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Let $H$ be a separable Hilbert space with an orthonormal basis $e_1,e_2,\dots$. Consider a linear functional $L: H\to \mathbb{R}$, which may be bounded or not, that obeys $L(e_i)=0$ for all $e_i$. Does that imply that $L(x)=0$ for all $x\in H$?

Artemy
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    No. It only implies it for continuous, equivalently bounded, linear functionals. However, everywhere defined unbounded linear functionals are rather exotic objects that can only be constructed using the axiom of choice via the so-called Hamel basis. – Conifold Sep 21 '20 at 04:00
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    Slightly more explicitly, consider the quotient of $H$ by the (non-closed) subspace spanned by the $e_i$. Since $H$ is uncountable-dimensional and the subspace is countable-dimensional this quotient is also uncountable-dimensional, and in particular nonzero. So the axiom of choice can be used to construct a nonzero linear functional on it, which pulls back to an unbounded nonzero linear functional on $H$ vanishing on the $e_i$. – Qiaochu Yuan Sep 21 '20 at 07:21
  • @QiaochuYuan and @ Conifold: these comments should be answers imo. – supinf Sep 21 '20 at 08:49
  • I'm happy to accept some combination of these comments as an answer. – Artemy Sep 22 '20 at 00:40

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Complete $\{ e_n \}$ to a Hamel basis $\{ f_{\alpha}\}_{\alpha\in\Lambda}$ for $H$ where $f_{\alpha_n} = e_{n}$. Then define a linear functional $\Phi$ on $\{ f_{\alpha} \}_{\alpha\in\Lambda}$ in such a way that $\Phi(f_{\alpha_n})=0$ but so that $\Phi \ne 0$. This is possible because $\{ e_n \}$ cannot be a Hamel basis.

Disintegrating By Parts
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