This is a problem from IMC training camp last year.

Given differentiable functions $f_i:\mathbb{R}\rightarrow\mathbb{R}, i=1,2,\ldots,n$ such that $\{f_1,f_2,\ldots,f_n\}$ is linearly independent. Show that there are $n-1$ linearly independent functions among $f_1',f_2',\ldots,f_n'$

  • $\begingroup$ Is the problem correctly stated? It seems to me that defining $f_i(x) = x^i$ gives a counterexample. $\endgroup$
    – bradhd
    Jul 3 '13 at 5:11
  • $\begingroup$ @Brad: At least $n-1$; it is also possible that there are $n$. $\endgroup$ Jul 3 '13 at 5:12
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    $\begingroup$ So the word "dependent" should be "independent"? $\endgroup$
    – bradhd
    Jul 3 '13 at 5:13
  • $\begingroup$ Otherwise, $f_k(x)=\sin kx$ would be an obvious counterexample $\endgroup$
    – leshik
    Jul 3 '13 at 5:15
  • $\begingroup$ @Brad: Oh yeah, I misread what may be a typo, my mistake. Hopefully Yoshua will confirm & edit. $\endgroup$ Jul 3 '13 at 5:46

The proof is by contradiction. The fact that each $n-1$ derivatives are linearly depend imply that nontrivial linear combination of any $n-1$ of the functions is constant. Now, without loss of generality we can assume that $\sum_{k=1}^{n-1}a_kf_k(x)=b_n$ and $a_1\ne 0$ and $\sum_{l=2}^nc_lf_l(x)=b_1.$ Since $f_1,...f_n$ are linearly independent we have $b_1,b_n\ne 0.$ Multiply the first equation by $b_1$ and the second one by $b_n$ and subtract them to get a nontrivial linear combination of $f_1,...f_n$ that adds up to $0.$ This provides a contradiction.


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