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Disclaimer: Though I don't need it anymore this is interesting in its own!

Is it true that if the orthogonal complement is trivial then the subset was dense: $$A^\bot=\{0\}\implies\overline{A}=X$$ Moreover does it hold that a dense subset remains dense in the completion: $$\overline{A}^X=X\implies\overline{A}^{\hat{X}}=\hat{X}$$ Of course the converses are both trivial and I suppose both fail in general. Besides for complete a.k.a. Hilbert spaces the first is just a consequence of the orthogonal decomposition for closed subspaces.

freishahiri
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  • What does $\perp$ mean in your question? –  Jun 12 '14 at 16:36
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    By the way, your recent edit makes nonsense out of Vladimir's answer. This is very rude to him. You should have asked a new question instead. –  Jun 12 '14 at 16:43
  • It is meant 'being orthogonal' within the direct sum of Hilbert spaces: $(x_0,Ax_0)\perp(x,A\restriction_\mathcal{D} x):\iff \langle x_0,x\rangle_X+\langle Ax_0,A\restriction_\mathcal{D}x\rangle_Y=0$ – freishahiri Jun 12 '14 at 16:43
  • Better look from this site: If $A$ is closed then $\mathcal{G}(A)$ closed in $X\times Y$ so also complete. Then $\mathcal{G}(A)=\overline{\mathcal{G}(A\restriction_\mathcal{D})}\oplus\mathcal{G}(A\restriction_\mathcal{D})^\perp$ w.r.t. $\mathcal{G}(A)$. Now if $\mathcal{G}(A\restriction_\mathcal{D})^\perp={0}$ w.r.t. $\mathcal{G}(A)$ then $\mathcal{G}(A)=\overline{\mathcal{G}(A\restriction_\mathcal{D})}$ so $\mathcal{D}$ was a core. The condition written out is what I mentioned in the question... – freishahiri Jun 12 '14 at 16:50
  • @Vladimir: Do u want me to open a new question rather? I mean the former question is interesting in its own so I wouldn't mind opening a new thread rather... – freishahiri Jun 12 '14 at 17:19
  • Yes if you wish open a new thread. But pretty much the answer will be the same; for a countrerexample, just take D to be the identity operator $L^2\to L^2$ on the domain $A$; then $A$ is not a core but if we consider the operator in $C_0^\infty$ then in the graph of the identity operator there will be no vector orthogonal to the graph of D restricted to $A$. – Vladimir Jun 12 '14 at 18:25

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The second is of course true. If $A$ is dense in $X$, then it is also dense in $\hat X$. Just approximate $x\in\hat X$ by $x_n\in X$, $||x-x_n||\le 2^{-n}$ and then approximate each $x_n$ by $y_n\in A$, $||y_n-x_n||\le 2^{-n}$.

The first is false: let $X=C_0^\infty(R)$, $\hat X=L^2(R)$, and $A=\{f\in X|\int_a^bf(x)dx=0\}$. Then $A^\perp=\{0\}$ in $X$, because $1_{[a,b]}\notin X$, but $A$ is clearly not dense.

Vladimir
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