "Let $H$ be the set of all..."?

2. Can anybody explain more comprehensively what does the last five lines in the solution mean? It will be very helpful if a contradictory example is presented.

3. Is it always true that if the interior of a closed set is convex then the set is also convex?

Thanks in advance.

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My try to show that $C$ and $H$ have the same boundary.

Since $C$ and $H$ are both closed therefore we can write $C= int(C)\cup bd(c), H=int(H)\cup bd(H)$ (where $int$ means interior and $bd$ means boundary). Now I try to show that $bd(C)=bd(H)$ by first showing that $bd(C)\subseteq bd(H)$ and then showing that $bd(H)\subseteq bd(C)$.

1. If $bd(C)\subseteq bd(H)$ then it means that any $x\in bd(C)$ should also be in $bd(H)$ i.e. $x\in bd(C)$ implies $x \in bd(H)$. If $x \in bd(C)$ and if it is not in $bd(H)$ then it should be either in the complement of $H$ or in the $int(H)$. The first possibility is not valid since any element of $C$ (and hence any element in $bd(C)$) must be in $H$ because $C \subseteq H$. Now the second possibility is that $x\in int(H)$. If this is true then it means that there is some $a^T$ such that $a^Tx<\sup_{y\in H}\{a^Ty\}$. This actually means that we can have a hyperplane with $a^Ty'<a^Ty$ and the halfspace associated with the $a^Ty'$ also contains $C$. But the intersection of this new halfspace (associated with $a^Ty'$) and the halfspace associated with $a^Ty$ is a set which is smaller than the halfspace associated with $a^Ty$. And therefore the intersection of all the other halfspaces that contains $C$ and this new halfspace (associated with $a^Ty'$) results in a smaller set as compared to the original intersection of all the other halfspaces and the halfspace associated with $a^Ty$. The new intersection also contains $C$. Therefore, we conclude that our original choice of $a^Ty$ can be replaced with $a^Ty'$ (while changing $y'$ until we have $y'=x$). Since the points that lie in between $y'$ and $y$ are not in the set $H$ therefore $y'=x$ is a boundary point of $H$.

2. If $bd(H)\subseteq bd(C)$ then it means that $x\in bd(H)$ implies $x\in bd(C)$. Again we can check the two possibilities i.e. i- is it possible that $x\in bd(H)$ is in interior of $C$ and ii- is it possible that $x\in bd(H)$ and $x\in C^c$. The first possibility is not valid since it means that $C \nsubseteq H$. We check the second possibility. Using the reasoning provided in point 1 it can be shown that this case is also invalid. So the only set that $x$ can lie in is $bd(C)$.

Therefore, we have $bd(H)\subseteq bd(C)$ and $bd(C)\subseteq bd(H)$ which means $bd(C)=bd(H)$. Please let me know what is wrong with the above reasoning.

Answer 1

This is Robert Israel's answer https://math.stackexchange.com/a/27290/27978 marginally expanded.

It is clear that $C \subset H$.

Suppose $ p \notin C$, choose $x \in C^\circ$ and let $t = \sup \{ s \in [0,1] | x+\lambda(p-x) \in C \ \forall \lambda \in [0,s]\}$.

Since $C^c$ and $C^\circ$ are open, it is clear that $t \in (0,1)$ and $b=x+t(p-x) \in \partial C$ and, by assumption, there is a supporting hyperplane $\eta$ passing through $b$.

To finish, we need to establish that $x,p$ are strictly on opposite sides of $\eta$, from which it will follow that $p \notin H$, and hence $C=H$.

I will abuse notation slightly by using $\eta$ to denote the linear functional corresponding to the separating hyperplane. In particular, choosing the sign of $\eta$ appropriately we see that $C$ is contained in the half space $\{z | \eta(z) \le \eta(b) \}$.

Since $B(x,r) \subset C$ for some $r>0$, we have $\eta(x) < \eta(b)$. Since $p = b + {1 -t \over t} (b-x)$, we see that $\eta(p) = \eta(b) +{1 -t \over t} \eta(b-x)> \eta(b)$ from which the result follows.

Addendum: Clarifications for questions in comments.

To show that $b \in \partial C$:

Let $l(\lambda) = x+\lambda(p-x)$. Note that $l(0) = x \in C^\circ$ and $l(1) = p \in C^c$. Both $C^\circ, C^c$ are open, hence there is some $\delta>0$ such that $l(\lambda) \in C^\circ$ for $\lambda \in [0,\delta)$ and similarly $l(\lambda) \in C^c$ for $\lambda \in (1-\delta,1]$. Define $t$ as above, then it is clear that $t \in [\delta,1-\delta]$.

By definition of $t$, we have $l(s) \in C$ for all $s < t$, hence $l(t) \in C$, since $C$ is closed. By definition of $t$, $l(s) \notin C$ for any $s>t$ (otherwise the definition of $t$ would be contradicted). Hence $b=l(t) \in \partial C$.

To show that $\eta(x) < \eta(b)$:

We have that the halfspace $H= \{z | \eta(z) \le \eta(b) \}$ and $C \subset H$. In particular, $\eta(x) \le \eta(b)$. Since $\eta$ is non zero, it maps open sets into open sets. Since $B(x,r) \subset C^\circ$ for some $r>0$ we see that $\eta(B(x,r)) \subset (-\infty, \eta(b)]$ and so $\eta(x) < \eta(b)$.