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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".

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is it clear that this holds for polynomials? –  Suvrit Sep 25 '13 at 19:45
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I think it is of the same difficulty for polynomials as for arbitrary functions, of course if the degrees of the polynmials are not taken into account. –  Alexandre Eremenko Sep 26 '13 at 4:13
    
Thank you, it is a very interesting question. I suspect it is related to Kobayashi hyperbolicity. (Un)Fortunately, my attempts provided no answer yet. I will keep trying. –  Sasha Anan'in Oct 4 '13 at 18:30
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Yes, the question arises in my research on hyperbolicity, more precisely in trying to understand Cartan's 1928 thesis, and the Second Fundamental theorem of Cartan, 1933. But it seems to be of independent interest. –  Alexandre Eremenko Oct 5 '13 at 14:10

1 Answer 1

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.

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Yes, but how does this help to solve the question? –  Alexandre Eremenko Oct 5 '13 at 14:11
    
The function $x\mapsto\det(u_1,\dots,u_{N-1},x)$ is continuous and $\mathbb H$ could be taken as $span(f_1,\dots,f_{N-1})$. –  Bazin Nov 3 '13 at 17:02

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