TITLE: Minimal area of Seifert surfaces
QUESTION [12 upvotes]: 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 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).

REPLY [10 votes]: In question (1), if you allow $g$ to vary, then this is answered positively by Hardt and Simon (see also). 
The answer to question (2) is no. Almgren and Thurston 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.