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Is there a theorem that says:

If $H$ is a Hilbert space and $U$ is any subspace then $$ H = U \oplus U^\bot$$ if and only if $U$ is closed?

My conjecture is yes.

I can easily prove that if $U$ is closed then $H = U \oplus U^\bot$. But it's not so clear whether if $H = U \oplus U^\bot$ then $U$ is closed. I mean, it seems to me that it should be true but it's not clear how I can show it. So maybe it's not even true.

  • Do you know that $U^\perp$ is always closed, and $(U^\perp)^{\perp} = \overline{U}$? So using the direction you have proved, you know that $H = \overline{U} \oplus U^\perp$. Perhaps then you can argue that $U = \overline{U}$ (sort of "cancelling" the $U^\perp$ from both sides). – Matthew C Feb 08 '15 at 04:33

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Argue as follows: Since any $v\in H$ can be written $v=u+u^\perp$ and $\| v\|^2 =\| u\|^2 + \| u^\perp\|^2$, we get that the mapping $v\mapsto u$ is a continuous projection with range $U$. The result follows from general considerations:

If $T$ is a bounded projection (i.e. $T=T^2$), then the range of $T$ is closed. Indeed take $y_n=Tx_n$ a sequence in the range converging to some $y$. Then $$T(y)=\lim Ty_n=\lim T(Tx_n)=\lim Tx_n=y.$$ Therefore $y$ is in the range.

Jose27
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