Let $M$ and $N$ be nonzero closed vector subspaces of a normed vector space $E$. Prove that if $M \neq N$, then $M^\perp \neq N^\perp$.
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2In general normed spaces, you don't have "orthogonality". You need the inner product structure. – Feb 04 '21 at 16:57
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Do you mean the annihilator $M^\perp = {\lambda \in M' : (\forall x\in M)\lambda(x) = 0}$? – GhostAmarth Feb 04 '21 at 17:01
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@GhostAmarth yes – Anderson Corrêa Porto Feb 04 '21 at 17:05
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1Does this answer your question? Uniqueness of annihilator subspace in infinite dimensional normed space – GhostAmarth Feb 04 '21 at 17:17
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Suppose $M^\perp = N^\perp$, then by double annihilator theorem $\overline{M} = \overline{N}$. Since $M$ and $N$ are closed, it's a contradiction.
If $X$ is a normed vector space and $U\subseteq X$, how do we show $(U^\perp)_\perp=\overline U$?
Timur Bakiev
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Although the question was answered in the post I linked, I thought I add some details.
Take w.l.o.g. $x\in M \setminus N$. Since $N$ is closed, $X/N$ is also a normed vector space and $q\colon X\to X/N, y \mapsto y+N$ is linear and continuous.
By Hahn-Banach, we can find a $\lambda \in (X/N)'$ such that $\lambda(x+N) = \|x+N\|_{X/N} \neq 0$, since $x\notin N$ (and therefore $x+N\neq N$).
Now $f=\lambda \circ q \in X'$ and $f|_N = 0$ but $f(x) \neq 0$, so $f\in N^\perp$ but $f\notin M^\perp$.
GhostAmarth
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