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How to prove that for any sequence $(f_n) \subset L^2[0,1]\setminus \{0\}$ there is a function $g \in L^2[0,1]$ such that

$\int f_n g dx \neq0\ \forall n\geq 1?$

I tried to use a weak limit of $sign(f_n)$ or by argument by contradiction but I didn't get the result.

2 Answers 2

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If you know the Baire category theorem, the result falls out at once: $L^2$ cannot be written as a countable union of proper subspaces $\{f_n\}^\perp$.

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The constructive solution in case where $f_n$ are linearly independent.

Let $g_n$ be the orthonormal sequence obtained by Gram-Schmidt process from $f_n$. Construct inductively a sequence $\alpha_n$ in the following way:

  • put $\alpha_0=1$
  • given $\alpha_n$ for $n, put $0<\alpha_N\leq 2^{-N}$ other than $-\sum_{n.

Put $g=\sum_n \alpha_ng_n$, then $\langle g\vert f_N\rangle=\alpha_N\langle g_N\vert f_N\rangle+\sum_{n


In general, let $e_n$ be an orthonormal sequence obtained from $f_n$ by Gram-Schmidt process. Construct inductively sequence $\alpha_n$:

  • $\alpha_0=1$
  • Given $\alpha_n$ for $n, put $0<\alpha_N<2^{-N}$ such that for any $f_k$ in the linear span of $e_0,\ldots,e_N$, but not in the span of $e_0,\ldots,e_{N-1}$ we have $\alpha_N\neq-\sum_{n. These are only countably many points, so we can choose such an $\alpha_N$.

Similarly to the above, $g=\sum_n\alpha_ne_n$ will be okay (because the $e_n$ comes from Gram-Schmidt, any $f_n$ will be spanned by finitely many of them).