Prove that for a convex set C in the Euclidean space E, a convex function has bounded level sets if and only if it satisfies the growth condition.

Question

Requiring the function $f$ to have bounded level sets is a "growth condition". Another example is the stronger condition \begin{equation} \liminf_{\left\|x\right\|\to\infty}\frac{f(x)}{\left\|x\right\|}>0, \end{equation} where we define $$\liminf_{\left\|x\right\|\to\infty}\frac{f(x)}{\left\|x\right\|}=\lim_{r\to +\infty}\inf\left\{\frac{f(x)}{\left\|x\right\|}\ \middle|\ x\in C\cap rB^c\right\}.$$

The proof is said to be outlined as follows:

1. Find a function with bounded level sets which does does not satisfy the growth condition.
2. Prove that any function satisfying the growth condition has bounded level sets.
3. Suppose the convex function $f:C\to\mathbb{R}$ has bounded level sets but the growth condition fails. Deduce the existence of a sequence $(x^m)$ in $C$ with $$f(x^m)\leq\frac{\left\|x^m\right\|}{m}\to +\infty.$$ For a fixed point $\bar{x}$ in $C$, derive a contradiction by considering the sequence $$\bar{x}+\frac{m}{\left\|x^m\right\|}(x^m -\bar{x}).$$ Hence complete the proof of the proposition.

Answer 1

If $C$ is bounded then the level sets are bounded and the growth condition vacuous, so I am presuming that $C$ is unbounded.

Let $L_\alpha = \{x \in C| f(x) \le \alpha \}$.

Suppose $\liminf_{\|x\|\to\infty}\frac{f(x)}{\|x\|} = r > 0$. Then for some $M$ if $x \in C$ and $\|x\| \ge M$ we have ${f(x) \over \|x\| } \ge {1 \over 2}r$ or $f(x) \ge {1 \over 2} r \|x\|$.

In particular, if $\|x\| \ge M'=\max(M, {2 \over r} (\alpha+1))$, then $f(x) > \alpha$ and so $L_\alpha \subset B(0,M')$.

Now suppose the level sets are bounded. Pick some $x_0$ then there is some $M$ such that $L_{f(x_0)+1} \subset B(0,M)$.

Choose some $x$ with $\|x-x_0 \| \ge M$ and let $x' = x_0+ M {x - x_0 \over \| x - x_0 \|}$. Then ${f(x)-f(x_0) \over \| x-x_0\||} \ge {f(x')-f(x_0) \over \|x'-x_0\|} \ge {1 \over M}$ and so \begin{eqnarray} {f(x) \over \|x\|} & \ge & {f(x) -f(x_0)\over \|x\|} + {f(x_0) \over \|x\|} \\ &=& {f(x) -f(x_0)\over \|x-x_0\|} { \|x-x_0\| \over \|x\|} + {f(x_0) \over \|x\|} \end{eqnarray} and so $\liminf_{\|x\| \to \infty} {f(x) \over \|x\|} \ge {1 \over M} > 0$.