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Suppose we have a bounded convex domain $\Omega \subset \mathbb{R}^n$. Can we approximate the domain from within by an increasing sequence of uniformly convex domains $\Omega_n$ converging to $\Omega$ in the Hausdorff distance?

By a uniformly convex $X$, I mean that the principal curvatures of $\partial X$ have a positive lower bound (take it to be smooth). An equivalent definition is that for every $\epsilon >0$, there exists $\delta > 0$ such that for all $x, y \in X$ with $\|x-y\| \geq \epsilon$ the distance from $(x+y)/2$ to the boundary is at least $\delta$.

I want to apply an approximation argument to a proof about convex domains.

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  • $\begingroup$ any references will also be helpful! $\endgroup$
    – abe.nong
    Oct 12, 2015 at 13:20

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This is true. There is a smooth convex function $u:\Omega\to\mathbb{R}$ such that $u(x)\to\infty$ as $x\to\partial\Omega$; see below. The sum $v(x)=u(x)+\|x\|^2$ has the same properties, and is strongly convex. Since $v$ is also locally Lipschitz, it follows that the level sets $\Omega_t = \{x:v(x)<t\}$ are uniformly convex domains. The convergence $\Omega_t\to\Omega$ follows from the fact that any compact set $K\subset \Omega$ is contained in $\Omega_t$ for $t>\sup_K v$.

The existence of $u$ as above requires a proof. It suffices to construct a convex $w:\Omega\to(-\infty,0)$ such that $w(x)\to0$ as $x\to\partial\Omega$; then $u=-1/w$ does the job since the function $\xi\to -1/\xi$ is convex and increasing on $(-\infty,0)$. Such $w$ is called a (smooth) convex exhaustion function for $\Omega$. Google search brought up Smooth exhaustion functions in convex domains by Zbigniew Blocki (free access), where the existence of $w$ is stated with a reference to Hörmander's book Notions of Convexity; so I suggest consulting that book as well.

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  • $\begingroup$ thanks. i recently saw a similar argument, but i was unsure how to show that the sublevel sets in fact had uniform positively bounded principal curvature. Also, do we need to use Sard's Lemma at all to show that these domains are in fact smooth? I heard that mentioned in a discussion with some others. $\endgroup$
    – abe.nong
    Oct 14, 2015 at 14:23
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    $\begingroup$ You don't need Sard's lemma because a strictly convex function has at most one critical point. To argue about the curvature, you can either go through the curvature of implicit curves, or consider the local behavior of $v$ on a hyperplane tangent to level set. On this hyperplane, the first order Taylor term are zero, but the quadratic term has a lower bound thanks to $|x|^2$. Hence, $v$ grows quadratically away from the tangency point on this plane, which shows how the level set curves away from this plane. $\endgroup$
    – user147263
    Oct 14, 2015 at 14:44

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