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Why is $(V^\perp)^\perp=\overline{V}$ when $V$ is a subspace of a Hilbert space $H$? I googled about this but only $V$ as a subspace of finite vector space. Any reference or proof or idea are welcome.

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    This is true iff $V$ is a closed subspace. Almost any book on FA has a proof. – Kavi Rama Murthy Mar 20 '20 at 00:23
  • It's only true if $V$ is closed. This is a good exercise to prove for yourself. If you've already tried, please explain what you tried and where you got stuck. – Nate Eldredge Mar 20 '20 at 00:23
  • @NateEldredge I edited the question. $(V^\perp)^\perp=\overline{V}$ not $(V^\perp)^\perp=V$. Thanks for your comment. –  Mar 20 '20 at 00:32
  • There could be some weird cases like $V=\emptyset$ but otherwise it is true – Maximilian Janisch Mar 20 '20 at 00:32
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    And see also https://math.stackexchange.com/questions/1051514/a-subspace-x-is-closed-iff-x-x-perp-perp?noredirect=1&lq=1, https://math.stackexchange.com/questions/1716551/prove-that-for-a-closed-subspace-w-of-a-vector-space-v-w-perp-perp?noredirect=1&lq=1, https://math.stackexchange.com/questions/1989671/if-m-is-a-closed-subspace-of-an-hilbert-space-h-then-m-perp-perp-m?noredirect=1&lq=1 – Nate Eldredge Mar 20 '20 at 00:36
  • @MaximilianJanisch the emptyset is not a subspace – Just dropped in Mar 21 '20 at 19:09
  • This is true even if $V$ is a dense subset in $\bar{V}$. The density argument relies on the continuity of inner product, as @egreg mentions in his answer in this post (https://math.stackexchange.com/questions/1043940/double-orthogonal-complement-is-equal-to-topological-closure). – Sam Wong Apr 11 '20 at 11:03

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