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.
  • $\begingroup$ What is the question? $\endgroup$ – copper.hat Jan 24 at 6:25
  • $\begingroup$ Prove that for a convex set C in the Euclidean space E, a convex function f has bounded level sets if and only if it satisfies the growth condition. $\endgroup$ – Danilo Similatan Jan 24 at 6:41
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    $\begingroup$ I am a little confused, if $C$ is bounded, for example, the growth condition is meaningless. Are there some relevant characteristics of $C$ or are you defining $f$ as $+\infty$ for $x \notin C$? $\endgroup$ – copper.hat Jan 24 at 7:21
  • $\begingroup$ Could you try to refer in the outline of the proof and give some hints individually? $\endgroup$ – Danilo Similatan Jan 24 at 8:09
  • $\begingroup$ books.google.co.jp/… $\endgroup$ – Danilo Similatan Jan 24 at 8:36

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 $\liminf_{\|x\| \to \infty} {f(x) \over \|x\|} \ge {1 \over M} > 0$.

  • $\begingroup$ In $M'$, is it $\frac{r}{2}$ or $\frac{2}{r}$? $\endgroup$ – Danilo Similatan Jan 24 at 8:12
  • $\begingroup$ I am confused how $f(x)\geq \alpha$. Could you provide some more details please? $\endgroup$ – Danilo Similatan Jan 24 at 9:06
  • $\begingroup$ Yes, sorry, I meant ${2 \over r}$. $\endgroup$ – copper.hat Jan 24 at 19:24
  • $\begingroup$ It seems two arguments show sufficiency but proof for necessity is lacking. $\endgroup$ – Danilo Similatan Jan 24 at 23:48
  • $\begingroup$ @DaniloSimilatan Thanks for catching that, I corrected the necessity. $\endgroup$ – copper.hat Jan 25 at 18:04

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