Let $L : \mathbb{R}^n \to \mathbb{R}$ be a convex smooth function such that its derivative $DL : \mathbb{R}^n \to (\mathbb{R}^n)^*$ is diffeomorphism. For each $R > 0$ define $$S_R := \{x \in \mathbb{R}^n : \|DL_x\| = R\}.$$ Then $S_k$ is compact. Why is that true?
I mean, using Heine-Borel it is enough to show that $S_k$ is closed and bounded and it is easy to see that $S_k$ is closed as a preimage of a continuous function. But why is $S_k$ bounded?
I can offer a proof, if we assume that $(D^{-1})'$ is continuous:
$D$ is a diffeomorphism, so for each $\lambda\in \mathbb R^{n*},$ there is an unique $x\in \mathbb R^n$ such that $\lambda=DL_x$. That is, the map $D^{-1}:DL_x\mapsto x$ is defined on all of $\mathbb R^{n*}$, is onto $\mathbb R^n$ and is differentiable.
For convenience, assume that $DL_0=0$ and fix $x\in S_R.$
Since $L$ is convex, $DL_{tx}$ exists for each $0\leq t\leq 1.$
By the mean value theorem,
$\tag1 D^{-1}(DL_x)-D^{-1}(DL_0) =\int^1_0 (D^{-1})'(DL_0+DL_{tx})DL_xdt$
and so
$\tag2 x=D^{-1}(DL_x)=\int^1_0 (D^{-1})'(DL_{tx})DL_xdt,\ $
which implies that
$\tag 3\|x\|\leq \sup\{\|(D^{-1})'(DL_{tx})\|:0\leq t\leq 1\}\cdot R.$
To finish, note that the map $t\mapsto DL_{tx}$ is continuous, so the image $\{DL_{tx}:0\leq t\leq 1\}$ is compact, and so, because $(D^{-1})'$ is continuous, the sup in $(3)$ is bounded.
This is enough to prove the claim.