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I am trying to prove $E$ is closed iff $E=(E^\perp)^\perp.$

Forward direction: Assume $E$ is closed, i.e. by definition, there exists a sequence $\{x_n\} \in E$ such that it converges to a point let $x \in X,$ which is also in $E.$

Now, from the definition of an orthogonal component, it is clear that $E\subset (E^\perp)^\perp.$

To prove $ (E^\perp)^\perp \subset E, $ I am assuming first $x \notin E, but \in (E^\perp)^\perp.$ i.e., $$\langle x,y\rangle =0 , \quad \forall y \in E^\perp$$

Thus, the vector $x$ has a non-zero projection onto $E^\perp,$ i.e. there exists one such $ y \in E^\perp$ s.t. $\langle x,y \rangle \ne 0.$ Which is a contradiction to our assumption $x \in (E^\perp)^\perp.$

Hence $x$ must be in $E.$ Thus, $ (E^\perp)^\perp \subset E. $

Finally, $(E^\perp)^\perp = E.$

Backward Direction: From the definition of orthogonal component $(E^\perp)^\perp.$ Thus, $(E^\perp)^\perp = E,$ implies $E$ closed.

I don't know whether the forward direction is rigorous enough. (If $X$ was Hilbert, then the Hilbert projection theorem straightforward gives us the proof.)

Funny Guy
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    The forward direction is false, see https://math.stackexchange.com/a/4978204/465145 – David Gao Oct 15 '24 at 23:44
  • It's a proof , the similar question is asking for a counter example. – Funny Guy Oct 15 '24 at 23:47
  • I mean, the question has a counterexample, so the proof is wrong and the question just have a negative answer. – David Gao Oct 15 '24 at 23:50
  • Also, where in your proof of the forward direction did you use the fact that $E$ is closed? – Sammy Black Oct 15 '24 at 23:52
  • Okay, let's say I am wrong in both approaches (forward). Which I know I am. Is it possible for anyone to guide me in the right direction? – Funny Guy Oct 16 '24 at 00:05
  • @FunnyGuy There is no right direction, the proposition you’re trying to prove is just false, as the counterexample demonstrates. You do know what counterexamples mean, right? – David Gao Oct 16 '24 at 00:11
  • @DavidGao I talked with my professor, and she said it is not false. – Funny Guy Oct 16 '24 at 00:14
  • @FunnyGuy Regardless of what your professor said, it is false, in that direction at least, and David's linked answer proves this. There is no proof of this assertion. Now, we could find issues with your proof, or help you prove the converse direction, but we can't prove a false statement true! – Theo Bendit Oct 16 '24 at 00:44
  • @TheoBendit that would be helpful!! Thank you! – Funny Guy Oct 16 '24 at 00:46

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Let's start with a proof of the backwards direction. It suffices to show that $F^\perp$ is closed for any subspace $F$, as this implies $(E^\perp)^\perp$ is closed.

We write: \begin{align*} F^\perp &= \{x \in X : \forall f \in F, \langle x, f \rangle = 0\} \\ &= \bigcap_{f \in F} \{x \in X : \langle x, f \rangle = 0\}. \end{align*} Note that $\{x \in X : \langle x, f \rangle = 0\}$ is the inverse image of the closed set $\{0\}$ under the continuous map $x \mapsto \langle x, f \rangle$, and so by continuity, the set must be closed. This makes $F^\perp$ the intersection of closed sets, which itself must be closed.

As for your forward direction, I'm with you until:

Thus, the vector $x$ has a non-zero projection onto $E^\perp$

It is true, in Hilbert spaces, that vectors admit projections onto closed subspaces. In general inner product spaces, not so much. See here.

Theo Bendit
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  • I was trying to say that if $x \notin E$, then it can not be orthogonal to any element $y\in E^\perp$. Implying $\langle x,y\rangle \ne 0.$ Is this wrong? – Funny Guy Oct 16 '24 at 01:13
  • @FunnyGuy That is wrong. Just read the example I linked to. $E^\perp$ can literally be just ${0}$. – David Gao Oct 16 '24 at 01:36
  • @DavidGao in your example you are in $l^2$ right? I mean your space is $l^2.$ Correct me if I am wrong. I am trying to understand this. – Funny Guy Oct 16 '24 at 01:51
  • @FunnyGuy No. The space is $c_{00}$. It’s the subspace of $\ell^2$ consisting of only finitely supported sequences. – David Gao Oct 16 '24 at 01:52