Level set of convex functions 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.
 A: 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 $L_1$.
