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I hesitate if the following claim is true:

Let $V$ be a normed vector space that is complete. For example, Hilbert space. And assume $\{v_1,...v_n\}$ is a subset of linearly independent vectors in $V$.
Assume also that for any $v_k$ we have a sequence of vectors in $V$ that converges to $v_k$, denote it $({w_{m}}^k)_m$ . Is that true that there exists large enough $m$ for which the subset of vectors $\{{w_m}^1,...,{w_m}^n\}$ is linearly independent?

Thank you in advance!

user376810
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    It holds in every normed space: If $X$ is a normed space and $v_1, \ldots, v_n$ are linearly independent, then there exists an $\varepsilon>0$ such that for every $w_1\in B(v_1, \varepsilon), \ldots, w_n\in B(v_n, \varepsilon)$, the set ${w_1, \ldots, w_n}$ is linearly independent as well. – tree detective Oct 09 '16 at 21:02
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    @treedetective: Do you have a reference for this result or can share some hints on it's proof? – el_tenedor Feb 05 '17 at 15:27
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    @el_tenedor The proof is a little bit lengthy and somehow technical, but not very difficult. You can find it in "Exercises in Functional Analysis", by Costara and Popa, page 9, exercise 35 (excellent book by the way). It provides two solutions, the first one by induction and the second one by contradiction. – tree detective Feb 05 '17 at 17:35
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    @treedetective: Thank you for this reference! – el_tenedor Feb 05 '17 at 18:30

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In a Hilbert space, you can consider the Grammian matrix $ \langle w_i^m, w_j^m\rangle$: if its determinant is zero for all $m$, it is also zero in the limit, as the scalar product and determinant are continuous.

For more general normed vector spaces, a finite dimensional subspace has a closed complement, hence there is a continuous projection onto $\mathrm{span}(v_1,\ldots,v_n)$. The coordinates of the projections of $w_m^{\,\,\,j}$ converge to the canonical basis of $\Bbb R^n$ and independence of the projections implies independence of the vectors themselves.

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Peter Franek
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  • Not each Banach space has a Schauder basis – Bananach Oct 09 '16 at 21:47
  • The approach with continuous projections works though, but the fact that there are continuous projections is not trivial I think – Bananach Oct 10 '16 at 13:53
  • @Bananach Thanks for the comments; is it all right now? – Peter Franek Oct 10 '16 at 14:33
  • Yes, that's what I had in mind. Only one minor thing, in the part on Hilbert spaces, I think you forgot to mention that you consider the determinant of the Gramian matrix (you do say that it is continuous) – Bananach Oct 10 '16 at 15:18
  • @PeterFranek: I do not understand your last sentence: What are the "coordinates" of the projections of $w_m^j$. Do you mean coordinates w.r.t the basis $v_1,\dots,v_n$? In this case I don't understand your argument. Why should the projections of the $w_m^j$ be eventually independent? – el_tenedor Feb 05 '17 at 16:06
  • @el_tenedor Yes, wrt to the basis $v_1,\ldots v_n$. If the coordinates are "very close" to the canonical bases vectors $(1,0,\ldots, 0), (0,1,\ldots,0),\ldots, (0,0,\ldots, 0,1)$, then they are independent. – Peter Franek Feb 05 '17 at 16:15
  • @PeterFranek: How would you make "very close" a rigorous argument? Maybe giving $\mathbb{R}^n$ the 2-norm and using this proof for Hilbert (!) spaces? http://math.stackexchange.com/questions/1823716/orthonormal-basis-with-specific-norm/1823731#1823731 – el_tenedor Feb 05 '17 at 16:41
  • @el_tenedor That's just the continuity of the determinant. – Peter Franek Feb 05 '17 at 16:42
  • @PeterFranek: Even better... Thank you for pointing this out. +1 – el_tenedor Feb 05 '17 at 16:46