Let $D\subset\mathbb R^2\subset\mathbb R^n$ be a unit planar disc in $\mathbb R^n$. Let $S$ be an orientable two-dimensional surface in $\mathbb R^n$ such that $\partial S=\partial D$. Of course, we have $area(S)\ge area(D)$. Assume that $area(S)< area(D)+\delta$ where $\delta>0$ is small.


Then $S$ is close to $D$ in the following sense: there is a 3-dimensional surface $F$ filling the gap between $S$ and $D$ such that $volume(F)<\varepsilon(\delta)$ where $\varepsilon(\delta)\to 0$ as $\delta\to 0$ ($n$ is fixed). "Filling the gap" means that $\partial F=S-D$.


This fact immediately follows from the compactness theorem for flat norms. But this proof is indirect and does not answer the following questions (I am especially interested in the second one):


1) Are there explicit upper bounds for $\varepsilon(\delta)$? How do they depend on $\delta$ and $n$?


2) Can $\varepsilon(\delta)$ be independent of $n$? Or, equivalently, does the above fact hold true in the Hilbert space?


In the unlikely event that 2-dimensional surfaces are somehow special, what about the same questions about $m$-dimensional surfaces, for a fixed $m$?


Remarks: "Surfaces" here are Lipschitz surfaces or rectifiable currents or whatever you prefer to see in this context. Rather than talking about the filling surface $F$, one could equivalently say that the integral flat norm of $S-D$ is less than $\varepsilon(\delta)$.



Estimating flat norm distance from a planar disc
There is [Almgren's isoperimetric inequality](http://www.ams.org/bull/1985-13-02/S0273-0979-1985-15393-5/S0273-0979-1985-15393-5.pdf):



> 
> Let $\Sigma$ be a $k$-surface in $\mathbb R^n$. Assume $vol \_k \Sigma \le vol\_k S^k$. Then one can fill $\Sigma$ by a $(k+1)$-surface with volume $\le vol\_{k+1} B^{k+1}$. (Here the "surfaces" might have singularities.)
> 
> 
> 


I will use it to show that there is an estimate $\epsilon(\delta)$ which does not depend on $n$.


Take $r$-nbhd $Z\_r$ of $D$.
Note\* that one can give an explicite estimate of $r$, independent of $n$ so that total area of $S$ outside of $Z\_r$ is very small. Moving a bit $r$, one can make the length of intersection curve $\gamma=\partial Z\_r\cap S$ sufficiently small. 
Use Almgren to fill $\gamma$ by a surface;
it breaks $S$ into two pieces $S=S\_1+S\_2$; 


* the surface $S\_1$ lies in $Z\_r$ and $\partial S\_1=\partial D$,
* the surface $S\_2$ has small area and $\partial S\_2=0$.


Fill both $S\_1$ and $S\_2$ separately:


* taking all segments from point on $S\_1$ to its projection on $D$ gives a filling of $S\_1-D$
* fill $S\_2$ using Almgren again.




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(\*)There is a map $\mathbb R^n\to D$ which decrease distances by some factor $k=k(r)<1$ outside of $Z\_r$ and $k(r)$ can be found explicitly.
So if an essential piece of $S$ is outside of $Z\_r$ then the area of $S$ is essentially bigger that $area(D)$.


Say, take $f(x)=$ "sum of maximal and minimal distance to the points in $D$".
This function is convex and it is constant on $D$. 
Take Sharafutdinov retruction for the level sets of this function. 


