The knot $K\subset \Bbb S^3$ is smoothly slice, but the disc $D\subset \Bbb D^4$ is only locally flat. Can $D$ be smoothed?

Question

Suppose I am given a smoothly slice knot $K\subset\Bbb S^3$. But I am only given a locally flat disc $D\subset \Bbb D^4$ with boundary $K$.

Question: Is there a smooth disc $D'\subset\Bbb D^4$ with boundary $K$ that is $\epsilon$-close to $D$ and approaches $D$ towards the boundary?

More precisely, I want $D'$ to be in $D_\epsilon\cap X$, where $D_\epsilon$ is an $\epsilon$-thickening of $D$ and

$$X:=\bigcup_{x\in D} B_x(d(x,K)),$$

$B_x(r)$ being a ball around $x$ of radius $r$, and $d(x,K)$ being the distance of $x$ from $K$.

Answer 1

Nice question! Here's an example to think about. Let J be a knot that is topologically but not smoothly slice, and let $D_J$ be a locally flat disc that is smooth near the boundary, with boundary J. Note that $-J$ (the usual concordance inverse) is also slice, with disk $-D_J$. Let $D = D_J\natural(-D_J)$, where the $\natural$ denotes boundary connected sum done along a band near the boundary. Let $K = J \# (-J)$; then it is smoothly slice by a standard argument.

At first glance, it looks hard for any slice disk $D'$ for $K$ to be close to $D$ without splitting into smooth slice disks for $J$ and $-J$. But the proof is not obvious because you are approximating in the $C^0$ topology whereas one would presumably need a stronger sense of approximation near the boundary.

As I said, something to think about.