I couldn't fit the whole question in the title, sorry! I'm trying the following bonus question from my Convex Analysis homework:
Let $X$ be a vector space over $\mathbb R$, and let $F\subset X$ be an absorbing convex set. Given a nonempty set $\Omega$ in $X$, define $$T_F(x;\Omega)=\inf\{t\geq 0\mid (x+tF)\cap\Omega\neq\emptyset\},x\in X.$$ (d) Given $x\in X$, consider the set $$P_F(x;\Omega)=\{y\in\Omega\mid T_F(x;\Omega)=p_F(y-x)\}.$$ Prove or disprove: If $X$ is a finite-dimensional topological vector space, $0\in\text{int}(F)$, and $\Omega$ is closed, then $P_F(x;\Omega)\neq\emptyset$ for all $x\in X$.
Just to give some additional context, in our class $p_F:X\to\mathbb R$ is the Minkowski function for $F$ given by $p_F(x)=\inf\{t\geq 0\mid x\in tF\}$ for all $x\in X$. In particular, this means $p_F(y-x)=\inf\{t\geq 0\mid y\in x+tF\}$. Also, we proved in part (c) that for all $x\in X$, $T_F(x;\Omega)=\inf_{y\in\Omega}p_F(y-x)$. Additionally, we have a corollary in our book that if $X$ is a TVS and $0\in\text{int}(F)$, with $F$ convex, we have that $p_F$ is continuous. So, by the assumptions of the problem, $p_F$ is continuous. My thought is that for each $x\in X$, we need to ensure $A_x:=\{p_F(y-x)\mid y\in \Omega\}$ achieves its minimum. This is because, if $p_F(y_0-x)=\min A_x$, then we would immediately have that $y_0\in P_F(x;\Omega)$ so that $P_F(x;\Omega)$ is nonempty. But then if $A_x$ does not achieve a minimum, surely $P_F(x;\Omega)$ is empty. This is because we would not be able to find $y\in\Omega$ such that $T_F(x;\Omega)=p_F(y-x)$. My hunch is that $A_x$ does achieve a minimum, but this is just a hunch, and I'm not sure how to show this if it's true. Can anyone help point me in the right direction? Thank you!
$\def\R{\mathbb R}$
Let $\{y_n\}\subset \Omega$ such that $p_F(y_n-x)\to T_F(x;\Omega)$. If $\Omega$ is bounded, then it is compact (since $X$ is finite-dimensional). So there is a cluster point $y\in \Omega$. As $p_F$ is continuous, $p_F(y-x)=T_F(x;\Omega)$.
But, if $\Omega$ is unbounded, the above does not work. Let $X=\R^2$, $x=(0,0)$, $$F=\big\{(a,b):\ |a|<1\big\},\qquad\qquad \Omega=\Big\{\Big(s,\frac1s\Big):\ s>0\Big\}.$$ We have $\Omega=g^{-1}(\{1\})\cap[0,\infty)$, where $g$ is the continuous function $g(s,t)=st $; so $\Omega$ is closed. And $F$ is open, convex, absorbing, and $0\in F$.
For $t\geq0$, the point $(s,\frac1s)$ is in $tF$ if and only if $s\leq t $. So $p_F((s,\frac1s))=s $. It follows that $T_F(x;\Omega)=0$ and $p_F(y)>0$ for all $y\in \Omega$.