Let $K$ be a knot smooth knot in a 3-manifold $M$ and fix a metric on $M$. Let $F$ be a orientable surface of genus $g$ with one boundary component. Then we can consider the family of all maps $\mathscr{F} = \{ \phi: (F, \partial F) \to (M,K) : \phi \text{ is an embedding} \}$. By pulling back the metric we can talk about $\text{Area}(\phi)$ so we can ask if some element of $\mathscr{F}$ achieves the minimum area amongst all elements of $\mathscr{F}$. I suppose that I am mainly interested in knots in $S^3$ with the usual metric, if that simplifies things. 


(1) Is this minimum achieved? 


I have seen [this paper](https://www.sciencedirect.com/science/article/pii/0166864184900038) where Proposition 1 might address my question - however, they allow for piecewise smooth maps. Is there any reason why the area minimizing surfaces would need to be smooth? This seems intuitively clear, but I am not familiar with how to prove such things. 


(2) If the minimum is achieved, is it achieved by a genus minimizing surface? (I.E. the genus of $F$ is the genus of the knot). 



Minimal area of Seifert surfaces
In question (1), if you allow $g$ to vary, then this is answered positively by [Hardt and Simon](https://mathscinet.ams.org/mathscinet-getitem?mr=554379) (see [also](http://www.ams.org/journals/bull/1979-01-01/S0273-0979-1979-14581-6/home.html)). 


The answer to question (2) is no. [Almgren and Thurston](https://mathscinet.ams.org/mathscinet-getitem?mr=442833) construct unknots which do not bound an embedded disk in their convex hull. If $\mathscr{F}$ (fixing $g=0$) contained a minimal area member, then it would have to be a minimal surface. However, a minimal surface bounding $K$ must lie in the convex hull of $K$, a contradiction. 


