How to prove that if the level set of f for some level α is non-empty and bounded, then the level set of f for all levels are bounded

Question

Consider a convex function f, if the level set of f for some level α is non-empty and bounded, then the level sets of f for all levels are bounded.

how to prove this question?

(f doesn't need to be closed)

Answer 1

This is not true in infinite dimensional spaces. For example, consider $f\colon \ell^2\to\mathbb R$ defined by $f(x)=\sup \{|x_n|\}$. This is a convex function, and $f^{-1}(0)=\{0\}$ is nonempty and bounded. But $f^{-1}(1)$ is an unbounded set, since a sequence with supremum $1$ can have arbitrarily large $\ell^2$ norm.

So, suppose the space is finite dimensional. Without loss of generality, the bounded level set is $A=f^{-1}(0)$ and $0\in A$. For sufficiently large $R$, the sphere $S=\{x : |x|=R\}$ surrounds $A$. Let $m=\min_S f$; note that $m>0$.

Take $x$ of norm $|x|>R$, and consider the restriction of $f$ to the line segment joining $x$ to $0$. We know it attains values $0$ and $\ge m$ at certain points; beyond the latter point, the graph of $f$ lies above its secant line. Hence $$f(x)\ge \frac{m}{R}|x|$$ which implies that $f$ has bounded level sets.