Does $\overline{\text{span}\{x_n\mid n\geq 1\}}\neq H$ imply $x_1\notin \overline{\text{span}\{x_n\mid n\geq 2\}}$? Let $\{x_n\}_{n\geq 1}$ be a linearly independent family in a Hilbert space $H$.
Assume that $\overline{\text{span}\{x_n\mid n\geq 1\}}\neq H$.

Is it enough to conclude that $x_1\notin \overline{\text{span}\{x_n\mid n\geq 2\}}$?

This fact is used here (p. 142) for $H=L^2(0,T)$ and $x_n=e^{-n^2 \pi^2 t }$. Is it a general result or depends on this particular setting?
 A: I will try to answer the question about those notes. The title question of course has counterexamples.
It looks to me like that statement is indeed a non sequitur. The version of Müntz's theorem quoted as "Theorem 2.6.3 (Münz)" [sic] does not say anything about the closed linear span except that it's not the whole space. A sequence in a Hilbert space is minimal if and only if it has a biorthogonal sequence - the thing that section aims to demonstrate - so this is quite a significant point.
However, the usual proof of Müntz's theorem does show that $x_m$ is not in the closed linear span of $\{x_n\mid n\neq m\}$. One direction of Müntz's theorem "away from the origin", e.g. in $L^2[e^{-\pi^2T},1]$, in fact proves that the distance of the monomial $s^{m^2}$ from $\operatorname{span}\{s^{n^2}\mid n\neq m\}$ is bounded away from zero. By a change of variables $s=e^{-t}$, which is (bi-)Lipschitz on this range, this gives a lower bound on the distance in $L^2[0,T]$ of $e^{-m^2\pi^2t}$ from the span of $\{e^{-n^2\pi^2t}\mid n\neq m\}$. Sorry, I don't have a reference to hand.
A: It's not general. Let $y_n(k)=1$ if $k=n$, 0 otherwise. Let $z_n=y_{n+1}$ for $n\ge 1$ and let $x_n=z_n$ for $n>1$ and $x_1=\sum_{n>1} \frac1n x_n$. This is in $\ell_2(\mathbb N)$ which is isomorphic  to $L_2$.
