Let $f:\mathbb R^n \to\mathbb R \cup\{+\infty\}$ be a proper convex function, assume that there exists $c\in\mathbb R$ such that the $c$-level set $L_{\leq c}=\{x\in R^n: f(x)\leq c\}$ is nonempty and bounded. Prove that all level sets of $f$ are also bounded.
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If $f$ is closed then this is straightforward. – copper.hat Mar 04 '14 at 21:58
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I believe you need $f$ to be closed (equivalent to lower semi-continuous for a proper function).
Define $f:\mathbb R^2 \to\mathbb R \cup\{+\infty\}$ as follows: $f(x,y) = \begin{cases} 0, & x=0, y=0\\ y, & y >0 \\ +\infty, & \text{otherwise}\end{cases}$
$f$ is proper, convex, $L_{\leq 0} = \{ (0,0) \}$ which is bounded and $ \mathbb{R} \times \{ 1\} \subset L_{\leq 1} $ which is unbounded.
Now suppose $f$ is closed. In particular, this means the level sets are closed.
I will proceed by contradiction: Suppose $c_1<c_2$ are such that $L_1=L_{\leq c_1}$ is bounded, but $L_2=L_{\leq c_2}$ is not. (note that $L_1 \subset L_2$.) Then there is a non-zero $d \in 0^+ L_2$, the recession cone of $L_2$ (for example, see Rockafellar, "Convex Analysis", Theorem 8.4). In particular, if $x_1 \in L_1$, we have $x_1+t d \in L_2$ for all $t \ge 0$. The function $\phi(t) = f(x_1+t d)$ must be nonincreasing (if not, then $x_1+td \notin L_2$ for some $t>0$), in which case we must have $x_1+t d \in L_1$ for all $t \ge 0$, which contradicts boundedness of $L_1$.
copper.hat
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Sorry, I should have made it clear; the Theorem quoted above states that a closed convex set if unbounded iff its recession cone consists of the zero vector alone. The main use above is to ensure that if $d$ is a direction of recession of $L_2$, then $x_0+td \in L_2$ for all $t \ge 0$ (and all $x_0 \in L_2$). In the 'counterexample' above, you can see that $e_1$ is a direction of recession for $L_{\leq 1}$ (well, more correctly $\operatorname{ri} L_{\leq 1}$), but $(0,0)+ t e_1 \notin L_{\leq 1}$ for any $t$. – copper.hat Mar 06 '14 at 04:47
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Minor point: I have only shown that closed is sufficient to prove your statement, and provided a non-closed counterexample. It is not necessary for $f$ to be closed for the statement to be true as the example $f(x) = 0$ for $|x|<1$ and $+\infty$ otherwise shows. – copper.hat Mar 06 '14 at 05:06
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No, because $0^+(\operatorname{ri} C) = 0^+(\bar{C})$ if $C$ is non-empty. (And if $C$ is open then $\operatorname{ri} C = C$.) – copper.hat Mar 06 '14 at 06:31
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Can you show me: is there a nonempty (non closed) unbouded convex set C but $o^+(C)={0}$. Thank you very much! – Richkent Mar 06 '14 at 06:39
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Take $C=(\mathbb{R} \times (0,1)) \cup { (0,0) }$. Then the only directions of recession are $\pm e_1$, but it doesn't 'work' at $(0,0)$, so $0^+ C = {(0,0)}$. Note that $0^+(\operatorname{ri} C) = 0^+(\bar{C}) = \operatorname{sp} { e_1 }$. – copper.hat Mar 06 '14 at 06:51
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I mean an open horizontal slab with the point $(0,0)$ added in (so it is neither open nor closed). Oh, I see a point of confusion, the first $(0,1)$ means the open interval, the second $(0,0)$ means the origin in 2-space. – copper.hat Mar 06 '14 at 07:07
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Btw, if you are happy with an answer it is customary to accept it after some period of time. – copper.hat Mar 06 '14 at 07:28