A tame topological knot (for the purpose of this question) is a topological embedding of $S^1 \times D^2$ into $\Bbb R^3$.
Tame topological knots are known to be isotopic to smooth knots. This question aims to quantify "how far one must travel" to smooth the knot.
Q: Does a tame topological knot have an "almost" $C^\infty$ Seifert surface?
Here "almost $C^\infty$" Seifert surface means there is a compact, orientable topological $2$-manifold $\Sigma$ (with boundary) embedded in $\Bbb R^3$ such that its boundary is the image of $S^1 \times \{0\}$ under our embedding $S^1 \times D^2 \to \Bbb R^3$. Moreover, the interior of $\Sigma$ is a $C^\infty$ submanifold of $\Bbb R^3$.
i.e. this is a strong manifestation of the statement that smooth knots are dense in tame topological knots.
I would guess the answer to this question is known, and I imagine it would have likely happened in the 60's through mid 70's, but I do not recall hearing a statement of this form.
For some context, Moise's paper "Affine structures in 3-manifolds" Theorem 6 says there is a homeomorphism of $\Bbb R^3$ supported in an arbitrarily small neighbourhood of the knot, which smooths the knot. So by this theorem, there is a surface $\Sigma \subset \Bbb R^3$ whose boundary is the knot, and outside of a small neighbourhood of the knot, is smooth. The issue is that we do not have any control over this surface as the neighbourhood varies, so there isn't a direct way to just take a limit.
I suppose one could really hit the problem over the head with the 3-manifold theory hammer, and argue that the smoothings of the knot are unique up to a small smooth isotopy, so that the smooth Seifert surfaces we can assume are nested (outside of the neighbourhoods). This would allow for a limit process.
But I'm aiming for a more organic argument than this. Perhaps there isn't one in the literature?
