TITLE: Almost linearly dependent functions
QUESTION [18 upvotes]: Everyone knows that analytic functions $f_1,\ldots,f_n$ are linearly dependent if and only if their Wronski determinant is identically equal to zero. There are several proofs of this,
one in Polya-Szego, vol. 2 Ch. 7, probl. 60.
Is this result stable? More precisely, is the following true:

For every $\epsilon>0$ there exists $\delta>0$ with this property: if analytic
  functions $f_1,\ldots,f_n$ in the unit disc satisfy 
  $|W(f_1,\ldots,f_n)(z)|<\delta \| f\|^n$ for $|z|<1$, then there exists a unit vector
  $a=(a_1,\ldots,a_n)$ such that $|a_1f_1+\ldots+a_nf_n|<\epsilon\| f\|$ for $|z|<1/2$.
  Here $\| f\|=\sqrt{|f_1|^2+\ldots+|f_n|^2}$.

If this is true, I would like to know something about $\delta$ as a function of
$\epsilon$. This is easy to prove when $n=2$ with $\delta\approx\epsilon$. It is also
easy to prove for formal power series that if the Wronskian is small in "p-adic norm",
then there is a linear combination that is small in "p-adic norm". I can also prove the converse statement for every $n$: if the functions are almost linearly dependent, then their Wronskian is small.
EDIT. I would be glad to see any other statement meaning that "if the Wronskian is small,
then there is a linear combination which is small".

REPLY [2 votes]: A simple remark. Let $\mathbb H$ be an Euclidean space with dimension $N$ and let $u_1,\dots, u_{N-1}$ be an orthonormal set of vectors. Let $x$ be a vector in $\mathbb H$: it belongs to the hyperplane $V$ generated by $u_1,\dots, u_{N-1}$ if and only if
$$
u_1\wedge\dots\wedge u_{N-1}\wedge x=0,\quad\text{i.e. }\det(u_1,\dots,u_{N-1}, x)=0
$$
Now, for $x\in \mathbb H$,$\vert\det(u_1,\dots,u_{N-1}, x)\vert$ is the Euclidean distance from $x$ to $V$: the smallness of that determinant is simply the smallness of the distance to the hyperplane.