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Suppose $V$ is a real or complex inner product space, $U \subseteq V$ is a vector subspace. s.t. $\overline U \neq V$. Are the following two conditions equivalent?

  1. $\overline U = V$ (here I am using the topology induced by the norm induced by the inner product)
  2. $U^\perp = \{0\}$

The negation of 2 implies the negation of 1, because if $w \in U^\perp$ with $w \neq 0$, then $\forall u \in U$ $\|w - u\|^2 = \langle w-u, w-u \rangle = \langle w, w \rangle + \langle u, u \rangle = \|w\|^2 + \|u\|^2 \geq \|w\|^2 > 0$. But I can't figure out if 2 implies 1.

CrabMan
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  • Can you use the fact that $\big(U^{\perp}\big)^{\perp}=\overline{U}$? –  Dec 14 '21 at 15:19
  • @MatthewPilling https://math.stackexchange.com/questions/1478710/a-perp-perp-a-for-a-closed-subspace-a?noredirect=1&lq=1#comment3010927_1478710 suggests that what you wrote might not hold in an incomplete space and that the answer to my question is false. I don't know a counterexample though. – CrabMan Dec 14 '21 at 16:18
  • The answer in https://math.stackexchange.com/questions/1023061/an-inner-product-space-and-its-proper-closed-subspace-with-trivial-orthogonal-co also works as a negative answer to my question. – CrabMan Dec 14 '21 at 17:05

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