Concurrent normals conjecture It is conjectured that if $K$ is a convex body in $n$-dimensional Euclidean space, then there exists a point in the interior of $K$ which is the point of concurrency of normals from $2n$ points on the boundary of $K$. This has been proved for $n=2$ and $3$ by E. Heil. For $n=4$, a proof appeared (under a smoothness assumption on the boundary) in
Pardon, John, Concurrent normals to convex bodies and spaces of Morse functions, Math. Ann. 352, No. 1, 55-71 (2012). ZBL1242.52006.
but a reviewer’s remark in zbMATH says that in this paper, the proof of the first theorem "does not seem to be quite correct". So my question is:
What is the current status of this conjecture for $n=4$?
 A: Let me address the specific complaint of that review.  The situation is the following.  Our (bounded, open) convex set is denoted $K\subseteq\mathbb R^n$ with closure $\overline K$, and we consider the "distance to $p$" function $d_p:\partial K\to\mathbb R$ for $p\in\overline K$.  Let $V\subseteq\overline K$ be the set of $p\in\overline K$ for which $d_p$ has exactly one local minimum.  I claimed in my paper that "it is clear that $V$ is closed".  As the reviewer correctly points out, this is false (counterexample: $K$ the unit ball).  But here is a corrected version: "if $d_p$ has finitely many local minima for every $p\in\overline K$, then $V$ is closed".  Indeed, if $d_p$ has finitely many local minima, then if we perturb $p$ the number of local minima can only increase.  This corrected version of the statement is sufficient for the proof given in the paper, since if $d_p$ has infinitely many local minima for some $p$, this exactly means that there are infinitely many normals to $\partial K$ which are concurrent at $p$.  So the reviewer's remark doesn't invalidate the argument.
I have to admit, this is not a well written paper, and I would be surprised if it did not contain a number of other equally badly presented arguments.  I can say that I reread it a couple of years ago and was more or less convinced by the proof.  However, I have not discussed it in detail with anyone.
