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Let $H$ be a Hilbert space and $H_1, \cdots, H_n$ be closed subspaces of $H$.

$\mathbf{Question}:$ Is it always true that the orthogonal complement $(H_1\cap\cdots\cap H_n)^\bot$ of the intersection $H_1\cap\cdots\cap H_n$ in $H$ is equal to the sum $\sum_{j=1}^n H_j^\bot$ of the orthogonal complements of $H_1,\cdots, H_n$?

If the answer to the question is positive? Is the same statement still true if $n=+\infty$? (in this case, the sum $\sum_{j=1}^n H_j^\bot$ should be replaced by the closure of it.)

In some settings, the calculations of each $H_j^\bot$ are easy, while the calculation of $(H_1\cap\cdots\cap H_n)^\bot$ is much more difficult.

Thanks!

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    $\begingroup$ In general it is not true even for $n=2$, because $(H_1\cap H_2)^{\perp}$ is always closed, while the sum of two closed subspaces may fail to be closed. (For instance, in $H:=X\times X$ take the graph of the zero operator as $H_1^{\perp}$ and the graph of a dense, non-surjective bounded operator $T$ on $X$ as $H_2^{\perp}$: then $H_1^{\perp} + H_2^{\perp}= X\times T(X)$ is a dense not closed subspace). $\endgroup$
    – Pietro Majer
    May 5, 2015 at 8:37
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    $\begingroup$ On the other hand, the theorem of bipolars implies that $(H_1\cap \cdots \cap H_n)^\perp$ is always the closure of the sums of the orthogonal complements. $\endgroup$
    – Jochen Wengenroth
    May 6, 2015 at 6:51
  • $\begingroup$ @ Jochen Wengenroth This sounds interesting. Could you tell me a reference about it? Thanks a lot! Is the same statement still true in the case that $n=\infty$? $\endgroup$
    – Entaou
    May 6, 2015 at 16:41
  • $\begingroup$ @Pietro Majer This is an interesting construction. On the other hand, by the comment of Jochen Wengenroth, the answer to the question is yes if $\sum H^\bot_j$ is closed. $\endgroup$
    – Entaou
    May 6, 2015 at 16:44
  • $\begingroup$ Yes, it's easy that $(K_1+K_2)^\perp =K_1^\perp \cap K_2^\perp$ holds for linear subspaces, and you can generalize by induction to n subspaces. Then if you take another orthogonal, you get closures (and you can also rename $H_j:=K_j^\perp$) $\endgroup$
    – Pietro Majer
    May 6, 2015 at 16:59

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