0

I would like to show that $(U^{\perp \perp})^{\perp \perp} = U^{\perp \perp}$. I know that if $U$ is a finite dimensional subspace then the equality holds as $U^{\perp \perp} = U$. However I'm not sure how to prove the case in which $U$ is infinite dimensional, specifically, I can't see how $(U^{\perp \perp})^{\perp \perp} \subset U^{\perp \perp}$.

Any help would be appreciated!

hikari30
  • 15
  • A subspace of what? – martini Feb 17 '22 at 07:11
  • 1
    This may be relevant. https://math.stackexchange.com/questions/352469/orthogonal-complement-of-a-hilbert-space –  Feb 17 '22 at 07:12
  • @martini It's a subspace of an inner product space, sorry I forgot to include that. – hikari30 Feb 17 '22 at 07:21
  • As pointed out by @TheHype, separable Hilbert spaces seem the paradigmatic example of what you're asking about, especially now that you've added "inner product space". – eigengrau Feb 17 '22 at 07:26
  • 2
    We have in general, that $U^{\perp \perp} = \overline{U}$ (where the overline means the closure). Thus, we get $$ (U^{\perp \perp})^{\perp \perp} = (\overline{U})^{\perp \perp} = \overline{(\overline{U})} = \overline{U} = U^{\perp \perp}. $$ Here we used that the closure of the closure is again the closure. – Severin Schraven Feb 17 '22 at 07:30
  • @SeverinSchraven Thank you! I'm also curious if there is any way to prove this without using topology? – hikari30 Feb 17 '22 at 18:30
  • I must admit that I do not know whether it is possible to show it without topology. If I was forced to guess, I would say it should be possible to do so. Probably it would be a bit messy as there are many duals floating around. – Severin Schraven Feb 17 '22 at 18:39

0 Answers0