Minimum of sublinear function

Question

Let $f:X\rightarrow\mathbb{R}$ be a sub-linear functional on a Banach space (or even just the euclidean space) which is bounded from below on a hyper-plane (or a non-bounded convex set). Must it attain its minimum on this set?

Answer 1

On a general Banach space, even linear functionals don't necessarily attain their infimum on a bounded closed convex set. A typical example is $$ L(f) = \int_0^{1/2} f(x)\,dx - \int_{1/2}^1 f(x)\,dx $$ on the space of continuous functions $C[0,1]$ with the uniform norm. If $B$ is the closed unit ball in this space, then $\inf_B L = -1$ but this infimum is not attained.

Furthermore, on the closed hyperplane $M=\{f: L(f)=1\}$ the sublinear function $\phi(x)=\|x\|$ does not attain its infimum (which is $1$). These two facts are directly related.

Euclidean case

If the closed convex set is allowed to be unbounded, counterexamples are still easy to find, such as $f(x) = x$ on the set $\{(x,y)\in\mathbb R^2 : x,y>0, xy\ge 1\}$. The infimum is $0$ and is not attained.

Here is a sublinear functional not attaining its minimum on a line in $\mathbb R^2$: let $\Omega = \{(x,y)\in\mathbb{R}^2: y> x^2-1\}$ and let $f$ be the Minkowski functional of $\Omega$, that is $f(x,y) = \inf\{t>0: (x/t, y/t)\in \Omega\}$. Since $\Omega$ is convex, $f$ is sublinear. It is strictly positive on the line $x=1$, but $\lim_{y\to\infty} f(1,y)=0$ so the infimum is $0$, not attained.

One the set is assumed bounded and closed (hence compact), one can appeal to the continuity of $f$.